Optimal. Leaf size=122 \[ \frac{b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac{3 b (a+b)^2 \tanh (c+d x)}{d}+\frac{(a+b)^3}{4 d (1-\tanh (c+d x))}-\frac{(a+b)^3}{4 d (\tanh (c+d x)+1)}-\frac{1}{2} x (a+b)^2 (a+7 b)+\frac{b^3 \tanh ^5(c+d x)}{5 d} \]
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Rubi [A] time = 0.185146, antiderivative size = 139, normalized size of antiderivative = 1.14, 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 = {3663, 467, 528, 388, 206} \[ \frac{b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac{\sinh (c+d x) \cosh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac{1}{2} x (a+b)^2 (a+7 b) \]
Antiderivative was successfully verified.
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Rule 3663
Rule 467
Rule 528
Rule 388
Rule 206
Rubi steps
\begin{align*} \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right )^3}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^2 \left (a+7 b x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right ) \left (-a (5 a+7 b)-b (33 a+35 b) x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{10 d}\\ &=\frac{b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{a \left (15 a^2+54 a b+35 b^2\right )+b \left (81 a^2+190 a b+105 b^2\right ) x^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{30 d}\\ &=\frac{b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac{b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac{\left ((a+b)^2 (a+7 b)\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac{1}{2} (a+b)^2 (a+7 b) x+\frac{b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac{b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}\\ \end{align*}
Mathematica [A] time = 2.15865, size = 95, normalized size = 0.78 \[ \frac{4 b \tanh (c+d x) \left (45 a^2-b (15 a+16 b) \text{sech}^2(c+d x)+105 a b+3 b^2 \text{sech}^4(c+d x)+58 b^2\right )-30 (a+7 b) (a+b)^2 (c+d x)+15 (a+b)^3 \sinh (2 (c+d x))}{60 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.053, size = 180, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \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 ( 1/2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-5/2\,dx-5/2\,c+5/2\,\tanh \left ( dx+c \right ) +5/6\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3} \right ) +{b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{7\,dx}{2}}-{\frac{7\,c}{2}}+{\frac{7\,\tanh \left ( dx+c \right ) }{2}}+{\frac{7\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{6}}+{\frac{7\, \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{10}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.1, size = 509, normalized size = 4.17 \begin{align*} -\frac{1}{8} \, a^{3}{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{1}{120} \, b^{3}{\left (\frac{420 \,{\left (d x + c\right )}}{d} + \frac{15 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{1003 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3350 \, e^{\left (-4 \, d x - 4 \, c\right )} + 5590 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3915 \, e^{\left (-8 \, d x - 8 \, c\right )} + 1455 \, e^{\left (-10 \, d x - 10 \, c\right )} + 15}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 10 \, e^{\left (-8 \, d x - 8 \, c\right )} + 5 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )}\right )}}\right )} - \frac{1}{8} \, a b^{2}{\left (\frac{60 \,{\left (d x + c\right )}}{d} + \frac{3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\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 )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.10247, size = 1854, normalized size = 15.2 \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 [B] time = 2.1757, size = 533, normalized size = 4.37 \begin{align*} -\frac{60 \,{\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x - 15 \,{\left (2 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 18 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 30 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 14 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - 15 \,{\left (a^{3} e^{\left (2 \, d x + 16 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, d x + 16 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, d x + 16 \, c\right )} + b^{3} e^{\left (2 \, d x + 16 \, c\right )}\right )} e^{\left (-14 \, c\right )} + \frac{16 \,{\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 135 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 450 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 240 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 600 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 340 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 390 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 200 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 105 \, a b^{2} + 58 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{120 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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