Integrand size = 29, antiderivative size = 50 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {g^2 x}{e^2}+\frac {(e f+d g)^2}{e^3 (d-e x)}+\frac {2 g (e f+d g) \log (d-e x)}{e^3} \]
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Time = 0.04 (sec) , antiderivative size = 50, 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} \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {(d g+e f)^2}{e^3 (d-e x)}+\frac {2 g (d g+e f) \log (d-e x)}{e^3}+\frac {g^2 x}{e^2} \]
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Rule 45
Rule 862
Rubi steps \begin{align*} \text {integral}& = \int \frac {(f+g x)^2}{(d-e x)^2} \, dx \\ & = \int \left (\frac {g^2}{e^2}+\frac {(e f+d g)^2}{e^2 (-d+e x)^2}+\frac {2 g (e f+d g)}{e^2 (-d+e x)}\right ) \, dx \\ & = \frac {g^2 x}{e^2}+\frac {(e f+d g)^2}{e^3 (d-e x)}+\frac {2 g (e f+d g) \log (d-e x)}{e^3} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.92 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {e g^2 x+\frac {(e f+d g)^2}{d-e x}+2 g (e f+d g) \log (d-e x)}{e^3} \]
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Time = 0.42 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.26
method | result | size |
default | \(\frac {g^{2} x}{e^{2}}+\frac {2 g \left (d g +e f \right ) \ln \left (-e x +d \right )}{e^{3}}+\frac {d^{2} g^{2}+2 d e f g +e^{2} f^{2}}{e^{3} \left (-e x +d \right )}\) | \(63\) |
risch | \(\frac {g^{2} x}{e^{2}}+\frac {2 g^{2} \ln \left (-e x +d \right ) d}{e^{3}}+\frac {2 g \ln \left (-e x +d \right ) f}{e^{2}}+\frac {d^{2} g^{2}}{e^{3} \left (-e x +d \right )}+\frac {2 d f g}{e^{2} \left (-e x +d \right )}+\frac {f^{2}}{e \left (-e x +d \right )}\) | \(89\) |
norman | \(\frac {\frac {d \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3}}+\frac {\left (2 d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right ) x}{e^{2}}-g^{2} x^{3}}{-e^{2} x^{2}+d^{2}}+\frac {2 g \left (d g +e f \right ) \ln \left (-e x +d \right )}{e^{3}}\) | \(99\) |
parallelrisch | \(\frac {2 \ln \left (e x -d \right ) x d e \,g^{2}+2 \ln \left (e x -d \right ) x \,e^{2} f g +g^{2} x^{2} e^{2}-2 \ln \left (e x -d \right ) d^{2} g^{2}-2 \ln \left (e x -d \right ) d e f g -2 d^{2} g^{2}-2 d e f g -e^{2} f^{2}}{e^{3} \left (e x -d \right )}\) | \(109\) |
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Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.90 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {e^{2} g^{2} x^{2} - d e g^{2} x - e^{2} f^{2} - 2 \, d e f g - d^{2} g^{2} - 2 \, {\left (d e f g + d^{2} g^{2} - {\left (e^{2} f g + d e g^{2}\right )} x\right )} \log \left (e x - d\right )}{e^{4} x - d e^{3}} \]
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Time = 0.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.22 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {- d^{2} g^{2} - 2 d e f g - e^{2} f^{2}}{- d e^{3} + e^{4} x} + \frac {g^{2} x}{e^{2}} + \frac {2 g \left (d g + e f\right ) \log {\left (- d + e x \right )}}{e^{3}} \]
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Time = 0.19 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.38 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {g^{2} x}{e^{2}} - \frac {e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}}{e^{4} x - d e^{3}} + \frac {2 \, {\left (e f g + d g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]
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Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.36 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {g^{2} x}{e^{2}} + \frac {2 \, {\left (e f g + d g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {e^{2} f^{2} + 2 \, d e f g + d^{2} g^{2}}{{\left (e x - d\right )} e^{3}} \]
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Time = 11.90 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.44 \[ \int \frac {(d+e x)^2 (f+g x)^2}{\left (d^2-e^2 x^2\right )^2} \, dx=\frac {d^2\,g^2+2\,d\,e\,f\,g+e^2\,f^2}{e\,\left (d\,e^2-e^3\,x\right )}+\frac {g^2\,x}{e^2}+\frac {\ln \left (e\,x-d\right )\,\left (2\,d\,g^2+2\,e\,f\,g\right )}{e^3} \]
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