Integrand size = 20, antiderivative size = 119 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {16 e^{\coth ^{-1}(a x)}}{35 a c^4}-\frac {e^{\coth ^{-1}(a x)} (1-6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}-\frac {2 e^{\coth ^{-1}(a x)} (1-4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}-\frac {8 e^{\coth ^{-1}(a x)} (1-2 a x)}{35 a c^4 \left (1-a^2 x^2\right )} \]
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Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6320, 6318} \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {(1-6 a x) e^{\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}-\frac {8 (1-2 a x) e^{\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )}-\frac {2 (1-4 a x) e^{\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {16 e^{\coth ^{-1}(a x)}}{35 a c^4} \]
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Rule 6318
Rule 6320
Rubi steps \begin{align*} \text {integral}& = -\frac {e^{\coth ^{-1}(a x)} (1-6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}+\frac {6 \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^3} \, dx}{7 c} \\ & = -\frac {e^{\coth ^{-1}(a x)} (1-6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}-\frac {2 e^{\coth ^{-1}(a x)} (1-4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {24 \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^2} \, dx}{35 c^2} \\ & = -\frac {e^{\coth ^{-1}(a x)} (1-6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}-\frac {2 e^{\coth ^{-1}(a x)} (1-4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}-\frac {8 e^{\coth ^{-1}(a x)} (1-2 a x)}{35 a c^4 \left (1-a^2 x^2\right )}+\frac {16 \int \frac {e^{\coth ^{-1}(a x)}}{c-a^2 c x^2} \, dx}{35 c^3} \\ & = \frac {16 e^{\coth ^{-1}(a x)}}{35 a c^4}-\frac {e^{\coth ^{-1}(a x)} (1-6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}-\frac {2 e^{\coth ^{-1}(a x)} (1-4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}-\frac {8 e^{\coth ^{-1}(a x)} (1-2 a x)}{35 a c^4 \left (1-a^2 x^2\right )} \\ \end{align*}
Time = 0.46 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-5-30 a x+30 a^2 x^2+40 a^3 x^3-40 a^4 x^4-16 a^5 x^5+16 a^6 x^6\right )}{35 c^4 (-1+a x)^4 (1+a x)^3} \]
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Time = 0.48 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.68
method | result | size |
gosper | \(\frac {16 a^{6} x^{6}-16 a^{5} x^{5}-40 a^{4} x^{4}+40 a^{3} x^{3}+30 a^{2} x^{2}-30 a x -5}{35 \left (a^{2} x^{2}-1\right )^{3} c^{4} \sqrt {\frac {a x -1}{a x +1}}\, a}\) | \(81\) |
default | \(\frac {16 a^{6} x^{6}-16 a^{5} x^{5}-40 a^{4} x^{4}+40 a^{3} x^{3}+30 a^{2} x^{2}-30 a x -5}{35 \sqrt {\frac {a x -1}{a x +1}}\, c^{4} \left (a x -1\right )^{3} \left (a x +1\right )^{3} a}\) | \(84\) |
trager | \(\frac {\left (16 a^{6} x^{6}-16 a^{5} x^{5}-40 a^{4} x^{4}+40 a^{3} x^{3}+30 a^{2} x^{2}-30 a x -5\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{35 a \,c^{4} \left (a x +1\right )^{2} \left (a x -1\right )^{4}}\) | \(86\) |
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Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {{\left (16 \, a^{6} x^{6} - 16 \, a^{5} x^{5} - 40 \, a^{4} x^{4} + 40 \, a^{3} x^{3} + 30 \, a^{2} x^{2} - 30 \, a x - 5\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{7} c^{4} x^{6} - 2 \, a^{6} c^{4} x^{5} - a^{5} c^{4} x^{4} + 4 \, a^{4} c^{4} x^{3} - a^{3} c^{4} x^{2} - 2 \, a^{2} c^{4} x + a c^{4}\right )}} \]
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\[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\int \frac {1}{a^{8} x^{8} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - 4 a^{6} x^{6} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} + 6 a^{4} x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - 4 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} + \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx}{c^{4}} \]
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Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {1}{2240} \, a {\left (\frac {7 \, {\left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 10 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 75 \, \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{a^{2} c^{4}} + \frac {\frac {42 \, {\left (a x - 1\right )}}{a x + 1} - \frac {175 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {700 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 5}{a^{2} c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}}}\right )} \]
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\[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\int { \frac {1}{{\left (a^{2} c x^{2} - c\right )}^{4} \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]
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Time = 0.07 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {15\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{64\,a\,c^4}-\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{32\,a\,c^4}+\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{320\,a\,c^4}-\frac {\frac {5\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {20\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}-\frac {6\,\left (a\,x-1\right )}{5\,\left (a\,x+1\right )}+\frac {1}{7}}{64\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}} \]
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