Optimal. Leaf size=132 \[ -\frac{3 (a+8 b) \log (1-\tanh (c+d x))}{16 d}+\frac{3 (a-8 b) \log (\tanh (c+d x)+1)}{16 d}+\frac{\sinh ^4(c+d x) (a \tanh (c+d x)+b)}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}-\frac{3 a \tanh (c+d x)}{8 d}-\frac{3 b \tanh ^2(c+d x)}{2 d} \]
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Rubi [A] time = 0.172493, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3663, 1804, 801, 633, 31} \[ -\frac{3 (a+8 b) \log (1-\tanh (c+d x))}{16 d}+\frac{3 (a-8 b) \log (\tanh (c+d x)+1)}{16 d}+\frac{\sinh ^4(c+d x) (a \tanh (c+d x)+b)}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}-\frac{3 a \tanh (c+d x)}{8 d}-\frac{3 b \tanh ^2(c+d x)}{2 d} \]
Antiderivative was successfully verified.
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Rule 3663
Rule 1804
Rule 801
Rule 633
Rule 31
Rubi steps
\begin{align*} \int \sinh ^4(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^3\right )}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (-4 b-a x-4 b x^2\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac{\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{x^2 (3 a+24 b x)}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}+\frac{\operatorname{Subst}\left (\int \left (-3 a-24 b x+\frac{3 (a+8 b x)}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 a \tanh (c+d x)}{8 d}-\frac{3 b \tanh ^2(c+d x)}{2 d}+\frac{\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}+\frac{3 \operatorname{Subst}\left (\int \frac{a+8 b x}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 a \tanh (c+d x)}{8 d}-\frac{3 b \tanh ^2(c+d x)}{2 d}+\frac{\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}-\frac{(3 (a-8 b)) \operatorname{Subst}\left (\int \frac{1}{-1-x} \, dx,x,\tanh (c+d x)\right )}{16 d}+\frac{(3 (a+8 b)) \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=-\frac{3 (a+8 b) \log (1-\tanh (c+d x))}{16 d}+\frac{3 (a-8 b) \log (1+\tanh (c+d x))}{16 d}-\frac{3 a \tanh (c+d x)}{8 d}-\frac{3 b \tanh ^2(c+d x)}{2 d}+\frac{\sinh ^4(c+d x) (b+a \tanh (c+d x))}{4 d}-\frac{\sinh ^2(c+d x) \tanh (c+d x) (a+8 b \tanh (c+d x))}{8 d}\\ \end{align*}
Mathematica [A] time = 0.190275, size = 92, normalized size = 0.7 \[ \frac{3 a (c+d x)}{8 d}-\frac{a \sinh (2 (c+d x))}{4 d}+\frac{a \sinh (4 (c+d x))}{32 d}+\frac{b \left (\sinh ^4(c+d x)-4 \sinh ^2(c+d x)+2 \text{sech}^2(c+d x)+12 \log (\cosh (c+d x))\right )}{4 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.047, size = 122, normalized size = 0.9 \begin{align*}{\frac{a\cosh \left ( dx+c \right ) \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4\,d}}-{\frac{3\,a\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{8\,d}}+{\frac{3\,ax}{8}}+{\frac{3\,ac}{8\,d}}+{\frac{b \left ( \sinh \left ( dx+c \right ) \right ) ^{6}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,b \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}}+3\,{\frac{b\ln \left ( \cosh \left ( dx+c \right ) \right ) }{d}}-{\frac{3\,b \left ( \tanh \left ( dx+c \right ) \right ) ^{2}}{2\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.60844, size = 262, normalized size = 1.98 \begin{align*} \frac{1}{64} \, a{\left (24 \, x + \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac{8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac{1}{64} \, b{\left (\frac{192 \,{\left (d x + c\right )}}{d} - \frac{20 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} + \frac{192 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} - \frac{18 \, e^{\left (-2 \, d x - 2 \, c\right )} + 39 \, e^{\left (-4 \, d x - 4 \, c\right )} - 108 \, e^{\left (-6 \, d x - 6 \, c\right )} - 1}{d{\left (e^{\left (-4 \, d x - 4 \, c\right )} + 2 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.49029, size = 4209, normalized size = 31.89 \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 = 1.37402, size = 278, normalized size = 2.11 \begin{align*} \frac{24 \,{\left (a - 8 \, b\right )} d x +{\left (a e^{\left (4 \, d x + 24 \, c\right )} + b e^{\left (4 \, d x + 24 \, c\right )} - 8 \, a e^{\left (2 \, d x + 22 \, c\right )} - 20 \, b e^{\left (2 \, d x + 22 \, c\right )}\right )} e^{\left (-20 \, c\right )} + 192 \, b \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - \frac{{\left (9 \, a e^{\left (8 \, d x + 8 \, c\right )} + 72 \, b e^{\left (8 \, d x + 8 \, c\right )} + 10 \, a e^{\left (6 \, d x + 6 \, c\right )} + 36 \, b e^{\left (6 \, d x + 6 \, c\right )} - 6 \, a e^{\left (4 \, d x + 4 \, c\right )} + 111 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 18 \, b e^{\left (2 \, d x + 2 \, c\right )} + a - b\right )} e^{\left (-4 \, c\right )}}{{\left (e^{\left (2 \, d x\right )} + e^{\left (4 \, d x + 2 \, c\right )}\right )}^{2}}}{64 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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