Optimal. Leaf size=168 \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e}+\frac{b \text{PolyLog}\left (2,\frac{2}{-c-d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right )}{2 d e}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{-c-d x+1}-1\right )}{2 d e}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e} \]
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Rubi [A] time = 0.302744, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {6107, 12, 5914, 6052, 5948, 6058, 6610} \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e}+\frac{b \text{PolyLog}\left (2,\frac{2}{-c-d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right )}{2 d e}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{-c-d x+1}-1\right )}{2 d e}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e} \]
Antiderivative was successfully verified.
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Rule 6107
Rule 12
Rule 5914
Rule 6052
Rule 5948
Rule 6058
Rule 6610
Rubi steps
\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{c e+d e x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac{2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c-d x}\right )}{d e}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (1-\frac{2}{1-x}\right ) \left (a+b \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac{2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c-d x}\right )}{d e}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right ) \log \left (2-\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right ) \log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac{2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c-d x}\right )}{d e}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d e}+\frac{b \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c-d x}\right )}{d e}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-1+\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e}\\ &=\frac{2 \left (a+b \tanh ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-c-d x}\right )}{d e}-\frac{b \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d e}+\frac{b \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (-1+\frac{2}{1-c-d x}\right )}{d e}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1-c-d x}\right )}{2 d e}-\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1-c-d x}\right )}{2 d e}\\ \end{align*}
Mathematica [C] time = 0.368635, size = 424, normalized size = 2.52 \[ \frac{-\frac{1}{4} a b \left (4 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c+d x)}\right )+4 \text{PolyLog}\left (2,-e^{2 \tanh ^{-1}(c+d x)}\right )-4 i \pi \log \left (\frac{2}{\sqrt{1-(c+d x)^2}}\right )-8 \tanh ^{-1}(c+d x)^2-4 i \pi \tanh ^{-1}(c+d x)-8 \tanh ^{-1}(c+d x) \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )+8 \tanh ^{-1}(c+d x) \log \left (e^{2 \tanh ^{-1}(c+d x)}+1\right )-8 \log \left (\frac{2}{\sqrt{1-(c+d x)^2}}\right ) \tanh ^{-1}(c+d x)+8 \log \left (\frac{2 i (c+d x)}{\sqrt{1-(c+d x)^2}}\right ) \tanh ^{-1}(c+d x)+4 i \pi \log \left (e^{2 \tanh ^{-1}(c+d x)}+1\right )+\pi ^2\right )+b^2 \left (\tanh ^{-1}(c+d x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )+\tanh ^{-1}(c+d x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c+d x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c+d x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c+d x)}\right )-\frac{2}{3} \tanh ^{-1}(c+d x)^3-\tanh ^{-1}(c+d x)^2 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )+\tanh ^{-1}(c+d x)^2 \log \left (1-e^{2 \tanh ^{-1}(c+d x)}\right )+\frac{i \pi ^3}{24}\right )+a^2 \log (c+d x)+2 a b \left (-\log \left (\frac{1}{\sqrt{1-(c+d x)^2}}\right )+\log \left (\frac{i (c+d x)}{\sqrt{1-(c+d x)^2}}\right )\right ) \tanh ^{-1}(c+d x)}{d e} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.316, size = 893, normalized size = 5.3 \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{a^{2} \log \left (d e x + c e\right )}{d e} + \int \frac{b^{2}{\left (\log \left (d x + c + 1\right ) - \log \left (-d x - c + 1\right )\right )}^{2}}{4 \,{\left (d e x + c e\right )}} + \frac{a b{\left (\log \left (d x + c + 1\right ) - \log \left (-d x - c + 1\right )\right )}}{d e x + c e}\,{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 (d x + c\right )^{2} + 2 \, a b \operatorname{artanh}\left (d x + c\right ) + a^{2}}{d e x + c e}, 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*} \frac{\int \frac{a^{2}}{c + d x}\, dx + \int \frac{b^{2} \operatorname{atanh}^{2}{\left (c + d x \right )}}{c + d x}\, dx + \int \frac{2 a b \operatorname{atanh}{\left (c + d x \right )}}{c + d x}\, dx}{e} \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 (d x + c\right ) + a\right )}^{2}}{d e x + c e}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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