Optimal. Leaf size=87 \[ -\frac{1}{2} b^2 c \text{PolyLog}\left (2,\frac{2}{c x^2+1}-1\right )+\frac{1}{2} c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{2 x^2}+b c \log \left (2-\frac{2}{c x^2+1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \]
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Rubi [B] time = 0.63159, antiderivative size = 237, normalized size of antiderivative = 2.72, number of steps used = 24, number of rules used = 13, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.812, Rules used = {6099, 2454, 2397, 2392, 2391, 2395, 36, 29, 31, 2439, 2416, 2394, 2393} \[ -\frac{1}{2} b^2 c \text{PolyLog}\left (2,-c x^2\right )+\frac{1}{2} b^2 c \text{PolyLog}\left (2,c x^2\right )+\frac{1}{4} b^2 c \text{PolyLog}\left (2,\frac{1}{2} \left (1-c x^2\right )\right )-\frac{1}{4} b^2 c \text{PolyLog}\left (2,\frac{1}{2} \left (c x^2+1\right )\right )-\frac{1}{4} b c \log \left (\frac{1}{2} \left (c x^2+1\right )\right ) \left (2 a-b \log \left (1-c x^2\right )\right )-\frac{b \log \left (c x^2+1\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{4 x^2}-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}+2 a b c \log (x)-\frac{b^2 \left (c x^2+1\right ) \log ^2\left (c x^2+1\right )}{8 x^2}-\frac{1}{4} b^2 c \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (c x^2+1\right ) \]
Warning: Unable to verify antiderivative.
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Rule 6099
Rule 2454
Rule 2397
Rule 2392
Rule 2391
Rule 2395
Rule 36
Rule 29
Rule 31
Rule 2439
Rule 2416
Rule 2394
Rule 2393
Rubi steps
\begin{align*} \int \frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{x^3} \, dx &=\int \left (\frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{4 x^3}-\frac{b \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{2 x^3}+\frac{b^2 \log ^2\left (1+c x^2\right )}{4 x^3}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{x^3} \, dx-\frac{1}{2} b \int \frac{\left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{x^3} \, dx+\frac{1}{4} b^2 \int \frac{\log ^2\left (1+c x^2\right )}{x^3} \, dx\\ &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{(2 a-b \log (1-c x))^2}{x^2} \, dx,x,x^2\right )-\frac{1}{4} b \operatorname{Subst}\left (\int \frac{(-2 a+b \log (1-c x)) \log (1+c x)}{x^2} \, dx,x,x^2\right )+\frac{1}{8} b^2 \operatorname{Subst}\left (\int \frac{\log ^2(1+c x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac{b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}+\frac{1}{4} (b c) \operatorname{Subst}\left (\int \frac{2 a-b \log (1-c x)}{x} \, dx,x,x^2\right )-\frac{1}{4} (b c) \operatorname{Subst}\left (\int \frac{-2 a+b \log (1-c x)}{x (1+c x)} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+c x)}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+c x)}{x (1-c x)} \, dx,x,x^2\right )\\ &=a b c \log (x)-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac{b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac{1}{4} b^2 c \text{Li}_2\left (-c x^2\right )-\frac{1}{4} (b c) \operatorname{Subst}\left (\int \left (\frac{-2 a+b \log (1-c x)}{x}-\frac{c (-2 a+b \log (1-c x))}{1+c x}\right ) \, dx,x,x^2\right )-\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{\log (1+c x)}{x}-\frac{c \log (1+c x)}{-1+c x}\right ) \, dx,x,x^2\right )\\ &=a b c \log (x)-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac{b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac{1}{4} b^2 c \text{Li}_2\left (-c x^2\right )+\frac{1}{4} b^2 c \text{Li}_2\left (c x^2\right )-\frac{1}{4} (b c) \operatorname{Subst}\left (\int \frac{-2 a+b \log (1-c x)}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+c x)}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (b c^2\right ) \operatorname{Subst}\left (\int \frac{-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^2\right )-\frac{1}{4} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+c x)}{-1+c x} \, dx,x,x^2\right )\\ &=2 a b c \log (x)-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac{1}{4} b c \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+c x^2\right )\right )-\frac{1}{4} b^2 c \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac{b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac{1}{2} b^2 c \text{Li}_2\left (-c x^2\right )+\frac{1}{4} b^2 c \text{Li}_2\left (c x^2\right )-\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^2\right )+\frac{1}{4} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )\\ &=2 a b c \log (x)-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac{1}{4} b c \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+c x^2\right )\right )-\frac{1}{4} b^2 c \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac{b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac{1}{2} b^2 c \text{Li}_2\left (-c x^2\right )+\frac{1}{2} b^2 c \text{Li}_2\left (c x^2\right )-\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1-c x^2\right )+\frac{1}{4} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1+c x^2\right )\\ &=2 a b c \log (x)-\frac{\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac{1}{4} b c \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+c x^2\right )\right )-\frac{1}{4} b^2 c \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac{b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac{1}{2} b^2 c \text{Li}_2\left (-c x^2\right )+\frac{1}{2} b^2 c \text{Li}_2\left (c x^2\right )+\frac{1}{4} b^2 c \text{Li}_2\left (\frac{1}{2} \left (1-c x^2\right )\right )-\frac{1}{4} b^2 c \text{Li}_2\left (\frac{1}{2} \left (1+c x^2\right )\right )\\ \end{align*}
Mathematica [A] time = 0.156982, size = 119, normalized size = 1.37 \[ \frac{1}{2} b^2 c \left (\tanh ^{-1}\left (c x^2\right ) \left (-\frac{\tanh ^{-1}\left (c x^2\right )}{c x^2}+\tanh ^{-1}\left (c x^2\right )+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )-\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )-\frac{a^2}{2 x^2}+a b c \left (-\frac{1}{2} \log \left (1-c^2 x^4\right )+\log \left (c x^2\right )-\frac{\tanh ^{-1}\left (c x^2\right )}{c x^2}\right ) \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 180., size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\it Artanh} \left ( c{x}^{2} \right ) \right ) ^{2}}{{x}^{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \,{\left (c{\left (\log \left (c^{2} x^{4} - 1\right ) - \log \left (x^{4}\right )\right )} + \frac{2 \, \operatorname{artanh}\left (c x^{2}\right )}{x^{2}}\right )} a b - \frac{1}{8} \, b^{2}{\left (\frac{\log \left (-c x^{2} + 1\right )^{2}}{x^{2}} + 2 \, \int -\frac{{\left (c x^{2} - 1\right )} \log \left (c x^{2} + 1\right )^{2} + 2 \,{\left (c x^{2} -{\left (c x^{2} - 1\right )} \log \left (c x^{2} + 1\right )\right )} \log \left (-c x^{2} + 1\right )}{c x^{5} - x^{3}}\,{d x}\right )} - \frac{a^{2}}{2 \, x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\left (c x^{2} \right )}\right )^{2}}{x^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x^{2}\right ) + a\right )}^{2}}{x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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