Optimal. Leaf size=334 \[ -\frac {i a f \text {Li}_2\left (-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac {i a f \text {Li}_2\left (i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac {b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}+\frac {b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}-\frac {b f \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )}+\frac {b (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}+\frac {b (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}-\frac {b (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}+\frac {2 a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )} \]
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Rubi [A] time = 0.60, antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5573, 5561, 2190, 2279, 2391, 6742, 4180, 3718} \[ -\frac {i a f \text {PolyLog}\left (2,-i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac {i a f \text {PolyLog}\left (2,i e^{c+d x}\right )}{d^2 \left (a^2+b^2\right )}+\frac {b f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}\right )}{d^2 \left (a^2+b^2\right )}+\frac {b f \text {PolyLog}\left (2,-\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}\right )}{d^2 \left (a^2+b^2\right )}-\frac {b f \text {PolyLog}\left (2,-e^{2 (c+d x)}\right )}{2 d^2 \left (a^2+b^2\right )}+\frac {b (e+f x) \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )}{d \left (a^2+b^2\right )}+\frac {b (e+f x) \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )}{d \left (a^2+b^2\right )}-\frac {b (e+f x) \log \left (e^{2 (c+d x)}+1\right )}{d \left (a^2+b^2\right )}+\frac {2 a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{d \left (a^2+b^2\right )} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2391
Rule 3718
Rule 4180
Rule 5561
Rule 5573
Rule 6742
Rubi steps
\begin {align*} \int \frac {(e+f x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx &=\frac {\int (e+f x) \text {sech}(c+d x) (a-b \sinh (c+d x)) \, dx}{a^2+b^2}+\frac {b^2 \int \frac {(e+f x) \cosh (c+d x)}{a+b \sinh (c+d x)} \, dx}{a^2+b^2}\\ &=-\frac {b (e+f x)^2}{2 \left (a^2+b^2\right ) f}+\frac {\int (a (e+f x) \text {sech}(c+d x)-b (e+f x) \tanh (c+d x)) \, dx}{a^2+b^2}+\frac {b^2 \int \frac {e^{c+d x} (e+f x)}{a-\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}+\frac {b^2 \int \frac {e^{c+d x} (e+f x)}{a+\sqrt {a^2+b^2}+b e^{c+d x}} \, dx}{a^2+b^2}\\ &=-\frac {b (e+f x)^2}{2 \left (a^2+b^2\right ) f}+\frac {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 ) d}+\frac {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 ) d}+\frac {a \int (e+f x) \text {sech}(c+d x) \, dx}{a^2+b^2}-\frac {b \int (e+f x) \tanh (c+d x) \, dx}{a^2+b^2}-\frac {(b f) \int \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 {(b f) \int \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 (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac {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 ) d}+\frac {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 ) d}-\frac {(2 b) \int \frac {e^{2 (c+d x)} (e+f x)}{1+e^{2 (c+d x)}} \, dx}{a^2+b^2}-\frac {(b f) \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 ) d^2}-\frac {(b f) \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 ) d^2}-\frac {(i a f) \int \log \left (1-i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d}+\frac {(i a f) \int \log \left (1+i e^{c+d x}\right ) \, dx}{\left (a^2+b^2\right ) d}\\ &=\frac {2 a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac {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 ) d}+\frac {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 ) d}-\frac {b (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}+\frac {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 ) d^2}+\frac {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 ) d^2}-\frac {(i a f) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac {(i a f) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac {(b f) \int \log \left (1+e^{2 (c+d x)}\right ) \, dx}{\left (a^2+b^2\right ) d}\\ &=\frac {2 a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac {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 ) d}+\frac {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 ) d}-\frac {b (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {i a f \text {Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac {i a f \text {Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac {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 ) d^2}+\frac {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 ) d^2}+\frac {(b f) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}\\ &=\frac {2 a (e+f x) \tan ^{-1}\left (e^{c+d x}\right )}{\left (a^2+b^2\right ) d}+\frac {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 ) d}+\frac {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 ) d}-\frac {b (e+f x) \log \left (1+e^{2 (c+d x)}\right )}{\left (a^2+b^2\right ) d}-\frac {i a f \text {Li}_2\left (-i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac {i a f \text {Li}_2\left (i e^{c+d x}\right )}{\left (a^2+b^2\right ) d^2}+\frac {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 ) d^2}+\frac {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 ) d^2}-\frac {b f \text {Li}_2\left (-e^{2 (c+d x)}\right )}{2 \left (a^2+b^2\right ) d^2}\\ \end {align*}
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Mathematica [A] time = 2.61, size = 439, normalized size = 1.31 \[ \frac {2 b f \text {Li}_2\left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}-a}\right )+2 b f \text {Li}_2\left (-\frac {b e^{c+d x}}{a+\sqrt {a^2+b^2}}\right )+2 b c f \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+2 b c f \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )+2 b d f x \log \left (\frac {b e^{c+d x}}{a-\sqrt {a^2+b^2}}+1\right )+2 b d f x \log \left (\frac {b e^{c+d x}}{\sqrt {a^2+b^2}+a}+1\right )+2 b d e \log (a+b \sinh (c+d x))-2 b c f \log (a+b \sinh (c+d x))+4 a d e \tan ^{-1}(\sinh (c+d x)+\cosh (c+d x))-2 i a f \text {Li}_2(-i (\cosh (c+d x)+\sinh (c+d x)))+2 i a f \text {Li}_2(i (\cosh (c+d x)+\sinh (c+d x)))+4 a d f x \tan ^{-1}(\sinh (c+d x)+\cosh (c+d x))-2 b c^2 f-2 b d e \log (\sinh (2 (c+d x))+\cosh (2 (c+d x))+1)+2 b c d e-b f \text {Li}_2(-\cosh (2 (c+d x))-\sinh (2 (c+d x)))-2 b c d f x-2 b d f x \log (\sinh (2 (c+d x))+\cosh (2 (c+d x))+1)+2 b d^2 e x}{2 d^2 \left (a^2+b^2\right )} \]
Warning: Unable to verify antiderivative.
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fricas [A] time = 0.82, size = 588, normalized size = 1.76 \[ \frac {b f {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + b f {\rm Li}_2\left (\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b} + 1\right ) + {\left (i \, a f - b f\right )} {\rm Li}_2\left (i \, \cosh \left (d x + c\right ) + i \, \sinh \left (d x + c\right )\right ) + {\left (-i \, a f - b f\right )} {\rm Li}_2\left (-i \, \cosh \left (d x + c\right ) - i \, \sinh \left (d x + c\right )\right ) + {\left (b d e - b c f\right )} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) + 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) + {\left (b d e - b c f\right )} \log \left (2 \, b \cosh \left (d x + c\right ) + 2 \, b \sinh \left (d x + c\right ) - 2 \, b \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} + 2 \, a\right ) + {\left (b d f x + b c f\right )} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) + {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) + {\left (b d f x + b c f\right )} \log \left (-\frac {a \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right ) - {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \sqrt {\frac {a^{2} + b^{2}}{b^{2}}} - b}{b}\right ) + {\left (i \, a d e - b d e - i \, a c f + b c f\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + i\right ) + {\left (-i \, a d e - b d e + i \, a c f + b c f\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) - i\right ) + {\left (-i \, a d f x - b d f x - i \, a c f - b c f\right )} \log \left (i \, \cosh \left (d x + c\right ) + i \, \sinh \left (d x + c\right ) + 1\right ) + {\left (i \, a d f x - b d f x + i \, a c f - b c f\right )} \log \left (-i \, \cosh \left (d x + c\right ) - i \, \sinh \left (d x + c\right ) + 1\right )}{{\left (a^{2} + b^{2}\right )} d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )} \operatorname {sech}\left (d x + c\right )}{b \sinh \left (d x + c\right ) + a}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.28, size = 954, normalized size = 2.86 \[ \frac {2 e b \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}{d \left (2 a^{2}+2 b^{2}\right )}-\frac {2 e b \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{d \left (2 a^{2}+2 b^{2}\right )}+\frac {4 e a \arctan \left ({\mathrm e}^{d x +c}\right )}{d \left (2 a^{2}+2 b^{2}\right )}+\frac {2 f b \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \left (2 a^{2}+2 b^{2}\right )}+\frac {2 f b \ln \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} \left (2 a^{2}+2 b^{2}\right )}+\frac {2 f b \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) x}{d \left (2 a^{2}+2 b^{2}\right )}+\frac {2 f b \ln \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right ) c}{d^{2} \left (2 a^{2}+2 b^{2}\right )}+\frac {2 f b \dilog \left (\frac {-b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}-a}{-a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} \left (2 a^{2}+2 b^{2}\right )}+\frac {2 f b \dilog \left (\frac {b \,{\mathrm e}^{d x +c}+\sqrt {a^{2}+b^{2}}+a}{a +\sqrt {a^{2}+b^{2}}}\right )}{d^{2} \left (2 a^{2}+2 b^{2}\right )}-\frac {2 f \ln \left (1+i {\mathrm e}^{d x +c}\right ) b x}{d \left (2 a^{2}+2 b^{2}\right )}-\frac {2 f \ln \left (1+i {\mathrm e}^{d x +c}\right ) b c}{d^{2} \left (2 a^{2}+2 b^{2}\right )}-\frac {2 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) a x}{d \left (2 a^{2}+2 b^{2}\right )}-\frac {2 f \ln \left (1-i {\mathrm e}^{d x +c}\right ) b x}{d \left (2 a^{2}+2 b^{2}\right )}-\frac {2 f \ln \left (1-i {\mathrm e}^{d x +c}\right ) b c}{d^{2} \left (2 a^{2}+2 b^{2}\right )}+\frac {2 i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) a x}{d \left (2 a^{2}+2 b^{2}\right )}+\frac {2 i f \dilog \left (1-i {\mathrm e}^{d x +c}\right ) a}{d^{2} \left (2 a^{2}+2 b^{2}\right )}-\frac {2 f \dilog \left (1+i {\mathrm e}^{d x +c}\right ) b}{d^{2} \left (2 a^{2}+2 b^{2}\right )}-\frac {2 i f \dilog \left (1+i {\mathrm e}^{d x +c}\right ) a}{d^{2} \left (2 a^{2}+2 b^{2}\right )}-\frac {2 f \dilog \left (1-i {\mathrm e}^{d x +c}\right ) b}{d^{2} \left (2 a^{2}+2 b^{2}\right )}-\frac {2 i f \ln \left (1+i {\mathrm e}^{d x +c}\right ) a c}{d^{2} \left (2 a^{2}+2 b^{2}\right )}+\frac {2 i f \ln \left (1-i {\mathrm e}^{d x +c}\right ) a c}{d^{2} \left (2 a^{2}+2 b^{2}\right )}-\frac {2 f c b \ln \left (b \,{\mathrm e}^{2 d x +2 c}+2 a \,{\mathrm e}^{d x +c}-b \right )}{d^{2} \left (2 a^{2}+2 b^{2}\right )}+\frac {2 f c b \ln \left (1+{\mathrm e}^{2 d x +2 c}\right )}{d^{2} \left (2 a^{2}+2 b^{2}\right )}-\frac {4 f c a \arctan \left ({\mathrm e}^{d x +c}\right )}{d^{2} \left (2 a^{2}+2 b^{2}\right )} \]
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 \[ -e {\left (\frac {2 \, a \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac {b \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac {b \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d}\right )} + 2 \, f \int \frac {2 \, x}{{\left (b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a\right )} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}\,{d x} \]
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 \[ \int \frac {e+f\,x}{\mathrm {cosh}\left (c+d\,x\right )\,\left (a+b\,\mathrm {sinh}\left (c+d\,x\right )\right )} \,d x \]
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 \[ \int \frac {\left (e + f x\right ) \operatorname {sech}{\left (c + d x \right )}}{a + b \sinh {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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