Optimal. Leaf size=121 \[ -\frac {1}{32 d \left (a^5 \tanh (c+d x)+a^5\right )}+\frac {x}{32 a^5}-\frac {1}{32 a d \left (a^2 \tanh (c+d x)+a^2\right )^2}-\frac {1}{24 a^2 d (a \tanh (c+d x)+a)^3}-\frac {1}{16 a d (a \tanh (c+d x)+a)^4}-\frac {1}{10 d (a \tanh (c+d x)+a)^5} \]
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Rubi [A] time = 0.08, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3479, 8} \[ -\frac {1}{32 d \left (a^5 \tanh (c+d x)+a^5\right )}-\frac {1}{32 a d \left (a^2 \tanh (c+d x)+a^2\right )^2}-\frac {1}{24 a^2 d (a \tanh (c+d x)+a)^3}+\frac {x}{32 a^5}-\frac {1}{16 a d (a \tanh (c+d x)+a)^4}-\frac {1}{10 d (a \tanh (c+d x)+a)^5} \]
Antiderivative was successfully verified.
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Rule 8
Rule 3479
Rubi steps
\begin {align*} \int \frac {1}{(a+a \tanh (c+d x))^5} \, dx &=-\frac {1}{10 d (a+a \tanh (c+d x))^5}+\frac {\int \frac {1}{(a+a \tanh (c+d x))^4} \, dx}{2 a}\\ &=-\frac {1}{10 d (a+a \tanh (c+d x))^5}-\frac {1}{16 a d (a+a \tanh (c+d x))^4}+\frac {\int \frac {1}{(a+a \tanh (c+d x))^3} \, dx}{4 a^2}\\ &=-\frac {1}{10 d (a+a \tanh (c+d x))^5}-\frac {1}{16 a d (a+a \tanh (c+d x))^4}-\frac {1}{24 a^2 d (a+a \tanh (c+d x))^3}+\frac {\int \frac {1}{(a+a \tanh (c+d x))^2} \, dx}{8 a^3}\\ &=-\frac {1}{10 d (a+a \tanh (c+d x))^5}-\frac {1}{16 a d (a+a \tanh (c+d x))^4}-\frac {1}{24 a^2 d (a+a \tanh (c+d x))^3}-\frac {1}{32 a^3 d (a+a \tanh (c+d x))^2}+\frac {\int \frac {1}{a+a \tanh (c+d x)} \, dx}{16 a^4}\\ &=-\frac {1}{10 d (a+a \tanh (c+d x))^5}-\frac {1}{16 a d (a+a \tanh (c+d x))^4}-\frac {1}{24 a^2 d (a+a \tanh (c+d x))^3}-\frac {1}{32 a^3 d (a+a \tanh (c+d x))^2}-\frac {1}{32 d \left (a^5+a^5 \tanh (c+d x)\right )}+\frac {\int 1 \, dx}{32 a^5}\\ &=\frac {x}{32 a^5}-\frac {1}{10 d (a+a \tanh (c+d x))^5}-\frac {1}{16 a d (a+a \tanh (c+d x))^4}-\frac {1}{24 a^2 d (a+a \tanh (c+d x))^3}-\frac {1}{32 a^3 d (a+a \tanh (c+d x))^2}-\frac {1}{32 d \left (a^5+a^5 \tanh (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.30, size = 109, normalized size = 0.90 \[ \frac {\text {sech}^5(c+d x) (-100 \sinh (c+d x)-225 \sinh (3 (c+d x))+120 d x \sinh (5 (c+d x))+12 \sinh (5 (c+d x))-500 \cosh (c+d x)-375 \cosh (3 (c+d x))+120 d x \cosh (5 (c+d x))-12 \cosh (5 (c+d x)))}{3840 a^5 d (\tanh (c+d x)+1)^5} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.51, size = 287, normalized size = 2.37 \[ \frac {12 \, {\left (10 \, d x - 1\right )} \cosh \left (d x + c\right )^{5} + 60 \, {\left (10 \, d x - 1\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 12 \, {\left (10 \, d x + 1\right )} \sinh \left (d x + c\right )^{5} + 15 \, {\left (8 \, {\left (10 \, d x + 1\right )} \cosh \left (d x + c\right )^{2} - 15\right )} \sinh \left (d x + c\right )^{3} - 375 \, \cosh \left (d x + c\right )^{3} + 15 \, {\left (8 \, {\left (10 \, d x - 1\right )} \cosh \left (d x + c\right )^{3} - 75 \, \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 5 \, {\left (12 \, {\left (10 \, d x + 1\right )} \cosh \left (d x + c\right )^{4} - 135 \, \cosh \left (d x + c\right )^{2} - 20\right )} \sinh \left (d x + c\right ) - 500 \, \cosh \left (d x + c\right )}{3840 \, {\left (a^{5} d \cosh \left (d x + c\right )^{5} + 5 \, a^{5} d \cosh \left (d x + c\right )^{4} \sinh \left (d x + c\right ) + 10 \, a^{5} d \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right )^{2} + 10 \, a^{5} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{3} + 5 \, a^{5} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + a^{5} d \sinh \left (d x + c\right )^{5}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.12, size = 75, normalized size = 0.62 \[ -\frac {\frac {{\left (300 \, e^{\left (8 \, d x + 8 \, c\right )} + 300 \, e^{\left (6 \, d x + 6 \, c\right )} + 200 \, e^{\left (4 \, d x + 4 \, c\right )} + 75 \, e^{\left (2 \, d x + 2 \, c\right )} + 12\right )} e^{\left (-10 \, d x - 10 \, c\right )}}{a^{5}} - \frac {120 \, {\left (d x + c\right )}}{a^{5}}}{3840 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.10, size = 126, normalized size = 1.04 \[ -\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{64 d \,a^{5}}-\frac {1}{10 d \,a^{5} \left (1+\tanh \left (d x +c \right )\right )^{5}}-\frac {1}{16 d \,a^{5} \left (1+\tanh \left (d x +c \right )\right )^{4}}-\frac {1}{24 d \,a^{5} \left (1+\tanh \left (d x +c \right )\right )^{3}}-\frac {1}{32 d \,a^{5} \left (1+\tanh \left (d x +c \right )\right )^{2}}-\frac {1}{32 d \,a^{5} \left (1+\tanh \left (d x +c \right )\right )}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{64 d \,a^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.34, size = 78, normalized size = 0.64 \[ \frac {d x + c}{32 \, a^{5} d} - \frac {300 \, e^{\left (-2 \, d x - 2 \, c\right )} + 300 \, e^{\left (-4 \, d x - 4 \, c\right )} + 200 \, e^{\left (-6 \, d x - 6 \, c\right )} + 75 \, e^{\left (-8 \, d x - 8 \, c\right )} + 12 \, e^{\left (-10 \, d x - 10 \, c\right )}}{3840 \, a^{5} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.16, size = 92, normalized size = 0.76 \[ \frac {x}{32\,a^5}-\frac {5\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{64\,a^5\,d}-\frac {5\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,a^5\,d}-\frac {5\,{\mathrm {e}}^{-6\,c-6\,d\,x}}{96\,a^5\,d}-\frac {5\,{\mathrm {e}}^{-8\,c-8\,d\,x}}{256\,a^5\,d}-\frac {{\mathrm {e}}^{-10\,c-10\,d\,x}}{320\,a^5\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.66, size = 1018, normalized size = 8.41 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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