Optimal. Leaf size=197 \[ \frac{e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt{\tanh ^2(a c+b c x)}}-\frac{3 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{b c \sqrt{\tanh ^2(a c+b c x)}}+\frac{3 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right ) \sqrt{\tanh ^2(a c+b c x)}}-\frac{2 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt{\tanh ^2(a c+b c x)}} \]
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Rubi [A] time = 0.86523, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, 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 6720
Rule 2282
Rule 390
Rule 1158
Rule 12
Rule 288
Rule 207
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*}
Mathematica [C] time = 7.5859, size = 334, normalized size = 1.7 \[ -\frac{e^{-5 c (a+b x)} \tanh ^3(c (a+b x)) \left (256 e^{8 c (a+b x)} \left (e^{2 c (a+b x)}+1\right )^3 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2,2\right \},\left \{1,1,1,1,\frac{11}{2}\right \},e^{2 c (a+b x)}\right )+384 e^{8 c (a+b x)} \left (5 e^{2 c (a+b x)}+7\right ) \left (e^{2 c (a+b x)}+1\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2\right \},\left \{1,1,1,\frac{11}{2}\right \},e^{2 c (a+b x)}\right )-21 \left (507305 e^{2 c (a+b x)}+173916 e^{4 c (a+b x)}-154296 e^{6 c (a+b x)}-73885 e^{8 c (a+b x)}+4887 e^{10 c (a+b x)}+252105\right )-\frac{315 \left (-28218 e^{2 c (a+b x)}+1173 e^{4 c (a+b x)}+17748 e^{6 c (a+b x)}+4299 e^{8 c (a+b x)}-1434 e^{10 c (a+b x)}+7 e^{12 c (a+b x)}-16807\right ) \tanh ^{-1}\left (\sqrt{e^{2 c (a+b x)}}\right )}{\sqrt{e^{2 c (a+b x)}}}\right )}{60480 b c \tanh ^2(c (a+b x))^{3/2}} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.193, size = 298, normalized size = 1.5 \begin{align*}{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ){{\rm e}^{c \left ( bx+a \right ) }}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) cb}{\frac{1}{\sqrt{{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}}}}}}-{\frac{{{\rm e}^{c \left ( bx+a \right ) }} \left ( 3\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) }{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) cb}{\frac{1}{\sqrt{{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}}}}}}+{\frac{ \left ( 3\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-3 \right ) \ln \left ({{\rm e}^{c \left ( bx+a \right ) }}-1 \right ) }{ \left ( 2+2\,{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) cb}{\frac{1}{\sqrt{{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}}}}}}-{\frac{ \left ( 3\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-3 \right ) \ln \left ({{\rm e}^{c \left ( bx+a \right ) }}+1 \right ) }{ \left ( 2+2\,{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) cb}{\frac{1}{\sqrt{{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.75246, size = 151, normalized size = 0.77 \begin{align*} -\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{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.00662, size = 1574, normalized size = 7.99 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.50692, size = 217, normalized size = 1.1 \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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