Optimal. Leaf size=262 \[ -\frac {3 i a^2 \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{2 c}-\frac {3 i a^2 \text {Li}_4\left (\frac {2}{1-i a x}-1\right )}{4 c}+\frac {3 i a^2 \text {Li}_2\left (\frac {2}{1-i a x}-1\right ) \tan ^{-1}(a x)^2}{2 c}-\frac {3 a^2 \text {Li}_3\left (\frac {2}{1-i a x}-1\right ) \tan ^{-1}(a x)}{2 c}+\frac {i a^2 \tan ^{-1}(a x)^4}{4 c}-\frac {a^2 \tan ^{-1}(a x)^3}{2 c}-\frac {3 i a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {a^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c}+\frac {3 a^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c}-\frac {\tan ^{-1}(a x)^3}{2 c x^2}-\frac {3 a \tan ^{-1}(a x)^2}{2 c x} \]
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Rubi [A] time = 0.51, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {4918, 4852, 4924, 4868, 2447, 4884, 4992, 4996, 6610} \[ -\frac {3 i a^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {3 i a^2 \text {PolyLog}\left (4,-1+\frac {2}{1-i a x}\right )}{4 c}+\frac {3 i a^2 \tan ^{-1}(a x)^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {3 a^2 \tan ^{-1}(a x) \text {PolyLog}\left (3,-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {i a^2 \tan ^{-1}(a x)^4}{4 c}-\frac {a^2 \tan ^{-1}(a x)^3}{2 c}-\frac {3 i a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {a^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)^3}{c}+\frac {3 a^2 \log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c}-\frac {\tan ^{-1}(a x)^3}{2 c x^2}-\frac {3 a \tan ^{-1}(a x)^2}{2 c x} \]
Antiderivative was successfully verified.
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Rule 2447
Rule 4852
Rule 4868
Rule 4884
Rule 4918
Rule 4924
Rule 4992
Rule 4996
Rule 6610
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a x)^3}{x^3 \left (c+a^2 c x^2\right )} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)^3}{x \left (c+a^2 c x^2\right )} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)^3}{x^3} \, dx}{c}\\ &=-\frac {\tan ^{-1}(a x)^3}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^4}{4 c}+\frac {(3 a) \int \frac {\tan ^{-1}(a x)^2}{x^2 \left (1+a^2 x^2\right )} \, dx}{2 c}-\frac {\left (i a^2\right ) \int \frac {\tan ^{-1}(a x)^3}{x (i+a x)} \, dx}{c}\\ &=-\frac {\tan ^{-1}(a x)^3}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^4}{4 c}-\frac {a^2 \tan ^{-1}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c}+\frac {(3 a) \int \frac {\tan ^{-1}(a x)^2}{x^2} \, dx}{2 c}-\frac {\left (3 a^3\right ) \int \frac {\tan ^{-1}(a x)^2}{1+a^2 x^2} \, dx}{2 c}+\frac {\left (3 a^3\right ) \int \frac {\tan ^{-1}(a x)^2 \log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {3 a \tan ^{-1}(a x)^2}{2 c x}-\frac {a^2 \tan ^{-1}(a x)^3}{2 c}-\frac {\tan ^{-1}(a x)^3}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^4}{4 c}-\frac {a^2 \tan ^{-1}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c}+\frac {3 i a^2 \tan ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {\left (3 a^2\right ) \int \frac {\tan ^{-1}(a x)}{x \left (1+a^2 x^2\right )} \, dx}{c}-\frac {\left (3 i a^3\right ) \int \frac {\tan ^{-1}(a x) \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {3 i a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {3 a \tan ^{-1}(a x)^2}{2 c x}-\frac {a^2 \tan ^{-1}(a x)^3}{2 c}-\frac {\tan ^{-1}(a x)^3}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^4}{4 c}-\frac {a^2 \tan ^{-1}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c}+\frac {3 i a^2 \tan ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {3 a^2 \tan ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {\left (3 i a^2\right ) \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c}+\frac {\left (3 a^3\right ) \int \frac {\text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{2 c}\\ &=-\frac {3 i a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {3 a \tan ^{-1}(a x)^2}{2 c x}-\frac {a^2 \tan ^{-1}(a x)^3}{2 c}-\frac {\tan ^{-1}(a x)^3}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^4}{4 c}+\frac {3 a^2 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {a^2 \tan ^{-1}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c}+\frac {3 i a^2 \tan ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {3 a^2 \tan ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {3 i a^2 \text {Li}_4\left (-1+\frac {2}{1-i a x}\right )}{4 c}-\frac {\left (3 a^3\right ) \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {3 i a^2 \tan ^{-1}(a x)^2}{2 c}-\frac {3 a \tan ^{-1}(a x)^2}{2 c x}-\frac {a^2 \tan ^{-1}(a x)^3}{2 c}-\frac {\tan ^{-1}(a x)^3}{2 c x^2}+\frac {i a^2 \tan ^{-1}(a x)^4}{4 c}+\frac {3 a^2 \tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {a^2 \tan ^{-1}(a x)^3 \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {3 i a^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c}+\frac {3 i a^2 \tan ^{-1}(a x)^2 \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {3 a^2 \tan ^{-1}(a x) \text {Li}_3\left (-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {3 i a^2 \text {Li}_4\left (-1+\frac {2}{1-i a x}\right )}{4 c}\\ \end {align*}
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Mathematica [A] time = 0.41, size = 189, normalized size = 0.72 \[ \frac {i a^2 \left (\frac {32 i \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^3}{a^2 x^2}-96 \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{-2 i \tan ^{-1}(a x)}\right )+96 i \tan ^{-1}(a x) \text {Li}_3\left (e^{-2 i \tan ^{-1}(a x)}\right )-96 \text {Li}_2\left (e^{2 i \tan ^{-1}(a x)}\right )+48 \text {Li}_4\left (e^{-2 i \tan ^{-1}(a x)}\right )-16 \tan ^{-1}(a x)^4+\frac {96 i \tan ^{-1}(a x)^2}{a x}-96 \tan ^{-1}(a x)^2+64 i \tan ^{-1}(a x)^3 \log \left (1-e^{-2 i \tan ^{-1}(a x)}\right )-192 i \tan ^{-1}(a x) \log \left (1-e^{2 i \tan ^{-1}(a x)}\right )+\pi ^4\right )}{64 c} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )^{3}}{a^{2} c x^{5} + c x^{3}}, x\right ) \]
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 [B] time = 11.51, size = 479, normalized size = 1.83 \[ \frac {3 i a^{2} \arctan \left (a x \right )^{2} \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}-\frac {a^{2} \arctan \left (a x \right )^{3}}{2 c}+\frac {3 i a^{2} \arctan \left (a x \right )^{2} \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}-\frac {3 a \arctan \left (a x \right )^{2}}{2 c x}-\frac {\arctan \left (a x \right )^{3}}{2 c \,x^{2}}-\frac {3 i a^{2} \arctan \left (a x \right )^{2}}{2 c}-\frac {a^{2} \arctan \left (a x \right )^{3} \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}-\frac {6 a^{2} \arctan \left (a x \right ) \polylog \left (3, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}+\frac {i a^{2} \arctan \left (a x \right )^{4}}{4 c}-\frac {a^{2} \arctan \left (a x \right )^{3} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}-\frac {6 i a^{2} \polylog \left (4, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}-\frac {6 a^{2} \arctan \left (a x \right ) \polylog \left (3, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}-\frac {3 i a^{2} \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}-\frac {3 i a^{2} \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}+\frac {3 a^{2} \arctan \left (a x \right ) \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}-\frac {6 i a^{2} \polylog \left (4, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c}+\frac {3 a^{2} \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )}{c} \]
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 \[ \int \frac {\arctan \left (a x\right )^{3}}{{\left (a^{2} c x^{2} + c\right )} x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^3}{x^3\,\left (c\,a^2\,x^2+c\right )} \,d x \]
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 \[ \frac {\int \frac {\operatorname {atan}^{3}{\left (a x \right )}}{a^{2} x^{5} + x^{3}}\, dx}{c} \]
Verification of antiderivative is not currently implemented for this CAS.
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