Optimal. Leaf size=760 \[ \frac {a^2 b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )^2}+\frac {a^2 b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )^2}-\frac {a^2 b f \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )^2}-\frac {a^2 f \tanh (c+d x)}{2 b d^2 \left (a^2+b^2\right )}+\frac {a^2 b (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )^2}+\frac {a^2 b (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )^2}-\frac {a^2 b (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )^2}+\frac {a^2 (e+f x) \text {sech}^2(c+d x)}{2 b d \left (a^2+b^2\right )}-\frac {i a^3 f \text {Li}_2\left (-i e^{c+d x}\right )}{2 b^2 d^2 \left (a^2+b^2\right )}-\frac {i a^3 f \text {Li}_2\left (-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )^2}+\frac {i a^3 f \text {Li}_2\left (i e^{c+d x}\right )}{2 b^2 d^2 \left (a^2+b^2\right )}+\frac {i a^3 f \text {Li}_2\left (i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )^2}+\frac {a^3 f \text {sech}(c+d x)}{2 b^2 d^2 \left (a^2+b^2\right )}+\frac {a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d \left (a^2+b^2\right )}+\frac {2 a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )^2}+\frac {a^3 (e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 b^2 d \left (a^2+b^2\right )}+\frac {i a f \text {Li}_2\left (-i e^{c+d x}\right )}{2 b^2 d^2}-\frac {i a f \text {Li}_2\left (i e^{c+d x}\right )}{2 b^2 d^2}-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}-\frac {a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d}-\frac {a (e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 b^2 d}+\frac {f \tanh (c+d x)}{2 b d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d} \]
[Out]
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Rubi [A] time = 1.15, antiderivative size = 760, normalized size of antiderivative = 1.00, number of steps used = 42, number of rules used = 13, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.406, Rules used = {5583, 5451, 3767, 8, 4185, 4180, 2279, 2391, 5573, 5561, 2190, 6742, 3718} \[ -\frac {i a^3 f \text {PolyLog}\left (2,-i e^{c+d x}\right )}{2 b^2 d^2 \left (a^2+b^2\right )}-\frac {i a^3 f \text {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )^2}+\frac {i a^3 f \text {PolyLog}\left (2,i e^{c+d x}\right )}{2 b^2 d^2 \left (a^2+b^2\right )}+\frac {i a^3 f \text {PolyLog}\left (2,i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )^2}+\frac {a^2 b f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )^2}+\frac {a^2 b f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )^2}-\frac {a^2 b f \text {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )^2}+\frac {i a f \text {PolyLog}\left (2,-i e^{c+d x}\right )}{2 b^2 d^2}-\frac {i a f \text {PolyLog}\left (2,i e^{c+d x}\right )}{2 b^2 d^2}-\frac {a^2 f \tanh (c+d x)}{2 b d^2 \left (a^2+b^2\right )}+\frac {a^3 f \text {sech}(c+d x)}{2 b^2 d^2 \left (a^2+b^2\right )}+\frac {a^2 b (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )^2}+\frac {a^2 b (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )^2}-\frac {a^2 b (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )^2}+\frac {a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d \left (a^2+b^2\right )}+\frac {2 a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )^2}+\frac {a^2 (e+f x) \text {sech}^2(c+d x)}{2 b d \left (a^2+b^2\right )}+\frac {a^3 (e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 b^2 d \left (a^2+b^2\right )}-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}-\frac {a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d}-\frac {a (e+f x) \tanh (c+d x) \text {sech}(c+d x)}{2 b^2 d}+\frac {f \tanh (c+d x)}{2 b d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 8
Rule 2190
Rule 2279
Rule 2391
Rule 3718
Rule 3767
Rule 4180
Rule 4185
Rule 5451
Rule 5561
Rule 5573
Rule 5583
Rule 6742
Rubi steps
\begin {align*} \int \frac {(e+f x) \text {sech}(c+d x) \tanh ^2(c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac {\int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x) \, dx}{b}-\frac {a \int \frac {(e+f x) \text {sech}^2(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx}{b}\\ &=-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d}-\frac {a \int (e+f x) \text {sech}^3(c+d x) \, dx}{b^2}+\frac {a^2 \int \frac {(e+f x) \text {sech}^3(c+d x)}{a+b \sinh (c+d x)} \, dx}{b^2}+\frac {f \int \text {sech}^2(c+d x) \, dx}{2 b d}\\ &=-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d}-\frac {a (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 d}-\frac {a \int (e+f x) \text {sech}(c+d x) \, dx}{2 b^2}+\frac {a^2 \int \frac {(e+f x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2+b^2}+\frac {a^2 \int (e+f x) \text {sech}^3(c+d x) (a-b \sinh (c+d x)) \, dx}{b^2 \left (a^2+b^2\right )}+\frac {(i f) \operatorname {Subst}(\int 1 \, dx,x,-i \tanh (c+d x))}{2 b d^2}\\ &=-\frac {a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d}-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d}+\frac {f \tanh (c+d x)}{2 b d^2}-\frac {a (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 d}+\frac {a^2 \int (e+f x) \text {sech}(c+d x) (a-b \sinh (c+d x)) \, dx}{\left (a^2+b^2\right )^2}+\frac {\left (a^2 b^2\right ) \int \frac {(e+f x) \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{\left (a^2+b^2\right )^2}+\frac {a^2 \int \left (a (e+f x) \text {sech}^3(c+d x)-b (e+f x) \text {sech}^2(c+d x) \tanh (c+d x)\right ) \, dx}{b^2 \left (a^2+b^2\right )}+\frac {(i a f) \int \log \left (1-i e^{c+d x}\right ) \, dx}{2 b^2 d}-\frac {(i a f) \int \log \left (1+i e^{c+d x}\right ) \, dx}{2 b^2 d}\\ &=-\frac {a^2 b (e+f x)^2}{2 \left (a^2+b^2\right )^2 f}-\frac {a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d}-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d}+\frac {f \tanh (c+d x)}{2 b d^2}-\frac {a (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 d}+\frac {a^2 \int (a (e+f x) \text {sech}(c+d x)-b (e+f x) \tanh (c+d x)) \, dx}{\left (a^2+b^2\right )^2}+\frac {\left (a^2 b^2\right ) \int \frac {e^{c+d x} (e+f x)}{a-\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{\left (a^2+b^2\right )^2}+\frac {\left (a^2 b^2\right ) \int \frac {e^{c+d x} (e+f x)}{a+\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{\left (a^2+b^2\right )^2}+\frac {a^3 \int (e+f x) \text {sech}^3(c+d x) \, dx}{b^2 \left (a^2+b^2\right )}-\frac {a^2 \int (e+f x) \text {sech}^2(c+d x) \tanh (c+d x) \, dx}{b \left (a^2+b^2\right )}+\frac {(i a f) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{c+d x}\right )}{2 b^2 d^2}-\frac {(i a f) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{c+d x}\right )}{2 b^2 d^2}\\ &=-\frac {a^2 b (e+f x)^2}{2 \left (a^2+b^2\right )^2 f}-\frac {a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}+\frac {i a f \text {Li}_2\left (-i e^{c+d x}\right )}{2 b^2 d^2}-\frac {i a f \text {Li}_2\left (i e^{c+d x}\right )}{2 b^2 d^2}-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}+\frac {a^3 f \text {sech}(c+d x)}{2 b^2 \left (a^2+b^2\right ) d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d}+\frac {a^2 (e+f x) \text {sech}^2(c+d x)}{2 b \left (a^2+b^2\right ) d}+\frac {f \tanh (c+d x)}{2 b d^2}-\frac {a (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 d}+\frac {a^3 (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 \left (a^2+b^2\right ) d}+\frac {a^3 \int (e+f x) \text {sech}(c+d x) \, dx}{\left (a^2+b^2\right )^2}-\frac {\left (a^2 b\right ) \int (e+f x) \tanh (c+d x) \, dx}{\left (a^2+b^2\right )^2}+\frac {a^3 \int (e+f x) \text {sech}(c+d x) \, dx}{2 b^2 \left (a^2+b^2\right )}-\frac {\left (a^2 b f\right ) \int \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right )^2 d}-\frac {\left (a^2 b f\right ) \int \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right ) \, dx}{\left (a^2+b^2\right )^2 d}-\frac {\left (a^2 f\right ) \int \text {sech}^2(c+d x) \, dx}{2 b \left (a^2+b^2\right ) d}\\ &=-\frac {a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d}+\frac {2 a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d}+\frac {a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 \left (a^2+b^2\right ) d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}+\frac {i a f \text {Li}_2\left (-i e^{c+d x}\right )}{2 b^2 d^2}-\frac {i a f \text {Li}_2\left (i e^{c+d x}\right )}{2 b^2 d^2}-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}+\frac {a^3 f \text {sech}(c+d x)}{2 b^2 \left (a^2+b^2\right ) d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d}+\frac {a^2 (e+f x) \text {sech}^2(c+d x)}{2 b \left (a^2+b^2\right ) d}+\frac {f \tanh (c+d x)}{2 b d^2}-\frac {a (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 d}+\frac {a^3 (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 \left (a^2+b^2\right ) d}-\frac {\left (2 a^2 b\right ) \int \frac {e^{2 (c+d x)} (e+f x)}{1+e^{2 (c+d x)}} \, dx}{\left (a^2+b^2\right )^2}-\frac {\left (a^2 b f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a-\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {\left (a^2 b f\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {b x}{a+\sqrt {a^2+b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {\left (i a^2 f\right ) \operatorname {Subst}(\int 1 \, dx,x,-i \tanh (c+d x))}{2 b \left (a^2+b^2\right ) d^2}-\frac {\left (i a^3 f\right ) \int \log \left (1-i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right )^2 d}+\frac {\left (i a^3 f\right ) \int \log \left (1+i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right )^2 d}-\frac {\left (i a^3 f\right ) \int \log \left (1-i e^{c+d x}\right ) \, dx}{2 b^2 \left (a^2+b^2\right ) d}+\frac {\left (i a^3 f\right ) \int \log \left (1+i e^{c+d x}\right ) \, dx}{2 b^2 \left (a^2+b^2\right ) d}\\ &=-\frac {a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d}+\frac {2 a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d}+\frac {a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 \left (a^2+b^2\right ) d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}-\frac {a^2 b (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}+\frac {i a f \text {Li}_2\left (-i e^{c+d x}\right )}{2 b^2 d^2}-\frac {i a f \text {Li}_2\left (i e^{c+d x}\right )}{2 b^2 d^2}+\frac {a^2 b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {a^2 b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}+\frac {a^3 f \text {sech}(c+d x)}{2 b^2 \left (a^2+b^2\right ) d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d}+\frac {a^2 (e+f x) \text {sech}^2(c+d x)}{2 b \left (a^2+b^2\right ) d}+\frac {f \tanh (c+d x)}{2 b d^2}-\frac {a^2 f \tanh (c+d x)}{2 b \left (a^2+b^2\right ) d^2}-\frac {a (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 d}+\frac {a^3 (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 \left (a^2+b^2\right ) d}-\frac {\left (i a^3 f\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {\left (i a^3 f\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {\left (i a^3 f\right ) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{c+d x}\right )}{2 b^2 \left (a^2+b^2\right ) d^2}+\frac {\left (i a^3 f\right ) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{c+d x}\right )}{2 b^2 \left (a^2+b^2\right ) d^2}+\frac {\left (a^2 b f\right ) \int \log \left (1+e^{2 (c+d x)}\right ) \, dx}{\left (a^2+b^2\right )^2 d}\\ &=-\frac {a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d}+\frac {2 a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d}+\frac {a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 \left (a^2+b^2\right ) d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}-\frac {a^2 b (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}+\frac {i a f \text {Li}_2\left (-i e^{c+d x}\right )}{2 b^2 d^2}-\frac {i a^3 f \text {Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {i a^3 f \text {Li}_2\left (-i e^{c+d x}\right )}{2 b^2 \left (a^2+b^2\right ) d^2}-\frac {i a f \text {Li}_2\left (i e^{c+d x}\right )}{2 b^2 d^2}+\frac {i a^3 f \text {Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {i a^3 f \text {Li}_2\left (i e^{c+d x}\right )}{2 b^2 \left (a^2+b^2\right ) d^2}+\frac {a^2 b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {a^2 b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}+\frac {a^3 f \text {sech}(c+d x)}{2 b^2 \left (a^2+b^2\right ) d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d}+\frac {a^2 (e+f x) \text {sech}^2(c+d x)}{2 b \left (a^2+b^2\right ) d}+\frac {f \tanh (c+d x)}{2 b d^2}-\frac {a^2 f \tanh (c+d x)}{2 b \left (a^2+b^2\right ) d^2}-\frac {a (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 d}+\frac {a^3 (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 \left (a^2+b^2\right ) d}+\frac {\left (a^2 b f\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d^2}\\ &=-\frac {a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 d}+\frac {2 a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d}+\frac {a^3 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{b^2 \left (a^2+b^2\right ) d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}+\frac {a^2 b (e+f x) \log \left (1+\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d}-\frac {a^2 b (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right )^2 d}+\frac {i a f \text {Li}_2\left (-i e^{c+d x}\right )}{2 b^2 d^2}-\frac {i a^3 f \text {Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {i a^3 f \text {Li}_2\left (-i e^{c+d x}\right )}{2 b^2 \left (a^2+b^2\right ) d^2}-\frac {i a f \text {Li}_2\left (i e^{c+d x}\right )}{2 b^2 d^2}+\frac {i a^3 f \text {Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {i a^3 f \text {Li}_2\left (i e^{c+d x}\right )}{2 b^2 \left (a^2+b^2\right ) d^2}+\frac {a^2 b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {a^2 b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^2 d^2}-\frac {a^2 b f \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right )^2 d^2}-\frac {a f \text {sech}(c+d x)}{2 b^2 d^2}+\frac {a^3 f \text {sech}(c+d x)}{2 b^2 \left (a^2+b^2\right ) d^2}-\frac {(e+f x) \text {sech}^2(c+d x)}{2 b d}+\frac {a^2 (e+f x) \text {sech}^2(c+d x)}{2 b \left (a^2+b^2\right ) d}+\frac {f \tanh (c+d x)}{2 b d^2}-\frac {a^2 f \tanh (c+d x)}{2 b \left (a^2+b^2\right ) d^2}-\frac {a (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 d}+\frac {a^3 (e+f x) \text {sech}(c+d x) \tanh (c+d x)}{2 b^2 \left (a^2+b^2\right ) d}\\ \end {align*}
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Mathematica [A] time = 8.06, size = 588, normalized size = 0.77 \[ \frac {a \left (-i f \left (a^2-b^2\right ) \text {Li}_2\left (-i e^{c+d x}\right )+i f \left (a^2-b^2\right ) \text {Li}_2\left (i e^{c+d x}\right )+2 a^2 d e \tan ^{-1}\left (e^{c+d x}\right )+i a^2 f (c+d x) \log \left (1-i e^{c+d x}\right )-i a^2 f (c+d x) \log \left (1+i e^{c+d x}\right )-2 a^2 c f \tan ^{-1}\left (e^{c+d x}\right )+2 a b d e (c+d x)-2 a b d e \log \left (e^{2 (c+d x)}+1\right )-a b f \text {Li}_2\left (-e^{2 (c+d x)}\right )+a b f (c+d x)^2-2 a b c f (c+d x)+2 a b c f \log \left (e^{2 (c+d x)}+1\right )-2 a b f (c+d x) \log \left (e^{2 (c+d x)}+1\right )-2 b^2 d e \tan ^{-1}\left (e^{c+d x}\right )-i b^2 f (c+d x) \log \left (1-i e^{c+d x}\right )+i b^2 f (c+d x) \log \left (1+i e^{c+d x}\right )+2 b^2 c f \tan ^{-1}\left (e^{c+d x}\right )\right )+2 a^2 b \left (f \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+f (c+d x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+f (c+d x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )+d e \log (a+b \sinh (c+d x))-c f \log (a+b \sinh (c+d x))-\frac {1}{2} f (c+d x)^2\right )-d \left (a^2+b^2\right ) (e+f x) \text {sech}^2(c+d x) (a \sinh (c+d x)+b)+f \left (a^2+b^2\right ) \text {sech}(c+d x) (b \sinh (c+d x)-a)}{2 d^2 \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.93, size = 4873, normalized size = 6.41 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.27, size = 2068, normalized size = 2.72 \[ \text {result too large to display} \]
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 \[ {\left (\frac {a^{2} b \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {a^{2} b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {{\left (a^{3} - a b^{2}\right )} \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {a e^{\left (-d x - c\right )} + 2 \, b e^{\left (-2 \, d x - 2 \, c\right )} - a e^{\left (-3 \, d x - 3 \, c\right )}}{{\left (a^{2} + b^{2} + 2 \, {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d}\right )} e - f {\left (\frac {{\left (a d x e^{\left (3 \, c\right )} + a e^{\left (3 \, c\right )}\right )} e^{\left (3 \, d x\right )} + {\left (2 \, b d x e^{\left (2 \, c\right )} + b e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} - {\left (a d x e^{c} - a e^{c}\right )} e^{\left (d x\right )} + b}{a^{2} d^{2} + b^{2} d^{2} + {\left (a^{2} d^{2} e^{\left (4 \, c\right )} + b^{2} d^{2} e^{\left (4 \, c\right )}\right )} e^{\left (4 \, d x\right )} + 2 \, {\left (a^{2} d^{2} e^{\left (2 \, c\right )} + b^{2} d^{2} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}} + 2 \, \int -\frac {a^{3} b x e^{\left (d x + c\right )} - a^{2} b^{2} x}{a^{4} b + 2 \, a^{2} b^{3} + b^{5} - {\left (a^{4} b e^{\left (2 \, c\right )} + 2 \, a^{2} b^{3} e^{\left (2 \, c\right )} + b^{5} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )} - 2 \, {\left (a^{5} e^{c} + 2 \, a^{3} b^{2} e^{c} + a b^{4} e^{c}\right )} e^{\left (d x\right )}}\,{d x} - 2 \, \int \frac {2 \, a^{2} b x + {\left (a^{3} e^{c} - a b^{2} e^{c}\right )} x e^{\left (d x\right )}}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + {\left (a^{4} e^{\left (2 \, c\right )} + 2 \, a^{2} b^{2} e^{\left (2 \, c\right )} + b^{4} e^{\left (2 \, c\right )}\right )} e^{\left (2 \, d x\right )}\right )}}\,{d x}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {tanh}\left (c+d\,x\right )}^2\,\left (e+f\,x\right )}{\mathrm {cosh}\left (c+d\,x\right )\,\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )} \,d x \]
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 \[ \int \frac {\left (e + f x\right ) \tanh ^{2}{\left (c + d x \right )} \operatorname {sech}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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