Integrand size = 27, antiderivative size = 119 \[ \int \frac {1}{x (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6} \]
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Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {871, 837, 12, 272, 65, 214} \[ \int \frac {1}{x (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}} \]
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Rule 12
Rule 65
Rule 214
Rule 272
Rule 837
Rule 871
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-5 d e^2+4 e^3 x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2} \\ & = \frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-15 d^3 e^4+8 d^2 e^5 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^4} \\ & = \frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {15 d^5 e^6}{x \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{10} e^6} \\ & = \frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^5} \\ & = \frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {d^2-e^2 x}} \, dx,x,x^2\right )}{2 d^5} \\ & = \frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\text {Subst}\left (\int \frac {1}{\frac {d^2}{e^2}-\frac {x^2}{e^2}} \, dx,x,\sqrt {d^2-e^2 x^2}\right )}{d^5 e^2} \\ & = \frac {5 d-4 e x}{15 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {1}{5 d^2 (d+e x) \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d-8 e x}{15 d^6 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^6} \\ \end{align*}
Time = 0.47 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (23 d^4+8 d^3 e x-27 d^2 e^2 x^2-7 d e^3 x^3+8 e^4 x^4\right )}{(d-e x)^2 (d+e x)^3}+30 \text {arctanh}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^6} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(254\) vs. \(2(105)=210\).
Time = 0.38 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.14
method | result | size |
default | \(\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d}-\frac {-\frac {1}{5 d e \left (x +\frac {d}{e}\right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}+\frac {4 e \left (-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{6 d^{2} e^{2} \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}-\frac {-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e}{3 e^{2} d^{4} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{5 d}}{d}\) | \(255\) |
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Leaf count of result is larger than twice the leaf count of optimal. 237 vs. \(2 (105) = 210\).
Time = 0.30 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.99 \[ \int \frac {1}{x (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {23 \, e^{5} x^{5} + 23 \, d e^{4} x^{4} - 46 \, d^{2} e^{3} x^{3} - 46 \, d^{3} e^{2} x^{2} + 23 \, d^{4} e x + 23 \, d^{5} + 15 \, {\left (e^{5} x^{5} + d e^{4} x^{4} - 2 \, d^{2} e^{3} x^{3} - 2 \, d^{3} e^{2} x^{2} + d^{4} e x + d^{5}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) + {\left (8 \, e^{4} x^{4} - 7 \, d e^{3} x^{3} - 27 \, d^{2} e^{2} x^{2} + 8 \, d^{3} e x + 23 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{6} e^{5} x^{5} + d^{7} e^{4} x^{4} - 2 \, d^{8} e^{3} x^{3} - 2 \, d^{9} e^{2} x^{2} + d^{10} e x + d^{11}\right )}} \]
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\[ \int \frac {1}{x (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \]
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\[ \int \frac {1}{x (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )} x} \,d x } \]
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\[ \int \frac {1}{x (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )} x} \,d x } \]
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Timed out. \[ \int \frac {1}{x (d+e x) \left (d^2-e^2 x^2\right )^{5/2}} \, dx=\int \frac {1}{x\,{\left (d^2-e^2\,x^2\right )}^{5/2}\,\left (d+e\,x\right )} \,d x \]
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