Calculating this difficult integral (equation of a circle)

[Mad_MechE]

I have tried numerous methods to calculate this integral and can't seem to figure it out.
I have tried to use maple and mathematica but i am not very strong using these programs. I was wondering if someone would be able to help me solve this integral. This equation stemmed from the equation of a circle... and i got the following function:

h = h0 + R - \sqrt{R+x}\sqrt{R-x}

i am trying to integrate the following:
\int\frac{1}{h(x)^2}dx

and

\int\frac{1}{h(x)^3}dx

Thanks for all your help

MT

 

[g_edgar]

Do you really need symbolic answers? For the first one, Maple says:
<br /> -1/2\, \left( 2\,i{R}^{3}{\it h0}\,{\rm arctanh} \left( {\frac {{R}^{2<br /> }+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R \right) <br /> \sqrt {{R}^{2}-{x}^{2}}}} \right) +2\,i{R}^{3}{\it h0}\,{\rm arctanh}<br /> \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{<br /> \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -4\,{R}^{<br /> 3}{\it h0}\,\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+<br /> 2\,R}}} \right) +i{R}^{2}{{\it h0}}^{2}{\rm arctanh} \left( {\frac {{R<br /> }^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R<br /> \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) +i{R}^{2}{{\it h0}}^{2}{<br /> \rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{<br /> \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}<br /> \right) -2\,{R}^{2}\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{<br /> \it h0}+2\,R}}} \right) {{\it h0}}^{2}-2\,{R}^{2}x\sqrt {{\it h0}}<br /> \sqrt {{\it h0}+2\,R}+i{R}^{2}{\rm arctanh} \left( {\frac {-{R}^{2}+i<br /> \sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R \right) <br /> \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}+i{R}^{2}{\rm arctanh}<br /> \left( {\frac {{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{<br /> \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}-2<br /> \,{R}^{2}\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,<br /> R}}} \right) {x}^{2}-4\,Rx{{\it h0}}^{3/2}\sqrt {{\it h0}+2\,R}-2\,R<br /> \sqrt {R-x}\sqrt {R+x}\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}x-2\,x{{<br /> \it h0}}^{5/2}\sqrt {{\it h0}+2\,R}-2\,\sqrt {R-x}\sqrt {R+x}{{\it h0}<br /> }^{3/2}\sqrt {{\it h0}+2\,R}x \right) \left( {\it h0}+2\,R \right) ^{<br /> -3/2}{{\it h0}}^{-3/2} \left( {{\it h0}}^{2}+2\,{\it h0}\,R+{x}^{2}<br /> \right) ^{-1}<br />

 

[g_edgar]

and for the second one, Maple says:
<br /> 1/4\, \left( -12\,i{R}^{4}{\it h0}\,{\rm arctanh} \left( {\frac {{R}^{<br /> 2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R<br /> \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}-12\,i{R}^{4}{\it h0<br /> }\,{\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {<br /> {\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}<br /> \right) {x}^{2}-18\,i{R}^{3}{{\it h0}}^{2}{\rm arctanh} \left( {<br /> \frac {{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it <br /> h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}-18\,i{R}^{3}{{<br /> \it h0}}^{2}{\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,<br /> R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}<br /> }}} \right) {x}^{2}-6\,i{R}^{2}{{\it h0}}^{3}{\rm arctanh} \left( {<br /> \frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it <br /> h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{2}-6\,i{R}^{2}{{<br /> \it h0}}^{3}{\rm arctanh} \left( {\frac {{R}^{2}+i\sqrt {{\it h0}+2\,R<br /> }\sqrt {{\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}<br /> }} \right) {x}^{2}-3\,i{R}^{2}{\it h0}\,{\rm arctanh} \left( {\frac {{<br /> R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R<br /> \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{4}-3\,i{R}^{2}{\it h0}<br /> \,{\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{<br /> \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}<br /> \right) {x}^{4}+6\,\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{9/2}\sqrt {{\it <br /> h0}+2\,R}x+2\,\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{5/2}\sqrt {{\it h0}+2<br /> \,R}{x}^{3}+4\,{{\it h0}}^{11/2}\sqrt {{\it h0}+2\,R}x+24\,{R}^{5}<br /> \arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}}<br /> \right) {{\it h0}}^{2}+30\,{R}^{3}\arctan \left( {\frac {x}{\sqrt {{<br /> \it h0}}\sqrt {{\it h0}+2\,R}}} \right) {{\it h0}}^{4}+48\,{R}^{4}<br /> \arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}}<br /> \right) {{\it h0}}^{3}+6\,{R}^{3}\arctan \left( {\frac {x}{\sqrt {{<br /> \it h0}}\sqrt {{\it h0}+2\,R}}} \right) {x}^{4}+6\,{R}^{2}\arctan<br /> \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}} \right) {{<br /> \it h0}}^{5}+20\,{R}^{4}{{\it h0}}^{3/2}\sqrt {{\it h0}+2\,R}x+6\,{R}^<br /> {3}\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}{x}^{3}+46\,{R}^{3}{{\it h0}}^<br /> {5/2}\sqrt {{\it h0}+2\,R}x+42\,{R}^{2}{{\it h0}}^{7/2}\sqrt {{\it h0}<br /> +2\,R}x+6\,{R}^{2}{{\it h0}}^{3/2}\sqrt {{\it h0}+2\,R}{x}^{3}+20\,{{<br /> \it h0}}^{9/2}\sqrt {{\it h0}+2\,R}xR+36\,{R}^{3}\arctan \left( {<br /> \frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}} \right) {{\it h0}}^{<br /> 2}{x}^{2}+6\,{R}^{2}\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {{<br /> \it h0}+2\,R}}} \right) {\it h0}\,{x}^{4}+12\,{R}^{2}\arctan \left( {<br /> \frac {x}{\sqrt {{\it h0}}\sqrt {{\it h0}+2\,R}}} \right) {{\it h0}}^{<br /> 3}{x}^{2}+24\,{R}^{4}\arctan \left( {\frac {x}{\sqrt {{\it h0}}\sqrt {<br /> {\it h0}+2\,R}}} \right) {\it h0}\,{x}^{2}-12\,i{R}^{5}{{\it h0}}^{2}{<br /> \rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{<br /> \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}<br /> \right) -12\,i{R}^{5}{{\it h0}}^{2}{\rm arctanh} \left( {\frac {{R}^{<br /> 2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R<br /> \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -24\,i{R}^{4}{{\it h0}}^{3}<br /> {\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{<br /> \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}<br /> \right) -24\,i{R}^{4}{{\it h0}}^{3}{\rm arctanh} \left( {\frac {{R}^{<br /> 2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R<br /> \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -15\,i{R}^{3}{{\it h0}}^{4}<br /> {\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{<br /> \it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}<br /> \right) -15\,i{R}^{3}{{\it h0}}^{4}{\rm arctanh} \left( {\frac {{R}^{<br /> 2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it h0}+R<br /> \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -3\,i{R}^{3}{\rm arctanh}<br /> \left( {\frac {{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{<br /> \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) {x}^{4}-3<br /> \,i{R}^{3}{\rm arctanh} \left( {\frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}<br /> \sqrt {{\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}<br /> } \right) {x}^{4}-3\,i{R}^{2}{{\it h0}}^{5}{\rm arctanh} \left( {<br /> \frac {-{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {{\it h0}}x}{ \left( {\it <br /> h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}} \right) -3\,i{R}^{2}{{\it h0}}<br /> ^{5}{\rm arctanh} \left( {\frac {{R}^{2}+i\sqrt {{\it h0}+2\,R}\sqrt {<br /> {\it h0}}x}{ \left( {\it h0}+R \right) \sqrt {{R}^{2}-{x}^{2}}}}<br /> \right) +20\,{R}^{3}\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{3/2}\sqrt {{<br /> \it h0}+2\,R}x+34\,{R}^{2}\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{5/2}\sqrt <br /> {{\it h0}+2\,R}x+24\,R\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{7/2}\sqrt {{<br /> \it h0}+2\,R}x+4\,R\sqrt {R+x}\sqrt {R-x}{{\it h0}}^{3/2}\sqrt {{\it <br /> h0}+2\,R}{x}^{3}+6\,{R}^{2}\sqrt {R+x}\sqrt {R-x}\sqrt {{\it h0}}<br /> \sqrt {{\it h0}+2\,R}{x}^{3} \right) \left( -x+i\sqrt {{\it h0}+2\,R}<br /> \sqrt {{\it h0}} \right) ^{-2} \left( x+i\sqrt {{\it h0}+2\,R}\sqrt {{<br /> \it h0}} \right) ^{-2} \left( {\it h0}+2\,R \right) ^{-5/2}{{\it h0}}^<br /> {-5/2}<br />
but this window doesn't show the whole thing in either case.

 

[Mad_MechE]

unfortunately i do need a symbolic answer that i can use in matlab... every time i used maple i couldn't get an output i could copy and paste into MATLAB to use and i have no way to validate the solution from maple... thank you for your help!