Integrand size = 22, antiderivative size = 53 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {x}{c}-\frac {2}{a c (1-a x)^2}+\frac {8}{a c (1-a x)}+\frac {5 \log (1-a x)}{a c} \]
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Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6302, 6266, 6264, 78} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {8}{a c (1-a x)}-\frac {2}{a c (1-a x)^2}+\frac {5 \log (1-a x)}{a c}+\frac {x}{c} \]
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Rule 78
Rule 6264
Rule 6266
Rule 6302
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{4 \text {arctanh}(a x)}}{c-\frac {c}{a x}} \, dx \\ & = -\frac {a \int \frac {e^{4 \text {arctanh}(a x)} x}{1-a x} \, dx}{c} \\ & = -\frac {a \int \frac {x (1+a x)^2}{(1-a x)^3} \, dx}{c} \\ & = -\frac {a \int \left (-\frac {1}{a}-\frac {4}{a (-1+a x)^3}-\frac {8}{a (-1+a x)^2}-\frac {5}{a (-1+a x)}\right ) \, dx}{c} \\ & = \frac {x}{c}-\frac {2}{a c (1-a x)^2}+\frac {8}{a c (1-a x)}+\frac {5 \log (1-a x)}{a c} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=-\frac {a \left (-\frac {x}{a}+\frac {2}{a^2 (1-a x)^2}-\frac {8}{a^2 (1-a x)}-\frac {5 \log (1-a x)}{a^2}\right )}{c} \]
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Time = 0.58 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {x}{c}+\frac {-8 c x +\frac {6 c}{a}}{c^{2} \left (a x -1\right )^{2}}+\frac {5 \ln \left (a x -1\right )}{a c}\) | \(43\) |
default | \(\frac {a \left (\frac {x}{a}-\frac {2}{a^{2} \left (a x -1\right )^{2}}-\frac {8}{a^{2} \left (a x -1\right )}+\frac {5 \ln \left (a x -1\right )}{a^{2}}\right )}{c}\) | \(47\) |
norman | \(\frac {\frac {a^{2} x^{3}}{c}-\frac {8 a \,x^{2}}{c}+\frac {5 x}{c}}{\left (a x -1\right )^{2}}+\frac {5 \ln \left (a x -1\right )}{a c}\) | \(50\) |
parallelrisch | \(\frac {a^{3} x^{3}+5 a^{2} \ln \left (a x -1\right ) x^{2}-8 a^{2} x^{2}-10 a \ln \left (a x -1\right ) x +5 a x +5 \ln \left (a x -1\right )}{\left (a x -1\right )^{2} c a}\) | \(67\) |
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Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.21 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {a^{3} x^{3} - 2 \, a^{2} x^{2} - 7 \, a x + 5 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 6}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} \]
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Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {- 8 a x + 6}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac {x}{c} + \frac {5 \log {\left (a x - 1 \right )}}{a c} \]
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Time = 0.20 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.92 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=-\frac {2 \, {\left (4 \, a x - 3\right )}}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} + \frac {x}{c} + \frac {5 \, \log \left (a x - 1\right )}{a c} \]
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Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.40 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {a x - 1}{a c} - \frac {5 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a c} - \frac {2 \, {\left (\frac {4 \, a^{3} c}{a x - 1} + \frac {a^{3} c}{{\left (a x - 1\right )}^{2}}\right )}}{a^{4} c^{2}} \]
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Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx=\frac {x}{c}-\frac {8\,x-\frac {6}{a}}{c\,a^2\,x^2-2\,c\,a\,x+c}+\frac {5\,\ln \left (a\,x-1\right )}{a\,c} \]
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