Integrand size = 29, antiderivative size = 61 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {(e f+d g)^2}{2 e^3 (d-e x)^2}-\frac {2 g (e f+d g)}{e^3 (d-e x)}-\frac {g^2 \log (d-e x)}{e^3} \]
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Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {862, 45} \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {2 g (d g+e f)}{e^3 (d-e x)}+\frac {(d g+e f)^2}{2 e^3 (d-e x)^2}-\frac {g^2 \log (d-e x)}{e^3} \]
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Rule 45
Rule 862
Rubi steps \begin{align*} \text {integral}& = \int \frac {(f+g x)^2}{(d-e x)^3} \, dx \\ & = \int \left (\frac {(e f+d g)^2}{e^2 (d-e x)^3}-\frac {2 g (e f+d g)}{e^2 (d-e x)^2}+\frac {g^2}{e^2 (d-e x)}\right ) \, dx \\ & = \frac {(e f+d g)^2}{2 e^3 (d-e x)^2}-\frac {2 g (e f+d g)}{e^3 (d-e x)}-\frac {g^2 \log (d-e x)}{e^3} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {\frac {(e f+d g) (-3 d g+e (f+4 g x))}{(d-e x)^2}-2 g^2 \log (d-e x)}{2 e^3} \]
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Time = 0.46 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13
method | result | size |
risch | \(\frac {\frac {2 g \left (d g +e f \right ) x}{e^{2}}-\frac {3 d^{2} g^{2}+2 d e f g -e^{2} f^{2}}{2 e^{3}}}{\left (-e x +d \right )^{2}}-\frac {g^{2} \ln \left (-e x +d \right )}{e^{3}}\) | \(69\) |
default | \(-\frac {g^{2} \ln \left (-e x +d \right )}{e^{3}}-\frac {2 g \left (d g +e f \right )}{e^{3} \left (-e x +d \right )}-\frac {-d^{2} g^{2}-2 d e f g -e^{2} f^{2}}{2 e^{3} \left (-e x +d \right )^{2}}\) | \(74\) |
parallelrisch | \(-\frac {2 \ln \left (e x -d \right ) x^{2} e^{2} g^{2}-4 \ln \left (e x -d \right ) x d e \,g^{2}+2 \ln \left (e x -d \right ) d^{2} g^{2}-4 x d e \,g^{2}-4 x \,e^{2} f g +3 d^{2} g^{2}+2 d e f g -e^{2} f^{2}}{2 e^{3} \left (e x -d \right )^{2}}\) | \(105\) |
norman | \(\frac {\left (2 d \,g^{2}+2 f g e \right ) x^{3}-\frac {d^{2} \left (3 d^{2} g^{2} e +2 d f g \,e^{2}-f^{2} e^{3}\right )}{2 e^{4}}+\frac {\left (5 d^{2} g^{2} e +6 d f g \,e^{2}+f^{2} e^{3}\right ) x^{2}}{2 e^{2}}-\frac {d \left (d^{2} g^{2}-e^{2} f^{2}\right ) x}{e^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {g^{2} \ln \left (-e x +d \right )}{e^{3}}\) | \(139\) |
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Time = 0.26 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.64 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {e^{2} f^{2} - 2 \, d e f g - 3 \, d^{2} g^{2} + 4 \, {\left (e^{2} f g + d e g^{2}\right )} x - 2 \, {\left (e^{2} g^{2} x^{2} - 2 \, d e g^{2} x + d^{2} g^{2}\right )} \log \left (e x - d\right )}{2 \, {\left (e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.36 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=- \frac {3 d^{2} g^{2} + 2 d e f g - e^{2} f^{2} + x \left (- 4 d e g^{2} - 4 e^{2} f g\right )}{2 d^{2} e^{3} - 4 d e^{4} x + 2 e^{5} x^{2}} - \frac {g^{2} \log {\left (- d + e x \right )}}{e^{3}} \]
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Time = 0.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.33 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {e^{2} f^{2} - 2 \, d e f g - 3 \, d^{2} g^{2} + 4 \, {\left (e^{2} f g + d e g^{2}\right )} x}{2 \, {\left (e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}\right )}} - \frac {g^{2} \log \left (e x - d\right )}{e^{3}} \]
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Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {g^{2} \log \left ({\left | e x - d \right |}\right )}{e^{3}} + \frac {4 \, {\left (e f g + d g^{2}\right )} x + \frac {e^{2} f^{2} - 2 \, d e f g - 3 \, d^{2} g^{2}}{e}}{2 \, {\left (e x - d\right )}^{2} e^{2}} \]
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Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.31 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {\frac {3\,d^2\,g^2+2\,d\,e\,f\,g-e^2\,f^2}{2\,e^3}-\frac {2\,g\,x\,\left (d\,g+e\,f\right )}{e^2}}{d^2-2\,d\,e\,x+e^2\,x^2}-\frac {g^2\,\ln \left (e\,x-d\right )}{e^3} \]
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