Optimal. Leaf size=150 \[ -\frac {x \left (7 c d^2-e (5 a e+b d)\right )}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e (5 a e+b d)+c d^2\right )}{16 d^{7/2} e^{5/2}}+\frac {x \left (e (5 a e+b d)+c d^2\right )}{16 d^3 e^2 \left (d+e x^2\right )}+\frac {x \left (a+\frac {d (c d-b e)}{e^2}\right )}{6 d \left (d+e x^2\right )^3} \]
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Rubi [A] time = 0.21, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1157, 385, 199, 205} \begin {gather*} \frac {x \left (e (5 a e+b d)+c d^2\right )}{16 d^3 e^2 \left (d+e x^2\right )}-\frac {x \left (7 c d^2-e (5 a e+b d)\right )}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e (5 a e+b d)+c d^2\right )}{16 d^{7/2} e^{5/2}}+\frac {x \left (a+\frac {d (c d-b e)}{e^2}\right )}{6 d \left (d+e x^2\right )^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 199
Rule 205
Rule 385
Rule 1157
Rubi steps
\begin {align*} \int \frac {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx &=\frac {\left (c d^2-b d e+a e^2\right ) x}{6 d e^2 \left (d+e x^2\right )^3}-\frac {\int \frac {-5 a+\frac {d (c d-b e)}{e^2}-\frac {6 c d x^2}{e}}{\left (d+e x^2\right )^3} \, dx}{6 d}\\ &=\frac {\left (c d^2-b d e+a e^2\right ) x}{6 d e^2 \left (d+e x^2\right )^3}-\frac {\left (7 c d^2-e (b d+5 a e)\right ) x}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\left (c d^2+e (b d+5 a e)\right ) \int \frac {1}{\left (d+e x^2\right )^2} \, dx}{8 d^2 e^2}\\ &=\frac {\left (c d^2-b d e+a e^2\right ) x}{6 d e^2 \left (d+e x^2\right )^3}-\frac {\left (7 c d^2-e (b d+5 a e)\right ) x}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\left (c d^2+e (b d+5 a e)\right ) x}{16 d^3 e^2 \left (d+e x^2\right )}+\frac {\left (c d^2+e (b d+5 a e)\right ) \int \frac {1}{d+e x^2} \, dx}{16 d^3 e^2}\\ &=\frac {\left (c d^2-b d e+a e^2\right ) x}{6 d e^2 \left (d+e x^2\right )^3}-\frac {\left (7 c d^2-e (b d+5 a e)\right ) x}{24 d^2 e^2 \left (d+e x^2\right )^2}+\frac {\left (c d^2+e (b d+5 a e)\right ) x}{16 d^3 e^2 \left (d+e x^2\right )}+\frac {\left (c d^2+e (b d+5 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{16 d^{7/2} e^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 142, normalized size = 0.95 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e (5 a e+b d)+c d^2\right )}{16 d^{7/2} e^{5/2}}+\frac {x \left (e \left (a e \left (33 d^2+40 d e x^2+15 e^2 x^4\right )+b d \left (-3 d^2+8 d e x^2+3 e^2 x^4\right )\right )+c d^2 \left (-3 d^2-8 d e x^2+3 e^2 x^4\right )\right )}{48 d^3 e^2 \left (d+e x^2\right )^3} \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 {a+b x^2+c x^4}{\left (d+e x^2\right )^4} \, dx \end {gather*}
Verification is not applicable to the result.
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fricas [A] time = 0.60, size = 530, normalized size = 3.53 \begin {gather*} \left [\frac {6 \, {\left (c d^{3} e^{3} + b d^{2} e^{4} + 5 \, a d e^{5}\right )} x^{5} - 16 \, {\left (c d^{4} e^{2} - b d^{3} e^{3} - 5 \, a d^{2} e^{4}\right )} x^{3} - 3 \, {\left ({\left (c d^{2} e^{3} + b d e^{4} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + b d^{4} e + 5 \, a d^{3} e^{2} + 3 \, {\left (c d^{3} e^{2} + b d^{2} e^{3} + 5 \, a d e^{4}\right )} x^{4} + 3 \, {\left (c d^{4} e + b d^{3} e^{2} + 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 + b d^{4} e^{2} - 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} + b d^{2} e^{4} + 5 \, a d e^{5}\right )} x^{5} - 8 \, {\left (c d^{4} e^{2} - b d^{3} e^{3} - 5 \, a d^{2} e^{4}\right )} x^{3} + 3 \, {\left ({\left (c d^{2} e^{3} + b d e^{4} + 5 \, a e^{5}\right )} x^{6} + c d^{5} + b d^{4} e + 5 \, a d^{3} e^{2} + 3 \, {\left (c d^{3} e^{2} + b d^{2} e^{3} + 5 \, a d e^{4}\right )} x^{4} + 3 \, {\left (c d^{4} e + b d^{3} e^{2} + 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 + b d^{4} e^{2} - 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 ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.16, size = 134, normalized size = 0.89 \begin {gather*} \frac {{\left (c d^{2} + b d e + 5 \, a e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {5}{2}\right )}}{16 \, d^{\frac {7}{2}}} + \frac {{\left (3 \, c d^{2} x^{5} e^{2} + 3 \, b d x^{5} e^{3} - 8 \, c d^{3} x^{3} e + 15 \, a x^{5} e^{4} + 8 \, b d^{2} x^{3} e^{2} - 3 \, c d^{4} x + 40 \, a d x^{3} e^{3} - 3 \, b d^{3} x e + 33 \, a d^{2} x e^{2}\right )} e^{\left (-2\right )}}{48 \, {\left (x^{2} e + d\right )}^{3} d^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 158, normalized size = 1.05 \begin {gather*} \frac {5 a \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \sqrt {d e}\, d^{3}}+\frac {b \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \sqrt {d e}\, d^{2} e}+\frac {c \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \sqrt {d e}\, d \,e^{2}}+\frac {\frac {\left (5 a \,e^{2}+b d e +c \,d^{2}\right ) x^{5}}{16 d^{3}}+\frac {\left (5 a \,e^{2}+b d e -c \,d^{2}\right ) x^{3}}{6 d^{2} e}+\frac {\left (11 a \,e^{2}-b d e -c \,d^{2}\right ) x}{16 d \,e^{2}}}{\left (e \,x^{2}+d \right )^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.51, size = 162, normalized size = 1.08 \begin {gather*} \frac {3 \, {\left (c d^{2} e^{2} + b d e^{3} + 5 \, a e^{4}\right )} x^{5} - 8 \, {\left (c d^{3} e - b d^{2} e^{2} - 5 \, a d e^{3}\right )} x^{3} - 3 \, {\left (c d^{4} + b d^{3} e - 11 \, a d^{2} e^{2}\right )} x}{48 \, {\left (d^{3} e^{5} x^{6} + 3 \, d^{4} e^{4} x^{4} + 3 \, d^{5} e^{3} x^{2} + d^{6} e^{2}\right )}} + \frac {{\left (c d^{2} + b d e + 5 \, a e^{2}\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{16 \, \sqrt {d e} d^{3} e^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.51, size = 144, normalized size = 0.96 \begin {gather*} \frac {\frac {x^5\,\left (c\,d^2+b\,d\,e+5\,a\,e^2\right )}{16\,d^3}-\frac {x\,\left (c\,d^2+b\,d\,e-11\,a\,e^2\right )}{16\,d\,e^2}+\frac {x^3\,\left (-c\,d^2+b\,d\,e+5\,a\,e^2\right )}{6\,d^2\,e}}{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+b\,d\,e+5\,a\,e^2\right )}{16\,d^{7/2}\,e^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 4.41, size = 241, normalized size = 1.61 \begin {gather*} - \frac {\sqrt {- \frac {1}{d^{7} e^{5}}} \left (5 a e^{2} + b d e + 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}}} \left (5 a e^{2} + b d e + c d^{2}\right ) \log {\left (d^{4} e^{2} \sqrt {- \frac {1}{d^{7} e^{5}}} + x \right )}}{32} + \frac {x^{5} \left (15 a e^{4} + 3 b d e^{3} + 3 c d^{2} e^{2}\right ) + x^{3} \left (40 a d e^{3} + 8 b d^{2} e^{2} - 8 c d^{3} e\right ) + x \left (33 a d^{2} e^{2} - 3 b d^{3} e - 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}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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