Optimal. Leaf size=319 \[ \frac{e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt{\tanh ^2(a c+b c x)}}-\frac{15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt{\tanh ^2(a c+b c x)}}+\frac{25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt{\tanh ^2(a c+b c x)}}-\frac{55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt{\tanh ^2(a c+b c x)}}+\frac{26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\tanh ^2(a c+b c x)}}-\frac{4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\tanh ^2(a c+b c x)}} \]
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Rubi [A] time = 1.77048, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {6720, 2282, 390, 1814, 1157, 385, 207} \[ \frac{e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt{\tanh ^2(a c+b c x)}}-\frac{15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt{\tanh ^2(a c+b c x)}}+\frac{25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt{\tanh ^2(a c+b c x)}}-\frac{55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt{\tanh ^2(a c+b c x)}}+\frac{26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\tanh ^2(a c+b c x)}}-\frac{4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\tanh ^2(a c+b c x)}} \]
Antiderivative was successfully verified.
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Rule 6720
Rule 2282
Rule 390
Rule 1814
Rule 1157
Rule 385
Rule 207
Rubi steps
\begin{align*} \int \frac{e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx &=\frac{\tanh (a c+b c x) \int e^{c (a+b x)} \coth ^5(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 )^5}{\left (-1+x^2\right )^5} \, 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+10 x^4+5 x^8\right )}{\left (-1+x^2\right )^5}\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+10 x^4+5 x^8}{\left (-1+x^2\right )^5} \, 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{4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\tanh ^2(a c+b c x)}}+\frac{\tanh (a c+b c x) \operatorname{Subst}\left (\int \frac{8+120 x^2+40 x^4+40 x^6}{\left (-1+x^2\right )^4} \, dx,x,e^{c (a+b x)}\right )}{4 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{4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\tanh ^2(a c+b c x)}}+\frac{26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\tanh ^2(a c+b c x)}}+\frac{\tanh (a c+b c x) \operatorname{Subst}\left (\int \frac{160+480 x^2+240 x^4}{\left (-1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{24 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{4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\tanh ^2(a c+b c x)}}+\frac{26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\tanh ^2(a c+b c x)}}-\frac{55 e^{c (a+b x)} \tanh (a c+b c x)}{6 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{240+960 x^2}{\left (-1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{96 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{4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\tanh ^2(a c+b c x)}}+\frac{26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\tanh ^2(a c+b c x)}}-\frac{55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt{\tanh ^2(a c+b c x)}}+\frac{25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt{\tanh ^2(a c+b c x)}}+\frac{(15 \tanh (a c+b c x)) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{4 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{4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt{\tanh ^2(a c+b c x)}}+\frac{26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt{\tanh ^2(a c+b c x)}}-\frac{55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt{\tanh ^2(a c+b c x)}}+\frac{25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt{\tanh ^2(a c+b c x)}}-\frac{15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt{\tanh ^2(a c+b c x)}}\\ \end{align*}
Mathematica [A] time = 10.6569, size = 164, normalized size = 0.51 \[ \frac{\left (66 e^{c (a+b x)}-314 e^{3 c (a+b x)}+374 e^{5 c (a+b x)}-246 e^{7 c (a+b x)}+24 e^{9 c (a+b x)}+45 \left (e^{2 c (a+b x)}-1\right )^4 \log \left (1-e^{c (a+b x)}\right )-45 \left (e^{2 c (a+b x)}-1\right )^4 \log \left (e^{c (a+b x)}+1\right )\right ) \tanh (c (a+b x))}{24 b c \left (e^{2 c (a+b x)}-1\right )^4 \sqrt{\tanh ^2(c (a+b x))}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.197, size = 320, normalized size = 1. \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 ( 75\,{{\rm e}^{6\,c \left ( bx+a \right ) }}-115\,{{\rm e}^{4\,c \left ( bx+a \right ) }}+109\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-21 \right ) }{12\, \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{3} \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 ( 15\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-15 \right ) \ln \left ({{\rm e}^{c \left ( bx+a \right ) }}-1 \right ) }{ \left ( 8+8\,{{\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 ( 15\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-15 \right ) \ln \left ({{\rm e}^{c \left ( bx+a \right ) }}+1 \right ) }{ \left ( 8+8\,{{\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 = 1.68155, size = 225, normalized size = 0.71 \begin{align*} -\frac{15 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{8 \, b c} + \frac{15 \, \log \left (e^{\left (b c x + a c\right )} - 1\right )}{8 \, b c} + \frac{12 \, e^{\left (9 \, b c x + 9 \, a c\right )} - 123 \, e^{\left (7 \, b c x + 7 \, a c\right )} + 187 \, e^{\left (5 \, b c x + 5 \, a c\right )} - 157 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 33 \, e^{\left (b c x + a c\right )}}{12 \, b c{\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, 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 = 2.21561, size = 4176, normalized size = 13.09 \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 = 1.64773, size = 290, normalized size = 0.91 \begin{align*} \frac{24 \, e^{\left (b c x + a c\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 45 \, \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 ) + 45 \, \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 (75 \, e^{\left (7 \, b c x + 7 \, a c\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 115 \, e^{\left (5 \, b c x + 5 \, a c\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 109 \, 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 ) - 21 \, 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 )}^{4}}}{24 \, b c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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