Integrand size = 25, antiderivative size = 94 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^2 (d+e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d-e x)}{15 d e^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 x}{15 d^3 e^2 \sqrt {d^2-e^2 x^2}} \]
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Time = 0.03 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {810, 792, 197} \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^2 (d+e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d-e x)}{15 d e^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 x}{15 d^3 e^2 \sqrt {d^2-e^2 x^2}} \]
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Rule 197
Rule 792
Rule 810
Rubi steps \begin{align*} \text {integral}& = \frac {x^2 (d+e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {\int \frac {x \left (2 d^2 e-2 d e^2 x\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d^2 e^2} \\ & = \frac {x^2 (d+e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d-e x)}{15 d e^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d e^2} \\ & = \frac {x^2 (d+e x)}{5 d e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d-e x)}{15 d e^3 \left (d^2-e^2 x^2\right )^{3/2}}-\frac {2 x}{15 d^3 e^2 \sqrt {d^2-e^2 x^2}} \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-2 d^4+2 d^3 e x+3 d^2 e^2 x^2+2 d e^3 x^3-2 e^4 x^4\right )}{15 d^3 e^3 (d-e x)^3 (d+e x)^2} \]
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Time = 0.35 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82
method | result | size |
gosper | \(-\frac {\left (-e x +d \right ) \left (e x +d \right )^{2} \left (2 e^{4} x^{4}-2 d \,e^{3} x^{3}-3 d^{2} e^{2} x^{2}-2 d^{3} e x +2 d^{4}\right )}{15 d^{3} e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\) | \(77\) |
trager | \(-\frac {\left (2 e^{4} x^{4}-2 d \,e^{3} x^{3}-3 d^{2} e^{2} x^{2}-2 d^{3} e x +2 d^{4}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{3} e^{3} \left (-e x +d \right )^{3} \left (e x +d \right )^{2}}\) | \(79\) |
default | \(e \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+d \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )\) | \(147\) |
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Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (83) = 166\).
Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.84 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {2 \, e^{5} x^{5} - 2 \, d e^{4} x^{4} - 4 \, d^{2} e^{3} x^{3} + 4 \, d^{3} e^{2} x^{2} + 2 \, d^{4} e x - 2 \, d^{5} - {\left (2 \, e^{4} x^{4} - 2 \, d e^{3} x^{3} - 3 \, d^{2} e^{2} x^{2} - 2 \, d^{3} e x + 2 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{8} x^{5} - d^{4} e^{7} x^{4} - 2 \, d^{5} e^{6} x^{3} + 2 \, d^{6} e^{5} x^{2} + d^{7} e^{4} x - d^{8} e^{3}\right )}} \]
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Result contains complex when optimal does not.
Time = 7.11 (sec) , antiderivative size = 513, normalized size of antiderivative = 5.46 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=d \left (\begin {cases} - \frac {5 i d^{2} x^{3}}{15 d^{9} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} - 30 d^{7} e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} + 15 d^{5} e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} + \frac {2 i e^{2} x^{5}}{15 d^{9} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} - 30 d^{7} e^{2} x^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}} + 15 d^{5} e^{4} x^{4} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {5 d^{2} x^{3}}{15 d^{9} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} - 30 d^{7} e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} + 15 d^{5} e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} - \frac {2 e^{2} x^{5}}{15 d^{9} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} - 30 d^{7} e^{2} x^{2} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}} + 15 d^{5} e^{4} x^{4} \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}} & \text {otherwise} \end {cases}\right ) + e \left (\begin {cases} - \frac {2 d^{2}}{15 d^{4} e^{4} \sqrt {d^{2} - e^{2} x^{2}} - 30 d^{2} e^{6} x^{2} \sqrt {d^{2} - e^{2} x^{2}} + 15 e^{8} x^{4} \sqrt {d^{2} - e^{2} x^{2}}} + \frac {5 e^{2} x^{2}}{15 d^{4} e^{4} \sqrt {d^{2} - e^{2} x^{2}} - 30 d^{2} e^{6} x^{2} \sqrt {d^{2} - e^{2} x^{2}} + 15 e^{8} x^{4} \sqrt {d^{2} - e^{2} x^{2}}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \left (d^{2}\right )^{\frac {7}{2}}} & \text {otherwise} \end {cases}\right ) \]
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none
Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.19 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{2}} - \frac {2 \, d^{2}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e^{3}} - \frac {x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d e^{2}} - \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3} e^{2}} \]
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\[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int { \frac {{\left (e x + d\right )} x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} \,d x } \]
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Time = 11.44 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83 \[ \int \frac {x^2 (d+e x)}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (-2\,d^4+2\,d^3\,e\,x+3\,d^2\,e^2\,x^2+2\,d\,e^3\,x^3-2\,e^4\,x^4\right )}{15\,d^3\,e^3\,{\left (d+e\,x\right )}^2\,{\left (d-e\,x\right )}^3} \]
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