Optimal. Leaf size=439 \[ \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)^2+\frac {4 i c \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {ArcTan}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 c \sqrt {1+a^2 x^2} \text {ArcTan}(a x)^2 \tanh ^{-1}\left (e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i c \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {PolyLog}\left (2,-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i c \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {PolyLog}\left (2,e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 i c \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i c \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 c \sqrt {1+a^2 x^2} \text {PolyLog}\left (3,-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 c \sqrt {1+a^2 x^2} \text {PolyLog}\left (3,e^{i \text {ArcTan}(a x)}\right )}{\sqrt {c+a^2 c x^2}} \]
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Rubi [A]
time = 0.37, antiderivative size = 439, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5070, 5078,
5076, 4268, 2611, 2320, 6724, 5050, 5010, 5006} \begin {gather*} \frac {2 i c \sqrt {a^2 x^2+1} \text {ArcTan}(a x) \text {Li}_2\left (-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 i c \sqrt {a^2 x^2+1} \text {ArcTan}(a x) \text {Li}_2\left (e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 c \sqrt {a^2 x^2+1} \text {Li}_3\left (-e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\frac {2 c \sqrt {a^2 x^2+1} \text {Li}_3\left (e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}+\text {ArcTan}(a x)^2 \sqrt {a^2 c x^2+c}+\frac {4 i c \sqrt {a^2 x^2+1} \text {ArcTan}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right ) \text {ArcTan}(a x)}{\sqrt {a^2 c x^2+c}}-\frac {2 c \sqrt {a^2 x^2+1} \text {ArcTan}(a x)^2 \tanh ^{-1}\left (e^{i \text {ArcTan}(a x)}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 i c \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}+\frac {2 i c \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i 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 5006
Rule 5010
Rule 5050
Rule 5070
Rule 5076
Rule 5078
Rule 6724
Rubi steps
\begin {align*} \int \frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{x} \, dx &=c \int \frac {\tan ^{-1}(a x)^2}{x \sqrt {c+a^2 c x^2}} \, dx+\left (a^2 c\right ) \int \frac {x \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx\\ &=\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2-(2 a c) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx+\frac {\left (c \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}}\\ &=\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac {\left (c \sqrt {1+a^2 x^2}\right ) \text {Subst}\left (\int x^2 \csc (x) \, dx,x,\tan ^{-1}(a x)\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (2 a c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac {4 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 c \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 {2 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (2 c \sqrt {1+a^2 x^2}\right ) \text {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 (2 c \sqrt {1+a^2 x^2}\right ) \text {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}}\\ &=\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac {4 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 c \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 {2 i c \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 {2 i c \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 {2 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (2 i c \sqrt {1+a^2 x^2}\right ) \text {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 (2 i c \sqrt {1+a^2 x^2}\right ) \text {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}}\\ &=\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac {4 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 c \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 {2 i c \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 {2 i c \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 {2 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {\left (2 c \sqrt {1+a^2 x^2}\right ) \text {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 (2 c \sqrt {1+a^2 x^2}\right ) \text {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}}\\ &=\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac {4 i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 c \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 {2 i c \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 {2 i c \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 {2 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {2 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {2 c \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 {2 c \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 = 0.18, size = 250, normalized size = 0.57 \begin {gather*} \frac {\sqrt {c+a^2 c x^2} \left (\sqrt {1+a^2 x^2} \text {ArcTan}(a x)^2+\text {ArcTan}(a x)^2 \log \left (1-e^{i \text {ArcTan}(a x)}\right )-2 \text {ArcTan}(a x) \log \left (1-i e^{i \text {ArcTan}(a x)}\right )+2 \text {ArcTan}(a x) \log \left (1+i e^{i \text {ArcTan}(a x)}\right )-\text {ArcTan}(a x)^2 \log \left (1+e^{i \text {ArcTan}(a x)}\right )+2 i \text {ArcTan}(a x) \text {PolyLog}\left (2,-e^{i \text {ArcTan}(a x)}\right )-2 i \text {PolyLog}\left (2,-i e^{i \text {ArcTan}(a x)}\right )+2 i \text {PolyLog}\left (2,i e^{i \text {ArcTan}(a x)}\right )-2 i \text {ArcTan}(a x) \text {PolyLog}\left (2,e^{i \text {ArcTan}(a x)}\right )-2 \text {PolyLog}\left (3,-e^{i \text {ArcTan}(a x)}\right )+2 \text {PolyLog}\left (3,e^{i \text {ArcTan}(a x)}\right )\right )}{\sqrt {1+a^2 x^2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.44, size = 337, normalized size = 0.77
method | result | size |
default | \(\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \arctan \left (a x \right )^{2}-\frac {i \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i \arctan \left (a x \right )^{2} \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right )^{2} \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+2 i \arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+2 \arctan \left (a x \right ) \polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 \arctan \left (a x \right ) \polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+2 i \polylog \left (3, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \polylog \left (3, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+2 \dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-2 \dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{\sqrt {a^{2} x^{2}+1}}\) | \(337\) |
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 {\sqrt {c \left (a^{2} x^{2} + 1\right )} \operatorname {atan}^{2}{\left (a x \right )}}{x}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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 )}^2\,\sqrt {c\,a^2\,x^2+c}}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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