Optimal. Leaf size=80 \[ \frac {e^{2 a+2 b x}}{2 b}+\frac {6}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 \log \left (1-e^{2 a+2 b x}\right )}{b} \]
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Rubi [A] time = 0.06, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2282, 444, 43} \[ \frac {e^{2 a+2 b x}}{2 b}+\frac {6}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 \log \left (1-e^{2 a+2 b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 43
Rule 444
Rule 2282
Rubi steps
\begin {align*} \int e^{2 (a+b x)} \coth ^3(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x \left (1+x^2\right )^3}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(1+x)^3}{(-1+x)^3} \, dx,x,e^{2 a+2 b x}\right )}{2 b}\\ &=\frac {\operatorname {Subst}\left (\int \left (1+\frac {8}{(-1+x)^3}+\frac {12}{(-1+x)^2}+\frac {6}{-1+x}\right ) \, dx,x,e^{2 a+2 b x}\right )}{2 b}\\ &=\frac {e^{2 a+2 b x}}{2 b}-\frac {2}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {6}{b \left (1-e^{2 a+2 b x}\right )}+\frac {3 \log \left (1-e^{2 a+2 b x}\right )}{b}\\ \end {align*}
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Mathematica [A] time = 0.12, size = 60, normalized size = 0.75 \[ \frac {\frac {8-12 e^{2 (a+b x)}}{\left (e^{2 (a+b x)}-1\right )^2}+e^{2 (a+b x)}+6 \log \left (1-e^{2 (a+b x)}\right )}{2 b} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.48, size = 398, normalized size = 4.98 \[ \frac {\cosh \left (b x + a\right )^{6} + 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + \sinh \left (b x + a\right )^{6} + {\left (15 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right )^{4} - 2 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} - 2 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + {\left (15 \, \cosh \left (b x + a\right )^{4} - 12 \, \cosh \left (b x + a\right )^{2} - 11\right )} \sinh \left (b x + a\right )^{2} - 11 \, \cosh \left (b x + a\right )^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\frac {2 \, \sinh \left (b x + a\right )}{\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )}\right ) + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{5} - 4 \, \cosh \left (b x + a\right )^{3} - 11 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 8}{2 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 70, normalized size = 0.88 \[ -\frac {\frac {9 \, e^{\left (4 \, b x + 4 \, a\right )} - 6 \, e^{\left (2 \, b x + 2 \, a\right )} + 1}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - e^{\left (2 \, b x + 2 \, a\right )} - 6 \, \log \left ({\left | e^{\left (2 \, b x + 2 \, a\right )} - 1 \right |}\right )}{2 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.69, size = 70, normalized size = 0.88 \[ \frac {{\mathrm e}^{2 b x +2 a}}{2 b}-\frac {6 a}{b}-\frac {2 \left (3 \,{\mathrm e}^{2 b x +2 a}-2\right )}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{2 b x +2 a}-1\right )}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 106, normalized size = 1.32 \[ \frac {6 \, {\left (b x + a\right )}}{b} + \frac {3 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {3 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} - \frac {10 \, e^{\left (-2 \, b x - 2 \, a\right )} - 5 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1}{2 \, b {\left (e^{\left (-2 \, b x - 2 \, a\right )} - 2 \, e^{\left (-4 \, b x - 4 \, a\right )} + e^{\left (-6 \, b x - 6 \, a\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.07, size = 80, normalized size = 1.00 \[ \frac {3\,\ln \left ({\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1\right )}{b}-\frac {6}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )}-\frac {2}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}+\frac {{\mathrm {e}}^{2\,a+2\,b\,x}}{2\,b} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [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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