Optimal. Leaf size=188 \[ \frac {(e f+d g)^2}{32 d^4 e^3 (d-e x)^2}+\frac {f (e f+d g)}{8 d^5 e^2 (d-e x)}-\frac {(e f-d g)^2}{24 d^3 e^3 (d+e x)^3}-\frac {(e f-d g) (3 e f+d g)}{32 d^4 e^3 (d+e x)^2}-\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^3 (d+e x)}+\frac {\left (5 e^2 f^2+2 d e f g-d^2 g^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^6 e^3} \]
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Rubi [A]
time = 0.14, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {862, 90, 214}
\begin {gather*} \frac {f (d g+e f)}{8 d^5 e^2 (d-e x)}-\frac {(d g+3 e f) (e f-d g)}{32 d^4 e^3 (d+e x)^2}+\frac {(d g+e f)^2}{32 d^4 e^3 (d-e x)^2}-\frac {(e f-d g)^2}{24 d^3 e^3 (d+e x)^3}+\frac {\left (-d^2 g^2+2 d e f g+5 e^2 f^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^6 e^3}-\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^3 (d+e x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 90
Rule 214
Rule 862
Rubi steps
\begin {align*} \int \frac {(f+g x)^2}{(d+e x) \left (d^2-e^2 x^2\right )^3} \, dx &=\int \frac {(f+g x)^2}{(d-e x)^3 (d+e x)^4} \, dx\\ &=\int \left (\frac {(e f+d g)^2}{16 d^4 e^2 (d-e x)^3}+\frac {f (e f+d g)}{8 d^5 e (d-e x)^2}+\frac {(-e f+d g)^2}{8 d^3 e^2 (d+e x)^4}+\frac {(e f-d g) (3 e f+d g)}{16 d^4 e^2 (d+e x)^3}+\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^2 (d+e x)^2}+\frac {-5 e^2 f^2-2 d e f g+d^2 g^2}{16 d^5 e^2 \left (-d^2+e^2 x^2\right )}\right ) \, dx\\ &=\frac {(e f+d g)^2}{32 d^4 e^3 (d-e x)^2}+\frac {f (e f+d g)}{8 d^5 e^2 (d-e x)}-\frac {(e f-d g)^2}{24 d^3 e^3 (d+e x)^3}-\frac {(e f-d g) (3 e f+d g)}{32 d^4 e^3 (d+e x)^2}-\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^3 (d+e x)}-\frac {\left (5 e^2 f^2+2 d e f g-d^2 g^2\right ) \int \frac {1}{-d^2+e^2 x^2} \, dx}{16 d^5 e^2}\\ &=\frac {(e f+d g)^2}{32 d^4 e^3 (d-e x)^2}+\frac {f (e f+d g)}{8 d^5 e^2 (d-e x)}-\frac {(e f-d g)^2}{24 d^3 e^3 (d+e x)^3}-\frac {(e f-d g) (3 e f+d g)}{32 d^4 e^3 (d+e x)^2}-\frac {3 e^2 f^2-d^2 g^2}{16 d^5 e^3 (d+e x)}+\frac {\left (5 e^2 f^2+2 d e f g-d^2 g^2\right ) \tanh ^{-1}\left (\frac {e x}{d}\right )}{16 d^6 e^3}\\ \end {align*}
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Mathematica [A]
time = 0.11, size = 197, normalized size = 1.05 \begin {gather*} \frac {\frac {3 d^2 (e f+d g)^2}{(d-e x)^2}+\frac {12 d e f (e f+d g)}{d-e x}-\frac {4 d^3 (e f-d g)^2}{(d+e x)^3}+\frac {3 d^2 \left (-3 e^2 f^2+2 d e f g+d^2 g^2\right )}{(d+e x)^2}+\frac {6 d \left (-3 e^2 f^2+d^2 g^2\right )}{d+e x}+3 \left (-5 e^2 f^2-2 d e f g+d^2 g^2\right ) \log (d-e x)+3 \left (5 e^2 f^2+2 d e f g-d^2 g^2\right ) \log (d+e x)}{96 d^6 e^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.12, size = 245, normalized size = 1.30
method | result | size |
default | \(-\frac {-d^{2} g^{2}+3 e^{2} f^{2}}{16 e^{3} d^{5} \left (e x +d \right )}-\frac {-d^{2} g^{2}-2 d e f g +3 e^{2} f^{2}}{32 e^{3} d^{4} \left (e x +d \right )^{2}}+\frac {\left (-d^{2} g^{2}+2 d e f g +5 e^{2} f^{2}\right ) \ln \left (e x +d \right )}{32 e^{3} d^{6}}-\frac {d^{2} g^{2}-2 d e f g +e^{2} f^{2}}{24 e^{3} d^{3} \left (e x +d \right )^{3}}-\frac {-d^{2} g^{2}-2 d e f g -e^{2} f^{2}}{32 e^{3} d^{4} \left (-e x +d \right )^{2}}+\frac {\left (d^{2} g^{2}-2 d e f g -5 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{32 d^{6} e^{3}}+\frac {f \left (d g +e f \right )}{8 d^{5} e^{2} \left (-e x +d \right )}\) | \(245\) |
norman | \(\frac {\frac {\left (11 d^{2} g^{2}+26 d e f g -31 e^{2} f^{2}\right ) x^{3}}{48 d^{4}}+\frac {\left (d^{2} g^{2}+14 d e f g +3 e^{2} f^{2}\right ) x^{2}}{16 e \,d^{3}}-\frac {e \left (d^{2} g^{2}+22 d e f g +7 e^{2} f^{2}\right ) x^{4}}{48 d^{5}}-\frac {e^{2} \left (d^{2} g^{2}+4 d e f g -2 e^{2} f^{2}\right ) x^{5}}{12 d^{6}}+\frac {\left (d^{2} g^{2}-2 d e f g +11 e^{2} f^{2}\right ) x}{16 e^{2} d^{2}}}{\left (e x +d \right )^{3} \left (-e x +d \right )^{2}}+\frac {\left (d^{2} g^{2}-2 d e f g -5 e^{2} f^{2}\right ) \ln \left (-e x +d \right )}{32 d^{6} e^{3}}-\frac {\left (d^{2} g^{2}-2 d e f g -5 e^{2} f^{2}\right ) \ln \left (e x +d \right )}{32 d^{6} e^{3}}\) | \(251\) |
risch | \(\frac {\frac {\left (d^{2} g^{2}-2 d e f g -5 e^{2} f^{2}\right ) e \,x^{4}}{16 d^{5}}+\frac {\left (d^{2} g^{2}-2 d e f g -5 e^{2} f^{2}\right ) x^{3}}{16 d^{4}}-\frac {5 \left (d^{2} g^{2}-2 d e f g -5 e^{2} f^{2}\right ) x^{2}}{48 d^{3} e}+\frac {\left (7 d^{2} g^{2}+10 d e f g +25 e^{2} f^{2}\right ) x}{48 e^{2} d^{2}}+\frac {d^{2} g^{2}+4 d e f g -2 e^{2} f^{2}}{12 d \,e^{3}}}{\left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{2}}+\frac {\ln \left (e x -d \right ) g^{2}}{32 d^{4} e^{3}}-\frac {\ln \left (e x -d \right ) f g}{16 d^{5} e^{2}}-\frac {5 \ln \left (e x -d \right ) f^{2}}{32 d^{6} e}-\frac {\ln \left (-e x -d \right ) g^{2}}{32 d^{4} e^{3}}+\frac {\ln \left (-e x -d \right ) f g}{16 d^{5} e^{2}}+\frac {5 \ln \left (-e x -d \right ) f^{2}}{32 d^{6} e}\) | \(296\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.30, size = 286, normalized size = 1.52 \begin {gather*} \frac {4 \, d^{6} g^{2} + 16 \, d^{5} f g e - 8 \, d^{4} f^{2} e^{2} + 3 \, {\left (d^{2} g^{2} e^{4} - 2 \, d f g e^{5} - 5 \, f^{2} e^{6}\right )} x^{4} + 3 \, {\left (d^{3} g^{2} e^{3} - 2 \, d^{2} f g e^{4} - 5 \, d f^{2} e^{5}\right )} x^{3} - 5 \, {\left (d^{4} g^{2} e^{2} - 2 \, d^{3} f g e^{3} - 5 \, d^{2} f^{2} e^{4}\right )} x^{2} + {\left (7 \, d^{5} g^{2} e + 10 \, d^{4} f g e^{2} + 25 \, d^{3} f^{2} e^{3}\right )} x}{48 \, {\left (d^{5} x^{5} e^{8} + d^{6} x^{4} e^{7} - 2 \, d^{7} x^{3} e^{6} - 2 \, d^{8} x^{2} e^{5} + d^{9} x e^{4} + d^{10} e^{3}\right )}} - \frac {{\left (d^{2} g^{2} - 2 \, d f g e - 5 \, f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e + d\right )}{32 \, d^{6}} + \frac {{\left (d^{2} g^{2} - 2 \, d f g e - 5 \, f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left (x e - d\right )}{32 \, d^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 639 vs.
\(2 (180) = 360\).
time = 2.86, size = 639, normalized size = 3.40 \begin {gather*} \frac {8 \, d^{7} g^{2} - 30 \, d f^{2} x^{4} e^{6} - 6 \, {\left (2 \, d^{2} f g x^{4} + 5 \, d^{2} f^{2} x^{3}\right )} e^{5} + 2 \, {\left (3 \, d^{3} g^{2} x^{4} - 6 \, d^{3} f g x^{3} + 25 \, d^{3} f^{2} x^{2}\right )} e^{4} + 2 \, {\left (3 \, d^{4} g^{2} x^{3} + 10 \, d^{4} f g x^{2} + 25 \, d^{4} f^{2} x\right )} e^{3} - 2 \, {\left (5 \, d^{5} g^{2} x^{2} - 10 \, d^{5} f g x + 8 \, d^{5} f^{2}\right )} e^{2} + 2 \, {\left (7 \, d^{6} g^{2} x + 16 \, d^{6} f g\right )} e - 3 \, {\left (d^{7} g^{2} - 5 \, f^{2} x^{5} e^{7} - {\left (2 \, d f g x^{5} + 5 \, d f^{2} x^{4}\right )} e^{6} + {\left (d^{2} g^{2} x^{5} - 2 \, d^{2} f g x^{4} + 10 \, d^{2} f^{2} x^{3}\right )} e^{5} + {\left (d^{3} g^{2} x^{4} + 4 \, d^{3} f g x^{3} + 10 \, d^{3} f^{2} x^{2}\right )} e^{4} - {\left (2 \, d^{4} g^{2} x^{3} - 4 \, d^{4} f g x^{2} + 5 \, d^{4} f^{2} x\right )} e^{3} - {\left (2 \, d^{5} g^{2} x^{2} + 2 \, d^{5} f g x + 5 \, d^{5} f^{2}\right )} e^{2} + {\left (d^{6} g^{2} x - 2 \, d^{6} f g\right )} e\right )} \log \left (x e + d\right ) + 3 \, {\left (d^{7} g^{2} - 5 \, f^{2} x^{5} e^{7} - {\left (2 \, d f g x^{5} + 5 \, d f^{2} x^{4}\right )} e^{6} + {\left (d^{2} g^{2} x^{5} - 2 \, d^{2} f g x^{4} + 10 \, d^{2} f^{2} x^{3}\right )} e^{5} + {\left (d^{3} g^{2} x^{4} + 4 \, d^{3} f g x^{3} + 10 \, d^{3} f^{2} x^{2}\right )} e^{4} - {\left (2 \, d^{4} g^{2} x^{3} - 4 \, d^{4} f g x^{2} + 5 \, d^{4} f^{2} x\right )} e^{3} - {\left (2 \, d^{5} g^{2} x^{2} + 2 \, d^{5} f g x + 5 \, d^{5} f^{2}\right )} e^{2} + {\left (d^{6} g^{2} x - 2 \, d^{6} f g\right )} e\right )} \log \left (x e - d\right )}{96 \, {\left (d^{6} x^{5} e^{8} + d^{7} x^{4} e^{7} - 2 \, d^{8} x^{3} e^{6} - 2 \, d^{9} x^{2} e^{5} + d^{10} x e^{4} + d^{11} e^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.90, size = 321, normalized size = 1.71 \begin {gather*} - \frac {- 4 d^{6} g^{2} - 16 d^{5} e f g + 8 d^{4} e^{2} f^{2} + x^{4} \left (- 3 d^{2} e^{4} g^{2} + 6 d e^{5} f g + 15 e^{6} f^{2}\right ) + x^{3} \left (- 3 d^{3} e^{3} g^{2} + 6 d^{2} e^{4} f g + 15 d e^{5} f^{2}\right ) + x^{2} \cdot \left (5 d^{4} e^{2} g^{2} - 10 d^{3} e^{3} f g - 25 d^{2} e^{4} f^{2}\right ) + x \left (- 7 d^{5} e g^{2} - 10 d^{4} e^{2} f g - 25 d^{3} e^{3} f^{2}\right )}{48 d^{10} e^{3} + 48 d^{9} e^{4} x - 96 d^{8} e^{5} x^{2} - 96 d^{7} e^{6} x^{3} + 48 d^{6} e^{7} x^{4} + 48 d^{5} e^{8} x^{5}} + \frac {\left (d^{2} g^{2} - 2 d e f g - 5 e^{2} f^{2}\right ) \log {\left (- \frac {d}{e} + x \right )}}{32 d^{6} e^{3}} - \frac {\left (d^{2} g^{2} - 2 d e f g - 5 e^{2} f^{2}\right ) \log {\left (\frac {d}{e} + x \right )}}{32 d^{6} e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 2.03, size = 262, normalized size = 1.39 \begin {gather*} -\frac {{\left (d^{2} g^{2} - 2 \, d f g e - 5 \, f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right )}{32 \, d^{6}} + \frac {{\left (d^{2} g^{2} - 2 \, d f g e - 5 \, f^{2} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | x e - d \right |}\right )}{32 \, d^{6}} + \frac {{\left (4 \, d^{7} g^{2} + 16 \, d^{6} f g e - 8 \, d^{5} f^{2} e^{2} + 3 \, {\left (d^{3} g^{2} e^{4} - 2 \, d^{2} f g e^{5} - 5 \, d f^{2} e^{6}\right )} x^{4} + 3 \, {\left (d^{4} g^{2} e^{3} - 2 \, d^{3} f g e^{4} - 5 \, d^{2} f^{2} e^{5}\right )} x^{3} - 5 \, {\left (d^{5} g^{2} e^{2} - 2 \, d^{4} f g e^{3} - 5 \, d^{3} f^{2} e^{4}\right )} x^{2} + {\left (7 \, d^{6} g^{2} e + 10 \, d^{5} f g e^{2} + 25 \, d^{4} f^{2} e^{3}\right )} x\right )} e^{\left (-3\right )}}{48 \, {\left (x e + d\right )}^{3} {\left (x e - d\right )}^{2} d^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.68, size = 249, normalized size = 1.32 \begin {gather*} \frac {\frac {d^2\,g^2+4\,d\,e\,f\,g-2\,e^2\,f^2}{12\,d\,e^3}-\frac {x^3\,\left (-d^2\,g^2+2\,d\,e\,f\,g+5\,e^2\,f^2\right )}{16\,d^4}-\frac {e\,x^4\,\left (-d^2\,g^2+2\,d\,e\,f\,g+5\,e^2\,f^2\right )}{16\,d^5}+\frac {x\,\left (7\,d^2\,g^2+10\,d\,e\,f\,g+25\,e^2\,f^2\right )}{48\,d^2\,e^2}+\frac {5\,x^2\,\left (-d^2\,g^2+2\,d\,e\,f\,g+5\,e^2\,f^2\right )}{48\,d^3\,e}}{d^5+d^4\,e\,x-2\,d^3\,e^2\,x^2-2\,d^2\,e^3\,x^3+d\,e^4\,x^4+e^5\,x^5}+\frac {\mathrm {atanh}\left (\frac {e\,x}{d}\right )\,\left (-d^2\,g^2+2\,d\,e\,f\,g+5\,e^2\,f^2\right )}{16\,d^6\,e^3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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