3.2.80 \(\int \frac {(a+b x^2+c x^4)^2}{(d+e x^2)^2} \, dx\)

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.

[In]

Int[(a + b*x^2 + c*x^4)^2/(d + e*x^2)^2,x]

[Out]

((3*c^2*d^2 + b^2*e^2 - 2*c*e*(2*b*d - a*e))*x)/e^4 - (2*c*(c*d - b*e)*x^3)/(3*e^3) + (c^2*x^5)/(5*e^2) + ((c*
d^2 - b*d*e + a*e^2)^2*x)/(2*d*e^4*(d + e*x^2)) - ((c*d^2 - b*d*e + a*e^2)*(7*c*d^2 - e*(3*b*d + a*e))*ArcTan[
(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/2)*e^(9/2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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.

[In]

Integrate[(a + b*x^2 + c*x^4)^2/(d + e*x^2)^2,x]

[Out]

((3*c^2*d^2 + b^2*e^2 + 2*c*e*(-2*b*d + a*e))*x)/e^4 + (2*c*(-(c*d) + b*e)*x^3)/(3*e^3) + (c^2*x^5)/(5*e^2) +
((c*d^2 + e*(-(b*d) + a*e))^2*x)/(2*d*e^4*(d + e*x^2)) - ((7*c^2*d^4 + 2*c*d^2*e*(-5*b*d + 3*a*e) - e^2*(-3*b^
2*d^2 + 2*a*b*d*e + a^2*e^2))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(2*d^(3/2)*e^(9/2))

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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.

[In]

IntegrateAlgebraic[(a + b*x^2 + c*x^4)^2/(d + e*x^2)^2,x]

[Out]

IntegrateAlgebraic[(a + b*x^2 + c*x^4)^2/(d + e*x^2)^2, x]

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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.

[In]

integrate((c*x^4+b*x^2+a)^2/(e*x^2+d)^2,x, algorithm="fricas")

[Out]

[1/60*(12*c^2*d^2*e^4*x^7 - 4*(7*c^2*d^3*e^3 - 10*b*c*d^2*e^4)*x^5 + 20*(7*c^2*d^4*e^2 - 10*b*c*d^3*e^3 + 3*(b
^2 + 2*a*c)*d^2*e^4)*x^3 + 15*(7*c^2*d^5 - 10*b*c*d^4*e - 2*a*b*d^2*e^3 - a^2*d*e^4 + 3*(b^2 + 2*a*c)*d^3*e^2
+ (7*c^2*d^4*e - 10*b*c*d^3*e^2 - 2*a*b*d*e^4 - a^2*e^5 + 3*(b^2 + 2*a*c)*d^2*e^3)*x^2)*sqrt(-d*e)*log((e*x^2
- 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) + 30*(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*(b^2 + 2
*a*c)*d^3*e^3)*x)/(d^2*e^6*x^2 + d^3*e^5), 1/30*(6*c^2*d^2*e^4*x^7 - 2*(7*c^2*d^3*e^3 - 10*b*c*d^2*e^4)*x^5 +
10*(7*c^2*d^4*e^2 - 10*b*c*d^3*e^3 + 3*(b^2 + 2*a*c)*d^2*e^4)*x^3 - 15*(7*c^2*d^5 - 10*b*c*d^4*e - 2*a*b*d^2*e
^3 - a^2*d*e^4 + 3*(b^2 + 2*a*c)*d^3*e^2 + (7*c^2*d^4*e - 10*b*c*d^3*e^2 - 2*a*b*d*e^4 - a^2*e^5 + 3*(b^2 + 2*
a*c)*d^2*e^3)*x^2)*sqrt(d*e)*arctan(sqrt(d*e)*x/d) + 15*(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*(b^2 + 2*a*c)*d^3*e^3)*x)/(d^2*e^6*x^2 + d^3*e^5)]

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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.

[In]

integrate((c*x^4+b*x^2+a)^2/(e*x^2+d)^2,x, algorithm="giac")

[Out]

1/15*(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 + 3
0*a*c*x*e^8)*e^(-10) - 1/2*(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)*
arctan(x*e^(1/2)/sqrt(d))*e^(-9/2)/d^(3/2) + 1/2*(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)*e^(-4)/((x^2*e + d)*d)

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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.

[In]

int((c*x^4+b*x^2+a)^2/(e*x^2+d)^2,x)

[Out]

1/5*c^2/e^2*x^5+2/3/e^2*x^3*b*c-2/3*c^2*d/e^3*x^3+2*a*c/e^2*x+1/e^2*b^2*x-4/e^3*b*c*d*x+3*c^2*d^2/e^4*x+1/2/(e
*x^2+d)*a^2/d*x-1/e*x/(e*x^2+d)*a*b+1/(e*x^2+d)*a*c*d/e^2*x+1/2/e^2*d*x/(e*x^2+d)*b^2-1/e^3*d^2*x/(e*x^2+d)*b*
c+1/2/(e*x^2+d)*c^2*d^3/e^4*x+1/2/(d*e)^(1/2)*a^2/d*arctan(1/(d*e)^(1/2)*e*x)+1/e/(d*e)^(1/2)*arctan(1/(d*e)^(
1/2)*e*x)*a*b-3/(d*e)^(1/2)*a*c*d/e^2*arctan(1/(d*e)^(1/2)*e*x)-3/2/e^2*d/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x
)*b^2+5/e^3*d^2/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*b*c-7/2/(d*e)^(1/2)*c^2*d^3/e^4*arctan(1/(d*e)^(1/2)*e*x
)

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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.

[In]

integrate((c*x^4+b*x^2+a)^2/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

1/2*(c^2*d^4 - 2*b*c*d^3*e - 2*a*b*d*e^3 + a^2*e^4 + (b^2 + 2*a*c)*d^2*e^2)*x/(d*e^5*x^2 + d^2*e^4) + 1/15*(3*
c^2*e^2*x^5 - 10*(c^2*d*e - b*c*e^2)*x^3 + 15*(3*c^2*d^2 - 4*b*c*d*e + (b^2 + 2*a*c)*e^2)*x)/e^4 - 1/2*(7*c^2*
d^4 - 10*b*c*d^3*e - 2*a*b*d*e^3 - a^2*e^4 + 3*(b^2 + 2*a*c)*d^2*e^2)*arctan(e*x/sqrt(d*e))/(sqrt(d*e)*d*e^4)

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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.

[In]

int((a + b*x^2 + c*x^4)^2/(d + e*x^2)^2,x)

[Out]

x*((2*a*c + b^2)/e^2 + (2*d*((2*c^2*d)/e^3 - (2*b*c)/e^2))/e - (c^2*d^2)/e^4) - x^3*((2*c^2*d)/(3*e^3) - (2*b*
c)/(3*e^2)) + (c^2*x^5)/(5*e^2) + (x*(a^2*e^4 + c^2*d^4 + b^2*d^2*e^2 - 2*a*b*d*e^3 - 2*b*c*d^3*e + 2*a*c*d^2*
e^2))/(2*d*(d*e^4 + e^5*x^2)) + (atan((e^(1/2)*x*(a*e^2 + c*d^2 - b*d*e)*(a*e^2 - 7*c*d^2 + 3*b*d*e))/(d^(1/2)
*(a^2*e^4 - 7*c^2*d^4 - 3*b^2*d^2*e^2 + 2*a*b*d*e^3 + 10*b*c*d^3*e - 6*a*c*d^2*e^2)))*(a*e^2 + c*d^2 - b*d*e)*
(a*e^2 - 7*c*d^2 + 3*b*d*e))/(2*d^(3/2)*e^(9/2))

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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.

[In]

integrate((c*x**4+b*x**2+a)**2/(e*x**2+d)**2,x)

[Out]

c**2*x**5/(5*e**2) + x**3*(2*b*c/(3*e**2) - 2*c**2*d/(3*e**3)) + x*(2*a*c/e**2 + b**2/e**2 - 4*b*c*d/e**3 + 3*
c**2*d**2/e**4) + x*(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)/(
2*d**2*e**4 + 2*d*e**5*x**2) - sqrt(-1/(d**3*e**9))*(a*e**2 - b*d*e + c*d**2)*(a*e**2 + 3*b*d*e - 7*c*d**2)*lo
g(-d**2*e**4*sqrt(-1/(d**3*e**9))*(a*e**2 - b*d*e + c*d**2)*(a*e**2 + 3*b*d*e - 7*c*d**2)/(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)/4 + sqrt(-1/(d**3*e**9))*(a*e**
2 - b*d*e + c*d**2)*(a*e**2 + 3*b*d*e - 7*c*d**2)*log(d**2*e**4*sqrt(-1/(d**3*e**9))*(a*e**2 - b*d*e + c*d**2)
*(a*e**2 + 3*b*d*e - 7*c*d**2)/(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)/4

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