Optimal. Leaf size=182 \[ \frac{b \left (6 a^2-23 a b-8 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 a \left (a^2-12 a b+8 b^2\right ) \tanh (c+d x)}{8 d}+\frac{3}{8} a x \left (a^2-12 a b+8 b^2\right )-\frac{3 b^2 (5 a-16 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^2}{8 d}+\frac{\sinh ^3(c+d x) \cosh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^3}{4 d} \]
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Rubi [A] time = 0.231556, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4132, 467, 577, 570, 206} \[ \frac{b \left (6 a^2-23 a b-8 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 a \left (a^2-12 a b+8 b^2\right ) \tanh (c+d x)}{8 d}+\frac{3}{8} a x \left (a^2-12 a b+8 b^2\right )-\frac{3 b^2 (5 a-16 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^2}{8 d}+\frac{\sinh ^3(c+d x) \cosh (c+d x) \left (a-b \tanh ^2(c+d x)+b\right )^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 4132
Rule 467
Rule 577
Rule 570
Rule 206
Rubi steps
\begin{align*} \int \left (a+b \text{sech}^2(c+d x)\right )^3 \sinh ^4(c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b-b x^2\right )^3}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 (a+b)-9 b x^2\right ) \left (a+b-b x^2\right )^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=-\frac{3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b-b x^2\right ) \left (-3 (a-8 b) (a+b)+3 (5 a-16 b) b x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}-\frac{\operatorname{Subst}\left (\int \left (3 a \left (a^2-12 a b+8 b^2\right )-3 b \left (6 a^2-23 a b-8 b^2\right ) x^2+3 (5 a-16 b) b^2 x^4-\frac{3 \left (a^3-12 a^2 b+8 a b^2\right )}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 a \left (a^2-12 a b+8 b^2\right ) \tanh (c+d x)}{8 d}+\frac{b \left (6 a^2-23 a b-8 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 (5 a-16 b) b^2 \tanh ^5(c+d x)}{40 d}-\frac{3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}+\frac{\left (3 a \left (a^2-12 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{3}{8} a \left (a^2-12 a b+8 b^2\right ) x-\frac{3 a \left (a^2-12 a b+8 b^2\right ) \tanh (c+d x)}{8 d}+\frac{b \left (6 a^2-23 a b-8 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 (5 a-16 b) b^2 \tanh ^5(c+d x)}{40 d}-\frac{3 (a-2 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b-b \tanh ^2(c+d x)\right )^3}{4 d}\\ \end{align*}
Mathematica [B] time = 2.42398, size = 651, normalized size = 3.58 \[ \frac{\text{sech}(c) \text{sech}^5(c+d x) \left (a \cosh ^2(c+d x)+b\right )^3 \left (1200 a d x \left (a^2-12 a b+8 b^2\right ) \cosh (2 c+d x)+1200 a d x \left (a^2-12 a b+8 b^2\right ) \cosh (d x)-7080 a^2 b \sinh (2 c+d x)+8760 a^2 b \sinh (2 c+3 d x)-840 a^2 b \sinh (4 c+3 d x)+2520 a^2 b \sinh (4 c+5 d x)+600 a^2 b \sinh (6 c+5 d x)+120 a^2 b \sinh (6 c+7 d x)+120 a^2 b \sinh (8 c+7 d x)-7200 a^2 b d x \cosh (2 c+3 d x)-7200 a^2 b d x \cosh (4 c+3 d x)-1440 a^2 b d x \cosh (4 c+5 d x)-1440 a^2 b d x \cosh (6 c+5 d x)+12120 a^2 b \sinh (d x)-180 a^3 \sinh (2 c+d x)-310 a^3 \sinh (2 c+3 d x)-310 a^3 \sinh (4 c+3 d x)-150 a^3 \sinh (4 c+5 d x)-150 a^3 \sinh (6 c+5 d x)-15 a^3 \sinh (6 c+7 d x)-15 a^3 \sinh (8 c+7 d x)+5 a^3 \sinh (8 c+9 d x)+5 a^3 \sinh (10 c+9 d x)+600 a^3 d x \cosh (2 c+3 d x)+600 a^3 d x \cosh (4 c+3 d x)+120 a^3 d x \cosh (4 c+5 d x)+120 a^3 d x \cosh (6 c+5 d x)-180 a^3 \sinh (d x)+11520 a b^2 \sinh (2 c+d x)-8960 a b^2 \sinh (2 c+3 d x)+3840 a b^2 \sinh (4 c+3 d x)-2560 a b^2 \sinh (4 c+5 d x)+4800 a b^2 d x \cosh (2 c+3 d x)+4800 a b^2 d x \cosh (4 c+3 d x)+960 a b^2 d x \cosh (4 c+5 d x)+960 a b^2 d x \cosh (6 c+5 d x)-14080 a b^2 \sinh (d x)-640 b^3 \sinh (4 c+3 d x)+128 b^3 \sinh (4 c+5 d x)+1280 b^3 \sinh (d x)\right )}{1280 d (a \cosh (2 (c+d x))+a+2 b)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.046, size = 182, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8}} \right ) \cosh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +3\,{a}^{2}b \left ( 1/2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{\cosh \left ( dx+c \right ) }}-3/2\,dx-3/2\,c+3/2\,\tanh \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ( dx+c-\tanh \left ( dx+c \right ) -1/3\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3} \right ) +{b}^{3} \left ( -{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{3\,\tanh \left ( dx+c \right ) }{8} \left ({\frac{8}{15}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) } \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.07936, size = 570, normalized size = 3.13 \begin{align*} \frac{1}{64} \, a^{3}{\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 )} + a b^{2}{\left (3 \, x + \frac{3 \, c}{d} - \frac{4 \,{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} - \frac{3}{8} \, a^{2} b{\left (\frac{12 \,{\left (d x + c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} + \frac{2}{5} \, b^{3}{\left (\frac{10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{5 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{1}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.64733, size = 1858, normalized size = 10.21 \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 = 1.18007, size = 474, normalized size = 2.6 \begin{align*} \frac{3 \,{\left (a^{3} - 12 \, a^{2} b + 8 \, a b^{2}\right )}{\left (d x + c\right )}}{8 \, d} - \frac{{\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 216 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 144 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + a^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac{a^{3} d e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, d^{2}} - \frac{2 \,{\left (15 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} - 30 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 5 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 110 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 70 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} b - 20 \, a b^{2} + b^{3}\right )}}{5 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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