Optimal. Leaf size=73 \[ \frac {x}{8 a^3}-\frac {1}{6 d (a+a \tanh (c+d x))^3}-\frac {1}{8 a d (a+a \tanh (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \tanh (c+d x)\right )} \]
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Rubi [A]
time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3560, 8}
\begin {gather*} -\frac {1}{8 d \left (a^3 \tanh (c+d x)+a^3\right )}+\frac {x}{8 a^3}-\frac {1}{8 a d (a \tanh (c+d x)+a)^2}-\frac {1}{6 d (a \tanh (c+d x)+a)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 8
Rule 3560
Rubi steps
\begin {align*} \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx &=-\frac {1}{6 d (a+a \tanh (c+d x))^3}+\frac {\int \frac {1}{(a+a \tanh (c+d x))^2} \, dx}{2 a}\\ &=-\frac {1}{6 d (a+a \tanh (c+d x))^3}-\frac {1}{8 a d (a+a \tanh (c+d x))^2}+\frac {\int \frac {1}{a+a \tanh (c+d x)} \, dx}{4 a^2}\\ &=-\frac {1}{6 d (a+a \tanh (c+d x))^3}-\frac {1}{8 a d (a+a \tanh (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \tanh (c+d x)\right )}+\frac {\int 1 \, dx}{8 a^3}\\ &=\frac {x}{8 a^3}-\frac {1}{6 d (a+a \tanh (c+d x))^3}-\frac {1}{8 a d (a+a \tanh (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \tanh (c+d x)\right )}\\ \end {align*}
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Mathematica [A]
time = 0.18, size = 83, normalized size = 1.14 \begin {gather*} \frac {\text {sech}^3(c+d x) (-27 \cosh (c+d x)+2 (-1+6 d x) \cosh (3 (c+d x))-9 \sinh (c+d x)+2 \sinh (3 (c+d x))+12 d x \sinh (3 (c+d x)))}{96 a^3 d (1+\tanh (c+d x))^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.66, size = 67, normalized size = 0.92
method | result | size |
risch | \(\frac {x}{8 a^{3}}-\frac {3 \,{\mathrm e}^{-2 d x -2 c}}{16 a^{3} d}-\frac {3 \,{\mathrm e}^{-4 d x -4 c}}{32 a^{3} d}-\frac {{\mathrm e}^{-6 d x -6 c}}{48 a^{3} d}\) | \(59\) |
derivativedivides | \(\frac {-\frac {1}{6 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{16}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{16}}{d \,a^{3}}\) | \(67\) |
default | \(\frac {-\frac {1}{6 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{16}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{16}}{d \,a^{3}}\) | \(67\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.27, size = 56, normalized size = 0.77 \begin {gather*} \frac {d x + c}{8 \, a^{3} d} - \frac {18 \, e^{\left (-2 \, d x - 2 \, c\right )} + 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2 \, e^{\left (-6 \, d x - 6 \, c\right )}}{96 \, a^{3} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 160 vs.
\(2 (65) = 130\).
time = 0.44, size = 160, normalized size = 2.19 \begin {gather*} \frac {2 \, {\left (6 \, d x - 1\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (6 \, d x - 1\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 2 \, {\left (6 \, d x + 1\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (2 \, {\left (6 \, d x + 1\right )} \cosh \left (d x + c\right )^{2} - 3\right )} \sinh \left (d x + c\right ) - 27 \, \cosh \left (d x + c\right )}{96 \, {\left (a^{3} d \cosh \left (d x + c\right )^{3} + 3 \, a^{3} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, a^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + a^{3} d \sinh \left (d x + c\right )^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 430 vs.
\(2 (60) = 120\).
time = 0.77, size = 430, normalized size = 5.89 \begin {gather*} \begin {cases} \frac {3 d x \tanh ^{3}{\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} + \frac {9 d x \tanh ^{2}{\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} + \frac {9 d x \tanh {\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} + \frac {3 d x}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} - \frac {3 \tanh ^{2}{\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} - \frac {9 \tanh {\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} - \frac {10}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x}{\left (a \tanh {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.42, size = 53, normalized size = 0.73 \begin {gather*} -\frac {\frac {{\left (18 \, e^{\left (4 \, d x + 4 \, c\right )} + 9 \, e^{\left (2 \, d x + 2 \, c\right )} + 2\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{a^{3}} - \frac {12 \, {\left (d x + c\right )}}{a^{3}}}{96 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.10, size = 58, normalized size = 0.79 \begin {gather*} \frac {x}{8\,a^3}-\frac {3\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{16\,a^3\,d}-\frac {3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{32\,a^3\,d}-\frac {{\mathrm {e}}^{-6\,c-6\,d\,x}}{48\,a^3\,d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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