Integrand size = 17, antiderivative size = 123 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {\left (a+\frac {c d^2}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}+\frac {\left (\frac {5 a}{d^2}-\frac {7 c}{e^2}\right ) x}{24 \left (d+e x^2\right )^2}+\frac {\left (\frac {5 a}{d^2}+\frac {c}{e^2}\right ) x}{16 d \left (d+e x^2\right )}+\frac {\left (c d^2+5 a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}} \]
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Time = 0.07 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1172, 393, 205, 211} \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {\left (5 a e^2+c d^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}}+\frac {x \left (\frac {5 a}{d^2}+\frac {c}{e^2}\right )}{16 d \left (d+e x^2\right )}+\frac {x \left (\frac {5 a}{d^2}-\frac {7 c}{e^2}\right )}{24 \left (d+e x^2\right )^2}+\frac {x \left (a+\frac {c d^2}{e^2}\right )}{6 d \left (d+e x^2\right )^3} \]
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Rule 205
Rule 211
Rule 393
Rule 1172
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a+\frac {c d^2}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}-\frac {\int \frac {-5 a+\frac {c d^2}{e^2}-\frac {6 c d x^2}{e}}{\left (d+e x^2\right )^3} \, dx}{6 d} \\ & = \frac {\left (a+\frac {c d^2}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}+\frac {\left (\frac {5 a}{d^2}-\frac {7 c}{e^2}\right ) x}{24 \left (d+e x^2\right )^2}+\frac {1}{8} \left (\frac {5 a}{d^2}+\frac {c}{e^2}\right ) \int \frac {1}{\left (d+e x^2\right )^2} \, dx \\ & = \frac {\left (a+\frac {c d^2}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}+\frac {\left (\frac {5 a}{d^2}-\frac {7 c}{e^2}\right ) x}{24 \left (d+e x^2\right )^2}+\frac {\left (\frac {5 a}{d^2}+\frac {c}{e^2}\right ) x}{16 d \left (d+e x^2\right )}+\frac {\left (\frac {5 a}{d^2}+\frac {c}{e^2}\right ) \int \frac {1}{d+e x^2} \, dx}{16 d} \\ & = \frac {\left (a+\frac {c d^2}{e^2}\right ) x}{6 d \left (d+e x^2\right )^3}+\frac {\left (\frac {5 a}{d^2}-\frac {7 c}{e^2}\right ) x}{24 \left (d+e x^2\right )^2}+\frac {\left (\frac {5 a}{d^2}+\frac {c}{e^2}\right ) x}{16 d \left (d+e x^2\right )}+\frac {\left (c d^2+5 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}} \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.92 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {x \left (c d^2 \left (-3 d^2-8 d e x^2+3 e^2 x^4\right )+a e^2 \left (33 d^2+40 d e x^2+15 e^2 x^4\right )\right )}{48 d^3 e^2 \left (d+e x^2\right )^3}+\frac {\left (c d^2+5 a e^2\right ) \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}} \]
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Time = 0.25 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.92
method | result | size |
default | \(\frac {\frac {\left (5 a \,e^{2}+c \,d^{2}\right ) x^{5}}{16 d^{3}}+\frac {\left (5 a \,e^{2}-c \,d^{2}\right ) x^{3}}{6 d^{2} e}+\frac {\left (11 a \,e^{2}-c \,d^{2}\right ) x}{16 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{3}}+\frac {\left (5 a \,e^{2}+c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {e d}}\right )}{16 d^{3} e^{2} \sqrt {e d}}\) | \(113\) |
risch | \(\frac {\frac {\left (5 a \,e^{2}+c \,d^{2}\right ) x^{5}}{16 d^{3}}+\frac {\left (5 a \,e^{2}-c \,d^{2}\right ) x^{3}}{6 d^{2} e}+\frac {\left (11 a \,e^{2}-c \,d^{2}\right ) x}{16 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{3}}-\frac {5 \ln \left (e x +\sqrt {-e d}\right ) a}{32 \sqrt {-e d}\, d^{3}}-\frac {\ln \left (e x +\sqrt {-e d}\right ) c}{32 \sqrt {-e d}\, e^{2} d}+\frac {5 \ln \left (-e x +\sqrt {-e d}\right ) a}{32 \sqrt {-e d}\, d^{3}}+\frac {\ln \left (-e x +\sqrt {-e d}\right ) c}{32 \sqrt {-e d}\, e^{2} d}\) | \(179\) |
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Time = 0.26 (sec) , antiderivative size = 424, normalized size of antiderivative = 3.45 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\left [\frac {6 \, {\left (c d^{3} e^{3} + 5 \, a d e^{5}\right )} x^{5} - 16 \, {\left (c d^{4} e^{2} - 5 \, a d^{2} e^{4}\right )} x^{3} - 3 \, {\left ({\left (c d^{2} e^{3} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + 5 \, a d^{3} e^{2} + 3 \, {\left (c d^{3} e^{2} + 5 \, a d e^{4}\right )} x^{4} + 3 \, {\left (c d^{4} e + 5 \, a d^{2} e^{3}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) - 6 \, {\left (c d^{5} e - 11 \, a d^{3} e^{3}\right )} x}{96 \, {\left (d^{4} e^{6} x^{6} + 3 \, d^{5} e^{5} x^{4} + 3 \, d^{6} e^{4} x^{2} + d^{7} e^{3}\right )}}, \frac {3 \, {\left (c d^{3} e^{3} + 5 \, a d e^{5}\right )} x^{5} - 8 \, {\left (c d^{4} e^{2} - 5 \, a d^{2} e^{4}\right )} x^{3} + 3 \, {\left ({\left (c d^{2} e^{3} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + 5 \, a d^{3} e^{2} + 3 \, {\left (c d^{3} e^{2} + 5 \, a d e^{4}\right )} x^{4} + 3 \, {\left (c d^{4} e + 5 \, a d^{2} e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - 3 \, {\left (c d^{5} e - 11 \, a d^{3} e^{3}\right )} x}{48 \, {\left (d^{4} e^{6} x^{6} + 3 \, d^{5} e^{5} x^{4} + 3 \, d^{6} e^{4} x^{2} + d^{7} e^{3}\right )}}\right ] \]
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Time = 0.44 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.66 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=- \frac {\sqrt {- \frac {1}{d^{7} e^{5}}} \cdot \left (5 a e^{2} + c d^{2}\right ) \log {\left (- d^{4} e^{2} \sqrt {- \frac {1}{d^{7} e^{5}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{d^{7} e^{5}}} \cdot \left (5 a e^{2} + c d^{2}\right ) \log {\left (d^{4} e^{2} \sqrt {- \frac {1}{d^{7} e^{5}}} + x \right )}}{32} + \frac {x^{5} \cdot \left (15 a e^{4} + 3 c d^{2} e^{2}\right ) + x^{3} \cdot \left (40 a d e^{3} - 8 c d^{3} e\right ) + x \left (33 a d^{2} e^{2} - 3 c d^{4}\right )}{48 d^{6} e^{2} + 144 d^{5} e^{3} x^{2} + 144 d^{4} e^{4} x^{4} + 48 d^{3} e^{5} x^{6}} \]
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Exception generated. \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.32 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.90 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {{\left (c d^{2} + 5 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3} e^{2}} + \frac {3 \, c d^{2} e^{2} x^{5} + 15 \, a e^{4} x^{5} - 8 \, c d^{3} e x^{3} + 40 \, a d e^{3} x^{3} - 3 \, c d^{4} x + 33 \, a d^{2} e^{2} x}{48 \, {\left (e x^{2} + d\right )}^{3} d^{3} e^{2}} \]
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Time = 13.97 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.05 \[ \int \frac {a+c x^4}{\left (d+e x^2\right )^4} \, dx=\frac {\frac {x^5\,\left (c\,d^2+5\,a\,e^2\right )}{16\,d^3}+\frac {x^3\,\left (5\,a\,e^2-c\,d^2\right )}{6\,d^2\,e}+\frac {x\,\left (11\,a\,e^2-c\,d^2\right )}{16\,d\,e^2}}{d^3+3\,d^2\,e\,x^2+3\,d\,e^2\,x^4+e^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (c\,d^2+5\,a\,e^2\right )}{16\,d^{7/2}\,e^{5/2}} \]
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