Optimal. Leaf size=260 \[ \frac {6 i a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 i a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 a \sqrt {a^2 x^2+1} \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {6 a \sqrt {a^2 x^2+1} \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3}{c x}-\frac {6 a \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}} \]
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Rubi [A] time = 0.37, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {4944, 4958, 4956, 4183, 2531, 2282, 6589} \[ \frac {6 i a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 i a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 a \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {6 a \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^3}{c x}-\frac {6 a \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 c x^2+c}} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 2531
Rule 4183
Rule 4944
Rule 4956
Rule 4958
Rule 6589
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a x)^3}{x^2 \sqrt {c+a^2 c x^2}} \, dx &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c x}+(3 a) \int \frac {\tan ^{-1}(a x)^2}{x \sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c x}+\frac {\left (3 a \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^2}{x \sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c x}+\frac {\left (3 a \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \csc (x) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c x}-\frac {6 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 a \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1-e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (6 a \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1+e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c x}-\frac {6 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 i a \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (6 i a \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c x}-\frac {6 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 a \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (6 a \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^3}{c x}-\frac {6 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 i a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 a \sqrt {1+a^2 x^2} \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 a \sqrt {1+a^2 x^2} \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ \end {align*}
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Mathematica [A] time = 0.34, size = 174, normalized size = 0.67 \[ -\frac {a \sqrt {a^2 x^2+1} \left (\frac {\sqrt {a^2 x^2+1} \tan ^{-1}(a x)^3}{a x}-6 i \tan ^{-1}(a x) \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )+6 i \tan ^{-1}(a x) \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )+6 \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )-6 \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )-3 \tan ^{-1}(a x)^2 \log \left (1-e^{i \tan ^{-1}(a x)}\right )+3 \tan ^{-1}(a x)^2 \log \left (1+e^{i \tan ^{-1}(a x)}\right )\right )}{\sqrt {c \left (a^2 x^2+1\right )}} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{3}}{a^{2} c x^{4} + c x^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.90, size = 230, normalized size = 0.88 \[ -\frac {\arctan \left (a x \right )^{3} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{c x}+\frac {3 a \left (\arctan \left (a x \right )^{2} \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \arctan \left (a x \right )-\arctan \left (a x \right )^{2} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+2 i \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \arctan \left (a x \right )+2 \polylog \left (3, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 \polylog \left (3, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c} \]
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 \[ \int \frac {\arctan \left (a x\right )^{3}}{\sqrt {a^{2} c x^{2} + c} x^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^3}{x^2\,\sqrt {c\,a^2\,x^2+c}} \,d x \]
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 \[ \int \frac {\operatorname {atan}^{3}{\left (a x \right )}}{x^{2} \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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