Integrand size = 22, antiderivative size = 37 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {c^2 (1-a x)^4}{2 a}-\frac {c^2 (1-a x)^5}{5 a} \]
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Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6302, 6275, 45} \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {c^2 (1-a x)^4}{2 a}-\frac {c^2 (1-a x)^5}{5 a} \]
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Rule 45
Rule 6275
Rule 6302
Rubi steps \begin{align*} \text {integral}& = -\int e^{-2 \text {arctanh}(a x)} \left (c-a^2 c x^2\right )^2 \, dx \\ & = -\left (c^2 \int (1-a x)^3 (1+a x) \, dx\right ) \\ & = -\left (c^2 \int \left (2 (1-a x)^3-(1-a x)^4\right ) \, dx\right ) \\ & = \frac {c^2 (1-a x)^4}{2 a}-\frac {c^2 (1-a x)^5}{5 a} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {1}{10} c^2 x \left (-10+10 a x-5 a^3 x^3+2 a^4 x^4\right ) \]
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Time = 0.63 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78
method | result | size |
gosper | \(\frac {c^{2} x \left (2 a^{4} x^{4}-5 a^{3} x^{3}+10 a x -10\right )}{10}\) | \(29\) |
default | \(c^{2} \left (\frac {1}{5} a^{4} x^{5}-\frac {1}{2} a^{3} x^{4}+a \,x^{2}-x \right )\) | \(30\) |
norman | \(a \,c^{2} x^{2}-c^{2} x -\frac {1}{2} a^{3} c^{2} x^{4}+\frac {1}{5} a^{4} c^{2} x^{5}\) | \(38\) |
risch | \(a \,c^{2} x^{2}-c^{2} x -\frac {1}{2} a^{3} c^{2} x^{4}+\frac {1}{5} a^{4} c^{2} x^{5}\) | \(38\) |
parallelrisch | \(a \,c^{2} x^{2}-c^{2} x -\frac {1}{2} a^{3} c^{2} x^{4}+\frac {1}{5} a^{4} c^{2} x^{5}\) | \(38\) |
meijerg | \(\frac {c^{2} \left (\frac {a x \left (12 a^{4} x^{4}-15 a^{3} x^{3}+20 a^{2} x^{2}-30 a x +60\right )}{60}-\ln \left (a x +1\right )\right )}{a}-\frac {c^{2} \left (-\frac {a x \left (-15 a^{3} x^{3}+20 a^{2} x^{2}-30 a x +60\right )}{60}+\ln \left (a x +1\right )\right )}{a}-\frac {2 c^{2} \left (\frac {a x \left (4 a^{2} x^{2}-6 a x +12\right )}{12}-\ln \left (a x +1\right )\right )}{a}+\frac {2 c^{2} \left (-\frac {a x \left (-3 a x +6\right )}{6}+\ln \left (a x +1\right )\right )}{a}+\frac {c^{2} \left (a x -\ln \left (a x +1\right )\right )}{a}-\frac {c^{2} \ln \left (a x +1\right )}{a}\) | \(186\) |
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Time = 0.24 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {1}{5} \, a^{4} c^{2} x^{5} - \frac {1}{2} \, a^{3} c^{2} x^{4} + a c^{2} x^{2} - c^{2} x \]
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Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.97 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {a^{4} c^{2} x^{5}}{5} - \frac {a^{3} c^{2} x^{4}}{2} + a c^{2} x^{2} - c^{2} x \]
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Time = 0.20 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {1}{5} \, a^{4} c^{2} x^{5} - \frac {1}{2} \, a^{3} c^{2} x^{4} + a c^{2} x^{2} - c^{2} x \]
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Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {1}{5} \, a^{4} c^{2} x^{5} - \frac {1}{2} \, a^{3} c^{2} x^{4} + a c^{2} x^{2} - c^{2} x \]
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Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int e^{-2 \coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^2 \, dx=\frac {a^4\,c^2\,x^5}{5}-\frac {a^3\,c^2\,x^4}{2}+a\,c^2\,x^2-c^2\,x \]
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