Optimal. Leaf size=293 \[ -\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{c^2 x}+\frac {2 i a \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {2 i a \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}}+\frac {2 a^2 x}{c \sqrt {a^2 c x^2+c}}-\frac {a^2 x \tan ^{-1}(a x)^2}{c \sqrt {a^2 c x^2+c}}-\frac {2 a \tan ^{-1}(a x)}{c \sqrt {a^2 c x^2+c}}-\frac {4 a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}} \]
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Rubi [A] time = 0.43, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4966, 4944, 4958, 4954, 4898, 191} \[ \frac {2 i a \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {2 i a \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}}-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{c^2 x}+\frac {2 a^2 x}{c \sqrt {a^2 c x^2+c}}-\frac {a^2 x \tan ^{-1}(a x)^2}{c \sqrt {a^2 c x^2+c}}-\frac {2 a \tan ^{-1}(a x)}{c \sqrt {a^2 c x^2+c}}-\frac {4 a \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {a^2 c x^2+c}} \]
Antiderivative was successfully verified.
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Rule 191
Rule 4898
Rule 4944
Rule 4954
Rule 4958
Rule 4966
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a x)^2}{x^2 \left (c+a^2 c x^2\right )^{3/2}} \, dx &=-\left (a^2 \int \frac {\tan ^{-1}(a x)^2}{\left (c+a^2 c x^2\right )^{3/2}} \, dx\right )+\frac {\int \frac {\tan ^{-1}(a x)^2}{x^2 \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=-\frac {2 a \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{c^2 x}+\left (2 a^2\right ) \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx+\frac {(2 a) \int \frac {\tan ^{-1}(a x)}{x \sqrt {c+a^2 c x^2}} \, dx}{c}\\ &=\frac {2 a^2 x}{c \sqrt {c+a^2 c x^2}}-\frac {2 a \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{c^2 x}+\frac {\left (2 a \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx}{c \sqrt {c+a^2 c x^2}}\\ &=\frac {2 a^2 x}{c \sqrt {c+a^2 c x^2}}-\frac {2 a \tan ^{-1}(a x)}{c \sqrt {c+a^2 c x^2}}-\frac {a^2 x \tan ^{-1}(a x)^2}{c \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{c^2 x}-\frac {4 a \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}}+\frac {2 i a \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}}-\frac {2 i a \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{c \sqrt {c+a^2 c x^2}}\\ \end {align*}
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Mathematica [A] time = 1.12, size = 226, normalized size = 0.77 \[ \frac {a \left (4 i \sqrt {a^2 x^2+1} \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )-4 i \sqrt {a^2 x^2+1} \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )+4 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \log \left (1-e^{i \tan ^{-1}(a x)}\right )-4 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \log \left (1+e^{i \tan ^{-1}(a x)}\right )-\frac {2 \left (a^2 x^2+1\right ) \tan ^{-1}(a x)^2 \sin ^2\left (\frac {1}{2} \tan ^{-1}(a x)\right )}{a x}+4 a x-2 a x \tan ^{-1}(a x)^2-4 \tan ^{-1}(a x)-\frac {1}{2} a x \tan ^{-1}(a x)^2 \csc ^2\left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )}{2 c \sqrt {a^2 c x^2+c}} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{2}}{a^{4} c^{2} x^{6} + 2 \, a^{2} c^{2} x^{4} + c^{2} x^{2}}, x\right ) \]
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 = 0.76, size = 279, normalized size = 0.95 \[ -\frac {a \left (\arctan \left (a x \right )^{2}-2+2 i \arctan \left (a x \right )\right ) \left (a x -i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right ) c^{2}}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (a x +i\right ) \left (\arctan \left (a x \right )^{2}-2-2 i \arctan \left (a x \right )\right ) a}{2 \left (a^{2} x^{2}+1\right ) c^{2}}-\frac {\arctan \left (a x \right )^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{x \,c^{2}}-\frac {2 i a \left (-i \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+i \ln \left (1-\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) \arctan \left (a x \right )+\polylog \left (2, \frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-\polylog \left (2, -\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, c^{2}} \]
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 \[ \int \frac {\arctan \left (a x\right )^{2}}{{\left (a^{2} c x^{2} + c\right )}^{\frac {3}{2}} x^{2}}\,{d x} \]
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 \[ \int \frac {{\mathrm {atan}\left (a\,x\right )}^2}{x^2\,{\left (c\,a^2\,x^2+c\right )}^{3/2}} \,d x \]
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 \[ \int \frac {\operatorname {atan}^{2}{\left (a x \right )}}{x^{2} \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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