3.875 \(\int \frac {x^7}{(a+b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=119 \[ \frac {3 a b \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}+\frac {3 b x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {x^6 \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2} \]

[Out]

-1/4*x^6*(2*c*x^2+b)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)^2+3/4*b*x^2*(b*x^2+2*a)/(-4*a*c+b^2)^2/(c*x^4+b*x^2+a)+3*a*b
*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 = 0.10, antiderivative size = 119, 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, 728, 722, 618, 206} \[ -\frac {x^6 \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {3 b x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {3 a b \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^7/(a + b*x^2 + c*x^4)^3,x]

[Out]

-(x^6*(b + 2*c*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) + (3*b*x^2*(2*a + b*x^2))/(4*(b^2 - 4*a*c)^2*(a +
b*x^2 + c*x^4)) + (3*a*b*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 722

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[(2*(2*p + 3)*(c*d
^2 - b*d*e + a*e^2))/((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 2)*(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] && EqQ[m +
2*p + 2, 0] && LtQ[p, -1]

Rule 728

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

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^7}{\left (a+b x^2+c x^4\right )^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^3}{\left (a+b x+c x^2\right )^3} \, dx,x,x^2\right )\\ &=-\frac {x^6 \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {x^2}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )}{4 \left (b^2-4 a c\right )}\\ &=-\frac {x^6 \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {3 b x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {(3 a b) \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^6 \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {3 b x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {(3 a b) \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^6 \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}+\frac {3 b x^2 \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {3 a b \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 = 0.19, size = 137, normalized size = 1.15 \[ -\frac {8 a^3 c+a^2 \left (b^2+10 b c x^2+16 c^2 x^4\right )+a b x^2 \left (2 b^2+b c x^2+6 c^2 x^4\right )+b^4 x^4}{4 c \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )^2}-\frac {3 a b \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/(a + b*x^2 + c*x^4)^3,x]

[Out]

-1/4*(8*a^3*c + b^4*x^4 + a*b*x^2*(2*b^2 + b*c*x^2 + 6*c^2*x^4) + a^2*(b^2 + 10*b*c*x^2 + 16*c^2*x^4))/(c*(b^2
- 4*a*c)^2*(a + b*x^2 + c*x^4)^2) - (3*a*b*ArcTan[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(5/2)

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fricas [B]  time = 1.11, size = 892, normalized size = 7.50 \[ \left [-\frac {6 \, {\left (a b^{3} c^{2} - 4 \, a^{2} b c^{3}\right )} x^{6} + a^{2} b^{4} + 4 \, a^{3} b^{2} c - 32 \, a^{4} c^{2} + {\left (b^{6} - 3 \, a b^{4} c + 12 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} x^{4} + 2 \, {\left (a b^{5} + a^{2} b^{3} c - 20 \, a^{3} b c^{2}\right )} x^{2} - 6 \, {\left (a b c^{3} x^{8} + 2 \, a b^{2} c^{2} x^{6} + 2 \, a^{2} b^{2} c x^{2} + a^{3} b c + {\left (a b^{3} c + 2 \, a^{2} b c^{2}\right )} x^{4}\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 (a^{2} b^{6} c - 12 \, a^{3} b^{4} c^{2} + 48 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4} + {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} x^{8} + 2 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} x^{6} + {\left (b^{8} c - 10 \, a b^{6} c^{2} + 24 \, a^{2} b^{4} c^{3} + 32 \, a^{3} b^{2} c^{4} - 128 \, a^{4} c^{5}\right )} x^{4} + 2 \, {\left (a b^{7} c - 12 \, a^{2} b^{5} c^{2} + 48 \, a^{3} b^{3} c^{3} - 64 \, a^{4} b c^{4}\right )} x^{2}\right )}}, -\frac {6 \, {\left (a b^{3} c^{2} - 4 \, a^{2} b c^{3}\right )} x^{6} + a^{2} b^{4} + 4 \, a^{3} b^{2} c - 32 \, a^{4} c^{2} + {\left (b^{6} - 3 \, a b^{4} c + 12 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} x^{4} + 2 \, {\left (a b^{5} + a^{2} b^{3} c - 20 \, a^{3} b c^{2}\right )} x^{2} - 12 \, {\left (a b c^{3} x^{8} + 2 \, a b^{2} c^{2} x^{6} + 2 \, a^{2} b^{2} c x^{2} + a^{3} b c + {\left (a b^{3} c + 2 \, a^{2} b c^{2}\right )} x^{4}\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 (a^{2} b^{6} c - 12 \, a^{3} b^{4} c^{2} + 48 \, a^{4} b^{2} c^{3} - 64 \, a^{5} c^{4} + {\left (b^{6} c^{3} - 12 \, a b^{4} c^{4} + 48 \, a^{2} b^{2} c^{5} - 64 \, a^{3} c^{6}\right )} x^{8} + 2 \, {\left (b^{7} c^{2} - 12 \, a b^{5} c^{3} + 48 \, a^{2} b^{3} c^{4} - 64 \, a^{3} b c^{5}\right )} x^{6} + {\left (b^{8} c - 10 \, a b^{6} c^{2} + 24 \, a^{2} b^{4} c^{3} + 32 \, a^{3} b^{2} c^{4} - 128 \, a^{4} c^{5}\right )} x^{4} + 2 \, {\left (a b^{7} c - 12 \, a^{2} b^{5} c^{2} + 48 \, a^{3} b^{3} c^{3} - 64 \, a^{4} b c^{4}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(c*x^4+b*x^2+a)^3,x, algorithm="fricas")

[Out]

[-1/4*(6*(a*b^3*c^2 - 4*a^2*b*c^3)*x^6 + a^2*b^4 + 4*a^3*b^2*c - 32*a^4*c^2 + (b^6 - 3*a*b^4*c + 12*a^2*b^2*c^
2 - 64*a^3*c^3)*x^4 + 2*(a*b^5 + a^2*b^3*c - 20*a^3*b*c^2)*x^2 - 6*(a*b*c^3*x^8 + 2*a*b^2*c^2*x^6 + 2*a^2*b^2*
c*x^2 + a^3*b*c + (a*b^3*c + 2*a^2*b*c^2)*x^4)*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)))/(a^2*b^6*c - 12*a^3*b^4*c^2 + 48*a^4*b^2*c^3 - 64*a^5*c^4
+ (b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^8 + 2*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^2*b^3*c^4 - 6
4*a^3*b*c^5)*x^6 + (b^8*c - 10*a*b^6*c^2 + 24*a^2*b^4*c^3 + 32*a^3*b^2*c^4 - 128*a^4*c^5)*x^4 + 2*(a*b^7*c - 1
2*a^2*b^5*c^2 + 48*a^3*b^3*c^3 - 64*a^4*b*c^4)*x^2), -1/4*(6*(a*b^3*c^2 - 4*a^2*b*c^3)*x^6 + a^2*b^4 + 4*a^3*b
^2*c - 32*a^4*c^2 + (b^6 - 3*a*b^4*c + 12*a^2*b^2*c^2 - 64*a^3*c^3)*x^4 + 2*(a*b^5 + a^2*b^3*c - 20*a^3*b*c^2)
*x^2 - 12*(a*b*c^3*x^8 + 2*a*b^2*c^2*x^6 + 2*a^2*b^2*c*x^2 + a^3*b*c + (a*b^3*c + 2*a^2*b*c^2)*x^4)*sqrt(-b^2
+ 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)/(b^2 - 4*a*c)))/(a^2*b^6*c - 12*a^3*b^4*c^2 + 48*a^4*b^2*c^3
- 64*a^5*c^4 + (b^6*c^3 - 12*a*b^4*c^4 + 48*a^2*b^2*c^5 - 64*a^3*c^6)*x^8 + 2*(b^7*c^2 - 12*a*b^5*c^3 + 48*a^
2*b^3*c^4 - 64*a^3*b*c^5)*x^6 + (b^8*c - 10*a*b^6*c^2 + 24*a^2*b^4*c^3 + 32*a^3*b^2*c^4 - 128*a^4*c^5)*x^4 + 2
*(a*b^7*c - 12*a^2*b^5*c^2 + 48*a^3*b^3*c^3 - 64*a^4*b*c^4)*x^2)]

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giac [A]  time = 1.77, size = 171, normalized size = 1.44 \[ -\frac {3 \, a b \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {6 \, a b c^{2} x^{6} + b^{4} x^{4} + a b^{2} c x^{4} + 16 \, a^{2} c^{2} x^{4} + 2 \, a b^{3} x^{2} + 10 \, a^{2} b c x^{2} + a^{2} b^{2} + 8 \, a^{3} c}{4 \, {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} {\left (c x^{4} + b x^{2} + a\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(c*x^4+b*x^2+a)^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.02, size = 230, normalized size = 1.93 \[ -\frac {3 a b \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}+\frac {-\frac {3 a b c \,x^{6}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}-\frac {\left (5 a c +b^{2}\right ) a b \,x^{2}}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) c}-\frac {\left (16 a^{2} c^{2}+a \,b^{2} c +b^{4}\right ) x^{4}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) c}-\frac {\left (8 a c +b^{2}\right ) a^{2}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) c}}{2 \left (c \,x^{4}+b \,x^{2}+a \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(c*x^4+b*x^2+a)^3,x)

[Out]

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

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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^7/(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 = 4.44, size = 423, normalized size = 3.55 \[ -\frac {\frac {x^2\,\left (5\,c\,a^2\,b+a\,b^3\right )}{2\,c\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x^4\,\left (16\,a^2\,c^2+a\,b^2\,c+b^4\right )}{4\,c\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {a\,\left (8\,c\,a^2+a\,b^2\right )}{4\,c\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {3\,a\,b\,c\,x^6}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^4\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^8+2\,a\,b\,x^2+2\,b\,c\,x^6}-\frac {3\,a\,b\,\mathrm {atan}\left (\frac {\left (x^2\,\left (\frac {9\,a\,b^2\,c^2}{{\left (4\,a\,c-b^2\right )}^{9/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {9\,a\,b^3\,\left (32\,a^2\,b\,c^4-16\,a\,b^3\,c^3+2\,b^5\,c^2\right )}{2\,{\left (4\,a\,c-b^2\right )}^{15/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )+\frac {18\,a^2\,b^3\,c^2}{{\left (4\,a\,c-b^2\right )}^{15/2}}\right )\,\left (b^4\,{\left (4\,a\,c-b^2\right )}^5+16\,a^2\,c^2\,{\left (4\,a\,c-b^2\right )}^5-8\,a\,b^2\,c\,{\left (4\,a\,c-b^2\right )}^5\right )}{18\,a^2\,b^2\,c^2}\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(a + b*x^2 + c*x^4)^3,x)

[Out]

- ((x^2*(a*b^3 + 5*a^2*b*c))/(2*c*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (x^4*(b^4 + 16*a^2*c^2 + a*b^2*c))/(4*c*(b
^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (a*(a*b^2 + 8*a^2*c))/(4*c*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (3*a*b*c*x^6)/(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) - (3*a*b*atan(((x
^2*((9*a*b^2*c^2)/((4*a*c - b^2)^(9/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (9*a*b^3*(2*b^5*c^2 - 16*a*b^3*c^3 +
32*a^2*b*c^4))/(2*(4*a*c - b^2)^(15/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))) + (18*a^2*b^3*c^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))/(18*a^2*b^2*c^2)))/(4*a*c -
b^2)^(5/2)

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sympy [B]  time = 3.81, size = 524, normalized size = 4.40 \[ \frac {3 a b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x^{2} + \frac {- 192 a^{4} b c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 144 a^{3} b^{3} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 36 a^{2} b^{5} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 a b^{7} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 a b^{2}}{6 a b c} \right )}}{2} - \frac {3 a b \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x^{2} + \frac {192 a^{4} b c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 144 a^{3} b^{3} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 36 a^{2} b^{5} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 3 a b^{7} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 a b^{2}}{6 a b c} \right )}}{2} + \frac {- 8 a^{3} c - a^{2} b^{2} - 6 a b c^{2} x^{6} + x^{4} \left (- 16 a^{2} c^{2} - a b^{2} c - b^{4}\right ) + x^{2} \left (- 10 a^{2} b c - 2 a b^{3}\right )}{64 a^{4} c^{3} - 32 a^{3} b^{2} c^{2} + 4 a^{2} b^{4} c + x^{8} \left (64 a^{2} c^{5} - 32 a b^{2} c^{4} + 4 b^{4} c^{3}\right ) + x^{6} \left (128 a^{2} b c^{4} - 64 a b^{3} c^{3} + 8 b^{5} c^{2}\right ) + x^{4} \left (128 a^{3} c^{4} - 24 a b^{4} c^{2} + 4 b^{6} c\right ) + x^{2} \left (128 a^{3} b c^{3} - 64 a^{2} b^{3} c^{2} + 8 a b^{5} c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(c*x**4+b*x**2+a)**3,x)

[Out]

3*a*b*sqrt(-1/(4*a*c - b**2)**5)*log(x**2 + (-192*a**4*b*c**3*sqrt(-1/(4*a*c - b**2)**5) + 144*a**3*b**3*c**2*
sqrt(-1/(4*a*c - b**2)**5) - 36*a**2*b**5*c*sqrt(-1/(4*a*c - b**2)**5) + 3*a*b**7*sqrt(-1/(4*a*c - b**2)**5) +
3*a*b**2)/(6*a*b*c))/2 - 3*a*b*sqrt(-1/(4*a*c - b**2)**5)*log(x**2 + (192*a**4*b*c**3*sqrt(-1/(4*a*c - b**2)*
*5) - 144*a**3*b**3*c**2*sqrt(-1/(4*a*c - b**2)**5) + 36*a**2*b**5*c*sqrt(-1/(4*a*c - b**2)**5) - 3*a*b**7*sqr
t(-1/(4*a*c - b**2)**5) + 3*a*b**2)/(6*a*b*c))/2 + (-8*a**3*c - a**2*b**2 - 6*a*b*c**2*x**6 + x**4*(-16*a**2*c
**2 - a*b**2*c - b**4) + x**2*(-10*a**2*b*c - 2*a*b**3))/(64*a**4*c**3 - 32*a**3*b**2*c**2 + 4*a**2*b**4*c + x
**8*(64*a**2*c**5 - 32*a*b**2*c**4 + 4*b**4*c**3) + x**6*(128*a**2*b*c**4 - 64*a*b**3*c**3 + 8*b**5*c**2) + x*
*4*(128*a**3*c**4 - 24*a*b**4*c**2 + 4*b**6*c) + x**2*(128*a**3*b*c**3 - 64*a**2*b**3*c**2 + 8*a*b**5*c))

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