3.876 \(\int \frac {x^5}{(a+b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=130 \[ -\frac {\left (2 a c+b^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}+\frac {x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {x^2 \left (2 a c+b^2\right )+3 a b}{2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )} \]

[Out]

1/4*x^2*(b*x^2+2*a)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)^2+1/2*(3*a*b+x^2*(2*a*c+b^2))/(-4*a*c+b^2)^2/(c*x^4+b*x^2+a)-
(2*a*c+b^2)*arctanh((2*c*x^2+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(5/2)

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Rubi [A]  time = 0.13, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1114, 738, 638, 618, 206} \[ \frac {x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {x^2 \left (2 a c+b^2\right )+3 a b}{2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {\left (2 a c+b^2\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^2 + c*x^4)^3,x]

[Out]

(x^2*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (3*a*b + (b^2 + 2*a*c)*x^2)/(2*(b^2 - 4*a*c)^2*(
a + b*x^2 + c*x^4)) - ((b^2 + 2*a*c)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(b^2 - 4*a*c)^(5/2)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), 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] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {x^5}{\left (a+b x^2+c x^4\right )^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^2}{\left (a+b x+c x^2\right )^3} \, dx,x,x^2\right )\\ &=\frac {x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {\operatorname {Subst}\left (\int \frac {2 a-2 b x}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )}{4 \left (b^2-4 a c\right )}\\ &=\frac {x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {3 a b+\left (b^2+2 a c\right ) x^2}{2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {\left (b^2+2 a c\right ) \operatorname {Subst}\left (\int \frac {1}{a+b x+c x^2} \, dx,x,x^2\right )}{2 \left (b^2-4 a c\right )^2}\\ &=\frac {x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {3 a b+\left (b^2+2 a c\right ) x^2}{2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {\left (b^2+2 a c\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{\left (b^2-4 a c\right )^2}\\ &=\frac {x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {3 a b+\left (b^2+2 a c\right ) x^2}{2 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {\left (b^2+2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 145, normalized size = 1.12 \[ \frac {1}{4} \left (\frac {4 \left (2 a c+b^2\right ) \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}}+\frac {\left (2 a c+b^2\right ) \left (b+2 c x^2\right )}{c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {a \left (b-2 c x^2\right )+b^2 x^2}{c \left (4 a c-b^2\right ) \left (a+b x^2+c x^4\right )^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^2 + c*x^4)^3,x]

[Out]

(((b^2 + 2*a*c)*(b + 2*c*x^2))/(c*(b^2 - 4*a*c)^2*(a + b*x^2 + c*x^4)) + (b^2*x^2 + a*(b - 2*c*x^2))/(c*(-b^2
+ 4*a*c)*(a + b*x^2 + c*x^4)^2) + (4*(b^2 + 2*a*c)*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5
/2))/4

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fricas [B]  time = 0.85, size = 907, normalized size = 6.98 \[ \left [\frac {2 \, {\left (b^{4} c - 2 \, a b^{2} c^{2} - 8 \, a^{2} c^{3}\right )} x^{6} + 6 \, a^{2} b^{3} - 24 \, a^{3} b c + 3 \, {\left (b^{5} - 2 \, a b^{3} c - 8 \, a^{2} b c^{2}\right )} x^{4} + 2 \, {\left (5 \, a b^{4} - 22 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} x^{2} + 2 \, {\left ({\left (b^{2} c^{2} + 2 \, a c^{3}\right )} x^{8} + 2 \, {\left (b^{3} c + 2 \, a b c^{2}\right )} x^{6} + {\left (b^{4} + 4 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} + 2 \, a^{3} c + 2 \, {\left (a b^{3} + 2 \, a^{2} b c\right )} x^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right )}{4 \, {\left ({\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} x^{8} + a^{2} b^{6} - 12 \, a^{3} b^{4} c + 48 \, a^{4} b^{2} c^{2} - 64 \, a^{5} c^{3} + 2 \, {\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} x^{6} + {\left (b^{8} - 10 \, a b^{6} c + 24 \, a^{2} b^{4} c^{2} + 32 \, a^{3} b^{2} c^{3} - 128 \, a^{4} c^{4}\right )} x^{4} + 2 \, {\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} x^{2}\right )}}, \frac {2 \, {\left (b^{4} c - 2 \, a b^{2} c^{2} - 8 \, a^{2} c^{3}\right )} x^{6} + 6 \, a^{2} b^{3} - 24 \, a^{3} b c + 3 \, {\left (b^{5} - 2 \, a b^{3} c - 8 \, a^{2} b c^{2}\right )} x^{4} + 2 \, {\left (5 \, a b^{4} - 22 \, a^{2} b^{2} c + 8 \, a^{3} c^{2}\right )} x^{2} - 4 \, {\left ({\left (b^{2} c^{2} + 2 \, a c^{3}\right )} x^{8} + 2 \, {\left (b^{3} c + 2 \, a b c^{2}\right )} x^{6} + {\left (b^{4} + 4 \, a b^{2} c + 4 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} + 2 \, a^{3} c + 2 \, {\left (a b^{3} + 2 \, a^{2} b c\right )} x^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{4 \, {\left ({\left (b^{6} c^{2} - 12 \, a b^{4} c^{3} + 48 \, a^{2} b^{2} c^{4} - 64 \, a^{3} c^{5}\right )} x^{8} + a^{2} b^{6} - 12 \, a^{3} b^{4} c + 48 \, a^{4} b^{2} c^{2} - 64 \, a^{5} c^{3} + 2 \, {\left (b^{7} c - 12 \, a b^{5} c^{2} + 48 \, a^{2} b^{3} c^{3} - 64 \, a^{3} b c^{4}\right )} x^{6} + {\left (b^{8} - 10 \, a b^{6} c + 24 \, a^{2} b^{4} c^{2} + 32 \, a^{3} b^{2} c^{3} - 128 \, a^{4} c^{4}\right )} x^{4} + 2 \, {\left (a b^{7} - 12 \, a^{2} b^{5} c + 48 \, a^{3} b^{3} c^{2} - 64 \, a^{4} b c^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/4*(2*(b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)*x^6 + 6*a^2*b^3 - 24*a^3*b*c + 3*(b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*x^4
+ 2*(5*a*b^4 - 22*a^2*b^2*c + 8*a^3*c^2)*x^2 + 2*((b^2*c^2 + 2*a*c^3)*x^8 + 2*(b^3*c + 2*a*b*c^2)*x^6 + (b^4
+ 4*a*b^2*c + 4*a^2*c^2)*x^4 + a^2*b^2 + 2*a^3*c + 2*(a*b^3 + 2*a^2*b*c)*x^2)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^4
+ 2*b*c*x^2 + b^2 - 2*a*c - (2*c*x^2 + b)*sqrt(b^2 - 4*a*c))/(c*x^4 + b*x^2 + a)))/((b^6*c^2 - 12*a*b^4*c^3 +
48*a^2*b^2*c^4 - 64*a^3*c^5)*x^8 + a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + 2*(b^7*c - 12*a*b^5
*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^6 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)
*x^4 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x^2), 1/4*(2*(b^4*c - 2*a*b^2*c^2 - 8*a^2*c^3)
*x^6 + 6*a^2*b^3 - 24*a^3*b*c + 3*(b^5 - 2*a*b^3*c - 8*a^2*b*c^2)*x^4 + 2*(5*a*b^4 - 22*a^2*b^2*c + 8*a^3*c^2)
*x^2 - 4*((b^2*c^2 + 2*a*c^3)*x^8 + 2*(b^3*c + 2*a*b*c^2)*x^6 + (b^4 + 4*a*b^2*c + 4*a^2*c^2)*x^4 + a^2*b^2 +
2*a^3*c + 2*(a*b^3 + 2*a^2*b*c)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)
))/((b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^8 + a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*
a^5*c^3 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^6 + (b^8 - 10*a*b^6*c + 24*a^2*b^4*c^2 +
32*a^3*b^2*c^3 - 128*a^4*c^4)*x^4 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x^2)]

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giac [A]  time = 1.81, size = 161, normalized size = 1.24 \[ \frac {{\left (b^{2} + 2 \, a c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {2 \, b^{2} c x^{6} + 4 \, a c^{2} x^{6} + 3 \, b^{3} x^{4} + 6 \, a b c x^{4} + 10 \, a b^{2} x^{2} - 4 \, a^{2} c x^{2} + 6 \, a^{2} b}{4 \, {\left (c x^{4} + b x^{2} + a\right )}^{2} {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2+a)^3,x, algorithm="giac")

[Out]

(b^2 + 2*a*c)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(-b^2 + 4*a*c)) + 1
/4*(2*b^2*c*x^6 + 4*a*c^2*x^6 + 3*b^3*x^4 + 6*a*b*c*x^4 + 10*a*b^2*x^2 - 4*a^2*c*x^2 + 6*a^2*b)/((c*x^4 + b*x^
2 + a)^2*(b^4 - 8*a*b^2*c + 16*a^2*c^2))

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maple [B]  time = 0.02, size = 270, normalized size = 2.08 \[ \frac {2 a c \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}+\frac {b^{2} \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}+\frac {\frac {\left (2 a c +b^{2}\right ) c \,x^{6}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}+\frac {3 \left (2 a c +b^{2}\right ) b \,x^{4}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}+\frac {3 a^{2} b}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {\left (2 a c -5 b^{2}\right ) a \,x^{2}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}}{2 \left (c \,x^{4}+b \,x^{2}+a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^4+b*x^2+a)^3,x)

[Out]

1/2*((2*a*c+b^2)*c/(16*a^2*c^2-8*a*b^2*c+b^4)*x^6+3/2*b*(2*a*c+b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^4-a*(2*a*c-5*
b^2)/(16*a^2*c^2-8*a*b^2*c+b^4)*x^2+3*a^2*b/(16*a^2*c^2-8*a*b^2*c+b^4))/(c*x^4+b*x^2+a)^2+2/(16*a^2*c^2-8*a*b^
2*c+b^4)/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a*c+1/(16*a^2*c^2-8*a*b^2*c+b^4)/(4*a*c-b^2)^
(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 4.46, size = 460, normalized size = 3.54 \[ \frac {\frac {3\,a^2\,b}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x^2\,\left (5\,a\,b^2-2\,a^2\,c\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {3\,b\,x^4\,\left (b^2+2\,a\,c\right )}{4\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {c\,x^6\,\left (b^2+2\,a\,c\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^4\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^8+2\,a\,b\,x^2+2\,b\,c\,x^6}+\frac {\mathrm {atan}\left (\frac {\left (x^2\,\left (\frac {\left (b^2+2\,a\,c\right )\,\left (b^2\,c^2+2\,a\,c^3\right )}{a\,{\left (4\,a\,c-b^2\right )}^{9/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {b\,{\left (b^2+2\,a\,c\right )}^2\,\left (32\,a^2\,b\,c^4-16\,a\,b^3\,c^3+2\,b^5\,c^2\right )}{2\,a\,{\left (4\,a\,c-b^2\right )}^{15/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )+\frac {2\,b\,c^2\,{\left (b^2+2\,a\,c\right )}^2}{{\left (4\,a\,c-b^2\right )}^{15/2}}\right )\,\left (b^4\,{\left (4\,a\,c-b^2\right )}^5+16\,a^2\,c^2\,{\left (4\,a\,c-b^2\right )}^5-8\,a\,b^2\,c\,{\left (4\,a\,c-b^2\right )}^5\right )}{8\,a^2\,c^4+8\,a\,b^2\,c^3+2\,b^4\,c^2}\right )\,\left (b^2+2\,a\,c\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^2 + c*x^4)^3,x)

[Out]

((3*a^2*b)/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (x^2*(5*a*b^2 - 2*a^2*c))/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) +
(3*b*x^4*(2*a*c + b^2))/(4*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (c*x^6*(2*a*c + b^2))/(2*(b^4 + 16*a^2*c^2 - 8*a
*b^2*c)))/(x^4*(2*a*c + b^2) + a^2 + c^2*x^8 + 2*a*b*x^2 + 2*b*c*x^6) + (atan(((x^2*(((2*a*c + b^2)*(2*a*c^3 +
b^2*c^2))/(a*(4*a*c - b^2)^(9/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (b*(2*a*c + b^2)^2*(2*b^5*c^2 - 16*a*b^3*c
^3 + 32*a^2*b*c^4))/(2*a*(4*a*c - b^2)^(15/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))) + (2*b*c^2*(2*a*c + b^2)^2)/(4*
a*c - b^2)^(15/2))*(b^4*(4*a*c - b^2)^5 + 16*a^2*c^2*(4*a*c - b^2)^5 - 8*a*b^2*c*(4*a*c - b^2)^5))/(8*a^2*c^4
+ 2*b^4*c^2 + 8*a*b^2*c^3))*(2*a*c + b^2))/(4*a*c - b^2)^(5/2)

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sympy [B]  time = 5.41, size = 580, normalized size = 4.46 \[ - \frac {\sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) \log {\left (x^{2} + \frac {- 64 a^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + 48 a^{2} b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) - 12 a b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + 2 a b c + b^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + b^{3}}{4 a c^{2} + 2 b^{2} c} \right )}}{2} + \frac {\sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) \log {\left (x^{2} + \frac {64 a^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) - 48 a^{2} b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + 12 a b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + 2 a b c - b^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (2 a c + b^{2}\right ) + b^{3}}{4 a c^{2} + 2 b^{2} c} \right )}}{2} + \frac {6 a^{2} b + x^{6} \left (4 a c^{2} + 2 b^{2} c\right ) + x^{4} \left (6 a b c + 3 b^{3}\right ) + x^{2} \left (- 4 a^{2} c + 10 a b^{2}\right )}{64 a^{4} c^{2} - 32 a^{3} b^{2} c + 4 a^{2} b^{4} + x^{8} \left (64 a^{2} c^{4} - 32 a b^{2} c^{3} + 4 b^{4} c^{2}\right ) + x^{6} \left (128 a^{2} b c^{3} - 64 a b^{3} c^{2} + 8 b^{5} c\right ) + x^{4} \left (128 a^{3} c^{3} - 24 a b^{4} c + 4 b^{6}\right ) + x^{2} \left (128 a^{3} b c^{2} - 64 a^{2} b^{3} c + 8 a b^{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**4+b*x**2+a)**3,x)

[Out]

-sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2)*log(x**2 + (-64*a**3*c**3*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2)
+ 48*a**2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) - 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c
+ b**2) + 2*a*b*c + b**6*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + b**3)/(4*a*c**2 + 2*b**2*c))/2 + sqrt(-1
/(4*a*c - b**2)**5)*(2*a*c + b**2)*log(x**2 + (64*a**3*c**3*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) - 48*a**
2*b**2*c**2*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + 12*a*b**4*c*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2)
+ 2*a*b*c - b**6*sqrt(-1/(4*a*c - b**2)**5)*(2*a*c + b**2) + b**3)/(4*a*c**2 + 2*b**2*c))/2 + (6*a**2*b + x**6
*(4*a*c**2 + 2*b**2*c) + x**4*(6*a*b*c + 3*b**3) + x**2*(-4*a**2*c + 10*a*b**2))/(64*a**4*c**2 - 32*a**3*b**2*
c + 4*a**2*b**4 + x**8*(64*a**2*c**4 - 32*a*b**2*c**3 + 4*b**4*c**2) + x**6*(128*a**2*b*c**3 - 64*a*b**3*c**2
+ 8*b**5*c) + x**4*(128*a**3*c**3 - 24*a*b**4*c + 4*b**6) + x**2*(128*a**3*b*c**2 - 64*a**2*b**3*c + 8*a*b**5)
)

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