Integrand size = 22, antiderivative size = 127 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {16 e^{-\coth ^{-1}(a x)}}{35 a c^4}+\frac {e^{-\coth ^{-1}(a x)} (1+6 a x)}{35 a c^4 \left (1-a^2 x^2\right )^3}+\frac {2 e^{-\coth ^{-1}(a x)} (1+4 a x)}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {8 e^{-\coth ^{-1}(a x)} (1+2 a x)}{35 a c^4 \left (1-a^2 x^2\right )} \]
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Time = 0.11 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6320, 6318} \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {8 (2 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )}+\frac {2 (4 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^2}+\frac {(6 a x+1) e^{-\coth ^{-1}(a x)}}{35 a c^4 \left (1-a^2 x^2\right )^3}-\frac {16 e^{-\coth ^{-1}(a x)}}{35 a c^4} \]
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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 = 0.47 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.63 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (-5+30 a x+30 a^2 x^2-40 a^3 x^3-40 a^4 x^4+16 a^5 x^5+16 a^6 x^6\right )}{35 (-1+a x)^3 (c+a c x)^4} \]
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Time = 0.51 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (16 a^{6} x^{6}+16 a^{5} x^{5}-40 a^{4} x^{4}-40 a^{3} x^{3}+30 a^{2} x^{2}+30 a x -5\right )}{35 \left (a^{2} x^{2}-1\right )^{3} c^{4} a}\) | \(81\) |
default | \(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (16 a^{6} x^{6}+16 a^{5} x^{5}-40 a^{4} x^{4}-40 a^{3} x^{3}+30 a^{2} x^{2}+30 a x -5\right )}{35 c^{4} \left (a x +1\right )^{3} \left (a x -1\right )^{3} a}\) | \(84\) |
trager | \(-\frac {\left (16 a^{6} x^{6}+16 a^{5} x^{5}-40 a^{4} x^{4}-40 a^{3} x^{3}+30 a^{2} x^{2}+30 a x -5\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{35 a \,c^{4} \left (a x -1\right )^{3} \left (a x +1\right )^{3}}\) | \(86\) |
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Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=-\frac {{\left (16 \, a^{6} x^{6} + 16 \, a^{5} x^{5} - 40 \, a^{4} x^{4} - 40 \, a^{3} x^{3} + 30 \, a^{2} x^{2} + 30 \, a x - 5\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{7} c^{4} x^{6} - 3 \, a^{5} c^{4} x^{4} + 3 \, a^{3} c^{4} x^{2} - a c^{4}\right )}} \]
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\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {\int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{8} x^{8} - 4 a^{6} x^{6} + 6 a^{4} x^{4} - 4 a^{2} x^{2} + 1}\, dx}{c^{4}} \]
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Time = 0.20 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {1}{2240} \, a {\left (\frac {5 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 42 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 175 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 700 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{4}} + \frac {7 \, {\left (\frac {10 \, {\left (a x - 1\right )}}{a x + 1} - \frac {75 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 1\right )}}{a^{2} c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}}\right )} \]
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\[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\int { \frac {\sqrt {\frac {a x - 1}{a x + 1}}}{{\left (a^{2} c x^{2} - c\right )}^{4}} \,d x } \]
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Time = 4.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\left (c-a^2 c x^2\right )^4} \, dx=\frac {5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{64\,a\,c^4}-\frac {5\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{16\,a\,c^4}-\frac {3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{160\,a\,c^4}+\frac {{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{448\,a\,c^4}-\frac {\frac {15\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,\left (a\,x-1\right )}{a\,x+1}+\frac {1}{5}}{64\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]
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