Optimal. Leaf size=51 \[ -\frac {1}{4 d \left (a^2 \tanh (c+d x)+a^2\right )}+\frac {x}{4 a^2}-\frac {1}{4 d (a \tanh (c+d x)+a)^2} \]
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Rubi [A] time = 0.03, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3479, 8} \[ -\frac {1}{4 d \left (a^2 \tanh (c+d x)+a^2\right )}+\frac {x}{4 a^2}-\frac {1}{4 d (a \tanh (c+d x)+a)^2} \]
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))^2} \, dx &=-\frac {1}{4 d (a+a \tanh (c+d x))^2}+\frac {\int \frac {1}{a+a \tanh (c+d x)} \, dx}{2 a}\\ &=-\frac {1}{4 d (a+a \tanh (c+d x))^2}-\frac {1}{4 d \left (a^2+a^2 \tanh (c+d x)\right )}+\frac {\int 1 \, dx}{4 a^2}\\ &=\frac {x}{4 a^2}-\frac {1}{4 d (a+a \tanh (c+d x))^2}-\frac {1}{4 d \left (a^2+a^2 \tanh (c+d x)\right )}\\ \end {align*}
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Mathematica [A] time = 0.18, size = 60, normalized size = 1.18 \[ \frac {\text {sech}^2(c+d x) ((4 d x+1) \sinh (2 (c+d x))+(4 d x-1) \cosh (2 (c+d x))-4)}{16 a^2 d (\tanh (c+d x)+1)^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.47, size = 101, normalized size = 1.98 \[ \frac {{\left (4 \, d x - 1\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (4 \, d x + 1\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (4 \, d x - 1\right )} \sinh \left (d x + c\right )^{2} - 4}{16 \, {\left (a^{2} d \cosh \left (d x + c\right )^{2} + 2 \, a^{2} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a^{2} d \sinh \left (d x + c\right )^{2}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.14, size = 42, normalized size = 0.82 \[ -\frac {\frac {{\left (4 \, e^{\left (2 \, d x + 2 \, c\right )} + 1\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{a^{2}} - \frac {4 \, {\left (d x + c\right )}}{a^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 72, normalized size = 1.41 \[ -\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{8 d \,a^{2}}-\frac {1}{4 d \,a^{2} \left (1+\tanh \left (d x +c \right )\right )^{2}}-\frac {1}{4 d \,a^{2} \left (1+\tanh \left (d x +c \right )\right )}+\frac {\ln \left (1+\tanh \left (d x +c \right )\right )}{8 d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.31, size = 43, normalized size = 0.84 \[ \frac {d x + c}{4 \, a^{2} d} - \frac {4 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}}{16 \, a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.09, size = 41, normalized size = 0.80 \[ \frac {x}{4\,a^2}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{4\,a^2\,d}-\frac {{\mathrm {e}}^{-4\,c-4\,d\,x}}{16\,a^2\,d} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.92, size = 223, normalized size = 4.37 \[ \begin {cases} \frac {d x \tanh ^{2}{\left (c + d x \right )}}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh {\left (c + d x \right )} + 4 a^{2} d} + \frac {2 d x \tanh {\left (c + d x \right )}}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh {\left (c + d x \right )} + 4 a^{2} d} + \frac {d x}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh {\left (c + d x \right )} + 4 a^{2} d} - \frac {\tanh {\left (c + d x \right )}}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh {\left (c + d x \right )} + 4 a^{2} d} - \frac {2}{4 a^{2} d \tanh ^{2}{\left (c + d x \right )} + 8 a^{2} d \tanh {\left (c + d x \right )} + 4 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x}{\left (a \tanh {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
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