Optimal. Leaf size=85 \[ \frac {2 e^{a+b x}}{b}-\frac {2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {5 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A]
time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2320, 12, 474,
466, 396, 213} \begin {gather*} \frac {2 e^{a+b x}}{b}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac {5 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 213
Rule 396
Rule 466
Rule 474
Rule 2320
Rubi steps
\begin {align*} \int e^{2 (a+b x)} \coth ^2(a+b x) \text {csch}(a+b x) \, dx &=\frac {\text {Subst}\left (\int \frac {2 x^2 \left (1+x^2\right )^2}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {2 \text {Subst}\left (\int \frac {x^2 \left (1+x^2\right )^2}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac {2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {\text {Subst}\left (\int \frac {x^2 \left (8+4 x^2\right )}{\left (-1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=-\frac {2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {\text {Subst}\left (\int \frac {-12-8 x^2}{-1+x^2} \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac {2 e^{a+b x}}{b}-\frac {2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}+\frac {5 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {2 e^{a+b x}}{b}-\frac {2 e^{3 a+3 b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {5 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 3.34, size = 247, normalized size = 2.91 \begin {gather*} -\frac {e^{-3 (a+b x)} \left (-21 \left (56595+62725 e^{2 (a+b x)}-12071 e^{4 (a+b x)}-19353 e^{6 (a+b x)}+768 e^{8 (a+b x)}\right )+\frac {315 \left (3773+2924 e^{2 (a+b x)}-2534 e^{4 (a+b x)}-1548 e^{6 (a+b x)}+297 e^{8 (a+b x)}\right ) \tanh ^{-1}\left (\sqrt {e^{2 (a+b x)}}\right )}{\sqrt {e^{2 (a+b x)}}}+128 e^{8 (a+b x)} \left (9+16 e^{2 (a+b x)}+7 e^{4 (a+b x)}\right ) \, _4F_3\left (2,2,2,\frac {5}{2};1,1,\frac {11}{2};e^{2 (a+b x)}\right )+128 e^{8 (a+b x)} \left (1+e^{2 (a+b x)}\right )^2 \, _5F_4\left (2,2,2,2,\frac {5}{2};1,1,1,\frac {11}{2};e^{2 (a+b x)}\right )\right )}{10080 b} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 2.83, size = 78, normalized size = 0.92
method | result | size |
risch | \(\frac {2 \,{\mathrm e}^{b x +a}}{b}-\frac {{\mathrm e}^{b x +a} \left (5 \,{\mathrm e}^{2 b x +2 a}-3\right )}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}+\frac {5 \ln \left ({\mathrm e}^{b x +a}-1\right )}{2 b}-\frac {5 \ln \left ({\mathrm e}^{b x +a}+1\right )}{2 b}\) | \(78\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.26, size = 96, normalized size = 1.13 \begin {gather*} -\frac {5 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{2 \, b} + \frac {5 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{2 \, b} - \frac {9 \, e^{\left (-2 \, b x - 2 \, a\right )} - 5 \, e^{\left (-4 \, b x - 4 \, a\right )} - 2}{b {\left (e^{\left (-b x - a\right )} - 2 \, e^{\left (-3 \, b x - 3 \, a\right )} + e^{\left (-5 \, b x - 5 \, a\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 459 vs.
\(2 (75) = 150\).
time = 0.35, size = 459, normalized size = 5.40 \begin {gather*} \frac {4 \, \cosh \left (b x + a\right )^{5} + 20 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 4 \, \sinh \left (b x + a\right )^{5} + 2 \, {\left (20 \, \cosh \left (b x + a\right )^{2} - 9\right )} \sinh \left (b x + a\right )^{3} - 18 \, \cosh \left (b x + a\right )^{3} + 2 \, {\left (20 \, \cosh \left (b x + a\right )^{3} - 27 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 5 \, {\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 (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 5 \, {\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 (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \, {\left (10 \, \cosh \left (b x + a\right )^{4} - 27 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right ) + 10 \, \cosh \left (b x + a\right )}{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 )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.41, size = 72, normalized size = 0.85 \begin {gather*} -\frac {\frac {2 \, {\left (5 \, e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - 4 \, e^{\left (b x + a\right )} + 5 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 5 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.90, size = 98, normalized size = 1.15 \begin {gather*} \frac {2\,{\mathrm {e}}^{a+b\,x}}{b}-\frac {5\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {5\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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