Integrand size = 27, antiderivative size = 117 \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \]
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Time = 0.10 (sec) , antiderivative size = 117, 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 = {1819, 837, 12, 272, 65, 214} \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5}+\frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}} \]
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Rule 12
Rule 65
Rule 214
Rule 272
Rule 837
Rule 1819
Rubi steps \begin{align*} \text {integral}& = \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {-5 d^2-8 d e x}{x \left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2} \\ & = \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {\int \frac {-15 d^4 e^2-16 d^3 e^3 x}{x \left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^6 e^2} \\ & = \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\int -\frac {15 d^6 e^4}{x \sqrt {d^2-e^2 x^2}} \, dx}{15 d^{10} e^4} \\ & = \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}+\frac {\int \frac {1}{x \sqrt {d^2-e^2 x^2}} \, dx}{d^4} \\ & = \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \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^4} \\ & = \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \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^4 e^2} \\ & = \frac {2 (d+e x)}{5 d \left (d^2-e^2 x^2\right )^{5/2}}+\frac {5 d+8 e x}{15 d^3 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {15 d+16 e x}{15 d^5 \sqrt {d^2-e^2 x^2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {d^2-e^2 x^2}}{d}\right )}{d^5} \\ \end{align*}
Time = 0.48 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\frac {\sqrt {d^2-e^2 x^2} \left (26 d^3-22 d^2 e x-17 d e^2 x^2+16 e^3 x^3\right )}{(d-e x)^3 (d+e x)}+30 \text {arctanh}\left (\frac {\sqrt {-e^2} x-\sqrt {d^2-e^2 x^2}}{d}\right )}{15 d^5} \]
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Time = 0.36 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.73
method | result | size |
default | \(\frac {1}{5 \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+d^{2} \left (\frac {1}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {1}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {\frac {1}{d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\ln \left (\frac {2 d^{2}+2 \sqrt {d^{2}}\, \sqrt {-e^{2} x^{2}+d^{2}}}{x}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}}{d^{2}}\right )+2 d e \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )\) | \(202\) |
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Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.44 \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {26 \, e^{4} x^{4} - 52 \, d e^{3} x^{3} + 52 \, d^{3} e x - 26 \, d^{4} + 15 \, {\left (e^{4} x^{4} - 2 \, d e^{3} x^{3} + 2 \, d^{3} e x - d^{4}\right )} \log \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{x}\right ) - {\left (16 \, e^{3} x^{3} - 17 \, d e^{2} x^{2} - 22 \, d^{2} e x + 26 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{5} e^{4} x^{4} - 2 \, d^{6} e^{3} x^{3} + 2 \, d^{8} e x - d^{9}\right )}} \]
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\[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (d + e x\right )^{2}}{x \left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \]
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Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.32 \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 \, e x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d} + \frac {2}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {8 \, e x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{3}} + \frac {1}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {16 \, e x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{5}} - \frac {\log \left (\frac {2 \, d^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d}{{\left | x \right |}}\right )}{d^{5}} + \frac {1}{\sqrt {-e^{2} x^{2} + d^{2}} d^{4}} \]
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\[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} x} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^2}{x \left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{x\,{\left (d^2-e^2\,x^2\right )}^{7/2}} \,d x \]
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