Optimal. Leaf size=84 \[ -\frac {a^2 \coth ^5(c+d x)}{5 d}+\frac {2 a^2 \coth ^3(c+d x)}{3 d}-\frac {a (a+2 b) \coth (c+d x)}{d}+\frac {b^2 \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac {b^2 x}{2} \]
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Rubi [A] time = 0.15, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3217, 1259, 1802, 207} \[ -\frac {a^2 \coth ^5(c+d x)}{5 d}+\frac {2 a^2 \coth ^3(c+d x)}{3 d}-\frac {a (a+2 b) \coth (c+d x)}{d}+\frac {b^2 \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac {b^2 x}{2} \]
Antiderivative was successfully verified.
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Rule 207
Rule 1259
Rule 1802
Rule 3217
Rubi steps
\begin {align*} \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a-2 a x^2+(a+b) x^4\right )^2}{x^6 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {-2 a^2+6 a^2 x^2-2 a (3 a+2 b) x^4+\left (2 a^2+4 a b+b^2\right ) x^6}{x^6 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac {\operatorname {Subst}\left (\int \left (-\frac {2 a^2}{x^6}+\frac {4 a^2}{x^4}-\frac {2 a (a+2 b)}{x^2}-\frac {b^2}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac {a (a+2 b) \coth (c+d x)}{d}+\frac {2 a^2 \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d}+\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac {b^2 x}{2}-\frac {a (a+2 b) \coth (c+d x)}{d}+\frac {2 a^2 \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d}+\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.83, size = 67, normalized size = 0.80 \[ \frac {15 b^2 (\sinh (2 (c+d x))-2 (c+d x))-4 a \coth (c+d x) \left (3 a \text {csch}^4(c+d x)-4 a \text {csch}^2(c+d x)+8 a+30 b\right )}{60 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.31, size = 457, normalized size = 5.44 \[ \frac {15 \, b^{2} \cosh \left (d x + c\right )^{7} + 105 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} - {\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} - 4 \, {\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (105 \, b^{2} \cosh \left (d x + c\right )^{3} - {\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 5 \, {\left (64 \, a^{2} + 144 \, a b + 27 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 20 \, {\left (15 \, b^{2} d x - 2 \, {\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{2} - 16 \, a^{2} - 60 \, a b\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (63 \, b^{2} \cosh \left (d x + c\right )^{5} - 2 \, {\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (64 \, a^{2} + 144 \, a b + 27 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 5 \, {\left (128 \, a^{2} + 96 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right ) - 20 \, {\left ({\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{4} + 30 \, b^{2} d x - 3 \, {\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{2} - 32 \, a^{2} - 120 \, a b\right )} \sinh \left (d x + c\right )}{120 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.26, size = 166, normalized size = 1.98 \[ -\frac {60 \, {\left (d x + c\right )} b^{2} - 15 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 15 \, {\left (2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + \frac {32 \, {\left (15 \, a b e^{\left (8 \, d x + 8 \, c\right )} - 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 40 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 20 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 60 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{2} + 15 \, a b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{120 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 74, normalized size = 0.88 \[ \frac {a^{2} \left (-\frac {8}{15}-\frac {\mathrm {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \mathrm {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )-2 a b \coth \left (d x +c \right )+b^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.34, size = 267, normalized size = 3.18 \[ -\frac {1}{8} \, b^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {16}{15} \, a^{2} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac {4 \, a b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.75, size = 397, normalized size = 4.73 \[ \frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (4\,a^2+3\,b\,a\right )}{5\,d}-\frac {4\,a\,b}{5\,d}-\frac {12\,a\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {b^2\,x}{2}-\frac {\frac {4\,\left (4\,a^2+3\,b\,a\right )}{15\,d}-\frac {8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {\frac {8\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (4\,a^2+3\,b\,a\right )}{5\,d}+\frac {4\,a\,b}{5\,d}-\frac {16\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}-\frac {16\,a\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {8\,a\,b}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]
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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