Optimal. Leaf size=113 \[ \frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A]
time = 0.05, antiderivative size = 113, 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 = {2320, 398,
1272, 1171, 393, 212} \begin {gather*} \frac {e^{a+b x}}{b}+\frac {5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \end {gather*}
Antiderivative was successfully verified.
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Rule 212
Rule 393
Rule 398
Rule 1171
Rule 1272
Rule 2320
Rubi steps
\begin {align*} \int e^{a+b x} \coth ^4(a+b x) \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^4}{\left (1-x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left (1+\frac {8 x^2 \left (1+x^4\right )}{\left (1-x^2\right )^4}\right ) \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}+\frac {8 \text {Subst}\left (\int \frac {x^2 \left (1+x^4\right )}{\left (1-x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}+\frac {4 \text {Subst}\left (\int \frac {-2-6 x^2-6 x^4}{\left (1-x^2\right )^3} \, dx,x,e^{a+b x}\right )}{3 b}\\ &=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}-\frac {\text {Subst}\left (\int \frac {-6-24 x^2}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{3 b}\\ &=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {5 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 [A]
time = 10.08, size = 115, normalized size = 1.02 \begin {gather*} \frac {-24 e^{a+b x}+50 e^{3 (a+b x)}-48 e^{5 (a+b x)}+6 e^{7 (a+b x)}+9 \left (-1+e^{2 (a+b x)}\right )^3 \log \left (1-e^{a+b x}\right )-9 \left (-1+e^{2 (a+b x)}\right )^3 \log \left (1+e^{a+b x}\right )}{6 b \left (-1+e^{2 (a+b x)}\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 1.28, size = 88, normalized size = 0.78
method | result | size |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}-\frac {{\mathrm e}^{b x +a} \left (15 \,{\mathrm e}^{4 b x +4 a}-16 \,{\mathrm e}^{2 b x +2 a}+9\right )}{3 b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{3}}+\frac {3 \ln \left ({\mathrm e}^{b x +a}-1\right )}{2 b}-\frac {3 \ln \left ({\mathrm e}^{b x +a}+1\right )}{2 b}\) | \(88\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.27, size = 110, normalized size = 0.97 \begin {gather*} \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 {15 \, e^{\left (5 \, b x + 5 \, a\right )} - 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a\right )} - 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\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. 796 vs.
\(2 (95) = 190\).
time = 0.35, size = 796, normalized size = 7.04 \begin {gather*} \frac {6 \, \cosh \left (b x + a\right )^{7} + 42 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{6} + 6 \, \sinh \left (b x + a\right )^{7} + 6 \, {\left (21 \, \cosh \left (b x + a\right )^{2} - 8\right )} \sinh \left (b x + a\right )^{5} - 48 \, \cosh \left (b x + a\right )^{5} + 30 \, {\left (7 \, \cosh \left (b x + a\right )^{3} - 8 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{4} + 10 \, {\left (21 \, \cosh \left (b x + a\right )^{4} - 48 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right )^{3} + 50 \, \cosh \left (b x + a\right )^{3} + 6 \, {\left (21 \, \cosh \left (b x + a\right )^{5} - 80 \, \cosh \left (b x + a\right )^{3} + 25 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 9 \, {\left (\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} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{4} - 3 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} - 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 3 \, \cosh \left (b x + a\right )^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{5} - 2 \, \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 ) + 9 \, {\left (\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} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{4} - 3 \, \cosh \left (b x + a\right )^{4} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} - 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, {\left (5 \, \cosh \left (b x + a\right )^{4} - 6 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 3 \, \cosh \left (b x + a\right )^{2} + 6 \, {\left (\cosh \left (b x + a\right )^{5} - 2 \, \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 ) + 6 \, {\left (7 \, \cosh \left (b x + a\right )^{6} - 40 \, \cosh \left (b x + a\right )^{4} + 25 \, \cosh \left (b x + a\right )^{2} - 4\right )} \sinh \left (b x + a\right ) - 24 \, \cosh \left (b x + a\right )}{6 \, {\left (b \cosh \left (b x + a\right )^{6} + 6 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + b \sinh \left (b x + a\right )^{6} - 3 \, b \cosh \left (b x + a\right )^{4} + 3 \, {\left (5 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{4} + 4 \, {\left (5 \, b \cosh \left (b x + a\right )^{3} - 3 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 3 \, b \cosh \left (b x + a\right )^{2} + 3 \, {\left (5 \, b \cosh \left (b x + a\right )^{4} - 6 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 6 \, {\left (b \cosh \left (b x + a\right )^{5} - 2 \, 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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{a} \int e^{b x} \coth ^{4}{\left (a + b x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.44, size = 83, normalized size = 0.73 \begin {gather*} -\frac {\frac {2 \, {\left (15 \, e^{\left (5 \, b x + 5 \, a\right )} - 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{3}} - 6 \, e^{\left (b x + a\right )} + 9 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 9 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{6 \, b} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.10, size = 160, normalized size = 1.42 \begin {gather*} \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 {\frac {4\,{\mathrm {e}}^{a+b\,x}}{3\,b}+\frac {4\,{\mathrm {e}}^{5\,a+5\,b\,x}}{3\,b}}{3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1}-\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 {11\,{\mathrm {e}}^{a+b\,x}}{3\,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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