Optimal. Leaf size=716 \[ \frac{2 i a^2 f (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac{2 i a^2 f (e+f x) \text{PolyLog}\left (2,i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac{2 a f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}-\frac{2 a f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}+\frac{a f (e+f x) \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{d^2 \left (a^2+b^2\right )}-\frac{2 i a^2 f^2 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{b d^3 \left (a^2+b^2\right )}+\frac{2 i a^2 f^2 \text{PolyLog}\left (3,i e^{c+d x}\right )}{b d^3 \left (a^2+b^2\right )}+\frac{2 a f^2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^3 \left (a^2+b^2\right )}+\frac{2 a f^2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^3 \left (a^2+b^2\right )}-\frac{a f^2 \text{PolyLog}\left (3,-e^{2 (c+d x)}\right )}{2 d^3 \left (a^2+b^2\right )}-\frac{2 i f (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2}+\frac{2 i f (e+f x) \text{PolyLog}\left (2,i e^{c+d x}\right )}{b d^2}+\frac{2 i f^2 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{b d^3}-\frac{2 i f^2 \text{PolyLog}\left (3,i e^{c+d x}\right )}{b d^3}-\frac{a (e+f x)^2 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}-\frac{a (e+f x)^2 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}+\frac{a (e+f x)^2 \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}-\frac{2 a^2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d \left (a^2+b^2\right )}+\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d} \]
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Rubi [A] time = 1.06746, antiderivative size = 716, normalized size of antiderivative = 1., number of steps used = 32, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {5567, 4180, 2531, 2282, 6589, 5573, 5561, 2190, 6742, 3718} \[ \frac{2 i a^2 f (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac{2 i a^2 f (e+f x) \text{PolyLog}\left (2,i e^{c+d x}\right )}{b d^2 \left (a^2+b^2\right )}-\frac{2 a f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}-\frac{2 a f (e+f x) \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}+\frac{a f (e+f x) \text{PolyLog}\left (2,-e^{2 (c+d x)}\right )}{d^2 \left (a^2+b^2\right )}-\frac{2 i a^2 f^2 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{b d^3 \left (a^2+b^2\right )}+\frac{2 i a^2 f^2 \text{PolyLog}\left (3,i e^{c+d x}\right )}{b d^3 \left (a^2+b^2\right )}+\frac{2 a f^2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{d^3 \left (a^2+b^2\right )}+\frac{2 a f^2 \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}\right )}{d^3 \left (a^2+b^2\right )}-\frac{a f^2 \text{PolyLog}\left (3,-e^{2 (c+d x)}\right )}{2 d^3 \left (a^2+b^2\right )}-\frac{2 i f (e+f x) \text{PolyLog}\left (2,-i e^{c+d x}\right )}{b d^2}+\frac{2 i f (e+f x) \text{PolyLog}\left (2,i e^{c+d x}\right )}{b d^2}+\frac{2 i f^2 \text{PolyLog}\left (3,-i e^{c+d x}\right )}{b d^3}-\frac{2 i f^2 \text{PolyLog}\left (3,i e^{c+d x}\right )}{b d^3}-\frac{a (e+f x)^2 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}-\frac{a (e+f x)^2 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}+\frac{a (e+f x)^2 \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}-\frac{2 a^2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d \left (a^2+b^2\right )}+\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d} \]
Antiderivative was successfully verified.
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Rule 5567
Rule 4180
Rule 2531
Rule 2282
Rule 6589
Rule 5573
Rule 5561
Rule 2190
Rule 6742
Rule 3718
Rubi steps
\begin{align*} \int \frac{(e+f x)^2 \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac{\int (e+f x)^2 \text{sech}(c+d x) \, dx}{b}-\frac{a \int \frac{(e+f x)^2 \text{sech}(c+d x)}{a+b \sinh (c+d x)} \, dx}{b}\\ &=\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{a \int (e+f x)^2 \text{sech}(c+d x) (a-b \sinh (c+d x)) \, dx}{b \left (a^2+b^2\right )}-\frac{(a b) \int \frac{(e+f x)^2 \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2+b^2}-\frac{(2 i f) \int (e+f x) \log \left (1-i e^{c+d x}\right ) \, dx}{b d}+\frac{(2 i f) \int (e+f x) \log \left (1+i e^{c+d x}\right ) \, dx}{b d}\\ &=\frac{a (e+f x)^3}{3 \left (a^2+b^2\right ) f}+\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{2 i f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{a \int \left (a (e+f x)^2 \text{sech}(c+d x)-b (e+f x)^2 \tanh (c+d x)\right ) \, dx}{b \left (a^2+b^2\right )}-\frac{(a b) \int \frac{e^{c+d x} (e+f x)^2}{a-\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}-\frac{(a b) \int \frac{e^{c+d x} (e+f x)^2}{a+\sqrt{a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}+\frac{\left (2 i f^2\right ) \int \text{Li}_2\left (-i e^{c+d x}\right ) \, dx}{b d^2}-\frac{\left (2 i f^2\right ) \int \text{Li}_2\left (i e^{c+d x}\right ) \, dx}{b d^2}\\ &=\frac{a (e+f x)^3}{3 \left (a^2+b^2\right ) f}+\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{2 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{2 i f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}+\frac{a \int (e+f x)^2 \tanh (c+d x) \, dx}{a^2+b^2}-\frac{a^2 \int (e+f x)^2 \text{sech}(c+d x) \, dx}{b \left (a^2+b^2\right )}+\frac{(2 a f) \int (e+f x) \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d}+\frac{(2 a f) \int (e+f x) \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d}+\frac{\left (2 i f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{c+d x}\right )}{b d^3}-\frac{\left (2 i f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{c+d x}\right )}{b d^3}\\ &=\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{2 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{2 i f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{2 i f^2 \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{2 i f^2 \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{(2 a) \int \frac{e^{2 (c+d x)} (e+f x)^2}{1+e^{2 (c+d x)}} \, dx}{a^2+b^2}+\frac{\left (2 i a^2 f\right ) \int (e+f x) \log \left (1-i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d}-\frac{\left (2 i a^2 f\right ) \int (e+f x) \log \left (1+i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d}+\frac{\left (2 a f^2\right ) \int \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^2}+\frac{\left (2 a f^2\right ) \int \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right ) d^2}\\ &=\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^2 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{2 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{2 i a^2 f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac{2 i f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{2 i a^2 f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{2 i f^2 \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{2 i f^2 \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}-\frac{(2 a f) \int (e+f x) \log \left (1+e^{2 (c+d x)}\right ) \, dx}{\left (a^2+b^2\right ) d}+\frac{\left (2 a f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{b x}{-a+\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}+\frac{\left (2 a f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{b x}{a+\sqrt{a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^3}-\frac{\left (2 i a^2 f^2\right ) \int \text{Li}_2\left (-i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^2}+\frac{\left (2 i a^2 f^2\right ) \int \text{Li}_2\left (i e^{c+d x}\right ) \, dx}{b \left (a^2+b^2\right ) d^2}\\ &=\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^2 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{2 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{2 i a^2 f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac{2 i f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{2 i a^2 f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{a f (e+f x) \text{Li}_2\left (-e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d^2}+\frac{2 i f^2 \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{2 i f^2 \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{2 a f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac{2 a f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac{\left (2 i a^2 f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac{\left (2 i a^2 f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac{\left (a f^2\right ) \int \text{Li}_2\left (-e^{2 (c+d x)}\right ) \, dx}{\left (a^2+b^2\right ) d^2}\\ &=\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^2 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{2 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{2 i a^2 f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac{2 i f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{2 i a^2 f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{a f (e+f x) \text{Li}_2\left (-e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d^2}+\frac{2 i f^2 \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{2 i a^2 f^2 \text{Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac{2 i f^2 \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{2 i a^2 f^2 \text{Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac{2 a f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac{2 a f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac{\left (a f^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^3}\\ &=\frac{2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b d}-\frac{2 a^2 (e+f x)^2 \tan ^{-1}\left (e^{c+d x}\right )}{b \left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}-\frac{a (e+f x)^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d}+\frac{a (e+f x)^2 \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac{2 i f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b d^2}+\frac{2 i a^2 f (e+f x) \text{Li}_2\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}+\frac{2 i f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b d^2}-\frac{2 i a^2 f (e+f x) \text{Li}_2\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}-\frac{2 a f (e+f x) \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^2}+\frac{a f (e+f x) \text{Li}_2\left (-e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d^2}+\frac{2 i f^2 \text{Li}_3\left (-i e^{c+d x}\right )}{b d^3}-\frac{2 i a^2 f^2 \text{Li}_3\left (-i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}-\frac{2 i f^2 \text{Li}_3\left (i e^{c+d x}\right )}{b d^3}+\frac{2 i a^2 f^2 \text{Li}_3\left (i e^{c+d x}\right )}{b \left (a^2+b^2\right ) d^3}+\frac{2 a f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}+\frac{2 a f^2 \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2+b^2}}\right )}{\left (a^2+b^2\right ) d^3}-\frac{a f^2 \text{Li}_3\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^3}\\ \end{align*}
Mathematica [B] time = 19.14, size = 1640, normalized size = 2.29 \[ \text{result too large to display} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.285, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{2}\tanh \left ( dx+c \right ) }{a+b\sinh \left ( dx+c \right ) }}\, 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*} -e^{2}{\left (\frac{2 \, b \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac{a \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac{a \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d}\right )} + \int \frac{2 \, f^{2} x^{2}{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left (b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} + \frac{4 \, e f x{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}}{{\left (b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.70121, size = 2743, normalized size = 3.83 \begin{align*} \text{result too large to display} \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 (e + f x\right )^{2} \tanh{\left (c + d x \right )}}{a + b \sinh{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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