Integrand size = 25, antiderivative size = 271 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {11 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac {5 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 \left (c-a^2 c x^2\right )^{5/2}} \]
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Time = 0.19 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6327, 6328, 90} \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {11 a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac {5 a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 \left (c-a^2 c x^2\right )^{5/2}} \]
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Rule 90
Rule 6327
Rule 6328
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^6} \, dx}{\left (c-a^2 c x^2\right )^{5/2}} \\ & = \frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {1}{x (-1+a x)^3 (1+a x)^2} \, dx}{\left (c-a^2 c x^2\right )^{5/2}} \\ & = \frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \left (-\frac {1}{x}+\frac {a}{4 (-1+a x)^3}-\frac {a}{2 (-1+a x)^2}+\frac {11 a}{16 (-1+a x)}+\frac {a}{8 (1+a x)^2}+\frac {5 a}{16 (1+a x)}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}} \\ & = -\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {11 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac {5 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 \left (c-a^2 c x^2\right )^{5/2}} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.32 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \left (-\frac {2}{(-1+a x)^2}+\frac {8}{-1+a x}-\frac {2}{1+a x}-16 \log (x)+11 \log (1-a x)+5 \log (1+a x)\right )}{16 \left (c-a^2 c x^2\right )^{5/2}} \]
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Time = 0.53 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.72
method | result | size |
default | \(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (5 a^{3} \ln \left (a x +1\right ) x^{3}-16 a^{3} \ln \left (x \right ) x^{3}+11 a^{3} \ln \left (a x -1\right ) x^{3}-5 a^{2} \ln \left (a x +1\right ) x^{2}+16 a^{2} \ln \left (x \right ) x^{2}-11 a^{2} \ln \left (a x -1\right ) x^{2}+6 a^{2} x^{2}-5 a \ln \left (a x +1\right ) x +16 a \ln \left (x \right ) x -11 a \ln \left (a x -1\right ) x +2 a x +5 \ln \left (a x +1\right )-16 \ln \left (x \right )+11 \ln \left (a x -1\right )-12\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} \left (a x +1\right )}\) | \(196\) |
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Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.54 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (6 \, a^{2} x^{2} + 2 \, a x + 5 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 11 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (x\right ) - 12\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{4} c^{3} x^{3} - a^{3} c^{3} x^{2} - a^{2} c^{3} x + a c^{3}\right )}} \]
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Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]
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\[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]
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Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {1}{x\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]
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