Optimal. Leaf size=197 \[ \frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}+\frac {3 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}-\frac {3 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}} \]
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Rubi [A] time = 0.87, 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 = {6720, 2282, 390, 1158, 12, 288, 207} \[ \frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {3 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}+\frac {3 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}} \]
Antiderivative was successfully verified.
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Rule 12
Rule 207
Rule 288
Rule 390
Rule 1158
Rule 2282
Rule 6720
Rubi steps
\begin {align*} \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{3/2}} \, dx &=\frac {\tanh (a c+b c x) \int e^{c (a+b x)} \coth ^3(a c+b c x) \, dx}{\sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {\tanh (a c+b c x) \operatorname {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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {\tanh (a c+b c x) \operatorname {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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}+\frac {(2 \tanh (a c+b c x)) \operatorname {Subst}\left (\int \frac {1+3 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {\tanh (a c+b c x) \operatorname {Subst}\left (\int \frac {12 x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{2 b c \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {(6 \tanh (a c+b c x)) \operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {3 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}+\frac {(3 \tanh (a c+b c x)) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {3 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {3 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}\\ \end {align*}
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Mathematica [C] time = 7.72, size = 334, normalized size = 1.70 \[ -\frac {e^{-5 c (a+b x)} \tanh ^3(c (a+b x)) \left (256 e^{8 c (a+b x)} \left (e^{2 c (a+b x)}+1\right )^3 \, _6F_5\left (\frac {3}{2},2,2,2,2,2;1,1,1,1,\frac {11}{2};e^{2 c (a+b x)}\right )+384 e^{8 c (a+b x)} \left (5 e^{2 c (a+b x)}+7\right ) \left (e^{2 c (a+b x)}+1\right )^2 \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {11}{2};e^{2 c (a+b x)}\right )-21 \left (507305 e^{2 c (a+b x)}+173916 e^{4 c (a+b x)}-154296 e^{6 c (a+b x)}-73885 e^{8 c (a+b x)}+4887 e^{10 c (a+b x)}+252105\right )-\frac {315 \left (-28218 e^{2 c (a+b x)}+1173 e^{4 c (a+b x)}+17748 e^{6 c (a+b x)}+4299 e^{8 c (a+b x)}-1434 e^{10 c (a+b x)}+7 e^{12 c (a+b x)}-16807\right ) \tanh ^{-1}\left (\sqrt {e^{2 c (a+b x)}}\right )}{\sqrt {e^{2 c (a+b x)}}}\right )}{60480 b c \tanh ^2(c (a+b x))^{3/2}} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 0.58, size = 613, normalized size = 3.11 \[ \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 )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 161, normalized size = 0.82 \[ \frac {2 \, e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 3 \, \log \left (e^{\left (b c x + a c\right )} + 1\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 3 \, \log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \frac {2 \, {\left (3 \, e^{\left (3 \, b c x + 3 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )\right )}}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{2}}}{2 \, b c} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.98, size = 298, normalized size = 1.51 \[ \frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) {\mathrm e}^{c \left (b x +a \right )}}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}-\frac {{\mathrm e}^{c \left (b x +a \right )} \left (3 \,{\mathrm e}^{2 c \left (b x +a \right )}-1\right )}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, c b}+\frac {3 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )}{2 \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}-\frac {3 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left (1+{\mathrm e}^{c \left (b x +a \right )}\right )}{2 \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.41, size = 112, normalized size = 0.57 \[ -\frac {3 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{2 \, b c} + \frac {3 \, \log \left (e^{\left (b c x + a c\right )} - 1\right )}{2 \, b c} + \frac {e^{\left (5 \, b c x + 5 \, a c\right )} - 5 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 2 \, e^{\left (b c x + a c\right )}}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} - 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]
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 \[ \int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left ({\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2\right )}^{3/2}} \,d x \]
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 \[ e^{a c} \int \frac {e^{b c x}}{\left (\tanh ^{2}{\left (a c + b c x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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