Optimal. Leaf size=166 \[ \frac {x \left (-2 c e (2 b d-a e)+b^2 e^2+3 c^2 d^2\right )}{e^4}+\frac {x \left (a e^2-b d e+c d^2\right )^2}{2 d e^4 \left (d+e x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a e^2-b d e+c d^2\right ) \left (7 c d^2-e (a e+3 b d)\right )}{2 d^{3/2} e^{9/2}}-\frac {2 c x^3 (c d-b e)}{3 e^3}+\frac {c^2 x^5}{5 e^2} \]
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Rubi [A] time = 0.30, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1157, 1810, 205} \begin {gather*} \frac {x \left (-2 c e (2 b d-a e)+b^2 e^2+3 c^2 d^2\right )}{e^4}+\frac {x \left (a e^2-b d e+c d^2\right )^2}{2 d e^4 \left (d+e x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a e^2-b d e+c d^2\right ) \left (7 c d^2-e (a e+3 b d)\right )}{2 d^{3/2} e^{9/2}}-\frac {2 c x^3 (c d-b e)}{3 e^3}+\frac {c^2 x^5}{5 e^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 205
Rule 1157
Rule 1810
Rubi steps
\begin {align*} \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^2} \, dx &=\frac {\left (c d^2-b d e+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac {\int \frac {\frac {c^2 d^4-2 c d^2 e (b d-a e)+e^2 \left (b^2 d^2-2 a b d e-a^2 e^2\right )}{e^4}-\frac {2 d \left (c^2 d^2+b^2 e^2-2 c e (b d-a e)\right ) x^2}{e^3}+\frac {2 c d (c d-2 b e) x^4}{e^2}-\frac {2 c^2 d x^6}{e}}{d+e x^2} \, dx}{2 d}\\ &=\frac {\left (c d^2-b d e+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac {\int \left (-\frac {2 d \left (3 c^2 d^2+b^2 e^2-2 c e (2 b d-a e)\right )}{e^4}+\frac {4 c d (c d-b e) x^2}{e^3}-\frac {2 c^2 d x^4}{e^2}+\frac {7 c^2 d^4-10 b c d^3 e+3 b^2 d^2 e^2+6 a c d^2 e^2-2 a b d e^3-a^2 e^4}{e^4 \left (d+e x^2\right )}\right ) \, dx}{2 d}\\ &=\frac {\left (3 c^2 d^2+b^2 e^2-2 c e (2 b d-a e)\right ) x}{e^4}-\frac {2 c (c d-b e) x^3}{3 e^3}+\frac {c^2 x^5}{5 e^2}+\frac {\left (c d^2-b d e+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac {\left (7 c^2 d^4-10 b c d^3 e+3 b^2 d^2 e^2+6 a c d^2 e^2-2 a b d e^3-a^2 e^4\right ) \int \frac {1}{d+e x^2} \, dx}{2 d e^4}\\ &=\frac {\left (3 c^2 d^2+b^2 e^2-2 c e (2 b d-a e)\right ) x}{e^4}-\frac {2 c (c d-b e) x^3}{3 e^3}+\frac {c^2 x^5}{5 e^2}+\frac {\left (c d^2-b d e+a e^2\right )^2 x}{2 d e^4 \left (d+e x^2\right )}-\frac {\left (7 c d^2-3 b d e-a e^2\right ) \left (c d^2-b d e+a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 d^{3/2} e^{9/2}}\\ \end {align*}
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Mathematica [A] time = 0.10, size = 183, normalized size = 1.10 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (-e^2 \left (a^2 e^2+2 a b d e-3 b^2 d^2\right )+2 c d^2 e (3 a e-5 b d)+7 c^2 d^4\right )}{2 d^{3/2} e^{9/2}}+\frac {x \left (2 c e (a e-2 b d)+b^2 e^2+3 c^2 d^2\right )}{e^4}+\frac {x \left (e (a e-b d)+c d^2\right )^2}{2 d e^4 \left (d+e x^2\right )}+\frac {2 c x^3 (b e-c d)}{3 e^3}+\frac {c^2 x^5}{5 e^2} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^2} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.91, size = 600, normalized size = 3.61 \begin {gather*} \left [\frac {12 \, c^{2} d^{2} e^{4} x^{7} - 4 \, {\left (7 \, c^{2} d^{3} e^{3} - 10 \, b c d^{2} e^{4}\right )} x^{5} + 20 \, {\left (7 \, c^{2} d^{4} e^{2} - 10 \, b c d^{3} e^{3} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{4}\right )} x^{3} + 15 \, {\left (7 \, c^{2} d^{5} - 10 \, b c d^{4} e - 2 \, a b d^{2} e^{3} - a^{2} d e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} + {\left (7 \, c^{2} d^{4} e - 10 \, b c d^{3} e^{2} - 2 \, a b d e^{4} - a^{2} e^{5} + 3 \, {\left (b^{2} + 2 \, a c\right )} 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 ) + 30 \, {\left (7 \, c^{2} d^{5} e - 10 \, b c d^{4} e^{2} - 2 \, a b d^{2} e^{4} + a^{2} d e^{5} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{3} e^{3}\right )} x}{60 \, {\left (d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}, \frac {6 \, c^{2} d^{2} e^{4} x^{7} - 2 \, {\left (7 \, c^{2} d^{3} e^{3} - 10 \, b c d^{2} e^{4}\right )} x^{5} + 10 \, {\left (7 \, c^{2} d^{4} e^{2} - 10 \, b c d^{3} e^{3} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{4}\right )} x^{3} - 15 \, {\left (7 \, c^{2} d^{5} - 10 \, b c d^{4} e - 2 \, a b d^{2} e^{3} - a^{2} d e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2} + {\left (7 \, c^{2} d^{4} e - 10 \, b c d^{3} e^{2} - 2 \, a b d e^{4} - a^{2} e^{5} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) + 15 \, {\left (7 \, c^{2} d^{5} e - 10 \, b c d^{4} e^{2} - 2 \, a b d^{2} e^{4} + a^{2} d e^{5} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{3} e^{3}\right )} x}{30 \, {\left (d^{2} e^{6} x^{2} + d^{3} e^{5}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 207, normalized size = 1.25 \begin {gather*} \frac {1}{15} \, {\left (3 \, c^{2} x^{5} e^{8} - 10 \, c^{2} d x^{3} e^{7} + 10 \, b c x^{3} e^{8} + 45 \, c^{2} d^{2} x e^{6} - 60 \, b c d x e^{7} + 15 \, b^{2} x e^{8} + 30 \, a c x e^{8}\right )} e^{\left (-10\right )} - \frac {{\left (7 \, c^{2} d^{4} - 10 \, b c d^{3} e + 3 \, b^{2} d^{2} e^{2} + 6 \, a c d^{2} e^{2} - 2 \, a b d e^{3} - a^{2} e^{4}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{2 \, d^{\frac {3}{2}}} + \frac {{\left (c^{2} d^{4} x - 2 \, b c d^{3} x e + b^{2} d^{2} x e^{2} + 2 \, a c d^{2} x e^{2} - 2 \, a b d x e^{3} + a^{2} x e^{4}\right )} e^{\left (-4\right )}}{2 \, {\left (x^{2} e + d\right )} d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.01, size = 320, normalized size = 1.93 \begin {gather*} \frac {c^{2} x^{5}}{5 e^{2}}+\frac {2 b c \,x^{3}}{3 e^{2}}-\frac {2 c^{2} d \,x^{3}}{3 e^{3}}+\frac {a^{2} x}{2 \left (e \,x^{2}+d \right ) d}+\frac {a^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, d}-\frac {a b x}{\left (e \,x^{2}+d \right ) e}+\frac {a b \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}\, e}+\frac {a c d x}{\left (e \,x^{2}+d \right ) e^{2}}-\frac {3 a c d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}\, e^{2}}+\frac {b^{2} d x}{2 \left (e \,x^{2}+d \right ) e^{2}}-\frac {3 b^{2} d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{2}}-\frac {b c \,d^{2} x}{\left (e \,x^{2}+d \right ) e^{3}}+\frac {5 b c \,d^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{\sqrt {d e}\, e^{3}}+\frac {c^{2} d^{3} x}{2 \left (e \,x^{2}+d \right ) e^{4}}-\frac {7 c^{2} d^{3} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}\, e^{4}}+\frac {2 a c x}{e^{2}}+\frac {b^{2} x}{e^{2}}-\frac {4 b c d x}{e^{3}}+\frac {3 c^{2} d^{2} x}{e^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.42, size = 205, normalized size = 1.23 \begin {gather*} \frac {{\left (c^{2} d^{4} - 2 \, b c d^{3} e - 2 \, a b d e^{3} + a^{2} e^{4} + {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2}\right )} x}{2 \, {\left (d e^{5} x^{2} + d^{2} e^{4}\right )}} + \frac {3 \, c^{2} e^{2} x^{5} - 10 \, {\left (c^{2} d e - b c e^{2}\right )} x^{3} + 15 \, {\left (3 \, c^{2} d^{2} - 4 \, b c d e + {\left (b^{2} + 2 \, a c\right )} e^{2}\right )} x}{15 \, e^{4}} - \frac {{\left (7 \, c^{2} d^{4} - 10 \, b c d^{3} e - 2 \, a b d e^{3} - a^{2} e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \, \sqrt {d e} d e^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.56, size = 293, normalized size = 1.77 \begin {gather*} x\,\left (\frac {b^2+2\,a\,c}{e^2}+\frac {2\,d\,\left (\frac {2\,c^2\,d}{e^3}-\frac {2\,b\,c}{e^2}\right )}{e}-\frac {c^2\,d^2}{e^4}\right )-x^3\,\left (\frac {2\,c^2\,d}{3\,e^3}-\frac {2\,b\,c}{3\,e^2}\right )+\frac {c^2\,x^5}{5\,e^2}+\frac {x\,\left (a^2\,e^4-2\,a\,b\,d\,e^3+2\,a\,c\,d^2\,e^2+b^2\,d^2\,e^2-2\,b\,c\,d^3\,e+c^2\,d^4\right )}{2\,d\,\left (e^5\,x^2+d\,e^4\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x\,\left (c\,d^2-b\,d\,e+a\,e^2\right )\,\left (-7\,c\,d^2+3\,b\,d\,e+a\,e^2\right )}{\sqrt {d}\,\left (a^2\,e^4+2\,a\,b\,d\,e^3-6\,a\,c\,d^2\,e^2-3\,b^2\,d^2\,e^2+10\,b\,c\,d^3\,e-7\,c^2\,d^4\right )}\right )\,\left (c\,d^2-b\,d\,e+a\,e^2\right )\,\left (-7\,c\,d^2+3\,b\,d\,e+a\,e^2\right )}{2\,d^{3/2}\,e^{9/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 3.79, size = 484, normalized size = 2.92 \begin {gather*} \frac {c^{2} x^{5}}{5 e^{2}} + x^{3} \left (\frac {2 b c}{3 e^{2}} - \frac {2 c^{2} d}{3 e^{3}}\right ) + x \left (\frac {2 a c}{e^{2}} + \frac {b^{2}}{e^{2}} - \frac {4 b c d}{e^{3}} + \frac {3 c^{2} d^{2}}{e^{4}}\right ) + \frac {x \left (a^{2} e^{4} - 2 a b d e^{3} + 2 a c d^{2} e^{2} + b^{2} d^{2} e^{2} - 2 b c d^{3} e + c^{2} d^{4}\right )}{2 d^{2} e^{4} + 2 d e^{5} x^{2}} - \frac {\sqrt {- \frac {1}{d^{3} e^{9}}} \left (a e^{2} - b d e + c d^{2}\right ) \left (a e^{2} + 3 b d e - 7 c d^{2}\right ) \log {\left (- \frac {d^{2} e^{4} \sqrt {- \frac {1}{d^{3} e^{9}}} \left (a e^{2} - b d e + c d^{2}\right ) \left (a e^{2} + 3 b d e - 7 c d^{2}\right )}{a^{2} e^{4} + 2 a b d e^{3} - 6 a c d^{2} e^{2} - 3 b^{2} d^{2} e^{2} + 10 b c d^{3} e - 7 c^{2} d^{4}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{d^{3} e^{9}}} \left (a e^{2} - b d e + c d^{2}\right ) \left (a e^{2} + 3 b d e - 7 c d^{2}\right ) \log {\left (\frac {d^{2} e^{4} \sqrt {- \frac {1}{d^{3} e^{9}}} \left (a e^{2} - b d e + c d^{2}\right ) \left (a e^{2} + 3 b d e - 7 c d^{2}\right )}{a^{2} e^{4} + 2 a b d e^{3} - 6 a c d^{2} e^{2} - 3 b^{2} d^{2} e^{2} + 10 b c d^{3} e - 7 c^{2} d^{4}} + x \right )}}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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