Optimal. Leaf size=327 \[ -\frac {2 \sqrt {1+a^2 x^2} \text {ArcTan}(a x)^3 \tanh ^{-1}\left (e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \text {ArcTan}(a x)^2 \text {PolyLog}\left (2,-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \text {ArcTan}(a x)^2 \text {PolyLog}\left (2,e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {PolyLog}\left (3,-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {PolyLog}\left (3,e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 i \sqrt {1+a^2 x^2} \text {PolyLog}\left (4,-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \text {PolyLog}\left (4,e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}} \]
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Rubi [A]
time = 0.21, antiderivative size = 327, normalized size of antiderivative = 1.00, number of steps
used = 11, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {5078, 5076,
4268, 2611, 6744, 2320, 6724} \begin {gather*} \frac {3 i \sqrt {a^2 x^2+1} \text {ArcTan}(a x)^2 \text {Li}_2\left (-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {3 i \sqrt {a^2 x^2+1} \text {ArcTan}(a x)^2 \text {Li}_2\left (e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 \sqrt {a^2 x^2+1} \text {ArcTan}(a x) \text {Li}_3\left (-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {6 \sqrt {a^2 x^2+1} \text {ArcTan}(a x) \text {Li}_3\left (e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {6 i \sqrt {a^2 x^2+1} \text {Li}_4\left (-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {6 i \sqrt {a^2 x^2+1} \text {Li}_4\left (e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 \sqrt {a^2 x^2+1} \text {ArcTan}(a x)^3 \tanh ^{-1}\left (e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}} \end {gather*}
Antiderivative was successfully verified.
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Rule 2320
Rule 2611
Rule 4268
Rule 5076
Rule 5078
Rule 6724
Rule 6744
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a x)^3}{x \sqrt {c+a^2 c x^2}} \, dx &=\frac {\sqrt {1+a^2 x^2} \int \frac {\tan ^{-1}(a x)^3}{x \sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=\frac {\sqrt {1+a^2 x^2} \text {Subst}\left (\int x^3 \csc (x) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int x^2 \log \left (1-e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (3 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int x^2 \log \left (1+e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 i \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int x \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 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int x \text {Li}_2\left (e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (6 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \text {Li}_3\left (-e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \text {Li}_3\left (e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (6 i \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {\left (6 i \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {2 \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^3 \tanh ^{-1}\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {3 i \sqrt {1+a^2 x^2} \tan ^{-1}(a x)^2 \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_3\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {6 i \sqrt {1+a^2 x^2} \text {Li}_4\left (-e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {6 i \sqrt {1+a^2 x^2} \text {Li}_4\left (e^{i \tan ^{-1}(a x)}\right )}{\sqrt {c+a^2 c x^2}}\\ \end {align*}
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Mathematica [A]
time = 0.16, size = 208, normalized size = 0.64 \begin {gather*} -\frac {i \sqrt {1+a^2 x^2} \left (\pi ^4-2 \text {ArcTan}(a x)^4+8 i \text {ArcTan}(a x)^3 \log \left (1-e^{-i \text {ArcTan}(a x)}\right )-8 i \text {ArcTan}(a x)^3 \log \left (1+e^{i \text {ArcTan}(a x)}\right )-24 \text {ArcTan}(a x)^2 \text {PolyLog}\left (2,e^{-i \text {ArcTan}(a x)}\right )-24 \text {ArcTan}(a x)^2 \text {PolyLog}\left (2,-e^{i \text {ArcTan}(a x)}\right )+48 i \text {ArcTan}(a x) \text {PolyLog}\left (3,e^{-i \text {ArcTan}(a x)}\right )-48 i \text {ArcTan}(a x) \text {PolyLog}\left (3,-e^{i \text {ArcTan}(a x)}\right )+48 \text {PolyLog}\left (4,e^{-i \text {ArcTan}(a x)}\right )+48 \text {PolyLog}\left (4,-e^{i \text {ArcTan}(a x)}\right )\right )}{8 \sqrt {c \left (1+a^2 x^2\right )}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 2.55, size = 261, normalized size = 0.80
method | result | size |
default | \(-\frac {i \left (i \arctan \left (a x \right )^{3} \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right )^{3} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+3 \arctan \left (a x \right )^{2} \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-3 \arctan \left (a x \right )^{2} \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+6 i \arctan \left (a x \right ) \polylog \left (3, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-6 i \arctan \left (a x \right ) \polylog \left (3, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-6 \polylog \left (4, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+6 \polylog \left (4, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c}\) | \(261\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\operatorname {atan}^{3}{\left (a x \right )}}{x \sqrt {c \left (a^{2} x^{2} + 1\right )}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {{\mathrm {atan}\left (a\,x\right )}^3}{x\,\sqrt {c\,a^2\,x^2+c}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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