Integrand size = 20, antiderivative size = 33 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c}{a^2 x}+c x-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a} \]
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Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6302, 6292, 6285, 90} \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c}{a^2 x}-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a}+c x \]
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Rule 90
Rule 6285
Rule 6292
Rule 6302
Rubi steps \begin{align*} \text {integral}& = \int e^{4 \text {arctanh}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx \\ & = -\frac {c \int \frac {e^{4 \text {arctanh}(a x)} \left (1-a^2 x^2\right )}{x^2} \, dx}{a^2} \\ & = -\frac {c \int \frac {(1+a x)^3}{x^2 (1-a x)} \, dx}{a^2} \\ & = -\frac {c \int \left (-a^2+\frac {1}{x^2}+\frac {4 a}{x}-\frac {8 a^2}{-1+a x}\right ) \, dx}{a^2} \\ & = \frac {c}{a^2 x}+c x-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {c}{a^2 x}+c x-\frac {4 c \log (x)}{a}+\frac {8 c \log (1-a x)}{a} \]
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Time = 0.64 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88
method | result | size |
default | \(\frac {c \left (a^{2} x +\frac {1}{x}-4 a \ln \left (x \right )+8 a \ln \left (a x -1\right )\right )}{a^{2}}\) | \(29\) |
risch | \(\frac {c}{a^{2} x}+c x -\frac {4 c \ln \left (x \right )}{a}+\frac {8 c \ln \left (-a x +1\right )}{a}\) | \(34\) |
parallelrisch | \(-\frac {-a^{2} c \,x^{2}+4 c \ln \left (x \right ) a x -8 c \ln \left (a x -1\right ) a x -c}{a^{2} x}\) | \(40\) |
norman | \(\frac {a^{2} c \,x^{3}-\frac {c}{a}}{x \left (a x -1\right ) a}-\frac {4 c \ln \left (x \right )}{a}+\frac {8 c \ln \left (a x -1\right )}{a}\) | \(51\) |
meijerg | \(-\frac {c \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a}+\frac {2 c \left (\frac {a x}{-a x +1}+\ln \left (-a x +1\right )\right )}{a}-\frac {2 c \left (\frac {2 a x}{-2 a x +2}-\ln \left (-a x +1\right )+1+\ln \left (x \right )+\ln \left (-a \right )\right )}{a}+\frac {c \left (-\frac {3 a x}{-3 a x +3}+2 \ln \left (-a x +1\right )-1-2 \ln \left (x \right )-2 \ln \left (-a \right )+\frac {1}{a x}\right )}{a}\) | \(141\) |
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Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=\frac {a^{2} c x^{2} + 8 \, a c x \log \left (a x - 1\right ) - 4 \, a c x \log \left (x\right ) + c}{a^{2} x} \]
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Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=c x + \frac {4 c \left (- \log {\left (x \right )} + 2 \log {\left (x - \frac {1}{a} \right )}\right )}{a} + \frac {c}{a^{2} x} \]
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Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=c x + \frac {8 \, c \log \left (a x - 1\right )}{a} - \frac {4 \, c \log \left (x\right )}{a} + \frac {c}{a^{2} x} \]
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Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.00 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=-\frac {4 \, c \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {4 \, c \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right )}{a} + \frac {{\left (a x - 1\right )} c}{a {\left (\frac {1}{a x - 1} + 1\right )}} \]
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Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int e^{4 \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx=c\,x+\frac {c}{a^2\,x}-\frac {4\,c\,\ln \left (x\right )}{a}+\frac {8\,c\,\ln \left (a\,x-1\right )}{a} \]
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