Integrand size = 18, antiderivative size = 53 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {4}{5 a c^4 (1-a x)^5}-\frac {1}{a c^4 (1-a x)^4}+\frac {1}{3 a c^4 (1-a x)^3} \]
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Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6302, 6264, 45} \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {1}{3 a c^4 (1-a x)^3}-\frac {1}{a c^4 (1-a x)^4}+\frac {4}{5 a c^4 (1-a x)^5} \]
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Rule 45
Rule 6264
Rule 6302
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{4 \text {arctanh}(a x)}}{(c-a c x)^4} \, dx \\ & = \frac {\int \frac {(1+a x)^2}{(1-a x)^6} \, dx}{c^4} \\ & = \frac {\int \left (\frac {4}{(-1+a x)^6}+\frac {4}{(-1+a x)^5}+\frac {1}{(-1+a x)^4}\right ) \, dx}{c^4} \\ & = \frac {4}{5 a c^4 (1-a x)^5}-\frac {1}{a c^4 (1-a x)^4}+\frac {1}{3 a c^4 (1-a x)^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.58 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {2+5 a x+5 a^2 x^2}{15 a c^4 (-1+a x)^5} \]
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Time = 0.54 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.51
method | result | size |
risch | \(\frac {-\frac {a \,x^{2}}{3}-\frac {x}{3}-\frac {2}{15 a}}{\left (a x -1\right )^{5} c^{4}}\) | \(27\) |
gosper | \(-\frac {5 a^{2} x^{2}+5 a x +2}{15 \left (a x -1\right )^{5} a \,c^{4}}\) | \(30\) |
default | \(\frac {-\frac {1}{a \left (a x -1\right )^{4}}-\frac {1}{3 a \left (a x -1\right )^{3}}-\frac {4}{5 a \left (a x -1\right )^{5}}}{c^{4}}\) | \(42\) |
parallelrisch | \(\frac {-2 a^{4} x^{5}+10 a^{3} x^{4}-20 a^{2} x^{3}+15 a \,x^{2}-15 x}{15 \left (a x -1\right )^{5} c^{4}}\) | \(47\) |
norman | \(\frac {\frac {a \,x^{2}}{c}-\frac {x}{c}-\frac {4 a^{2} x^{3}}{3 c}+\frac {2 a^{3} x^{4}}{3 c}-\frac {2 a^{4} x^{5}}{15 c}}{\left (a x -1\right )^{5} c^{3}}\) | \(60\) |
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Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.45 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {5 \, a^{2} x^{2} + 5 \, a x + 2}{15 \, {\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, a^{2} c^{4} x - a c^{4}\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.51 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {- 5 a^{2} x^{2} - 5 a x - 2}{15 a^{6} c^{4} x^{5} - 75 a^{5} c^{4} x^{4} + 150 a^{4} c^{4} x^{3} - 150 a^{3} c^{4} x^{2} + 75 a^{2} c^{4} x - 15 a c^{4}} \]
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Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.45 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {5 \, a^{2} x^{2} + 5 \, a x + 2}{15 \, {\left (a^{6} c^{4} x^{5} - 5 \, a^{5} c^{4} x^{4} + 10 \, a^{4} c^{4} x^{3} - 10 \, a^{3} c^{4} x^{2} + 5 \, a^{2} c^{4} x - a c^{4}\right )}} \]
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Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\frac {5}{{\left (a x - 1\right )}^{3} a} + \frac {15}{{\left (a x - 1\right )}^{4} a} + \frac {12}{{\left (a x - 1\right )}^{5} a}}{15 \, c^{4}} \]
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Time = 4.53 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.55 \[ \int \frac {e^{4 \coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {5\,a^2\,x^2+5\,a\,x+2}{15\,a\,c^4\,{\left (a\,x-1\right )}^5} \]
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