Optimal. Leaf size=197 \[ \frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {2 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {3 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )}-\frac {3 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c} \]
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Rubi [A]
time = 0.21, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6852, 2320,
398, 1172, 12, 294, 213} \begin {gather*} -\frac {3 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c}+\frac {e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c}+\frac {3 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )}-\frac {2 e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 213
Rule 294
Rule 398
Rule 1172
Rule 2320
Rule 6852
Rubi steps
\begin {align*} \int e^{c (a+b x)} \coth ^2(a c+b c x)^{3/2} \, dx &=\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \int e^{c (a+b x)} \coth ^3(a c+b c x) \, dx\\ &=\frac {\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{\left (-1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \text {Subst}\left (\int \left (1+\frac {2 \left (1+3 x^4\right )}{\left (-1+x^2\right )^3}\right ) \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}+\frac {\left (2 \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \text {Subst}\left (\int \frac {1+3 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {2 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {\left (\sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \text {Subst}\left (\int \frac {12 x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{2 b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {2 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {\left (6 \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \text {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {2 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {3 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )}+\frac {\left (3 \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)\right ) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {2 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2}+\frac {3 e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )}-\frac {3 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 2.75, size = 334, normalized size = 1.70 \begin {gather*} -\frac {e^{-5 c (a+b x)} \coth ^2(c (a+b x))^{3/2} \left (-21 \left (252105+507305 e^{2 c (a+b x)}+173916 e^{4 c (a+b x)}-154296 e^{6 c (a+b x)}-73885 e^{8 c (a+b x)}+4887 e^{10 c (a+b x)}\right )-\frac {315 \left (-16807-28218 e^{2 c (a+b x)}+1173 e^{4 c (a+b x)}+17748 e^{6 c (a+b x)}+4299 e^{8 c (a+b x)}-1434 e^{10 c (a+b x)}+7 e^{12 c (a+b x)}\right ) \tanh ^{-1}\left (\sqrt {e^{2 c (a+b x)}}\right )}{\sqrt {e^{2 c (a+b x)}}}+384 e^{8 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^2 \left (7+5 e^{2 c (a+b x)}\right ) \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};e^{2 c (a+b x)}\right )+256 e^{8 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^3 \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {11}{2};e^{2 c (a+b x)}\right )\right ) \tanh ^3(c (a+b x))}{60480 b c} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 5.13, size = 298, normalized size = 1.51
method | result | size |
risch | \(\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}-\frac {\sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )} \left (3 \,{\mathrm e}^{2 c \left (b x +a \right )}-1\right )}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) b c}-\frac {3 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}+1\right )}{2 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}+\frac {3 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )}{2 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}\) | \(298\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 112, normalized size = 0.57 \begin {gather*} -\frac {3 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{2 \, b c} + \frac {3 \, \log \left (e^{\left (b c x + a c\right )} - 1\right )}{2 \, b c} + \frac {e^{\left (5 \, b c x + 5 \, a c\right )} - 5 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 2 \, e^{\left (b c x + a c\right )}}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} - 2 \, e^{\left (2 \, b c x + 2 \, a c\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. 613 vs.
\(2 (179) = 358\).
time = 0.38, size = 613, normalized size = 3.11 \begin {gather*} \frac {2 \, \cosh \left (b c x + a c\right )^{5} + 10 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} + 2 \, \sinh \left (b c x + a c\right )^{5} + 10 \, {\left (2 \, \cosh \left (b c x + a c\right )^{2} - 1\right )} \sinh \left (b c x + a c\right )^{3} - 10 \, \cosh \left (b c x + a c\right )^{3} + 10 \, {\left (2 \, \cosh \left (b c x + a c\right )^{3} - 3 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 3 \, {\left (\cosh \left (b c x + a c\right )^{4} + 4 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + \sinh \left (b c x + a c\right )^{4} + 2 \, {\left (3 \, \cosh \left (b c x + a c\right )^{2} - 1\right )} \sinh \left (b c x + a c\right )^{2} - 2 \, \cosh \left (b c x + a c\right )^{2} + 4 \, {\left (\cosh \left (b c x + a c\right )^{3} - \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right ) + 1\right )} \log \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right ) + 1\right ) + 3 \, {\left (\cosh \left (b c x + a c\right )^{4} + 4 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + \sinh \left (b c x + a c\right )^{4} + 2 \, {\left (3 \, \cosh \left (b c x + a c\right )^{2} - 1\right )} \sinh \left (b c x + a c\right )^{2} - 2 \, \cosh \left (b c x + a c\right )^{2} + 4 \, {\left (\cosh \left (b c x + a c\right )^{3} - \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right ) + 1\right )} \log \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right ) - 1\right ) + 2 \, {\left (5 \, \cosh \left (b c x + a c\right )^{4} - 15 \, \cosh \left (b c x + a c\right )^{2} + 2\right )} \sinh \left (b c x + a c\right ) + 4 \, \cosh \left (b c x + a c\right )}{2 \, {\left (b c \cosh \left (b c x + a c\right )^{4} + 4 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + b c \sinh \left (b c x + a c\right )^{4} - 2 \, b c \cosh \left (b c x + a c\right )^{2} + 2 \, {\left (3 \, b c \cosh \left (b c x + a c\right )^{2} - b c\right )} \sinh \left (b c x + a c\right )^{2} + b c + 4 \, {\left (b c \cosh \left (b c x + a c\right )^{3} - b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )\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 = 155, normalized size = 0.79 \begin {gather*} \frac {\frac {2 \, e^{\left (b c x + a c\right )}}{\mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} - \frac {3 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{\mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} + \frac {3 \, \log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right )}{\mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} - \frac {2 \, {\left (3 \, e^{\left (3 \, b c x + 3 \, a c\right )} - e^{\left (b c x + a c\right )}\right )}}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{2} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}}{2 \, b c} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {coth}\left (a\,c+b\,c\,x\right )}^2\right )}^{3/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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