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 = 0.22, antiderivative size = 182, normalized size of antiderivative = 1.00, 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 206
Rule 467
Rule 570
Rule 577
Rule 3663
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*}
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Mathematica [A] time = 3.92, size = 125, normalized size = 0.69 \[ \frac {-32 b \tanh (c+d x) \left (15 a^2-b (5 a+7 b) \text {sech}^2(c+d x)+50 a b+b^2 \text {sech}^4(c+d x)+36 b^2\right )+60 \left (a^3+15 a^2 b+35 a b^2+21 b^3\right ) (c+d x)+5 (a+b)^3 \sinh (4 (c+d x))-40 (a+4 b) (a+b)^2 \sinh (2 (c+d x))}{160 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.55, size = 879, normalized size = 4.83 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.79, size = 507, normalized size = 2.79 \[ \frac {120 \, {\left (a^{3} + 15 \, a^{2} b + 35 \, a b^{2} + 21 \, b^{3}\right )} d x - 5 \, {\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 630 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 378 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 48 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 72 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 32 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 5 \, {\left (a^{3} e^{\left (4 \, d x + 36 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, d x + 36 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, d x + 36 \, c\right )} + b^{3} e^{\left (4 \, d x + 36 \, c\right )} - 8 \, a^{3} e^{\left (2 \, d x + 34 \, c\right )} - 48 \, a^{2} b e^{\left (2 \, d x + 34 \, c\right )} - 72 \, a b^{2} e^{\left (2 \, d x + 34 \, c\right )} - 32 \, b^{3} e^{\left (2 \, d x + 34 \, c\right )}\right )} e^{\left (-32 \, c\right )} + \frac {128 \, {\left (15 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 50 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 210 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 150 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 290 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 210 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 190 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 130 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} b + 50 \, a b^{2} + 36 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{320 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.36, size = 246, normalized size = 1.35 \[ \frac {a^{3} \left (\left (\frac {\left (\sinh ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\sinh ^{5}\left (d x +c \right )}{4 \cosh \left (d x +c \right )}-\frac {5 \left (\sinh ^{3}\left (d x +c \right )\right )}{8 \cosh \left (d x +c \right )}+\frac {15 d x}{8}+\frac {15 c}{8}-\frac {15 \tanh \left (d x +c \right )}{8}\right )+3 a \,b^{2} \left (\frac {\sinh ^{7}\left (d x +c \right )}{4 \cosh \left (d x +c \right )^{3}}-\frac {7 \left (\sinh ^{5}\left (d x +c \right )\right )}{8 \cosh \left (d x +c \right )^{3}}+\frac {35 d x}{8}+\frac {35 c}{8}-\frac {35 \tanh \left (d x +c \right )}{8}-\frac {35 \left (\tanh ^{3}\left (d x +c \right )\right )}{24}\right )+b^{3} \left (\frac {\sinh ^{9}\left (d x +c \right )}{4 \cosh \left (d x +c \right )^{5}}-\frac {9 \left (\sinh ^{7}\left (d x +c \right )\right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {63 d x}{8}+\frac {63 c}{8}-\frac {63 \tanh \left (d x +c \right )}{8}-\frac {21 \left (\tanh ^{3}\left (d x +c \right )\right )}{8}-\frac {63 \left (\tanh ^{5}\left (d x +c \right )\right )}{40}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.34, size = 480, normalized size = 2.64 \[ \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 )} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.39, size = 730, normalized size = 4.01 \[ \frac {\frac {2\,\left (3\,a^2\,b+12\,a\,b^2+10\,b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+9\,a\,b^2+5\,b^3\right )}{5\,d}+\frac {12\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b+8\,a\,b^2+6\,b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b+9\,a\,b^2+5\,b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (3\,a^2\,b+12\,a\,b^2+10\,b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}+\frac {\frac {2\,\left (3\,a^2\,b+9\,a\,b^2+5\,b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+12\,a\,b^2+10\,b^3\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+x\,\left (\frac {3\,a^3}{8}+\frac {45\,a^2\,b}{8}+\frac {105\,a\,b^2}{8}+\frac {63\,b^3}{8}\right )+\frac {\frac {2\,\left (3\,a^2\,b+9\,a\,b^2+5\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+8\,a\,b^2+6\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b+9\,a\,b^2+5\,b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b+12\,a\,b^2+10\,b^3\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+\frac {\frac {2\,\left (3\,a^2\,b+8\,a\,b^2+6\,b^3\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+9\,a\,b^2+5\,b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b+12\,a\,b^2+10\,b^3\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {2\,\left (3\,a^2\,b+12\,a\,b^2+10\,b^3\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}\,{\left (a+b\right )}^3}{64\,d}+\frac {{\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^3}{64\,d}+\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a+b\right )}^2\,\left (a+4\,b\right )}{8\,d}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2\,\left (a+4\,b\right )}{8\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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