Optimal. Leaf size=135 \[ \frac {\left (3 c d^2-e (2 b d-a e)\right ) x}{e^4}-\frac {(2 c d-b e) x^3}{3 e^3}+\frac {c x^5}{5 e^2}+\frac {d \left (c d^2-b d e+a e^2\right ) x}{2 e^4 \left (d+e x^2\right )}-\frac {\sqrt {d} \left (7 c d^2-e (5 b d-3 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 e^{9/2}} \]
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Rubi [A]
time = 0.11, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1271, 1824,
211} \begin {gather*} -\frac {\sqrt {d} \text {ArcTan}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (7 c d^2-e (5 b d-3 a e)\right )}{2 e^{9/2}}+\frac {x \left (3 c d^2-e (2 b d-a e)\right )}{e^4}+\frac {d x \left (a e^2-b d e+c d^2\right )}{2 e^4 \left (d+e x^2\right )}-\frac {x^3 (2 c d-b e)}{3 e^3}+\frac {c x^5}{5 e^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 211
Rule 1271
Rule 1824
Rubi steps
\begin {align*} \int \frac {x^4 \left (a+b x^2+c x^4\right )}{\left (d+e x^2\right )^2} \, dx &=\frac {d \left (c d^2-b d e+a e^2\right ) x}{2 e^4 \left (d+e x^2\right )}-\frac {\int \frac {d \left (c d^2-b d e+a e^2\right )-2 e \left (c d^2-b d e+a e^2\right ) x^2+2 e^2 (c d-b e) x^4-2 c e^3 x^6}{d+e x^2} \, dx}{2 e^4}\\ &=\frac {d \left (c d^2-b d e+a e^2\right ) x}{2 e^4 \left (d+e x^2\right )}-\frac {\int \left (-2 \left (3 c d^2-2 b d e+a e^2\right )+2 e (2 c d-b e) x^2-2 c e^2 x^4+\frac {7 c d^3-5 b d^2 e+3 a d e^2}{d+e x^2}\right ) \, dx}{2 e^4}\\ &=\frac {\left (3 c d^2-e (2 b d-a e)\right ) x}{e^4}-\frac {(2 c d-b e) x^3}{3 e^3}+\frac {c x^5}{5 e^2}+\frac {d \left (c d^2-b d e+a e^2\right ) x}{2 e^4 \left (d+e x^2\right )}-\frac {\left (d \left (7 c d^2-e (5 b d-3 a e)\right )\right ) \int \frac {1}{d+e x^2} \, dx}{2 e^4}\\ &=\frac {\left (3 c d^2-e (2 b d-a e)\right ) x}{e^4}-\frac {(2 c d-b e) x^3}{3 e^3}+\frac {c x^5}{5 e^2}+\frac {d \left (c d^2-b d e+a e^2\right ) x}{2 e^4 \left (d+e x^2\right )}-\frac {\sqrt {d} \left (7 c d^2-e (5 b d-3 a e)\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 e^{9/2}}\\ \end {align*}
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Mathematica [A]
time = 0.05, size = 133, normalized size = 0.99 \begin {gather*} \frac {\left (3 c d^2-2 b d e+a e^2\right ) x}{e^4}+\frac {(-2 c d+b e) x^3}{3 e^3}+\frac {c x^5}{5 e^2}+\frac {\left (c d^3-b d^2 e+a d e^2\right ) x}{2 e^4 \left (d+e x^2\right )}-\frac {\sqrt {d} \left (7 c d^2-5 b d e+3 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 e^{9/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.14, size = 122, normalized size = 0.90
method | result | size |
default | \(\frac {\frac {1}{5} c \,x^{5} e^{2}+\frac {1}{3} b \,e^{2} x^{3}-\frac {2}{3} c d e \,x^{3}+a \,e^{2} x -2 d e b x +3 c \,d^{2} x}{e^{4}}-\frac {d \left (\frac {\left (-\frac {1}{2} a \,e^{2}+\frac {1}{2} d e b -\frac {1}{2} c \,d^{2}\right ) x}{e \,x^{2}+d}+\frac {\left (3 a \,e^{2}-5 d e b +7 c \,d^{2}\right ) \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{2 \sqrt {d e}}\right )}{e^{4}}\) | \(122\) |
risch | \(\frac {c \,x^{5}}{5 e^{2}}+\frac {b \,x^{3}}{3 e^{2}}-\frac {2 c d \,x^{3}}{3 e^{3}}+\frac {a x}{e^{2}}-\frac {2 d b x}{e^{3}}+\frac {3 c \,d^{2} x}{e^{4}}+\frac {\left (\frac {1}{2} d \,e^{2} a -\frac {1}{2} d^{2} e b +\frac {1}{2} c \,d^{3}\right ) x}{e^{4} \left (e \,x^{2}+d \right )}+\frac {3 \sqrt {-d e}\, \ln \left (-\sqrt {-d e}\, x -d \right ) a}{4 e^{3}}-\frac {5 \sqrt {-d e}\, \ln \left (-\sqrt {-d e}\, x -d \right ) d b}{4 e^{4}}+\frac {7 \sqrt {-d e}\, \ln \left (-\sqrt {-d e}\, x -d \right ) c \,d^{2}}{4 e^{5}}-\frac {3 \sqrt {-d e}\, \ln \left (\sqrt {-d e}\, x -d \right ) a}{4 e^{3}}+\frac {5 \sqrt {-d e}\, \ln \left (\sqrt {-d e}\, x -d \right ) d b}{4 e^{4}}-\frac {7 \sqrt {-d e}\, \ln \left (\sqrt {-d e}\, x -d \right ) c \,d^{2}}{4 e^{5}}\) | \(250\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.50, size = 122, normalized size = 0.90 \begin {gather*} -\frac {{\left (7 \, c d^{3} - 5 \, b d^{2} e + 3 \, a d e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{2 \, \sqrt {d}} + \frac {1}{15} \, {\left (3 \, c x^{5} e^{2} - 5 \, {\left (2 \, c d e - b e^{2}\right )} x^{3} + 15 \, {\left (3 \, c d^{2} - 2 \, b d e + a e^{2}\right )} x\right )} e^{\left (-4\right )} + \frac {{\left (c d^{3} - b d^{2} e + a d e^{2}\right )} x}{2 \, {\left (x^{2} e^{5} + d e^{4}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.39, size = 341, normalized size = 2.53 \begin {gather*} \left [\frac {210 \, c d^{3} x + 15 \, {\left (7 \, c d^{3} + 3 \, a x^{2} e^{3} - {\left (5 \, b d x^{2} - 3 \, a d\right )} e^{2} + {\left (7 \, c d^{2} x^{2} - 5 \, b d^{2}\right )} e\right )} \sqrt {-d e^{\left (-1\right )}} \log \left (\frac {x^{2} e - 2 \, \sqrt {-d e^{\left (-1\right )}} x e - d}{x^{2} e + d}\right ) + 4 \, {\left (3 \, c x^{7} + 5 \, b x^{5} + 15 \, a x^{3}\right )} e^{3} - 2 \, {\left (14 \, c d x^{5} + 50 \, b d x^{3} - 45 \, a d x\right )} e^{2} + 10 \, {\left (14 \, c d^{2} x^{3} - 15 \, b d^{2} x\right )} e}{60 \, {\left (x^{2} e^{5} + d e^{4}\right )}}, \frac {105 \, c d^{3} x - 15 \, {\left (7 \, c d^{3} + 3 \, a x^{2} e^{3} - {\left (5 \, b d x^{2} - 3 \, a d\right )} e^{2} + {\left (7 \, c d^{2} x^{2} - 5 \, b d^{2}\right )} e\right )} \sqrt {d} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )} + 2 \, {\left (3 \, c x^{7} + 5 \, b x^{5} + 15 \, a x^{3}\right )} e^{3} - {\left (14 \, c d x^{5} + 50 \, b d x^{3} - 45 \, a d x\right )} e^{2} + 5 \, {\left (14 \, c d^{2} x^{3} - 15 \, b d^{2} x\right )} e}{30 \, {\left (x^{2} e^{5} + d e^{4}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.71, size = 189, normalized size = 1.40 \begin {gather*} \frac {c x^{5}}{5 e^{2}} + x^{3} \left (\frac {b}{3 e^{2}} - \frac {2 c d}{3 e^{3}}\right ) + x \left (\frac {a}{e^{2}} - \frac {2 b d}{e^{3}} + \frac {3 c d^{2}}{e^{4}}\right ) + \frac {x \left (a d e^{2} - b d^{2} e + c d^{3}\right )}{2 d e^{4} + 2 e^{5} x^{2}} + \frac {\sqrt {- \frac {d}{e^{9}}} \cdot \left (3 a e^{2} - 5 b d e + 7 c d^{2}\right ) \log {\left (- e^{4} \sqrt {- \frac {d}{e^{9}}} + x \right )}}{4} - \frac {\sqrt {- \frac {d}{e^{9}}} \cdot \left (3 a e^{2} - 5 b d e + 7 c d^{2}\right ) \log {\left (e^{4} \sqrt {- \frac {d}{e^{9}}} + x \right )}}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 4.16, size = 125, normalized size = 0.93 \begin {gather*} -\frac {{\left (7 \, c d^{3} - 5 \, b d^{2} e + 3 \, a d e^{2}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{2 \, \sqrt {d}} + \frac {1}{15} \, {\left (3 \, c x^{5} e^{8} - 10 \, c d x^{3} e^{7} + 5 \, b x^{3} e^{8} + 45 \, c d^{2} x e^{6} - 30 \, b d x e^{7} + 15 \, a x e^{8}\right )} e^{\left (-10\right )} + \frac {{\left (c d^{3} x - b d^{2} x e + a d x e^{2}\right )} e^{\left (-4\right )}}{2 \, {\left (x^{2} e + d\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.32, size = 179, normalized size = 1.33 \begin {gather*} x^3\,\left (\frac {b}{3\,e^2}-\frac {2\,c\,d}{3\,e^3}\right )-x\,\left (\frac {c\,d^2}{e^4}-\frac {a}{e^2}+\frac {2\,d\,\left (\frac {b}{e^2}-\frac {2\,c\,d}{e^3}\right )}{e}\right )+\frac {c\,x^5}{5\,e^2}+\frac {x\,\left (\frac {c\,d^3}{2}-\frac {b\,d^2\,e}{2}+\frac {a\,d\,e^2}{2}\right )}{e^5\,x^2+d\,e^4}-\frac {\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e}\,x\,\left (7\,c\,d^2-5\,b\,d\,e+3\,a\,e^2\right )}{7\,c\,d^3-5\,b\,d^2\,e+3\,a\,d\,e^2}\right )\,\left (7\,c\,d^2-5\,b\,d\,e+3\,a\,e^2\right )}{2\,e^{9/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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