Integrand size = 29, antiderivative size = 78 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {g (2 e f+3 d g) x}{e^2}+\frac {g^2 x^2}{2 e}+\frac {2 d (e f+d g)^2}{e^3 (d-e x)}+\frac {(e f+d g) (e f+5 d g) \log (d-e x)}{e^3} \]
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Time = 0.06 (sec) , antiderivative size = 78, 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, 78} \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {2 d (d g+e f)^2}{e^3 (d-e x)}+\frac {(5 d g+e f) (d g+e f) \log (d-e x)}{e^3}+\frac {g x (3 d g+2 e f)}{e^2}+\frac {g^2 x^2}{2 e} \]
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Rule 78
Rule 862
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x) (f+g x)^2}{(d-e x)^2} \, dx \\ & = \int \left (\frac {g (2 e f+3 d g)}{e^2}+\frac {g^2 x}{e}+\frac {(-e f-5 d g) (e f+d g)}{e^2 (d-e x)}+\frac {2 d (e f+d g)^2}{e^2 (-d+e x)^2}\right ) \, dx \\ & = \frac {g (2 e f+3 d g) x}{e^2}+\frac {g^2 x^2}{2 e}+\frac {2 d (e f+d g)^2}{e^3 (d-e x)}+\frac {(e f+d g) (e f+5 d g) \log (d-e x)}{e^3} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {2 e g (2 e f+3 d g) x+e^2 g^2 x^2+\frac {4 d (e f+d g)^2}{d-e x}+2 \left (e^2 f^2+6 d e f g+5 d^2 g^2\right ) \log (d-e x)}{2 e^3} \]
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Time = 0.43 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.19
method | result | size |
default | \(\frac {g \left (\frac {1}{2} e g \,x^{2}+3 d g x +2 e f x \right )}{e^{2}}+\frac {\left (5 d^{2} g^{2}+6 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}+\frac {2 d \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}\) | \(93\) |
risch | \(\frac {g^{2} x^{2}}{2 e}+\frac {3 g^{2} d x}{e^{2}}+\frac {2 g f x}{e}+\frac {5 \ln \left (-e x +d \right ) d^{2} g^{2}}{e^{3}}+\frac {6 \ln \left (-e x +d \right ) d f g}{e^{2}}+\frac {\ln \left (-e x +d \right ) f^{2}}{e}+\frac {2 d^{3} g^{2}}{e^{3} \left (-e x +d \right )}+\frac {4 d^{2} f g}{e^{2} \left (-e x +d \right )}+\frac {2 d \,f^{2}}{e \left (-e x +d \right )}\) | \(132\) |
norman | \(\frac {\frac {d \left (5 d^{2} g^{2}+6 d e f g +2 e^{2} f^{2}\right ) x}{e^{2}}+\frac {d^{2} \left (5 d^{2} g^{2}+8 d e f g +4 e^{2} f^{2}\right )}{2 e^{3}}-\frac {e \,g^{2} x^{4}}{2}-g \left (3 d g +2 e f \right ) x^{3}}{-e^{2} x^{2}+d^{2}}+\frac {\left (5 d^{2} g^{2}+6 d e f g +e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\) | \(135\) |
parallelrisch | \(\frac {g^{2} x^{3} e^{3}+10 \ln \left (e x -d \right ) x \,d^{2} e \,g^{2}+12 \ln \left (e x -d \right ) x d \,e^{2} f g +2 \ln \left (e x -d \right ) x \,e^{3} f^{2}+5 x^{2} d \,e^{2} g^{2}+4 x^{2} e^{3} f g -10 \ln \left (e x -d \right ) d^{3} g^{2}-12 \ln \left (e x -d \right ) d^{2} e f g -2 \ln \left (e x -d \right ) d \,e^{2} f^{2}-10 d^{3} g^{2}-12 d^{2} e f g -4 d \,e^{2} f^{2}}{2 e^{3} \left (e x -d \right )}\) | \(174\) |
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Leaf count of result is larger than twice the leaf count of optimal. 157 vs. \(2 (77) = 154\).
Time = 0.27 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.01 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {e^{3} g^{2} x^{3} - 4 \, d e^{2} f^{2} - 8 \, d^{2} e f g - 4 \, d^{3} g^{2} + {\left (4 \, e^{3} f g + 5 \, d e^{2} g^{2}\right )} x^{2} - 2 \, {\left (2 \, d e^{2} f g + 3 \, d^{2} e g^{2}\right )} x - 2 \, {\left (d e^{2} f^{2} + 6 \, d^{2} e f g + 5 \, d^{3} g^{2} - {\left (e^{3} f^{2} + 6 \, d e^{2} f g + 5 \, d^{2} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{2 \, {\left (e^{4} x - d e^{3}\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.21 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=x \left (\frac {3 d g^{2}}{e^{2}} + \frac {2 f g}{e}\right ) + \frac {- 2 d^{3} g^{2} - 4 d^{2} e f g - 2 d e^{2} f^{2}}{- d e^{3} + e^{4} x} + \frac {g^{2} x^{2}}{2 e} + \frac {\left (d g + e f\right ) \left (5 d g + e f\right ) \log {\left (- d + e x \right )}}{e^{3}} \]
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Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.33 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=-\frac {2 \, {\left (d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2}\right )}}{e^{4} x - d e^{3}} + \frac {e g^{2} x^{2} + 2 \, {\left (2 \, e f g + 3 \, d g^{2}\right )} x}{2 \, e^{2}} + \frac {{\left (e^{2} f^{2} + 6 \, d e f g + 5 \, d^{2} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]
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Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.38 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {{\left (e^{2} f^{2} + 6 \, d e f g + 5 \, d^{2} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} + \frac {e^{3} g^{2} x^{2} + 4 \, e^{3} f g x + 6 \, d e^{2} g^{2} x}{2 \, e^{4}} - \frac {2 \, {\left (d e^{2} f^{2} + 2 \, d^{2} e f g + d^{3} g^{2}\right )}}{{\left (e x - d\right )} e^{3}} \]
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Time = 12.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.49 \[ \int \frac {(d+e x)^3 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=x\,\left (\frac {d\,g^2+2\,e\,f\,g}{e^2}+\frac {2\,d\,g^2}{e^2}\right )+\frac {\ln \left (e\,x-d\right )\,\left (5\,d^2\,g^2+6\,d\,e\,f\,g+e^2\,f^2\right )}{e^3}+\frac {g^2\,x^2}{2\,e}+\frac {2\,\left (d^3\,g^2+2\,d^2\,e\,f\,g+d\,e^2\,f^2\right )}{e\,\left (d\,e^2-e^3\,x\right )} \]
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