
Yeah,it's a big black bug there...
Mathematica latest version won't do this
\int \frac{x+1}{e^{x}+2e^{2x}+3e^{3x}} \ dx
A 10 y.o. Maple returns proudly
\int \frac{x+1}{e^x+2e^{2x}+3e^{3x}}dx=\allowbreak -\frac {1}{e^x}\ln \left( e^x\right) -\frac {2}{e^x}-\ln ^2\left( e^x\right)+\allowbreak i\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \arctan \frac {1}{2}\sqrt{2}
+\frac {1}{8}i\sqrt{2}\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \ln \left( 1+2e^x+3e^{2x}\right) -\allowbreak \frac {1}{4}\sqrt{2}\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \arctan \left( \frac {3}{2}e^x\sqrt{2}+\frac {1}2\sqrt{2}\right)
+\frac {1}{4}\sqrt{2}\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \arctan \frac {1}{2}\sqrt{2}+\allowbreak \frac {1}{2}\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \ln \left( 1+2e^x+3e^{2x}\right)
-\allowbreak i\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \arctan \frac {1}{2}\sqrt{2}-\frac {1}{8}i\sqrt{2}\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \ln \left( 1+2e^x+3e^{2x}\right)
-\allowbreak \frac {1}{4}\sqrt{2}\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \arctan \left( \frac {3}{2}e^x\sqrt{2}+\frac {1}{2}\sqrt{2}\right) +\frac {1}{4}\sqrt{2}\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \arctan \frac {1}{2}\sqrt{2}
+\allowbreak \frac {1}{2}\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \ln \left( 1+2e^x+3e^{2x}\right) -i\ln \left( -\frac {1}{3}-\frac {1}{3}i\sqrt{2}\right) \arctan \left( \frac {3}{2}e^x\sqrt{2}+\frac {1}{2}\sqrt{2}\right)
-\frac {1}{4}i\sqrt{2}\ \mbox{dilog}\left( 3\frac{e^x}{-1+i\sqrt{2}}\right) - \ \mbox{dilog}\left( 3\frac{e^x}{-1+i\sqrt{2}}\right) +\allowbreak \frac {1}{4}i\sqrt{2} \ \mbox{dilog}\left( -3\frac{e^x}{1+i\sqrt{2}}\right) - \ \mbox{dilog}\left( -3\frac{e^x}{1+i\sqrt{2}}\right)
-2\ln \left( e^x\right) +\allowbreak \ln \left( 1+2e^x+3e^{2x}\right) -\frac {1}{2}\sqrt{2}\arctan \frac {1}{4}\left( 2+6e^x\right) \sqrt{2} <br />
+i\ln \left( -\frac {1}{3}+\frac {1}{3}i\sqrt{2}\right) \arctan \left( \frac {3}{2}e^x\sqrt{2}+\frac {1}{2}\sqrt{2}\right)+ C'
Daniel.