Integrand size = 29, antiderivative size = 81 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {g^2 x}{e^2}+\frac {d (e f+d g)^2}{e^3 (d-e x)^2}-\frac {(e f+d g) (e f+5 d g)}{e^3 (d-e x)}-\frac {2 g (e f+2 d g) \log (d-e x)}{e^3} \]
[Out]
Time = 0.07 (sec) , antiderivative size = 81, 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} \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {(d g+e f) (5 d g+e f)}{e^3 (d-e x)}+\frac {d (d g+e f)^2}{e^3 (d-e x)^2}-\frac {2 g (2 d g+e f) \log (d-e x)}{e^3}-\frac {g^2 x}{e^2} \]
[In]
[Out]
Rule 78
Rule 862
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x) (f+g x)^2}{(d-e x)^3} \, dx \\ & = \int \left (-\frac {g^2}{e^2}+\frac {(-e f-5 d g) (e f+d g)}{e^2 (d-e x)^2}-\frac {2 d (e f+d g)^2}{e^2 (-d+e x)^3}-\frac {2 g (e f+2 d g)}{e^2 (-d+e x)}\right ) \, dx \\ & = -\frac {g^2 x}{e^2}+\frac {d (e f+d g)^2}{e^3 (d-e x)^2}-\frac {(e f+d g) (e f+5 d g)}{e^3 (d-e x)}-\frac {2 g (e f+2 d g) \log (d-e x)}{e^3} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {-4 d^3 g^2+4 d^2 e g (-f+g x)+2 d e^2 g x (3 f+g x)+e^3 x \left (f^2-g^2 x^2\right )-2 g (e f+2 d g) (d-e x)^2 \log (d-e x)}{e^3 (d-e x)^2} \]
[In]
[Out]
Time = 0.43 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.19
method | result | size |
risch | \(-\frac {g^{2} x}{e^{2}}-\frac {\left (-5 d^{2} g^{2}-6 d e f g -e^{2} f^{2}\right ) x +\frac {4 d^{2} g \left (d g +e f \right )}{e}}{e^{2} \left (-e x +d \right )^{2}}-\frac {4 g^{2} \ln \left (-e x +d \right ) d}{e^{3}}-\frac {2 g \ln \left (-e x +d \right ) f}{e^{2}}\) | \(96\) |
default | \(-\frac {g^{2} x}{e^{2}}-\frac {2 g \left (2 d g +e f \right ) \ln \left (-e x +d \right )}{e^{3}}+\frac {-5 d^{2} g^{2}-6 d e f g -e^{2} f^{2}}{e^{3} \left (-e x +d \right )}+\frac {d \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )^{2}}\) | \(101\) |
norman | \(\frac {\left (7 d^{2} g^{2}+6 d e f g +e^{2} f^{2}\right ) x^{3}-\frac {d^{4} \left (4 d e \,g^{2}+4 e^{2} f g \right )}{e^{4}}-e^{2} g^{2} x^{5}+\frac {2 d \left (3 d^{2} g^{2} e +4 d f g \,e^{2}+f^{2} e^{3}\right ) x^{2}}{e^{2}}-\frac {d^{2} \left (4 d^{2} g^{2}+2 d e f g -e^{2} f^{2}\right ) x}{e^{2}}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {2 g \left (2 d g +e f \right ) \ln \left (-e x +d \right )}{e^{3}}\) | \(165\) |
parallelrisch | \(-\frac {4 \ln \left (e x -d \right ) x^{2} d \,e^{2} g^{2}+2 \ln \left (e x -d \right ) x^{2} e^{3} f g +g^{2} x^{3} e^{3}-8 \ln \left (e x -d \right ) x \,d^{2} e \,g^{2}-4 \ln \left (e x -d \right ) x d \,e^{2} f g +4 \ln \left (e x -d \right ) d^{3} g^{2}+2 \ln \left (e x -d \right ) d^{2} e f g -8 x \,d^{2} e \,g^{2}-6 x d \,e^{2} f g -x \,e^{3} f^{2}+6 d^{3} g^{2}+4 d^{2} e f g}{e^{3} \left (e x -d \right )^{2}}\) | \(175\) |
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.96 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {e^{3} g^{2} x^{3} - 2 \, d e^{2} g^{2} x^{2} + 4 \, d^{2} e f g + 4 \, d^{3} g^{2} - {\left (e^{3} f^{2} + 6 \, d e^{2} f g + 4 \, d^{2} e g^{2}\right )} x + 2 \, {\left (d^{2} e f g + 2 \, d^{3} g^{2} + {\left (e^{3} f g + 2 \, d e^{2} g^{2}\right )} x^{2} - 2 \, {\left (d e^{2} f g + 2 \, d^{2} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}} \]
[In]
[Out]
Time = 0.42 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=- \frac {4 d^{3} g^{2} + 4 d^{2} e f g + x \left (- 5 d^{2} e g^{2} - 6 d e^{2} f g - e^{3} f^{2}\right )}{d^{2} e^{3} - 2 d e^{4} x + e^{5} x^{2}} - \frac {g^{2} x}{e^{2}} - \frac {2 g \left (2 d g + e f\right ) \log {\left (- d + e x \right )}}{e^{3}} \]
[In]
[Out]
none
Time = 0.18 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.30 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {g^{2} x}{e^{2}} - \frac {4 \, d^{2} e f g + 4 \, d^{3} g^{2} - {\left (e^{3} f^{2} + 6 \, d e^{2} f g + 5 \, d^{2} e g^{2}\right )} x}{e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}} - \frac {2 \, {\left (e f g + 2 \, d g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {g^{2} x}{e^{2}} - \frac {2 \, {\left (e f g + 2 \, d g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {4 \, d^{2} e f g + 4 \, d^{3} g^{2} - {\left (e^{3} f^{2} + 6 \, d e^{2} f g + 5 \, d^{2} e g^{2}\right )} x}{{\left (e x - d\right )}^{2} e^{3}} \]
[In]
[Out]
Time = 11.74 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.32 \[ \int \frac {(d+e x)^4 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {\frac {4\,\left (d^3\,g^2+e\,f\,d^2\,g\right )}{e}-x\,\left (5\,d^2\,g^2+6\,d\,e\,f\,g+e^2\,f^2\right )}{d^2\,e^2-2\,d\,e^3\,x+e^4\,x^2}-\frac {g^2\,x}{e^2}-\frac {\ln \left (e\,x-d\right )\,\left (4\,d\,g^2+2\,e\,f\,g\right )}{e^3} \]
[In]
[Out]