Integrand size = 23, antiderivative size = 116 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2} \]
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Time = 0.02 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {794, 201, 223, 209} \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^6 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e} \]
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Rule 201
Rule 209
Rule 223
Rule 794
Rubi steps \begin{align*} \text {integral}& = -\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^2 \int \left (d^2-e^2 x^2\right )^{3/2} \, dx}{6 e} \\ & = \frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^4 \int \sqrt {d^2-e^2 x^2} \, dx}{8 e} \\ & = \frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{16 e} \\ & = \frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e} \\ & = \frac {d^4 x \sqrt {d^2-e^2 x^2}}{16 e}+\frac {d^2 x \left (d^2-e^2 x^2\right )^{3/2}}{24 e}-\frac {(6 d+5 e x) \left (d^2-e^2 x^2\right )^{5/2}}{30 e^2}+\frac {d^6 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{16 e^2} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {\sqrt {d^2-e^2 x^2} \left (48 d^5+15 d^4 e x-96 d^3 e^2 x^2-70 d^2 e^3 x^3+48 d e^4 x^4+40 e^5 x^5\right )+30 d^6 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{240 e^2} \]
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Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.93
method | result | size |
risch | \(-\frac {\left (40 e^{5} x^{5}+48 d \,e^{4} x^{4}-70 d^{2} e^{3} x^{3}-96 d^{3} e^{2} x^{2}+15 d^{4} e x +48 d^{5}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{240 e^{2}}+\frac {d^{6} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{16 e \sqrt {e^{2}}}\) | \(108\) |
default | \(e \left (-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6 e^{2}}+\frac {d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6 e^{2}}\right )-\frac {d \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{5 e^{2}}\) | \(126\) |
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Time = 0.35 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=-\frac {30 \, d^{6} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (40 \, e^{5} x^{5} + 48 \, d e^{4} x^{4} - 70 \, d^{2} e^{3} x^{3} - 96 \, d^{3} e^{2} x^{2} + 15 \, d^{4} e x + 48 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{240 \, e^{2}} \]
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Time = 0.55 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.36 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\begin {cases} \frac {d^{6} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 e} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{5}}{5 e^{2}} - \frac {d^{4} x}{16 e} + \frac {2 d^{3} x^{2}}{5} + \frac {7 d^{2} e x^{3}}{24} - \frac {d e^{2} x^{4}}{5} - \frac {e^{3} x^{5}}{6}\right ) & \text {for}\: e^{2} \neq 0 \\\left (\frac {d x^{2}}{2} + \frac {e x^{3}}{3}\right ) \left (d^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]
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Time = 0.29 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.98 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^{6} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{16 \, \sqrt {e^{2}} e} + \frac {\sqrt {-e^{2} x^{2} + d^{2}} d^{4} x}{16 \, e} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{2} x}{24 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} x}{6 \, e} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d}{5 \, e^{2}} \]
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Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.81 \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\frac {d^{6} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{16 \, e {\left | e \right |}} - \frac {1}{240} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {48 \, d^{5}}{e^{2}} + {\left (\frac {15 \, d^{4}}{e} - 2 \, {\left (48 \, d^{3} + {\left (35 \, d^{2} e - 4 \, {\left (5 \, e^{3} x + 6 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} \]
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Timed out. \[ \int x (d+e x) \left (d^2-e^2 x^2\right )^{3/2} \, dx=\int x\,{\left (d^2-e^2\,x^2\right )}^{3/2}\,\left (d+e\,x\right ) \,d x \]
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