Integrand size = 22, antiderivative size = 115 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {\left (a+\frac {d (c d-b e)}{e^2}\right ) x}{4 d \left (d+e x^2\right )^2}-\frac {\left (5 c d^2-e (b d+3 a e)\right ) x}{8 d^2 e^2 \left (d+e x^2\right )}+\frac {\left (3 c d^2+e (b d+3 a e)\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \]
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Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1171, 393, 211} \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {\arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e (3 a e+b d)+3 c d^2\right )}{8 d^{5/2} e^{5/2}}-\frac {x \left (5 c d^2-e (3 a e+b d)\right )}{8 d^2 e^2 \left (d+e x^2\right )}+\frac {x \left (a e^2-b d e+c d^2\right )}{4 d e^2 \left (d+e x^2\right )^2} \]
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Rule 211
Rule 393
Rule 1171
Rubi steps \begin{align*} \text {integral}& = \frac {\left (c d^2-b d e+a e^2\right ) x}{4 d e^2 \left (d+e x^2\right )^2}-\frac {\int \frac {-3 a+\frac {d (c d-b e)}{e^2}-\frac {4 c d x^2}{e}}{\left (d+e x^2\right )^2} \, dx}{4 d} \\ & = \frac {\left (c d^2-b d e+a e^2\right ) x}{4 d e^2 \left (d+e x^2\right )^2}-\frac {\left (5 c d^2-e (b d+3 a e)\right ) x}{8 d^2 e^2 \left (d+e x^2\right )}-\frac {\left (-\frac {4 c d^2}{e}+e \left (-3 a+\frac {d (c d-b e)}{e^2}\right )\right ) \int \frac {1}{d+e x^2} \, dx}{8 d^2 e} \\ & = \frac {\left (c d^2-b d e+a e^2\right ) x}{4 d e^2 \left (d+e x^2\right )^2}-\frac {\left (5 c d^2-e (b d+3 a e)\right ) x}{8 d^2 e^2 \left (d+e x^2\right )}+\frac {\left (3 c d^2+e (b d+3 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.96 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {x \left (-c d^2 \left (3 d+5 e x^2\right )+e \left (b d \left (-d+e x^2\right )+a e \left (5 d+3 e x^2\right )\right )\right )}{8 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\left (3 c d^2+e (b d+3 a e)\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{5/2}} \]
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Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.93
method | result | size |
default | \(\frac {\frac {\left (3 a \,e^{2}+b d e -5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-b d e -3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}+\frac {\left (3 a \,e^{2}+b d e +3 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{8 e^{2} d^{2} \sqrt {e d}}\) | \(107\) |
risch | \(\frac {\frac {\left (3 a \,e^{2}+b d e -5 c \,d^{2}\right ) x^{3}}{8 d^{2} e}+\frac {\left (5 a \,e^{2}-b d e -3 c \,d^{2}\right ) x}{8 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{2}}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) a}{16 \sqrt {-e d}\, d^{2}}-\frac {\ln \left (e x +\sqrt {-e d}\right ) b}{16 \sqrt {-e d}\, e d}-\frac {3 \ln \left (e x +\sqrt {-e d}\right ) c}{16 \sqrt {-e d}\, e^{2}}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) a}{16 \sqrt {-e d}\, d^{2}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) b}{16 \sqrt {-e d}\, e d}+\frac {3 \ln \left (-e x +\sqrt {-e d}\right ) c}{16 \sqrt {-e d}\, e^{2}}\) | \(215\) |
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none
Time = 0.27 (sec) , antiderivative size = 391, normalized size of antiderivative = 3.40 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\left [-\frac {2 \, {\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} x^{3} + {\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2} + {\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) + 2 \, {\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} x}{16 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}, -\frac {{\left (5 \, c d^{3} e^{2} - b d^{2} e^{3} - 3 \, a d e^{4}\right )} x^{3} - {\left (3 \, c d^{4} + b d^{3} e + 3 \, a d^{2} e^{2} + {\left (3 \, c d^{2} e^{2} + b d e^{3} + 3 \, a e^{4}\right )} x^{4} + 2 \, {\left (3 \, c d^{3} e + b d^{2} e^{2} + 3 \, a d e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + {\left (3 \, c d^{4} e + b d^{3} e^{2} - 5 \, a d^{2} e^{3}\right )} x}{8 \, {\left (d^{3} e^{5} x^{4} + 2 \, d^{4} e^{4} x^{2} + d^{5} e^{3}\right )}}\right ] \]
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Time = 0.66 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.70 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{d^{5} e^{5}}} \cdot \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log {\left (- d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{5} e^{5}}} \cdot \left (3 a e^{2} + b d e + 3 c d^{2}\right ) \log {\left (d^{3} e^{2} \sqrt {- \frac {1}{d^{5} e^{5}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a e^{3} + b d e^{2} - 5 c d^{2} e\right ) + x \left (5 a d e^{2} - b d^{2} e - 3 c d^{3}\right )}{8 d^{4} e^{2} + 16 d^{3} e^{3} x^{2} + 8 d^{2} e^{4} x^{4}} \]
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Exception generated. \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.95 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {{\left (3 \, c d^{2} + b d e + 3 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \, \sqrt {d e} d^{2} e^{2}} - \frac {5 \, c d^{2} e x^{3} - b d e^{2} x^{3} - 3 \, a e^{3} x^{3} + 3 \, c d^{3} x + b d^{2} e x - 5 \, a d e^{2} x}{8 \, {\left (e x^{2} + d\right )}^{2} d^{2} e^{2}} \]
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Time = 7.64 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97 \[ \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,c\,d^2+b\,d\,e+3\,a\,e^2\right )}{8\,d^{5/2}\,e^{5/2}}-\frac {\frac {x\,\left (3\,c\,d^2+b\,d\,e-5\,a\,e^2\right )}{8\,d\,e^2}-\frac {x^3\,\left (-5\,c\,d^2+b\,d\,e+3\,a\,e^2\right )}{8\,d^2\,e}}{d^2+2\,d\,e\,x^2+e^2\,x^4} \]
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