Integrand size = 27, antiderivative size = 148 \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {4 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \]
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Time = 0.16 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {1653, 1651, 673, 665, 677, 223, 209} \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {4 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}+\frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2} \]
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Rule 209
Rule 223
Rule 665
Rule 673
Rule 677
Rule 1651
Rule 1653
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac {\int \frac {\sqrt {d^2-e^2 x^2} \left (2 d^3 e^2+5 d^2 e^3 x+4 d e^4 x^2\right )}{(d+e x)^4} \, dx}{e^5} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac {\int \left (\frac {d^3 e^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^4}-\frac {3 d^2 e^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^3}+\frac {4 d e^2 \sqrt {d^2-e^2 x^2}}{(d+e x)^2}\right ) \, dx}{e^5} \\ & = -\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}-\frac {(4 d) \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^2} \, dx}{e^3}+\frac {\left (3 d^2\right ) \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{e^3}-\frac {d^3 \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx}{e^3} \\ & = \frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {d \left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {(4 d) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx}{e^3}-\frac {d^2 \int \frac {\sqrt {d^2-e^2 x^2}}{(d+e x)^3} \, dx}{5 e^3} \\ & = \frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {(4 d) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )}{e^3} \\ & = \frac {8 d \sqrt {d^2-e^2 x^2}}{e^4 (d+e x)}+\frac {d^2 \left (d^2-e^2 x^2\right )^{3/2}}{5 e^4 (d+e x)^4}-\frac {14 d \left (d^2-e^2 x^2\right )^{3/2}}{15 e^4 (d+e x)^3}-\frac {\left (d^2-e^2 x^2\right )^{3/2}}{e^4 (d+e x)^2}+\frac {4 d \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e^4} \\ \end{align*}
Time = 0.36 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.72 \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (94 d^3+222 d^2 e x+149 d e^2 x^2+15 e^3 x^3\right )}{15 e^4 (d+e x)^3}+\frac {4 d \sqrt {-e^2} \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{e^5} \]
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Time = 0.43 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.26
method | result | size |
risch | \(\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e^{4}}+\frac {4 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{3} \sqrt {e^{2}}}+\frac {104 d \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 e^{5} \left (x +\frac {d}{e}\right )}-\frac {31 d^{2} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{15 e^{6} \left (x +\frac {d}{e}\right )^{2}}+\frac {2 d^{3} \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{5 e^{7} \left (x +\frac {d}{e}\right )^{3}}\) | \(186\) |
default | \(\frac {\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}}{e^{4}}-\frac {d^{3} \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{5 d e \left (x +\frac {d}{e}\right )^{4}}-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{15 d^{2} \left (x +\frac {d}{e}\right )^{3}}\right )}{e^{7}}-\frac {3 d \left (-\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{d e \left (x +\frac {d}{e}\right )^{2}}-\frac {e \left (\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}+\frac {d e \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{\sqrt {e^{2}}}\right )}{d}\right )}{e^{5}}-\frac {d \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{e^{7} \left (x +\frac {d}{e}\right )^{3}}\) | \(349\) |
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Time = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.18 \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {94 \, d e^{3} x^{3} + 282 \, d^{2} e^{2} x^{2} + 282 \, d^{3} e x + 94 \, d^{4} - 120 \, {\left (d e^{3} x^{3} + 3 \, d^{2} e^{2} x^{2} + 3 \, d^{3} e x + d^{4}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (15 \, e^{3} x^{3} + 149 \, d e^{2} x^{2} + 222 \, d^{2} e x + 94 \, d^{3}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \]
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\[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int \frac {x^{3} \sqrt {- \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{4}}\, dx \]
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Timed out. \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\text {Timed out} \]
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Time = 0.31 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.43 \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\frac {4 \, d \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{e^{3} {\left | e \right |}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e^{4}} - \frac {2 \, {\left (79 \, d + \frac {335 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d}{e^{2} x} + \frac {505 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d}{e^{4} x^{2}} + \frac {285 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3} d}{e^{6} x^{3}} + \frac {60 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4} d}{e^{8} x^{4}}\right )}}{15 \, e^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} + 1\right )}^{5} {\left | e \right |}} \]
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Timed out. \[ \int \frac {x^3 \sqrt {d^2-e^2 x^2}}{(d+e x)^4} \, dx=\int \frac {x^3\,\sqrt {d^2-e^2\,x^2}}{{\left (d+e\,x\right )}^4} \,d x \]
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