How to calculate this type of integral, Thanks

[zhaojx84]

$$\int {z}^{2}\arcsin\left({\frac{a+\sqrt{392-{a}^{2}-2{z}^{2}}}{2 \sqrt{196-{z}^{2}}}}\right) dz$$
$$\int {z}^{2}\arcsin\left({\frac{a}{\sqrt{196-{z}^{2}}}}\right) dz$$

 

[Ackbach]

Hmm. For the first integral, I would sit down and cry. Wolfram Dev Platform doesn't do anything with it. The second integral is doable, but extremely messy. The Wolfram Langauge code

FullSimplify[Integrate[z^2 ArcSin[a/(Sqrt[196-z^2])],z] ]//TeXForm

yields
$$\frac{1}{6}\left(-a\sqrt{196-z^2}z\sqrt{\frac{a^2+z^2-196}{z^2-196}}+2744i
\left(\ln\left(\frac{3i\left(ia\sqrt{196-z^2}
\sqrt{\frac{a^2+z^2-196}{z^2-196}}+a^2-14(z+14)\right)}{686a^2
(z+14)}\right)-\ln\left(\frac{3\left(a\sqrt{196-z^2}
\sqrt{\frac{a^2+z^2-196}{z^2-196}}-ia^2-14i(z-14)\right)}{686a^2
(z-14)}\right)\right)-a\left(a^2-588\right)\tan
^{-1}\left(\frac{z}{\sqrt{196-z^2}\sqrt{\frac{a^2+z^2-196}{z^2-196}}}\right)+2
z^3\sin^{-1}\left(\frac{a}{\sqrt{196-z^2}}\right)\right).$$
I did tinker with this output just a hair (log to ln). This is not a very useable solution, but I think it's the best you're going to get for an indefinite integral.

 

[zhaojx84]

Hmm. For the first integral, I would sit down and cry. Wolfram Dev Platform doesn't do anything with it. The second integral is doable, but extremely messy. The Wolfram Langauge code

FullSimplify[Integrate[z^2 ArcSin[a/(Sqrt[196-z^2])],z] ]//TeXForm

yields
$$\frac{1}{6}\left(-a\sqrt{196-z^2}z\sqrt{\frac{a^2+z^2-196}{z^2-196}}+2744i
\left(\ln\left(\frac{3i\left(ia\sqrt{196-z^2}
\sqrt{\frac{a^2+z^2-196}{z^2-196}}+a^2-14(z+14)\right)}{686a^2
(z+14)}\right)-\ln\left(\frac{3\left(a\sqrt{196-z^2}
\sqrt{\frac{a^2+z^2-196}{z^2-196}}-ia^2-14i(z-14)\right)}{686a^2
(z-14)}\right)\right)-a\left(a^2-588\right)\tan
^{-1}\left(\frac{z}{\sqrt{196-z^2}\sqrt{\frac{a^2+z^2-196}{z^2-196}}}\right)+2
z^3\sin^{-1}\left(\frac{a}{\sqrt{196-z^2}}\right)\right).$$
I did tinker with this output just a hair (log to ln). This is not a very useable solution, but I think it's the best you're going to get for an indefinite integral.

Thank you very much. Regards.