Optimal. Leaf size=157 \[ \frac {i \text {Li}_2\left (1-\frac {2}{i a x+1}\right )}{a^6 c^2}+\frac {i \tan ^{-1}(a x)^2}{a^6 c^2}+\frac {3 \tan ^{-1}(a x)}{4 a^6 c^2}+\frac {2 \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^6 c^2}-\frac {x}{2 a^5 c^2}+\frac {x^2 \tan ^{-1}(a x)}{2 a^4 c^2}-\frac {\tan ^{-1}(a x)}{2 a^6 c^2 \left (a^2 x^2+1\right )}+\frac {x}{4 a^5 c^2 \left (a^2 x^2+1\right )} \]
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Rubi [A] time = 0.36, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 12, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4964, 4916, 4852, 321, 203, 4920, 4854, 2402, 2315, 4930, 199, 205} \[ \frac {i \text {PolyLog}\left (2,1-\frac {2}{1+i a x}\right )}{a^6 c^2}+\frac {x}{4 a^5 c^2 \left (a^2 x^2+1\right )}+\frac {x^2 \tan ^{-1}(a x)}{2 a^4 c^2}-\frac {\tan ^{-1}(a x)}{2 a^6 c^2 \left (a^2 x^2+1\right )}-\frac {x}{2 a^5 c^2}+\frac {i \tan ^{-1}(a x)^2}{a^6 c^2}+\frac {3 \tan ^{-1}(a x)}{4 a^6 c^2}+\frac {2 \log \left (\frac {2}{1+i a x}\right ) \tan ^{-1}(a x)}{a^6 c^2} \]
Antiderivative was successfully verified.
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Rule 199
Rule 203
Rule 205
Rule 321
Rule 2315
Rule 2402
Rule 4852
Rule 4854
Rule 4916
Rule 4920
Rule 4930
Rule 4964
Rubi steps
\begin {align*} \int \frac {x^5 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx &=-\frac {\int \frac {x^3 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{a^2}+\frac {\int \frac {x^3 \tan ^{-1}(a x)}{c+a^2 c x^2} \, dx}{a^2 c}\\ &=\frac {\int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^2} \, dx}{a^4}+\frac {\int x \tan ^{-1}(a x) \, dx}{a^4 c^2}-2 \frac {\int \frac {x \tan ^{-1}(a x)}{c+a^2 c x^2} \, dx}{a^4 c}\\ &=\frac {x^2 \tan ^{-1}(a x)}{2 a^4 c^2}-\frac {\tan ^{-1}(a x)}{2 a^6 c^2 \left (1+a^2 x^2\right )}+\frac {\int \frac {1}{\left (c+a^2 c x^2\right )^2} \, dx}{2 a^5}-2 \left (-\frac {i \tan ^{-1}(a x)^2}{2 a^6 c^2}-\frac {\int \frac {\tan ^{-1}(a x)}{i-a x} \, dx}{a^5 c^2}\right )-\frac {\int \frac {x^2}{1+a^2 x^2} \, dx}{2 a^3 c^2}\\ &=-\frac {x}{2 a^5 c^2}+\frac {x}{4 a^5 c^2 \left (1+a^2 x^2\right )}+\frac {x^2 \tan ^{-1}(a x)}{2 a^4 c^2}-\frac {\tan ^{-1}(a x)}{2 a^6 c^2 \left (1+a^2 x^2\right )}+\frac {\int \frac {1}{1+a^2 x^2} \, dx}{2 a^5 c^2}-2 \left (-\frac {i \tan ^{-1}(a x)^2}{2 a^6 c^2}-\frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^6 c^2}+\frac {\int \frac {\log \left (\frac {2}{1+i a x}\right )}{1+a^2 x^2} \, dx}{a^5 c^2}\right )+\frac {\int \frac {1}{c+a^2 c x^2} \, dx}{4 a^5 c}\\ &=-\frac {x}{2 a^5 c^2}+\frac {x}{4 a^5 c^2 \left (1+a^2 x^2\right )}+\frac {3 \tan ^{-1}(a x)}{4 a^6 c^2}+\frac {x^2 \tan ^{-1}(a x)}{2 a^4 c^2}-\frac {\tan ^{-1}(a x)}{2 a^6 c^2 \left (1+a^2 x^2\right )}-2 \left (-\frac {i \tan ^{-1}(a x)^2}{2 a^6 c^2}-\frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^6 c^2}-\frac {i \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i a x}\right )}{a^6 c^2}\right )\\ &=-\frac {x}{2 a^5 c^2}+\frac {x}{4 a^5 c^2 \left (1+a^2 x^2\right )}+\frac {3 \tan ^{-1}(a x)}{4 a^6 c^2}+\frac {x^2 \tan ^{-1}(a x)}{2 a^4 c^2}-\frac {\tan ^{-1}(a x)}{2 a^6 c^2 \left (1+a^2 x^2\right )}-2 \left (-\frac {i \tan ^{-1}(a x)^2}{2 a^6 c^2}-\frac {\tan ^{-1}(a x) \log \left (\frac {2}{1+i a x}\right )}{a^6 c^2}-\frac {i \text {Li}_2\left (1-\frac {2}{1+i a x}\right )}{2 a^6 c^2}\right )\\ \end {align*}
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Mathematica [A] time = 0.24, size = 90, normalized size = 0.57 \[ \frac {2 \tan ^{-1}(a x) \left (2 a^2 x^2+8 \log \left (1+e^{2 i \tan ^{-1}(a x)}\right )-\cos \left (2 \tan ^{-1}(a x)\right )+2\right )-8 i \text {Li}_2\left (-e^{2 i \tan ^{-1}(a x)}\right )-4 a x-8 i \tan ^{-1}(a x)^2+\sin \left (2 \tan ^{-1}(a x)\right )}{8 a^6 c^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{5} \arctan \left (a x\right )}{a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 281, normalized size = 1.79 \[ \frac {x^{2} \arctan \left (a x \right )}{2 a^{4} c^{2}}-\frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{a^{6} c^{2}}-\frac {\arctan \left (a x \right )}{2 a^{6} c^{2} \left (a^{2} x^{2}+1\right )}-\frac {x}{2 a^{5} c^{2}}+\frac {x}{4 a^{5} c^{2} \left (a^{2} x^{2}+1\right )}+\frac {3 \arctan \left (a x \right )}{4 a^{6} c^{2}}-\frac {i \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{2 a^{6} c^{2}}+\frac {i \ln \left (a x -i\right )^{2}}{4 a^{6} c^{2}}+\frac {i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{2 a^{6} c^{2}}+\frac {i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{2 a^{6} c^{2}}+\frac {i \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{2 a^{6} c^{2}}-\frac {i \ln \left (a x +i\right )^{2}}{4 a^{6} c^{2}}-\frac {i \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{2 a^{6} c^{2}}-\frac {i \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{2 a^{6} c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^5\,\mathrm {atan}\left (a\,x\right )}{{\left (c\,a^2\,x^2+c\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {x^{5} \operatorname {atan}{\left (a x \right )}}{a^{4} x^{4} + 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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