Optimal. Leaf size=73 \[ \frac {5 e^{a+b x}}{2 b}+\frac {e^{3 a+3 b x}}{6 b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {4 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.05, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2282, 12, 390, 385, 206} \[ \frac {5 e^{a+b x}}{2 b}+\frac {e^{3 a+3 b x}}{6 b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {4 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 12
Rule 206
Rule 385
Rule 390
Rule 2282
Rubi steps
\begin {align*} \int e^{2 (a+b x)} \cosh (a+b x) \coth ^2(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{2 \left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac {\operatorname {Subst}\left (\int \left (5+x^2-\frac {4 \left (1-3 x^2\right )}{\left (1-x^2\right )^2}\right ) \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac {5 e^{a+b x}}{2 b}+\frac {e^{3 a+3 b x}}{6 b}-\frac {2 \operatorname {Subst}\left (\int \frac {1-3 x^2}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {5 e^{a+b x}}{2 b}+\frac {e^{3 a+3 b x}}{6 b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {5 e^{a+b x}}{2 b}+\frac {e^{3 a+3 b x}}{6 b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {4 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}
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Mathematica [C] time = 0.96, size = 220, normalized size = 3.01 \[ \frac {e^{-5 (a+b x)} \left (256 e^{8 (a+b x)} \left (e^{2 (a+b x)}+1\right )^3 \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};e^{2 (a+b x)}\right )-21 \left (91925 e^{2 (a+b x)}+61158 e^{4 (a+b x)}-20166 e^{6 (a+b x)}-15061 e^{8 (a+b x)}+753 e^{10 (a+b x)}+36015\right )-\frac {315 \left (-5328 e^{2 (a+b x)}-1821 e^{4 (a+b x)}+3264 e^{6 (a+b x)}+1149 e^{8 (a+b x)}-240 e^{10 (a+b x)}+e^{12 (a+b x)}-2401\right ) \tanh ^{-1}\left (\sqrt {e^{2 (a+b x)}}\right )}{\sqrt {e^{2 (a+b x)}}}\right )}{60480 b} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.52, size = 272, normalized size = 3.73 \[ \frac {\cosh \left (b x + a\right )^{5} + 5 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{5} + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{2} + 7\right )} \sinh \left (b x + a\right )^{3} + 14 \, \cosh \left (b x + a\right )^{3} + 2 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + 21 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 12 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 12 \, {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + {\left (5 \, \cosh \left (b x + a\right )^{4} + 42 \, \cosh \left (b x + a\right )^{2} - 27\right )} \sinh \left (b x + a\right ) - 27 \, \cosh \left (b x + a\right )}{6 \, {\left (b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 75, normalized size = 1.03 \[ \frac {{\left (e^{\left (3 \, b x + 15 \, a\right )} + 15 \, e^{\left (b x + 13 \, a\right )}\right )} e^{\left (-12 \, a\right )} - \frac {12 \, e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - 12 \, \log \left (e^{\left (b x + a\right )} + 1\right ) + 12 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{6 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.64, size = 79, normalized size = 1.08 \[ \frac {{\mathrm e}^{3 b x +3 a}}{6 b}+\frac {5 \,{\mathrm e}^{b x +a}}{2 b}-\frac {2 \,{\mathrm e}^{b x +a}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}-\frac {2 \ln \left (1+{\mathrm e}^{b x +a}\right )}{b}+\frac {2 \ln \left ({\mathrm e}^{b x +a}-1\right )}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 87, normalized size = 1.19 \[ -\frac {2 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {2 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} + \frac {14 \, e^{\left (-2 \, b x - 2 \, a\right )} - 27 \, e^{\left (-4 \, b x - 4 \, a\right )} + 1}{6 \, b {\left (e^{\left (-3 \, b x - 3 \, a\right )} - e^{\left (-5 \, b x - 5 \, a\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.87, size = 77, normalized size = 1.05 \[ \frac {5\,{\mathrm {e}}^{a+b\,x}}{2\,b}-\frac {4\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}+\frac {{\mathrm {e}}^{3\,a+3\,b\,x}}{6\,b}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]
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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