Optimal. Leaf size=84 \[ \frac {a^3 \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac {3}{8} a x \left (a^2+4 a b+8 b^2\right )+\frac {3 a^2 (a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac {b^3 \tanh (c+d x)}{d} \]
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Rubi [A] time = 0.11, antiderivative size = 84, 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 = {4146, 390, 1157, 385, 206} \[ \frac {3}{8} a x \left (a^2+4 a b+8 b^2\right )+\frac {3 a^2 (a+4 b) \sinh (c+d x) \cosh (c+d x)}{8 d}+\frac {a^3 \sinh (c+d x) \cosh ^3(c+d x)}{4 d}+\frac {b^3 \tanh (c+d x)}{d} \]
Antiderivative was successfully verified.
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Rule 206
Rule 385
Rule 390
Rule 1157
Rule 4146
Rubi steps
\begin {align*} \int \cosh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b-b x^2\right )^3}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (b^3+\frac {a \left (a^2+3 a b+3 b^2\right )-3 a b (a+2 b) x^2+3 a b^2 x^4}{\left (1-x^2\right )^3}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {b^3 \tanh (c+d x)}{d}+\frac {\operatorname {Subst}\left (\int \frac {a \left (a^2+3 a b+3 b^2\right )-3 a b (a+2 b) x^2+3 a b^2 x^4}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac {a^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {b^3 \tanh (c+d x)}{d}-\frac {\operatorname {Subst}\left (\int \frac {-3 a (a+2 b)^2+12 a b^2 x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac {3 a^2 (a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {a^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {b^3 \tanh (c+d x)}{d}+\frac {\left (3 a \left (a^2+4 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+4 a b+8 b^2\right ) x+\frac {3 a^2 (a+4 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {a^3 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac {b^3 \tanh (c+d x)}{d}\\ \end {align*}
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Mathematica [A] time = 0.43, size = 70, normalized size = 0.83 \[ \frac {a^3 \sinh (4 (c+d x))+12 a \left (a^2+4 a b+8 b^2\right ) (c+d x)+8 a^2 (a+3 b) \sinh (2 (c+d x))+32 b^3 \tanh (c+d x)}{32 d} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.40, size = 153, normalized size = 1.82 \[ \frac {a^{3} \sinh \left (d x + c\right )^{5} + {\left (10 \, a^{3} \cosh \left (d x + c\right )^{2} + 9 \, a^{3} + 24 \, a^{2} b\right )} \sinh \left (d x + c\right )^{3} - 8 \, {\left (8 \, b^{3} - 3 \, {\left (a^{3} + 4 \, a^{2} b + 8 \, a b^{2}\right )} d x\right )} \cosh \left (d x + c\right ) + {\left (5 \, a^{3} \cosh \left (d x + c\right )^{4} + 8 \, a^{3} + 24 \, a^{2} b + 64 \, b^{3} + 9 \, {\left (3 \, a^{3} + 8 \, a^{2} b\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{64 \, d \cosh \left (d x + c\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.21, size = 177, normalized size = 2.11 \[ \frac {a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (a^{3} + 4 \, a^{2} b + 8 \, a b^{2}\right )} {\left (d x + c\right )} - \frac {128 \, b^{3}}{e^{\left (2 \, d x + 2 \, c\right )} + 1} - {\left (18 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 72 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 144 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 8 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + a^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.48, size = 93, normalized size = 1.11 \[ \frac {a^{3} \left (\left (\frac {\left (\cosh ^{3}\left (d x +c \right )\right )}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a \,b^{2} \left (d x +c \right )+b^{3} \tanh \left (d x +c \right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.32, size = 130, normalized size = 1.55 \[ \frac {1}{64} \, a^{3} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {3}{8} \, a^{2} b {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 3 \, a b^{2} x + \frac {2 \, b^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.53, size = 117, normalized size = 1.39 \[ \frac {3\,a\,x\,\left (a^2+4\,a\,b+8\,b^2\right )}{8}-\frac {2\,b^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {a^3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {a^3\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}-\frac {a^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a+3\,b\right )}{8\,d}+\frac {a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+3\,b\right )}{8\,d} \]
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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