Optimal. Leaf size=148 \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac{b \text{PolyLog}\left (2,\frac{2}{1-c x^n}-1\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x^n}\right )}{2 n}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-c x^n}-1\right )}{2 n}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{n} \]
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Rubi [A] time = 0.313463, antiderivative size = 148, normalized size of antiderivative = 1., 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} \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac{b \text{PolyLog}\left (2,\frac{2}{1-c x^n}-1\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x^n}\right )}{2 n}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-c x^n}-1\right )}{2 n}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{n} \]
Antiderivative was successfully verified.
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Rule 6095
Rule 5914
Rule 6052
Rule 5948
Rule 6058
Rule 6610
Rubi steps
\begin{align*} \int \frac{\left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx,x,x^n\right )}{n}\\ &=\frac{2 \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x^n}\right )}{n}-\frac{(4 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^n\right )}{n}\\ &=\frac{2 \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x^n}\right )}{n}+\frac{(2 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^n\right )}{n}-\frac{(2 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^n\right )}{n}\\ &=\frac{2 \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x^n}\right )}{n}-\frac{b \left (a+b \tanh ^{-1}\left (c x^n\right )\right ) \text{Li}_2\left (1-\frac{2}{1-c x^n}\right )}{n}+\frac{b \left (a+b \tanh ^{-1}\left (c x^n\right )\right ) \text{Li}_2\left (-1+\frac{2}{1-c x^n}\right )}{n}+\frac{\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^n\right )}{n}-\frac{\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^n\right )}{n}\\ &=\frac{2 \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c x^n}\right )}{n}-\frac{b \left (a+b \tanh ^{-1}\left (c x^n\right )\right ) \text{Li}_2\left (1-\frac{2}{1-c x^n}\right )}{n}+\frac{b \left (a+b \tanh ^{-1}\left (c x^n\right )\right ) \text{Li}_2\left (-1+\frac{2}{1-c x^n}\right )}{n}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1-c x^n}\right )}{2 n}-\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1-c x^n}\right )}{2 n}\\ \end{align*}
Mathematica [A] time = 0.128001, size = 146, normalized size = 0.99 \[ \frac{\frac{1}{2} b \left (2 \text{PolyLog}\left (2,\frac{c x^n+1}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )-2 \text{PolyLog}\left (2,\frac{c x^n+1}{c x^n-1}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )+b \left (\text{PolyLog}\left (3,\frac{c x^n+1}{c x^n-1}\right )-\text{PolyLog}\left (3,\frac{c x^n+1}{1-c x^n}\right )\right )\right )+2 \tanh ^{-1}\left (\frac{2}{c x^n-1}+1\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{n} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.33, size = 880, normalized size = 6. \begin{align*} \text{result too large to display} \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}{4} \, b^{2} \log \left (-c x^{n} + 1\right )^{2} \log \left (x\right ) + a^{2} \log \left (x\right ) - \int -\frac{{\left (b^{2} c x^{n} - b^{2}\right )} \log \left (c x^{n} + 1\right )^{2} + 4 \,{\left (a b c x^{n} - a b\right )} \log \left (c x^{n} + 1\right ) + 2 \,{\left (2 \, a b -{\left (b^{2} c n \log \left (x\right ) + 2 \, a b c\right )} x^{n} -{\left (b^{2} c x^{n} - b^{2}\right )} \log \left (c x^{n} + 1\right )\right )} \log \left (-c x^{n} + 1\right )}{4 \,{\left (c x x^{n} - x\right )}}\,{d x} \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^{n}\right )^{2} + 2 \, a b \operatorname{artanh}\left (c x^{n}\right ) + a^{2}}{x}, 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^{n} \right )}\right )^{2}}{x}\, 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^{n}\right ) + a\right )}^{2}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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