Optimal. Leaf size=77 \[ -\frac {\text {Li}_2\left (-e^{2 \text {sech}^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac {\text {sech}^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac {\text {sech}^{-1}\left (c e^{a+b x}\right ) \log \left (e^{2 \text {sech}^{-1}\left (c e^{a+b x}\right )}+1\right )}{b} \]
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Rubi [A] time = 0.10, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {2282, 6281, 5660, 3718, 2190, 2279, 2391} \[ -\frac {\text {PolyLog}\left (2,-e^{2 \text {sech}^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac {\text {sech}^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac {\text {sech}^{-1}\left (c e^{a+b x}\right ) \log \left (e^{2 \text {sech}^{-1}\left (c e^{a+b x}\right )}+1\right )}{b} \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2282
Rule 2391
Rule 3718
Rule 5660
Rule 6281
Rubi steps
\begin {align*} \int \text {sech}^{-1}\left (c e^{a+b x}\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\text {sech}^{-1}(c x)}{x} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\cosh ^{-1}\left (\frac {x}{c}\right )}{x} \, dx,x,e^{-a-b x}\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int x \tanh (x) \, dx,x,\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac {\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )^2}{2 b}-\frac {2 \operatorname {Subst}\left (\int \frac {e^{2 x} x}{1+e^{2 x}} \, dx,x,\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac {\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )^2}{2 b}-\frac {\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{b}+\frac {\operatorname {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac {\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )^2}{2 b}-\frac {\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{b}+\frac {\operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{2 b}\\ &=\frac {\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )^2}{2 b}-\frac {\cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right ) \log \left (1+e^{2 \cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{b}-\frac {\text {Li}_2\left (-e^{2 \cosh ^{-1}\left (\frac {e^{-a-b x}}{c}\right )}\right )}{2 b}\\ \end {align*}
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Mathematica [B] time = 1.17, size = 249, normalized size = 3.23 \[ x \text {sech}^{-1}\left (c e^{a+b x}\right )-\frac {\sqrt {\frac {1-c e^{a+b x}}{c e^{a+b x}+1}} \sqrt {c e^{a+b x}+1} \left (4 \text {Li}_2\left (\frac {1}{2} \left (1-\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )-\log ^2\left (c^2 e^{2 (a+b x)}\right )-2 \log ^2\left (\frac {1}{2} \left (\sqrt {1-c^2 e^{2 (a+b x)}}+1\right )\right )+4 \log \left (\frac {1}{2} \left (\sqrt {1-c^2 e^{2 (a+b x)}}+1\right )\right ) \log \left (c^2 e^{2 (a+b x)}\right )+\left (8 b x-4 \log \left (c^2 e^{2 (a+b x)}\right )\right ) \tanh ^{-1}\left (\sqrt {1-c^2 e^{2 (a+b x)}}\right )\right )}{8 b \sqrt {1-c e^{a+b x}}} \]
Antiderivative was successfully verified.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
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 \[ \int \operatorname {arsech}\left (c e^{\left (b x + a\right )}\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.15, size = 140, normalized size = 1.82 \[ \frac {\mathrm {arcsech}\left (c \,{\mathrm e}^{b x +a}\right )^{2}}{2 b}-\frac {\mathrm {arcsech}\left (c \,{\mathrm e}^{b x +a}\right ) \ln \left (1+\left (\frac {{\mathrm e}^{-b x -a}}{c}+\sqrt {\frac {{\mathrm e}^{-b x -a}}{c}-1}\, \sqrt {\frac {{\mathrm e}^{-b x -a}}{c}+1}\right )^{2}\right )}{b}-\frac {\polylog \left (2, -\left (\frac {{\mathrm e}^{-b x -a}}{c}+\sqrt {\frac {{\mathrm e}^{-b x -a}}{c}-1}\, \sqrt {\frac {{\mathrm e}^{-b x -a}}{c}+1}\right )^{2}\right )}{2 b} \]
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 \[ b c^{2} \int \frac {x e^{\left (2 \, b x + 2 \, a\right )}}{c^{2} e^{\left (2 \, b x + 2 \, a\right )} + {\left (c^{2} e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (\frac {1}{2} \, \log \left (c e^{\left (b x + a\right )} + 1\right ) + \frac {1}{2} \, \log \left (-c e^{\left (b x + a\right )} + 1\right )\right )} - 1}\,{d x} - \frac {1}{2} \, b x^{2} - {\left (a + \log \relax (c)\right )} x + x \log \left (\sqrt {c e^{\left (b x + a\right )} + 1} \sqrt {-c e^{\left (b x + a\right )} + 1} + 1\right ) - \frac {b x \log \left (c e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (-c e^{\left (b x + a\right )}\right )}{2 \, b} - \frac {b x \log \left (-c e^{\left (b x + a\right )} + 1\right ) + {\rm Li}_2\left (c e^{\left (b x + a\right )}\right )}{2 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \mathrm {acosh}\left (\frac {{\mathrm {e}}^{-a-b\,x}}{c}\right ) \,d x \]
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 \[ \int \operatorname {asech}{\left (c e^{a + b x} \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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