Optimal. Leaf size=182 \[ -\frac{b \left (6 a^2+35 a b+21 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 (a+b) \left (a^2+14 a b+21 b^2\right ) \tanh (c+d x)}{8 d}+\frac{3}{8} x (a+b) \left (a^2+14 a b+21 b^2\right )-\frac{3 b^2 (5 a+21 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\sinh ^3(c+d x) \cosh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d} \]
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Rubi [A] time = 0.22168, 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 = {3663, 467, 577, 570, 206} \[ -\frac{b \left (6 a^2+35 a b+21 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 (a+b) \left (a^2+14 a b+21 b^2\right ) \tanh (c+d x)}{8 d}+\frac{3}{8} x (a+b) \left (a^2+14 a b+21 b^2\right )-\frac{3 b^2 (5 a+21 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\sinh ^3(c+d x) \cosh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d} \]
Antiderivative was successfully verified.
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Rule 3663
Rule 467
Rule 577
Rule 570
Rule 206
Rubi steps
\begin{align*} \int \sinh ^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+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 \tanh ^2(c+d x)\right )^3}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right )^2 \left (3 a+9 b x^2\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right ) \left (-3 a (a+9 b)-3 b (5 a+21 b) x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}-\frac{\operatorname{Subst}\left (\int \left (3 (a+b) \left (a^2+14 a b+21 b^2\right )+3 b \left (6 a^2+35 a b+21 b^2\right ) x^2+3 b^2 (5 a+21 b) x^4-\frac{3 \left (a^3+15 a^2 b+35 a b^2+21 b^3\right )}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{3 (a+b) \left (a^2+14 a b+21 b^2\right ) \tanh (c+d x)}{8 d}-\frac{b \left (6 a^2+35 a b+21 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 b^2 (5 a+21 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}+\frac{\left (3 (a+b) \left (a^2+14 a b+21 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+b) \left (a^2+14 a b+21 b^2\right ) x-\frac{3 (a+b) \left (a^2+14 a b+21 b^2\right ) \tanh (c+d x)}{8 d}-\frac{b \left (6 a^2+35 a b+21 b^2\right ) \tanh ^3(c+d x)}{8 d}-\frac{3 b^2 (5 a+21 b) \tanh ^5(c+d x)}{40 d}-\frac{3 (a+3 b) \sinh ^2(c+d x) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{8 d}+\frac{\cosh (c+d x) \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{4 d}\\ \end{align*}
Mathematica [A] time = 3.97828, size = 125, normalized size = 0.69 \[ \frac{60 \left (15 a^2 b+a^3+35 a b^2+21 b^3\right ) (c+d x)-32 b \tanh (c+d x) \left (15 a^2-b (5 a+7 b) \text{sech}^2(c+d x)+50 a b+b^2 \text{sech}^4(c+d x)+36 b^2\right )+5 (a+b)^3 \sinh (4 (c+d x))-40 (a+4 b) (a+b)^2 \sinh (2 (c+d x))}{160 d} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.053, size = 246, normalized size = 1.4 \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 ( 1/4\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{\cosh \left ( dx+c \right ) }}-5/8\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{\cosh \left ( dx+c \right ) }}+{\frac{15\,dx}{8}}+{\frac{15\,c}{8}}-{\frac{15\,\tanh \left ( dx+c \right ) }{8}} \right ) +3\,a{b}^{2} \left ( 1/4\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-{\frac{7\, \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{8\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{35\,dx}{8}}+{\frac{35\,c}{8}}-{\frac{35\,\tanh \left ( dx+c \right ) }{8}}-{\frac{35\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{24}} \right ) +{b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{9}}{4\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{9\, \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{8\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{63\,dx}{8}}+{\frac{63\,c}{8}}-{\frac{63\,\tanh \left ( dx+c \right ) }{8}}-{\frac{21\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{8}}-{\frac{63\, \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{40}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.07719, size = 648, normalized size = 3.56 \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 )} + \frac{1}{320} \, b^{3}{\left (\frac{2520 \,{\left (d x + c\right )}}{d} + \frac{5 \,{\left (32 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}\right )}}{d} - \frac{135 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5358 \, e^{\left (-4 \, d x - 4 \, c\right )} + 18190 \, e^{\left (-6 \, d x - 6 \, c\right )} + 28455 \, e^{\left (-8 \, d x - 8 \, c\right )} + 19995 \, e^{\left (-10 \, d x - 10 \, c\right )} + 6560 \, e^{\left (-12 \, d x - 12 \, c\right )} - 5}{d{\left (e^{\left (-4 \, d x - 4 \, c\right )} + 5 \, e^{\left (-6 \, d x - 6 \, c\right )} + 10 \, e^{\left (-8 \, d x - 8 \, c\right )} + 10 \, e^{\left (-10 \, d x - 10 \, c\right )} + 5 \, e^{\left (-12 \, d x - 12 \, c\right )} + e^{\left (-14 \, d x - 14 \, c\right )}\right )}}\right )} + \frac{1}{64} \, a b^{2}{\left (\frac{840 \,{\left (d x + c\right )}}{d} + \frac{3 \,{\left (24 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}\right )}}{d} - \frac{63 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1487 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2517 \, e^{\left (-6 \, d x - 6 \, c\right )} + 1608 \, e^{\left (-8 \, d x - 8 \, c\right )} - 3}{d{\left (e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )}\right )}}\right )} + \frac{3}{64} \, a^{2} b{\left (\frac{120 \,{\left (d x + c\right )}}{d} + \frac{16 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )}}{d} - \frac{15 \, e^{\left (-2 \, d x - 2 \, c\right )} + 144 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d{\left (e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.07457, size = 2273, normalized size = 12.49 \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 = 2.74754, size = 684, normalized size = 3.76 \begin{align*} \frac{120 \,{\left (a^{3} + 15 \, a^{2} b + 35 \, a b^{2} + 21 \, b^{3}\right )} d x - 5 \,{\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 630 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 378 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 48 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 72 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 32 \, 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 (-4 \, d x - 4 \, c\right )} + 5 \,{\left (a^{3} e^{\left (4 \, d x + 36 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, d x + 36 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, d x + 36 \, c\right )} + b^{3} e^{\left (4 \, d x + 36 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 34 \, c\right )} - 48 \, a^{2} b e^{\left (2 \, d x + 34 \, c\right )} - 72 \, a b^{2} e^{\left (2 \, d x + 34 \, c\right )} - 32 \, b^{3} e^{\left (2 \, d x + 34 \, c\right )}\right )} e^{\left (-32 \, c\right )} + \frac{128 \,{\left (15 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 50 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 210 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 150 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 290 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 210 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 190 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 130 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} b + 50 \, a b^{2} + 36 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{320 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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