3.2.81 \(\int \frac {(a+b x^2+c x^4)^2}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=201 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e^2 \left (3 a^2 e^2+2 a b d e+3 b^2 d^2\right )-6 c d^2 e (5 b d-a e)+35 c^2 d^4\right )}{8 d^{5/2} e^{9/2}}-\frac {x \left (-3 a e^2-5 b d e+13 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{8 d^2 e^4 \left (d+e x^2\right )}+\frac {x \left (a e^2-b d e+c d^2\right )^2}{4 d e^4 \left (d+e x^2\right )^2}-\frac {c x (3 c d-2 b e)}{e^4}+\frac {c^2 x^3}{3 e^3} \]

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Rubi [A]  time = 0.42, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1157, 1814, 1153, 205} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e^2 \left (3 a^2 e^2+2 a b d e+3 b^2 d^2\right )-6 c d^2 e (5 b d-a e)+35 c^2 d^4\right )}{8 d^{5/2} e^{9/2}}-\frac {x \left (-3 a e^2-5 b d e+13 c d^2\right ) \left (a e^2-b d e+c d^2\right )}{8 d^2 e^4 \left (d+e x^2\right )}+\frac {x \left (a e^2-b d e+c d^2\right )^2}{4 d e^4 \left (d+e x^2\right )^2}-\frac {c x (3 c d-2 b e)}{e^4}+\frac {c^2 x^3}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2 + c*x^4)^2/(d + e*x^2)^3,x]

[Out]

-((c*(3*c*d - 2*b*e)*x)/e^4) + (c^2*x^3)/(3*e^3) + ((c*d^2 - b*d*e + a*e^2)^2*x)/(4*d*e^4*(d + e*x^2)^2) - ((1
3*c*d^2 - 5*b*d*e - 3*a*e^2)*(c*d^2 - b*d*e + a*e^2)*x)/(8*d^2*e^4*(d + e*x^2)) + ((35*c^2*d^4 - 6*c*d^2*e*(5*
b*d - a*e) + e^2*(3*b^2*d^2 + 2*a*b*d*e + 3*a^2*e^2))*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*d^(5/2)*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 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2+c x^4\right )^2}{\left (d+e x^2\right )^3} \, dx &=\frac {\left (c d^2-b d e+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}-\frac {\int \frac {\frac {\left (c d^2-b d e-a e^2\right ) \left (c d^2-b d e+3 a e^2\right )}{e^4}-\frac {4 d \left (c^2 d^2+b^2 e^2-2 c e (b d-a e)\right ) x^2}{e^3}+\frac {4 c d (c d-2 b e) x^4}{e^2}-\frac {4 c^2 d x^6}{e}}{\left (d+e x^2\right )^2} \, dx}{4 d}\\ &=\frac {\left (c d^2-b d e+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}-\frac {\left (13 c d^2-5 b d e-3 a e^2\right ) \left (c d^2-b d e+a e^2\right ) x}{8 d^2 e^4 \left (d+e x^2\right )}+\frac {\int \frac {\frac {11 c^2 d^4-2 c d^2 e (7 b d-3 a e)+e^2 \left (3 b^2 d^2+2 a b d e+3 a^2 e^2\right )}{e^4}-\frac {16 c d^2 (c d-b e) x^2}{e^3}+\frac {8 c^2 d^2 x^4}{e^2}}{d+e x^2} \, dx}{8 d^2}\\ &=\frac {\left (c d^2-b d e+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}-\frac {\left (13 c d^2-5 b d e-3 a e^2\right ) \left (c d^2-b d e+a e^2\right ) x}{8 d^2 e^4 \left (d+e x^2\right )}+\frac {\int \left (-\frac {8 c d^2 (3 c d-2 b e)}{e^4}+\frac {8 c^2 d^2 x^2}{e^3}+\frac {35 c^2 d^4-30 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+3 a^2 e^4}{e^4 \left (d+e x^2\right )}\right ) \, dx}{8 d^2}\\ &=-\frac {c (3 c d-2 b e) x}{e^4}+\frac {c^2 x^3}{3 e^3}+\frac {\left (c d^2-b d e+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}-\frac {\left (13 c d^2-5 b d e-3 a e^2\right ) \left (c d^2-b d e+a e^2\right ) x}{8 d^2 e^4 \left (d+e x^2\right )}+\frac {\left (35 c^2 d^4-6 c d^2 e (5 b d-a e)+e^2 \left (3 b^2 d^2+2 a b d e+3 a^2 e^2\right )\right ) \int \frac {1}{d+e x^2} \, dx}{8 d^2 e^4}\\ &=-\frac {c (3 c d-2 b e) x}{e^4}+\frac {c^2 x^3}{3 e^3}+\frac {\left (c d^2-b d e+a e^2\right )^2 x}{4 d e^4 \left (d+e x^2\right )^2}-\frac {\left (13 c d^2-5 b d e-3 a e^2\right ) \left (c d^2-b d e+a e^2\right ) x}{8 d^2 e^4 \left (d+e x^2\right )}+\frac {\left (35 c^2 d^4-6 c d^2 e (5 b d-a e)+e^2 \left (3 b^2 d^2+2 a b d e+3 a^2 e^2\right )\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 d^{5/2} e^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 217, normalized size = 1.08 \begin {gather*} -\frac {x \left (e^2 \left (-3 a^2 e^2-2 a b d e+5 b^2 d^2\right )-2 c d^2 e (9 b d-5 a e)+13 c^2 d^4\right )}{8 d^2 e^4 \left (d+e x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (e^2 \left (3 a^2 e^2+2 a b d e+3 b^2 d^2\right )+6 c d^2 e (a e-5 b d)+35 c^2 d^4\right )}{8 d^{5/2} e^{9/2}}+\frac {x \left (e (a e-b d)+c d^2\right )^2}{4 d e^4 \left (d+e x^2\right )^2}+\frac {c x (2 b e-3 c d)}{e^4}+\frac {c^2 x^3}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2 + c*x^4)^2/(d + e*x^2)^3,x]

[Out]

(c*(-3*c*d + 2*b*e)*x)/e^4 + (c^2*x^3)/(3*e^3) + ((c*d^2 + e*(-(b*d) + a*e))^2*x)/(4*d*e^4*(d + e*x^2)^2) - ((
13*c^2*d^4 - 2*c*d^2*e*(9*b*d - 5*a*e) + e^2*(5*b^2*d^2 - 2*a*b*d*e - 3*a^2*e^2))*x)/(8*d^2*e^4*(d + e*x^2)) +
((35*c^2*d^4 + 6*c*d^2*e*(-5*b*d + a*e) + e^2*(3*b^2*d^2 + 2*a*b*d*e + 3*a^2*e^2))*ArcTan[(Sqrt[e]*x)/Sqrt[d]
])/(8*d^(5/2)*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 )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x^2 + c*x^4)^2/(d + e*x^2)^3,x]

[Out]

IntegrateAlgebraic[(a + b*x^2 + c*x^4)^2/(d + e*x^2)^3, x]

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fricas [B]  time = 0.71, size = 794, normalized size = 3.95 \begin {gather*} \left [\frac {16 \, c^{2} d^{3} e^{4} x^{7} - 16 \, {\left (7 \, c^{2} d^{4} e^{3} - 6 \, b c d^{3} e^{4}\right )} x^{5} - 2 \, {\left (175 \, c^{2} d^{5} e^{2} - 150 \, b c d^{4} e^{3} - 6 \, a b d^{2} e^{5} - 9 \, a^{2} d e^{6} + 15 \, {\left (b^{2} + 2 \, a c\right )} d^{3} e^{4}\right )} x^{3} - 3 \, {\left (35 \, c^{2} d^{6} - 30 \, b c d^{5} e + 2 \, a b d^{3} e^{3} + 3 \, a^{2} d^{2} e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{4} e^{2} + {\left (35 \, c^{2} d^{4} e^{2} - 30 \, b c d^{3} e^{3} + 2 \, a b d e^{5} + 3 \, a^{2} e^{6} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{4}\right )} x^{4} + 2 \, {\left (35 \, c^{2} d^{5} e - 30 \, b c d^{4} e^{2} + 2 \, a b d^{2} e^{4} + 3 \, a^{2} d e^{5} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{3} 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 (35 \, c^{2} d^{6} e - 30 \, b c d^{5} e^{2} + 2 \, a b d^{3} e^{4} - 5 \, a^{2} d^{2} e^{5} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{4} e^{3}\right )} x}{48 \, {\left (d^{3} e^{7} x^{4} + 2 \, d^{4} e^{6} x^{2} + d^{5} e^{5}\right )}}, \frac {8 \, c^{2} d^{3} e^{4} x^{7} - 8 \, {\left (7 \, c^{2} d^{4} e^{3} - 6 \, b c d^{3} e^{4}\right )} x^{5} - {\left (175 \, c^{2} d^{5} e^{2} - 150 \, b c d^{4} e^{3} - 6 \, a b d^{2} e^{5} - 9 \, a^{2} d e^{6} + 15 \, {\left (b^{2} + 2 \, a c\right )} d^{3} e^{4}\right )} x^{3} + 3 \, {\left (35 \, c^{2} d^{6} - 30 \, b c d^{5} e + 2 \, a b d^{3} e^{3} + 3 \, a^{2} d^{2} e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{4} e^{2} + {\left (35 \, c^{2} d^{4} e^{2} - 30 \, b c d^{3} e^{3} + 2 \, a b d e^{5} + 3 \, a^{2} e^{6} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{4}\right )} x^{4} + 2 \, {\left (35 \, c^{2} d^{5} e - 30 \, b c d^{4} e^{2} + 2 \, a b d^{2} e^{4} + 3 \, a^{2} d e^{5} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{3} e^{3}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - 3 \, {\left (35 \, c^{2} d^{6} e - 30 \, b c d^{5} e^{2} + 2 \, a b d^{3} e^{4} - 5 \, a^{2} d^{2} e^{5} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{4} e^{3}\right )} x}{24 \, {\left (d^{3} e^{7} x^{4} + 2 \, d^{4} e^{6} x^{2} + d^{5} 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)^3,x, algorithm="fricas")

[Out]

[1/48*(16*c^2*d^3*e^4*x^7 - 16*(7*c^2*d^4*e^3 - 6*b*c*d^3*e^4)*x^5 - 2*(175*c^2*d^5*e^2 - 150*b*c*d^4*e^3 - 6*
a*b*d^2*e^5 - 9*a^2*d*e^6 + 15*(b^2 + 2*a*c)*d^3*e^4)*x^3 - 3*(35*c^2*d^6 - 30*b*c*d^5*e + 2*a*b*d^3*e^3 + 3*a
^2*d^2*e^4 + 3*(b^2 + 2*a*c)*d^4*e^2 + (35*c^2*d^4*e^2 - 30*b*c*d^3*e^3 + 2*a*b*d*e^5 + 3*a^2*e^6 + 3*(b^2 + 2
*a*c)*d^2*e^4)*x^4 + 2*(35*c^2*d^5*e - 30*b*c*d^4*e^2 + 2*a*b*d^2*e^4 + 3*a^2*d*e^5 + 3*(b^2 + 2*a*c)*d^3*e^3)
*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - 6*(35*c^2*d^6*e - 30*b*c*d^5*e^2 + 2*a*b*d^3*
e^4 - 5*a^2*d^2*e^5 + 3*(b^2 + 2*a*c)*d^4*e^3)*x)/(d^3*e^7*x^4 + 2*d^4*e^6*x^2 + d^5*e^5), 1/24*(8*c^2*d^3*e^4
*x^7 - 8*(7*c^2*d^4*e^3 - 6*b*c*d^3*e^4)*x^5 - (175*c^2*d^5*e^2 - 150*b*c*d^4*e^3 - 6*a*b*d^2*e^5 - 9*a^2*d*e^
6 + 15*(b^2 + 2*a*c)*d^3*e^4)*x^3 + 3*(35*c^2*d^6 - 30*b*c*d^5*e + 2*a*b*d^3*e^3 + 3*a^2*d^2*e^4 + 3*(b^2 + 2*
a*c)*d^4*e^2 + (35*c^2*d^4*e^2 - 30*b*c*d^3*e^3 + 2*a*b*d*e^5 + 3*a^2*e^6 + 3*(b^2 + 2*a*c)*d^2*e^4)*x^4 + 2*(
35*c^2*d^5*e - 30*b*c*d^4*e^2 + 2*a*b*d^2*e^4 + 3*a^2*d*e^5 + 3*(b^2 + 2*a*c)*d^3*e^3)*x^2)*sqrt(d*e)*arctan(s
qrt(d*e)*x/d) - 3*(35*c^2*d^6*e - 30*b*c*d^5*e^2 + 2*a*b*d^3*e^4 - 5*a^2*d^2*e^5 + 3*(b^2 + 2*a*c)*d^4*e^3)*x)
/(d^3*e^7*x^4 + 2*d^4*e^6*x^2 + d^5*e^5)]

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giac [A]  time = 0.18, size = 244, normalized size = 1.21 \begin {gather*} \frac {1}{3} \, {\left (c^{2} x^{3} e^{6} - 9 \, c^{2} d x e^{5} + 6 \, b c x e^{6}\right )} e^{\left (-9\right )} + \frac {{\left (35 \, c^{2} d^{4} - 30 \, 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} + 3 \, a^{2} e^{4}\right )} \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {9}{2}\right )}}{8 \, d^{\frac {5}{2}}} - \frac {{\left (13 \, c^{2} d^{4} x^{3} e - 18 \, b c d^{3} x^{3} e^{2} + 11 \, c^{2} d^{5} x + 5 \, b^{2} d^{2} x^{3} e^{3} + 10 \, a c d^{2} x^{3} e^{3} - 14 \, b c d^{4} x e - 2 \, a b d x^{3} e^{4} + 3 \, b^{2} d^{3} x e^{2} + 6 \, a c d^{3} x e^{2} - 3 \, a^{2} x^{3} e^{5} + 2 \, a b d^{2} x e^{3} - 5 \, a^{2} d x e^{4}\right )} e^{\left (-4\right )}}{8 \, {\left (x^{2} e + d\right )}^{2} d^{2}} \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)^3,x, algorithm="giac")

[Out]

1/3*(c^2*x^3*e^6 - 9*c^2*d*x*e^5 + 6*b*c*x*e^6)*e^(-9) + 1/8*(35*c^2*d^4 - 30*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 + 3*a^2*e^4)*arctan(x*e^(1/2)/sqrt(d))*e^(-9/2)/d^(5/2) - 1/8*(13*c^2*d^4*x^3*e - 18*b
*c*d^3*x^3*e^2 + 11*c^2*d^5*x + 5*b^2*d^2*x^3*e^3 + 10*a*c*d^2*x^3*e^3 - 14*b*c*d^4*x*e - 2*a*b*d*x^3*e^4 + 3*
b^2*d^3*x*e^2 + 6*a*c*d^3*x*e^2 - 3*a^2*x^3*e^5 + 2*a*b*d^2*x*e^3 - 5*a^2*d*x*e^4)*e^(-4)/((x^2*e + d)^2*d^2)

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maple [B]  time = 0.01, size = 402, normalized size = 2.00 \begin {gather*} \frac {3 a^{2} e \,x^{3}}{8 \left (e \,x^{2}+d \right )^{2} d^{2}}+\frac {a b \,x^{3}}{4 \left (e \,x^{2}+d \right )^{2} d}-\frac {5 a c \,x^{3}}{4 \left (e \,x^{2}+d \right )^{2} e}-\frac {5 b^{2} x^{3}}{8 \left (e \,x^{2}+d \right )^{2} e}+\frac {9 b c d \,x^{3}}{4 \left (e \,x^{2}+d \right )^{2} e^{2}}-\frac {13 c^{2} d^{2} x^{3}}{8 \left (e \,x^{2}+d \right )^{2} e^{3}}+\frac {5 a^{2} x}{8 \left (e \,x^{2}+d \right )^{2} d}-\frac {a b x}{4 \left (e \,x^{2}+d \right )^{2} e}-\frac {3 a c d x}{4 \left (e \,x^{2}+d \right )^{2} e^{2}}-\frac {3 b^{2} d x}{8 \left (e \,x^{2}+d \right )^{2} e^{2}}+\frac {7 b c \,d^{2} x}{4 \left (e \,x^{2}+d \right )^{2} e^{3}}-\frac {11 c^{2} d^{3} x}{8 \left (e \,x^{2}+d \right )^{2} e^{4}}+\frac {c^{2} x^{3}}{3 e^{3}}+\frac {3 a^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, d^{2}}+\frac {a b \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{4 \sqrt {d e}\, d e}+\frac {3 a c \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{4 \sqrt {d e}\, e^{2}}+\frac {3 b^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, e^{2}}-\frac {15 b c d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{4 \sqrt {d e}\, e^{3}}+\frac {35 c^{2} d^{2} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \sqrt {d e}\, e^{4}}+\frac {2 b c x}{e^{3}}-\frac {3 c^{2} d 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)^3,x)

[Out]

1/3*c^2/e^3*x^3+2*c/e^3*b*x-3*c^2*d/e^4*x+3/8/(e*x^2+d)^2*a^2/d^2*e*x^3+1/4/(e*x^2+d)^2/d*x^3*a*b-5/4/(e*x^2+d
)^2*a*c/e*x^3-5/8/e/(e*x^2+d)^2*x^3*b^2+9/4/e^2/(e*x^2+d)^2*x^3*b*c*d-13/8/(e*x^2+d)^2*c^2*d^2/e^3*x^3+5/8/(e*
x^2+d)^2*a^2/d*x-1/4/e/(e*x^2+d)^2*a*b*x-3/4/(e*x^2+d)^2*a*c*d/e^2*x-3/8/e^2/(e*x^2+d)^2*b^2*d*x+7/4/e^3/(e*x^
2+d)^2*b*c*d^2*x-11/8/(e*x^2+d)^2*c^2*d^3/e^4*x+3/8/(d*e)^(1/2)*a^2/d^2*arctan(1/(d*e)^(1/2)*e*x)+1/4/e/d/(d*e
)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*a*b+3/4/(d*e)^(1/2)*a*c/e^2*arctan(1/(d*e)^(1/2)*e*x)+3/8/e^2/(d*e)^(1/2)*ar
ctan(1/(d*e)^(1/2)*e*x)*b^2-15/4/e^3*d/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*b*c+35/8/(d*e)^(1/2)*c^2*d^2/e^4*
arctan(1/(d*e)^(1/2)*e*x)

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maxima [A]  time = 2.36, size = 245, normalized size = 1.22 \begin {gather*} -\frac {{\left (13 \, c^{2} d^{4} e - 18 \, b c d^{3} e^{2} - 2 \, a b d e^{4} - 3 \, a^{2} e^{5} + 5 \, {\left (b^{2} + 2 \, a c\right )} d^{2} e^{3}\right )} x^{3} + {\left (11 \, c^{2} d^{5} - 14 \, b c d^{4} e + 2 \, a b d^{2} e^{3} - 5 \, a^{2} d e^{4} + 3 \, {\left (b^{2} + 2 \, a c\right )} d^{3} e^{2}\right )} x}{8 \, {\left (d^{2} e^{6} x^{4} + 2 \, d^{3} e^{5} x^{2} + d^{4} e^{4}\right )}} + \frac {c^{2} e x^{3} - 3 \, {\left (3 \, c^{2} d - 2 \, b c e\right )} x}{3 \, e^{4}} + \frac {{\left (35 \, c^{2} d^{4} - 30 \, b c d^{3} e + 2 \, a b d e^{3} + 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 )}{8 \, \sqrt {d e} d^{2} 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)^3,x, algorithm="maxima")

[Out]

-1/8*((13*c^2*d^4*e - 18*b*c*d^3*e^2 - 2*a*b*d*e^4 - 3*a^2*e^5 + 5*(b^2 + 2*a*c)*d^2*e^3)*x^3 + (11*c^2*d^5 -
14*b*c*d^4*e + 2*a*b*d^2*e^3 - 5*a^2*d*e^4 + 3*(b^2 + 2*a*c)*d^3*e^2)*x)/(d^2*e^6*x^4 + 2*d^3*e^5*x^2 + d^4*e^
4) + 1/3*(c^2*e*x^3 - 3*(3*c^2*d - 2*b*c*e)*x)/e^4 + 1/8*(35*c^2*d^4 - 30*b*c*d^3*e + 2*a*b*d*e^3 + 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^2*e^4)

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mupad [B]  time = 0.12, size = 257, normalized size = 1.28 \begin {gather*} \frac {c^2\,x^3}{3\,e^3}-x\,\left (\frac {3\,c^2\,d}{e^4}-\frac {2\,b\,c}{e^3}\right )-\frac {\frac {x\,\left (-5\,a^2\,e^4+2\,a\,b\,d\,e^3+6\,a\,c\,d^2\,e^2+3\,b^2\,d^2\,e^2-14\,b\,c\,d^3\,e+11\,c^2\,d^4\right )}{8\,d}-\frac {x^3\,\left (3\,a^2\,e^5+2\,a\,b\,d\,e^4-10\,a\,c\,d^2\,e^3-5\,b^2\,d^2\,e^3+18\,b\,c\,d^3\,e^2-13\,c^2\,d^4\,e\right )}{8\,d^2}}{d^2\,e^4+2\,d\,e^5\,x^2+e^6\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )\,\left (3\,a^2\,e^4+2\,a\,b\,d\,e^3+6\,a\,c\,d^2\,e^2+3\,b^2\,d^2\,e^2-30\,b\,c\,d^3\,e+35\,c^2\,d^4\right )}{8\,d^{5/2}\,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)^3,x)

[Out]

(c^2*x^3)/(3*e^3) - x*((3*c^2*d)/e^4 - (2*b*c)/e^3) - ((x*(11*c^2*d^4 - 5*a^2*e^4 + 3*b^2*d^2*e^2 + 2*a*b*d*e^
3 - 14*b*c*d^3*e + 6*a*c*d^2*e^2))/(8*d) - (x^3*(3*a^2*e^5 - 13*c^2*d^4*e - 5*b^2*d^2*e^3 + 2*a*b*d*e^4 - 10*a
*c*d^2*e^3 + 18*b*c*d^3*e^2))/(8*d^2))/(d^2*e^4 + e^6*x^4 + 2*d*e^5*x^2) + (atan((e^(1/2)*x)/d^(1/2))*(3*a^2*e
^4 + 35*c^2*d^4 + 3*b^2*d^2*e^2 + 2*a*b*d*e^3 - 30*b*c*d^3*e + 6*a*c*d^2*e^2))/(8*d^(5/2)*e^(9/2))

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sympy [A]  time = 17.72, size = 398, normalized size = 1.98 \begin {gather*} \frac {c^{2} x^{3}}{3 e^{3}} + x \left (\frac {2 b c}{e^{3}} - \frac {3 c^{2} d}{e^{4}}\right ) - \frac {\sqrt {- \frac {1}{d^{5} e^{9}}} \left (3 a^{2} e^{4} + 2 a b d e^{3} + 6 a c d^{2} e^{2} + 3 b^{2} d^{2} e^{2} - 30 b c d^{3} e + 35 c^{2} d^{4}\right ) \log {\left (- d^{3} e^{4} \sqrt {- \frac {1}{d^{5} e^{9}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{d^{5} e^{9}}} \left (3 a^{2} e^{4} + 2 a b d e^{3} + 6 a c d^{2} e^{2} + 3 b^{2} d^{2} e^{2} - 30 b c d^{3} e + 35 c^{2} d^{4}\right ) \log {\left (d^{3} e^{4} \sqrt {- \frac {1}{d^{5} e^{9}}} + x \right )}}{16} + \frac {x^{3} \left (3 a^{2} e^{5} + 2 a b d e^{4} - 10 a c d^{2} e^{3} - 5 b^{2} d^{2} e^{3} + 18 b c d^{3} e^{2} - 13 c^{2} d^{4} e\right ) + x \left (5 a^{2} d e^{4} - 2 a b d^{2} e^{3} - 6 a c d^{3} e^{2} - 3 b^{2} d^{3} e^{2} + 14 b c d^{4} e - 11 c^{2} d^{5}\right )}{8 d^{4} e^{4} + 16 d^{3} e^{5} x^{2} + 8 d^{2} e^{6} x^{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)**3,x)

[Out]

c**2*x**3/(3*e**3) + x*(2*b*c/e**3 - 3*c**2*d/e**4) - sqrt(-1/(d**5*e**9))*(3*a**2*e**4 + 2*a*b*d*e**3 + 6*a*c
*d**2*e**2 + 3*b**2*d**2*e**2 - 30*b*c*d**3*e + 35*c**2*d**4)*log(-d**3*e**4*sqrt(-1/(d**5*e**9)) + x)/16 + sq
rt(-1/(d**5*e**9))*(3*a**2*e**4 + 2*a*b*d*e**3 + 6*a*c*d**2*e**2 + 3*b**2*d**2*e**2 - 30*b*c*d**3*e + 35*c**2*
d**4)*log(d**3*e**4*sqrt(-1/(d**5*e**9)) + x)/16 + (x**3*(3*a**2*e**5 + 2*a*b*d*e**4 - 10*a*c*d**2*e**3 - 5*b*
*2*d**2*e**3 + 18*b*c*d**3*e**2 - 13*c**2*d**4*e) + x*(5*a**2*d*e**4 - 2*a*b*d**2*e**3 - 6*a*c*d**3*e**2 - 3*b
**2*d**3*e**2 + 14*b*c*d**4*e - 11*c**2*d**5))/(8*d**4*e**4 + 16*d**3*e**5*x**2 + 8*d**2*e**6*x**4)

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