Optimal. Leaf size=137 \[ -\frac {1}{2} b \text {Li}_2\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac {1}{2} b \text {Li}_2\left (\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{1-c x^2}\right )-\frac {1}{4} b^2 \text {Li}_3\left (\frac {2}{1-c x^2}-1\right ) \]
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Rubi [A] time = 0.34, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6095, 5914, 6052, 5948, 6058, 6610} \[ -\frac {1}{2} b \text {PolyLog}\left (2,1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac {1}{2} b \text {PolyLog}\left (2,\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac {1}{4} b^2 \text {PolyLog}\left (3,1-\frac {2}{1-c x^2}\right )-\frac {1}{4} b^2 \text {PolyLog}\left (3,\frac {2}{1-c x^2}-1\right )+\tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \]
Antiderivative was successfully verified.
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Rule 5914
Rule 5948
Rule 6052
Rule 6058
Rule 6095
Rule 6610
Rubi steps
\begin {align*} \int \frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-(2 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )+(b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )-(b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {1}{2} \left (b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )-\frac {1}{2} \left (b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{1-c x^2}\right )-\frac {1}{4} b^2 \text {Li}_3\left (-1+\frac {2}{1-c x^2}\right )\\ \end {align*}
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Mathematica [A] time = 0.08, size = 183, normalized size = 1.34 \[ \frac {1}{2} \left (2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-4 b c \left (\frac {1}{2} \left (\frac {\text {Li}_2\left (\frac {-c x^2-1}{c x^2-1}\right ) \left (-a-b \tanh ^{-1}\left (c x^2\right )\right )}{2 c}+\frac {b \text {Li}_3\left (\frac {-c x^2-1}{c x^2-1}\right )}{4 c}\right )+\frac {1}{2} \left (-\frac {\text {Li}_2\left (\frac {c x^2+1}{c x^2-1}\right ) \left (-a-b \tanh ^{-1}\left (c x^2\right )\right )}{2 c}-\frac {b \text {Li}_3\left (\frac {c x^2+1}{c x^2-1}\right )}{4 c}\right )\right )\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (c x^{2}\right )^{2} + 2 \, a b \operatorname {artanh}\left (c x^{2}\right ) + a^{2}}{x}, x\right ) \]
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 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x^{2}\right ) + a\right )}^{2}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.22, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctanh \left (c \,x^{2}\right )\right )^{2}}{x}\, dx \]
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 \[ a^{2} \log \relax (x) + \int \frac {b^{2} {\left (\log \left (c x^{2} + 1\right ) - \log \left (-c x^{2} + 1\right )\right )}^{2}}{4 \, x} + \frac {a b {\left (\log \left (c x^{2} + 1\right ) - \log \left (-c x^{2} + 1\right )\right )}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^2}{x} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c x^{2} \right )}\right )^{2}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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