Optimal. Leaf size=179 \[ \frac {8 d^4 (d g+e f)^2}{e^3 (d-e x)^2}-\frac {32 d^3 (d g+e f) (2 d g+e f)}{e^3 (d-e x)}-\frac {d x \left (56 d^2 g^2+48 d e f g+7 e^2 f^2\right )}{e^2}-\frac {8 d^2 \left (13 d^2 g^2+14 d e f g+3 e^2 f^2\right ) \log (d-e x)}{e^3}-\frac {1}{3} g x^3 (7 d g+2 e f)-\frac {x^2 (2 d g+e f) (12 d g+e f)}{2 e}-\frac {1}{4} e g^2 x^4 \]
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Rubi [A] time = 0.24, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {848, 88} \[ -\frac {d x \left (56 d^2 g^2+48 d e f g+7 e^2 f^2\right )}{e^2}-\frac {8 d^2 \left (13 d^2 g^2+14 d e f g+3 e^2 f^2\right ) \log (d-e x)}{e^3}+\frac {8 d^4 (d g+e f)^2}{e^3 (d-e x)^2}-\frac {32 d^3 (d g+e f) (2 d g+e f)}{e^3 (d-e x)}-\frac {1}{3} g x^3 (7 d g+2 e f)-\frac {x^2 (2 d g+e f) (12 d g+e f)}{2 e}-\frac {1}{4} e g^2 x^4 \]
Antiderivative was successfully verified.
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Rule 88
Rule 848
Rubi steps
\begin {align*} \int \frac {(d+e x)^7 (f+g x)^2}{\left (d^2-e^2 x^2\right )^3} \, dx &=\int \frac {(d+e x)^4 (f+g x)^2}{(d-e x)^3} \, dx\\ &=\int \left (-\frac {d \left (7 e^2 f^2+48 d e f g+56 d^2 g^2\right )}{e^2}+\frac {(-e f-12 d g) (e f+2 d g) x}{e}-g (2 e f+7 d g) x^2-e g^2 x^3+\frac {32 d^3 (-e f-2 d g) (e f+d g)}{e^2 (d-e x)^2}-\frac {16 d^4 (e f+d g)^2}{e^2 (-d+e x)^3}-\frac {8 d^2 \left (3 e^2 f^2+14 d e f g+13 d^2 g^2\right )}{e^2 (-d+e x)}\right ) \, dx\\ &=-\frac {d \left (7 e^2 f^2+48 d e f g+56 d^2 g^2\right ) x}{e^2}-\frac {(e f+2 d g) (e f+12 d g) x^2}{2 e}-\frac {1}{3} g (2 e f+7 d g) x^3-\frac {1}{4} e g^2 x^4+\frac {8 d^4 (e f+d g)^2}{e^3 (d-e x)^2}-\frac {32 d^3 (e f+d g) (e f+2 d g)}{e^3 (d-e x)}-\frac {8 d^2 \left (3 e^2 f^2+14 d e f g+13 d^2 g^2\right ) \log (d-e x)}{e^3}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 193, normalized size = 1.08 \[ \frac {8 d^4 (d g+e f)^2}{e^3 (d-e x)^2}-\frac {x^2 \left (24 d^2 g^2+14 d e f g+e^2 f^2\right )}{2 e}-\frac {d x \left (56 d^2 g^2+48 d e f g+7 e^2 f^2\right )}{e^2}-\frac {8 d^2 \left (13 d^2 g^2+14 d e f g+3 e^2 f^2\right ) \log (d-e x)}{e^3}+\frac {32 d^3 \left (2 d^2 g^2+3 d e f g+e^2 f^2\right )}{e^3 (e x-d)}-\frac {1}{3} g x^3 (7 d g+2 e f)-\frac {1}{4} e g^2 x^4 \]
Antiderivative was successfully verified.
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fricas [A] time = 0.83, size = 336, normalized size = 1.88 \[ -\frac {3 \, e^{6} g^{2} x^{6} + 288 \, d^{4} e^{2} f^{2} + 960 \, d^{5} e f g + 672 \, d^{6} g^{2} + 2 \, {\left (4 \, e^{6} f g + 11 \, d e^{5} g^{2}\right )} x^{5} + {\left (6 \, e^{6} f^{2} + 68 \, d e^{5} f g + 91 \, d^{2} e^{4} g^{2}\right )} x^{4} + 4 \, {\left (18 \, d e^{5} f^{2} + 104 \, d^{2} e^{4} f g + 103 \, d^{3} e^{3} g^{2}\right )} x^{3} - 6 \, {\left (27 \, d^{2} e^{4} f^{2} + 178 \, d^{3} e^{3} f g + 200 \, d^{4} e^{2} g^{2}\right )} x^{2} - 12 \, {\left (25 \, d^{3} e^{3} f^{2} + 48 \, d^{4} e^{2} f g + 8 \, d^{5} e g^{2}\right )} x + 96 \, {\left (3 \, d^{4} e^{2} f^{2} + 14 \, d^{5} e f g + 13 \, d^{6} g^{2} + {\left (3 \, d^{2} e^{4} f^{2} + 14 \, d^{3} e^{3} f g + 13 \, d^{4} e^{2} g^{2}\right )} x^{2} - 2 \, {\left (3 \, d^{3} e^{3} f^{2} + 14 \, d^{4} e^{2} f g + 13 \, d^{5} e g^{2}\right )} x\right )} \log \left (e x - d\right )}{12 \, {\left (e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.19, size = 364, normalized size = 2.03 \[ -4 \, {\left (13 \, d^{4} g^{2} e^{7} + 14 \, d^{3} f g e^{8} + 3 \, d^{2} f^{2} e^{9}\right )} e^{\left (-10\right )} \log \left ({\left | x^{2} e^{2} - d^{2} \right |}\right ) - \frac {1}{12} \, {\left (3 \, g^{2} x^{4} e^{25} + 28 \, d g^{2} x^{3} e^{24} + 144 \, d^{2} g^{2} x^{2} e^{23} + 672 \, d^{3} g^{2} x e^{22} + 8 \, f g x^{3} e^{25} + 84 \, d f g x^{2} e^{24} + 576 \, d^{2} f g x e^{23} + 6 \, f^{2} x^{2} e^{25} + 84 \, d f^{2} x e^{24}\right )} e^{\left (-24\right )} - \frac {4 \, {\left (13 \, d^{5} g^{2} e^{6} + 14 \, d^{4} f g e^{7} + 3 \, d^{3} f^{2} e^{8}\right )} e^{\left (-9\right )} \log \left (\frac {{\left | 2 \, x e^{2} - 2 \, {\left | d \right |} e \right |}}{{\left | 2 \, x e^{2} + 2 \, {\left | d \right |} e \right |}}\right )}{{\left | d \right |}} - \frac {8 \, {\left (7 \, d^{8} g^{2} e^{7} + 10 \, d^{7} f g e^{8} + 3 \, d^{6} f^{2} e^{9} - 4 \, {\left (2 \, d^{5} g^{2} e^{10} + 3 \, d^{4} f g e^{11} + d^{3} f^{2} e^{12}\right )} x^{3} - {\left (9 \, d^{6} g^{2} e^{9} + 14 \, d^{5} f g e^{10} + 5 \, d^{4} f^{2} e^{11}\right )} x^{2} + 2 \, {\left (3 \, d^{7} g^{2} e^{8} + 4 \, d^{6} f g e^{9} + d^{5} f^{2} e^{10}\right )} x\right )} e^{\left (-10\right )}}{{\left (x^{2} e^{2} - d^{2}\right )}^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 263, normalized size = 1.47 \[ -\frac {e \,g^{2} x^{4}}{4}-\frac {7 d \,g^{2} x^{3}}{3}-\frac {2 e f g \,x^{3}}{3}+\frac {8 d^{6} g^{2}}{\left (e x -d \right )^{2} e^{3}}+\frac {16 d^{5} f g}{\left (e x -d \right )^{2} e^{2}}+\frac {8 d^{4} f^{2}}{\left (e x -d \right )^{2} e}-\frac {12 d^{2} g^{2} x^{2}}{e}-7 d f g \,x^{2}-\frac {e \,f^{2} x^{2}}{2}+\frac {64 d^{5} g^{2}}{\left (e x -d \right ) e^{3}}+\frac {96 d^{4} f g}{\left (e x -d \right ) e^{2}}-\frac {104 d^{4} g^{2} \ln \left (e x -d \right )}{e^{3}}+\frac {32 d^{3} f^{2}}{\left (e x -d \right ) e}-\frac {112 d^{3} f g \ln \left (e x -d \right )}{e^{2}}-\frac {56 d^{3} g^{2} x}{e^{2}}-\frac {24 d^{2} f^{2} \ln \left (e x -d \right )}{e}-\frac {48 d^{2} f g x}{e}-7 d \,f^{2} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.47, size = 227, normalized size = 1.27 \[ -\frac {8 \, {\left (3 \, d^{4} e^{2} f^{2} + 10 \, d^{5} e f g + 7 \, d^{6} g^{2} - 4 \, {\left (d^{3} e^{3} f^{2} + 3 \, d^{4} e^{2} f g + 2 \, d^{5} e g^{2}\right )} x\right )}}{e^{5} x^{2} - 2 \, d e^{4} x + d^{2} e^{3}} - \frac {3 \, e^{3} g^{2} x^{4} + 4 \, {\left (2 \, e^{3} f g + 7 \, d e^{2} g^{2}\right )} x^{3} + 6 \, {\left (e^{3} f^{2} + 14 \, d e^{2} f g + 24 \, d^{2} e g^{2}\right )} x^{2} + 12 \, {\left (7 \, d e^{2} f^{2} + 48 \, d^{2} e f g + 56 \, d^{3} g^{2}\right )} x}{12 \, e^{2}} - \frac {8 \, {\left (3 \, d^{2} e^{2} f^{2} + 14 \, d^{3} e f g + 13 \, d^{4} g^{2}\right )} \log \left (e x - d\right )}{e^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.14, size = 375, normalized size = 2.09 \[ \frac {x\,\left (64\,d^5\,g^2+96\,d^4\,e\,f\,g+32\,d^3\,e^2\,f^2\right )-\frac {8\,\left (7\,d^6\,g^2+10\,d^5\,e\,f\,g+3\,d^4\,e^2\,f^2\right )}{e}}{d^2\,e^2-2\,d\,e^3\,x+e^4\,x^2}-x^2\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{2\,e^3}-\frac {3\,d^2\,g^2}{2\,e}+\frac {3\,d\,\left (2\,g\,\left (2\,d\,g+e\,f\right )+3\,d\,g^2\right )}{2\,e}\right )-x\,\left (\frac {d^3\,g^2}{e^2}-\frac {3\,d^2\,\left (2\,g\,\left (2\,d\,g+e\,f\right )+3\,d\,g^2\right )}{e^2}+\frac {4\,d\,\left (d^2\,g^2+3\,d\,e\,f\,g+e^2\,f^2\right )}{e^2}+\frac {3\,d\,\left (\frac {6\,d^2\,e^2\,g^2+8\,d\,e^3\,f\,g+e^4\,f^2}{e^3}-\frac {3\,d^2\,g^2}{e}+\frac {3\,d\,\left (2\,g\,\left (2\,d\,g+e\,f\right )+3\,d\,g^2\right )}{e}\right )}{e}\right )-x^3\,\left (\frac {2\,g\,\left (2\,d\,g+e\,f\right )}{3}+d\,g^2\right )-\frac {\ln \left (e\,x-d\right )\,\left (104\,d^4\,g^2+112\,d^3\,e\,f\,g+24\,d^2\,e^2\,f^2\right )}{e^3}-\frac {e\,g^2\,x^4}{4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.55, size = 219, normalized size = 1.22 \[ - \frac {8 d^{2} \left (13 d^{2} g^{2} + 14 d e f g + 3 e^{2} f^{2}\right ) \log {\left (- d + e x \right )}}{e^{3}} - \frac {e g^{2} x^{4}}{4} - x^{3} \left (\frac {7 d g^{2}}{3} + \frac {2 e f g}{3}\right ) - x^{2} \left (\frac {12 d^{2} g^{2}}{e} + 7 d f g + \frac {e f^{2}}{2}\right ) - x \left (\frac {56 d^{3} g^{2}}{e^{2}} + \frac {48 d^{2} f g}{e} + 7 d f^{2}\right ) - \frac {56 d^{6} g^{2} + 80 d^{5} e f g + 24 d^{4} e^{2} f^{2} + x \left (- 64 d^{5} e g^{2} - 96 d^{4} e^{2} f g - 32 d^{3} e^{3} f^{2}\right )}{d^{2} e^{3} - 2 d e^{4} x + e^{5} x^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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