Optimal. Leaf size=81 \[ \frac {e^{a+b x}}{b}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.05, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2282, 390, 1158, 12, 288, 207} \[ \frac {e^{a+b x}}{b}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 12
Rule 207
Rule 288
Rule 390
Rule 1158
Rule 2282
Rubi steps
\begin {align*} \int e^{a+b x} \coth ^3(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\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 \left (1+\frac {2 \left (1+3 x^4\right )}{\left (-1+x^2\right )^3}\right ) \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}+\frac {2 \operatorname {Subst}\left (\int \frac {1+3 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {12 x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {6 \operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+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 {3 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+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 {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}
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Mathematica [C] time = 2.54, size = 286, normalized size = 3.53 \[ -\frac {e^{-5 (a+b x)} \left (256 e^{8 (a+b x)} \left (e^{2 (a+b x)}+1\right )^3 \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {11}{2};e^{2 (a+b x)}\right )+384 e^{8 (a+b x)} \left (5 e^{2 (a+b x)}+7\right ) \left (e^{2 (a+b x)}+1\right )^2 \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};e^{2 (a+b x)}\right )-21 \left (507305 e^{2 (a+b x)}+173916 e^{4 (a+b x)}-154296 e^{6 (a+b x)}-73885 e^{8 (a+b x)}+4887 e^{10 (a+b x)}+252105\right )-\frac {315 \left (-28218 e^{2 (a+b x)}+1173 e^{4 (a+b x)}+17748 e^{6 (a+b x)}+4299 e^{8 (a+b x)}-1434 e^{10 (a+b x)}+7 e^{12 (a+b x)}-16807\right ) \tanh ^{-1}\left (\sqrt {e^{2 (a+b x)}}\right )}{\sqrt {e^{2 (a+b x)}}}\right )}{60480 b} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.49, size = 459, normalized size = 5.67 \[ \frac {2 \, \cosh \left (b x + a\right )^{5} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 2 \, \sinh \left (b x + a\right )^{5} + 10 \, {\left (2 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{3} - 10 \, \cosh \left (b x + a\right )^{3} + 10 \, {\left (2 \, \cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 3 \, {\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 ) + 3 \, {\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 (5 \, \cosh \left (b x + a\right )^{4} - 15 \, \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right ) + 4 \, \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 )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 72, normalized size = 0.89 \[ -\frac {\frac {2 \, {\left (3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} - 2 \, e^{\left (b x + a\right )} + 3 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 3 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.38, size = 89, normalized size = 1.10 \[ \frac {\frac {\cosh ^{2}\left (b x +a \right )}{\sinh \left (b x +a \right )}-\frac {2}{\sinh \left (b x +a \right )}+\frac {\cosh ^{3}\left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}-\frac {3 \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}+\frac {3 \,\mathrm {csch}\left (b x +a \right ) \coth \left (b x +a \right )}{2}-3 \arctanh \left ({\mathrm e}^{b x +a}\right )}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.35, size = 88, normalized size = 1.09 \[ \frac {e^{\left (b x + a\right )}}{b} - \frac {3 \, \log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} + \frac {3 \, \log \left (e^{\left (b x + a\right )} - 1\right )}{2 \, b} - \frac {3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.79, size = 97, normalized size = 1.20 \[ \frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {3\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]
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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