Integrand size = 29, antiderivative size = 149 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {\left (e^2 f^2+12 d e f g+18 d^2 g^2\right ) x}{e^2}-\frac {g (e f+3 d g) x^2}{e}-\frac {g^2 x^3}{3}+\frac {4 d^3 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {4 d^2 (e f+d g) (3 e f+7 d g)}{e^3 (d-e x)}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right ) \log (d-e x)}{e^3} \]
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Time = 0.13 (sec) , antiderivative size = 149, 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, 90} \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {4 d^3 (d g+e f)^2}{e^3 (d-e x)^2}-\frac {4 d^2 (d g+e f) (7 d g+3 e f)}{e^3 (d-e x)}-\frac {x \left (18 d^2 g^2+12 d e f g+e^2 f^2\right )}{e^2}-\frac {2 d \left (19 d^2 g^2+18 d e f g+3 e^2 f^2\right ) \log (d-e x)}{e^3}-\frac {g x^2 (3 d g+e f)}{e}-\frac {g^2 x^3}{3} \]
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Rule 90
Rule 862
Rubi steps \begin{align*} \text {integral}& = \int \frac {(d+e x)^3 (f+g x)^2}{(d-e x)^3} \, dx \\ & = \int \left (\frac {-e^2 f^2-12 d e f g-18 d^2 g^2}{e^2}-\frac {2 g (e f+3 d g) x}{e}-g^2 x^2+\frac {4 d^2 (-3 e f-7 d g) (e f+d g)}{e^2 (d-e x)^2}-\frac {8 d^3 (e f+d g)^2}{e^2 (-d+e x)^3}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right )}{e^2 (-d+e x)}\right ) \, dx \\ & = -\frac {\left (e^2 f^2+12 d e f g+18 d^2 g^2\right ) x}{e^2}-\frac {g (e f+3 d g) x^2}{e}-\frac {g^2 x^3}{3}+\frac {4 d^3 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {4 d^2 (e f+d g) (3 e f+7 d g)}{e^3 (d-e x)}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right ) \log (d-e x)}{e^3} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.05 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {\left (e^2 f^2+12 d e f g+18 d^2 g^2\right ) x}{e^2}-\frac {g (e f+3 d g) x^2}{e}-\frac {g^2 x^3}{3}+\frac {4 d^3 (e f+d g)^2}{e^3 (d-e x)^2}+\frac {4 d^2 \left (3 e^2 f^2+10 d e f g+7 d^2 g^2\right )}{e^3 (-d+e x)}-\frac {2 d \left (3 e^2 f^2+18 d e f g+19 d^2 g^2\right ) \log (d-e x)}{e^3} \]
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Time = 0.44 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.17
method | result | size |
default | \(-\frac {\frac {1}{3} g^{2} x^{3} e^{2}+3 d e \,g^{2} x^{2}+e^{2} f g \,x^{2}+18 d^{2} g^{2} x +12 d e f g x +e^{2} f^{2} x}{e^{2}}-\frac {2 d \left (19 d^{2} g^{2}+18 d e f g +3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}-\frac {4 d^{2} \left (7 d^{2} g^{2}+10 d e f g +3 e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )}+\frac {4 d^{3} \left (d^{2} g^{2}+2 d e f g +e^{2} f^{2}\right )}{e^{3} \left (-e x +d \right )^{2}}\) | \(174\) |
risch | \(-\frac {g^{2} x^{3}}{3}-\frac {3 d \,g^{2} x^{2}}{e}-f g \,x^{2}-\frac {18 d^{2} g^{2} x}{e^{2}}-\frac {12 d f g x}{e}-f^{2} x -\frac {\left (-28 d^{4} g^{2}-40 f g e \,d^{3}-12 d^{2} e^{2} f^{2}\right ) x +\frac {8 d^{3} \left (3 d^{2} g^{2}+4 d e f g +e^{2} f^{2}\right )}{e}}{e^{2} \left (-e x +d \right )^{2}}-\frac {38 d^{3} \ln \left (-e x +d \right ) g^{2}}{e^{3}}-\frac {36 d^{2} \ln \left (-e x +d \right ) f g}{e^{2}}-\frac {6 d \ln \left (-e x +d \right ) f^{2}}{e}\) | \(181\) |
norman | \(\frac {\left (\frac {191}{3} d^{4} g^{2}+64 f g e \,d^{3}+14 d^{2} e^{2} f^{2}\right ) x^{3}+\left (-\frac {52}{3} d^{2} g^{2} e^{2}-12 d f g \,e^{3}-f^{2} e^{4}\right ) x^{5}+\frac {d^{2} \left (41 g^{2} e \,d^{3}+51 e^{2} f g \,d^{2}+16 e^{3} f^{2} d \right ) x^{2}}{e^{2}}-\frac {d^{4} \left (30 g^{2} e \,d^{3}+34 e^{2} f g \,d^{2}+8 e^{3} f^{2} d \right )}{e^{4}}-\frac {g^{2} e^{4} x^{7}}{3}-\frac {d^{4} \left (38 d^{2} g^{2}+36 d e f g +5 e^{2} f^{2}\right ) x}{e^{2}}-e^{3} g \left (3 d g +e f \right ) x^{6}}{\left (-e^{2} x^{2}+d^{2}\right )^{2}}-\frac {2 d \left (19 d^{2} g^{2}+18 d e f g +3 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{e^{3}}\) | \(254\) |
parallelrisch | \(-\frac {g^{2} e^{5} x^{5}+7 x^{4} d \,e^{4} g^{2}+3 x^{4} e^{5} f g +114 \ln \left (e x -d \right ) x^{2} d^{3} e^{2} g^{2}+108 \ln \left (e x -d \right ) x^{2} d^{2} e^{3} f g +18 \ln \left (e x -d \right ) x^{2} d \,e^{4} f^{2}+37 x^{3} d^{2} e^{3} g^{2}+30 x^{3} d \,e^{4} f g +3 x^{3} e^{5} f^{2}-228 \ln \left (e x -d \right ) x \,d^{4} e \,g^{2}-216 \ln \left (e x -d \right ) x \,d^{3} e^{2} f g -36 \ln \left (e x -d \right ) x \,d^{2} e^{3} f^{2}+114 \ln \left (e x -d \right ) d^{5} g^{2}+108 \ln \left (e x -d \right ) d^{4} e f g +18 \ln \left (e x -d \right ) d^{3} e^{2} f^{2}-228 x \,d^{4} e \,g^{2}-222 x \,d^{3} e^{2} f g -45 x \,d^{2} e^{3} f^{2}+171 g^{2} d^{5}+165 f g \,d^{4} e +30 f^{2} d^{3} e^{2}}{3 e^{3} \left (e x -d \right )^{2}}\) | \(315\) |
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Time = 0.36 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.97 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {e^{5} g^{2} x^{5} + 24 \, d^{3} e^{2} f^{2} + 96 \, d^{4} e f g + 72 \, d^{5} g^{2} + {\left (3 \, e^{5} f g + 7 \, d e^{4} g^{2}\right )} x^{4} + {\left (3 \, e^{5} f^{2} + 30 \, d e^{4} f g + 37 \, d^{2} e^{3} g^{2}\right )} x^{3} - 3 \, {\left (2 \, d e^{4} f^{2} + 23 \, d^{2} e^{3} f g + 33 \, d^{3} e^{2} g^{2}\right )} x^{2} - 3 \, {\left (11 \, d^{2} e^{3} f^{2} + 28 \, d^{3} e^{2} f g + 10 \, d^{4} e g^{2}\right )} x + 6 \, {\left (3 \, d^{3} e^{2} f^{2} + 18 \, d^{4} e f g + 19 \, d^{5} g^{2} + {\left (3 \, d e^{4} f^{2} + 18 \, d^{2} e^{3} f g + 19 \, d^{3} e^{2} g^{2}\right )} x^{2} - 2 \, {\left (3 \, d^{2} e^{3} f^{2} + 18 \, d^{3} e^{2} f g + 19 \, d^{4} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{3 \, {\left (e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]
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Time = 0.60 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.19 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=- \frac {2 d \left (19 d^{2} g^{2} + 18 d e f g + 3 e^{2} f^{2}\right ) \log {\left (- d + e x \right )}}{e^{3}} - \frac {g^{2} x^{3}}{3} - x^{2} \cdot \left (\frac {3 d g^{2}}{e} + f g\right ) - x \left (\frac {18 d^{2} g^{2}}{e^{2}} + \frac {12 d f g}{e} + f^{2}\right ) - \frac {24 d^{5} g^{2} + 32 d^{4} e f g + 8 d^{3} e^{2} f^{2} + x \left (- 28 d^{4} e g^{2} - 40 d^{3} e^{2} f g - 12 d^{2} e^{3} f^{2}\right )}{d^{2} e^{3} - 2 d e^{4} x + e^{5} x^{2}} \]
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Time = 0.19 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.26 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {4 \, {\left (2 \, d^{3} e^{2} f^{2} + 8 \, d^{4} e f g + 6 \, d^{5} g^{2} - {\left (3 \, d^{2} e^{3} f^{2} + 10 \, d^{3} e^{2} f g + 7 \, d^{4} e g^{2}\right )} x\right )}}{e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}} - \frac {e^{2} g^{2} x^{3} + 3 \, {\left (e^{2} f g + 3 \, d e g^{2}\right )} x^{2} + 3 \, {\left (e^{2} f^{2} + 12 \, d e f g + 18 \, d^{2} g^{2}\right )} x}{3 \, e^{2}} - \frac {2 \, {\left (3 \, d e^{2} f^{2} + 18 \, d^{2} e f g + 19 \, d^{3} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]
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Time = 0.27 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.24 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=-\frac {2 \, {\left (3 \, d e^{2} f^{2} + 18 \, d^{2} e f g + 19 \, d^{3} g^{2}\right )} \log \left ({\left | e x - d \right |}\right )}{e^{3}} - \frac {4 \, {\left (2 \, d^{3} e^{2} f^{2} + 8 \, d^{4} e f g + 6 \, d^{5} g^{2} - {\left (3 \, d^{2} e^{3} f^{2} + 10 \, d^{3} e^{2} f g + 7 \, d^{4} e g^{2}\right )} x\right )}}{{\left (e x - d\right )}^{2} e^{3}} - \frac {e^{9} g^{2} x^{3} + 3 \, e^{9} f g x^{2} + 9 \, d e^{8} g^{2} x^{2} + 3 \, e^{9} f^{2} x + 36 \, d e^{8} f g x + 54 \, d^{2} e^{7} g^{2} x}{3 \, e^{9}} \]
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Time = 0.11 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.61 \[ \int \frac {(d+e x)^6 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx=\frac {x\,\left (28\,d^4\,g^2+40\,d^3\,e\,f\,g+12\,d^2\,e^2\,f^2\right )-\frac {8\,\left (3\,d^5\,g^2+4\,d^4\,e\,f\,g+d^3\,e^2\,f^2\right )}{e}}{d^2\,e^2-2\,d\,e^3\,x+e^4\,x^2}-x\,\left (\frac {3\,d^2\,e\,g^2+6\,d\,e^2\,f\,g+e^3\,f^2}{e^3}+\frac {3\,d\,\left (\frac {g\,\left (3\,d\,g+2\,e\,f\right )}{e}+\frac {3\,d\,g^2}{e}\right )}{e}-\frac {3\,d^2\,g^2}{e^2}\right )-x^2\,\left (\frac {g\,\left (3\,d\,g+2\,e\,f\right )}{2\,e}+\frac {3\,d\,g^2}{2\,e}\right )-\frac {g^2\,x^3}{3}-\frac {\ln \left (e\,x-d\right )\,\left (38\,d^3\,g^2+36\,d^2\,e\,f\,g+6\,d\,e^2\,f^2\right )}{e^3} \]
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