Optimal. Leaf size=113 \[ \frac {e^{a+b x}}{b}+\frac {5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
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Rubi [A] time = 0.07, 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 = {2282, 390, 1258, 1157, 385, 206} \[ \frac {e^{a+b x}}{b}+\frac {5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]
Antiderivative was successfully verified.
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Rule 206
Rule 385
Rule 390
Rule 1157
Rule 1258
Rule 2282
Rubi steps
\begin {align*} \int e^{a+b x} \coth ^4(a+b x) \, dx &=\frac {\operatorname {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 {\operatorname {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 \operatorname {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 \operatorname {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 {\operatorname {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 \operatorname {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 = 10.12, size = 115, normalized size = 1.02 \[ \frac {-24 e^{a+b x}+50 e^{3 (a+b x)}-48 e^{5 (a+b x)}+6 e^{7 (a+b x)}+9 \left (e^{2 (a+b x)}-1\right )^3 \log \left (1-e^{a+b x}\right )-9 \left (e^{2 (a+b x)}-1\right )^3 \log \left (e^{a+b x}+1\right )}{6 b \left (e^{2 (a+b x)}-1\right )^3} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.52, size = 796, normalized size = 7.04 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 83, normalized size = 0.73 \[ -\frac {\frac {2 \, {\left (15 \, e^{\left (5 \, b x + 5 \, a\right )} - 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{3}} - 6 \, e^{\left (b x + a\right )} + 9 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 9 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{6 \, b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.27, size = 107, normalized size = 0.95 \[ \frac {\frac {\cosh ^{3}\left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}-\frac {3 \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}+\frac {3 \,\mathrm {csch}\left (b x +a \right ) \coth \left (b x +a \right )}{2}-3 \arctanh \left ({\mathrm e}^{b x +a}\right )+\frac {\cosh ^{4}\left (b x +a \right )}{\sinh \left (b x +a \right )^{3}}-\frac {4 \left (\cosh ^{2}\left (b x +a \right )\right )}{\sinh \left (b x +a \right )^{3}}+\frac {8}{3 \sinh \left (b x +a \right )^{3}}}{b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 110, normalized size = 0.97 \[ \frac {e^{\left (b x + a\right )}}{b} - \frac {3 \, \log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} + \frac {3 \, \log \left (e^{\left (b x + a\right )} - 1\right )}{2 \, b} - \frac {15 \, e^{\left (5 \, b x + 5 \, a\right )} - 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a\right )} - 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}} \]
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 \[ \frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {\frac {4\,{\mathrm {e}}^{a+b\,x}}{3\,b}+\frac {4\,{\mathrm {e}}^{5\,a+5\,b\,x}}{3\,b}}{3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {11\,{\mathrm {e}}^{a+b\,x}}{3\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]
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 \[ e^{a} \int e^{b x} \coth ^{4}{\left (a + b x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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