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

Optimal. Leaf size=113 \[ \frac {3 b c \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}}+\frac {2 a+b x^2}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 b \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )} \]

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Rubi [A]  time = 0.09, antiderivative size = 113, 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, 638, 614, 618, 206} \begin {gather*} \frac {2 a+b x^2}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 b \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {3 b c \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 {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b*x^2 + c*x^4)^3,x]

[Out]

(2*a + b*x^2)/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)^2) - (3*b*(b + 2*c*x^2))/(4*(b^2 - 4*a*c)^2*(a + b*x^2 + c*
x^4)) + (3*b*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 614

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

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 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^3}{\left (a+b x^2+c x^4\right )^3} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\left (a+b x+c x^2\right )^3} \, dx,x,x^2\right )\\ &=\frac {2 a+b x^2}{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 {1}{\left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )}{4 \left (b^2-4 a c\right )}\\ &=\frac {2 a+b x^2}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 b \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}-\frac {(3 b c) \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 {2 a+b x^2}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 b \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {(3 b c) \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 {2 a+b x^2}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )^2}-\frac {3 b \left (b+2 c x^2\right )}{4 \left (b^2-4 a c\right )^2 \left (a+b x^2+c x^4\right )}+\frac {3 b c \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 = 0.10, size = 114, normalized size = 1.01 \begin {gather*} \frac {-\frac {12 b c \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {\left (b^2-4 a c\right ) \left (2 a+b x^2\right )}{\left (a+b x^2+c x^4\right )^2}-\frac {3 b \left (b+2 c x^2\right )}{a+b x^2+c x^4}}{4 \left (b^2-4 a c\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((b^2 - 4*a*c)*(2*a + b*x^2))/(a + b*x^2 + c*x^4)^2 - (3*b*(b + 2*c*x^2))/(a + b*x^2 + c*x^4) - (12*b*c*ArcTa
n[(b + 2*c*x^2)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(4*(b^2 - 4*a*c)^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3}{\left (a+b x^2+c x^4\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 2.67, size = 808, normalized size = 7.15 \begin {gather*} \left [-\frac {6 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{6} + a b^{4} + 4 \, a^{2} b^{2} c - 32 \, a^{3} c^{2} + 9 \, {\left (b^{4} c - 4 \, a b^{2} c^{2}\right )} x^{4} + 2 \, {\left (b^{5} + a b^{3} c - 20 \, a^{2} b c^{2}\right )} x^{2} - 6 \, {\left (b c^{3} x^{8} + 2 \, b^{2} c^{2} x^{6} + 2 \, a b^{2} c x^{2} + {\left (b^{3} c + 2 \, a b c^{2}\right )} x^{4} + a^{2} b c\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 {6 \, {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{6} + a b^{4} + 4 \, a^{2} b^{2} c - 32 \, a^{3} c^{2} + 9 \, {\left (b^{4} c - 4 \, a b^{2} c^{2}\right )} x^{4} + 2 \, {\left (b^{5} + a b^{3} c - 20 \, a^{2} b c^{2}\right )} x^{2} - 12 \, {\left (b c^{3} x^{8} + 2 \, b^{2} c^{2} x^{6} + 2 \, a b^{2} c x^{2} + {\left (b^{3} c + 2 \, a b c^{2}\right )} x^{4} + a^{2} b c\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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(6*(b^3*c^2 - 4*a*b*c^3)*x^6 + a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2 + 9*(b^4*c - 4*a*b^2*c^2)*x^4 + 2*(b^5 +
a*b^3*c - 20*a^2*b*c^2)*x^2 - 6*(b*c^3*x^8 + 2*b^2*c^2*x^6 + 2*a*b^2*c*x^2 + (b^3*c + 2*a*b*c^2)*x^4 + a^2*b*
c)*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*
(6*(b^3*c^2 - 4*a*b*c^3)*x^6 + a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2 + 9*(b^4*c - 4*a*b^2*c^2)*x^4 + 2*(b^5 + a*b^3
*c - 20*a^2*b*c^2)*x^2 - 12*(b*c^3*x^8 + 2*b^2*c^2*x^6 + 2*a*b^2*c*x^2 + (b^3*c + 2*a*b*c^2)*x^4 + a^2*b*c)*sq
rt(-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 = 1.79, size = 143, normalized size = 1.27 \begin {gather*} -\frac {3 \, b c \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 \, b c^{2} x^{6} + 9 \, b^{2} c x^{4} + 2 \, b^{3} x^{2} + 10 \, a b c x^{2} + a b^{2} + 8 \, a^{2} c}{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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*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*(6*b
*c^2*x^6 + 9*b^2*c*x^4 + 2*b^3*x^2 + 10*a*b*c*x^2 + a*b^2 + 8*a^2*c)/((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 [A]  time = 0.01, size = 142, normalized size = 1.26 \begin {gather*} -\frac {3 b c \,x^{2}}{2 \left (4 a c -b^{2}\right )^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}-\frac {3 b c \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {5}{2}}}-\frac {3 b^{2}}{4 \left (4 a c -b^{2}\right )^{2} \left (c \,x^{4}+b \,x^{2}+a \right )}+\frac {-b \,x^{2}-2 a}{4 \left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{2}+a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-b*x^2-2*a)/(4*a*c-b^2)/(c*x^4+b*x^2+a)^2-3/2*b/(4*a*c-b^2)^2/(c*x^4+b*x^2+a)*x^2*c-3/4*b^2/(4*a*c-b^2)^2
/(c*x^4+b*x^2+a)-3*b/(4*a*c-b^2)^(5/2)*c*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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(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 = 4.39, size = 400, normalized size = 3.54 \begin {gather*} -\frac {\frac {8\,c\,a^2+a\,b^2}{4\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x^2\,\left (b^3+5\,a\,c\,b\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {9\,b^2\,c\,x^4}{4\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {3\,b\,c^2\,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\,b\,c\,\mathrm {atan}\left (\frac {\left (x^2\,\left (\frac {9\,b^2\,c^4}{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^3\,c^2\,\left (144\,a^2\,b\,c^4-72\,a\,b^3\,c^3+9\,b^5\,c^2\right )}{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 {18\,b^3\,c^4}{{\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\,b^2\,c^4}\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b*x^2 + c*x^4)^3,x)

[Out]

- ((a*b^2 + 8*a^2*c)/(4*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (x^2*(b^3 + 5*a*b*c))/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2
*c)) + (9*b^2*c*x^4)/(4*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (3*b*c^2*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*b*c*atan(((x^2*((9*b^2*c^4)/(a*(4*a*c - b^2)^(9
/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) + (b^3*c^2*(9*b^5*c^2 - 72*a*b^3*c^3 + 144*a^2*b*c^4))/(a*(4*a*c - b^2)^(1
5/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))) + (18*b^3*c^4)/(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*b^2*c^4)))/(4*a*c - b^2)^(5/2)

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sympy [B]  time = 3.04, size = 491, normalized size = 4.35 \begin {gather*} \frac {3 b c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x^{2} + \frac {- 192 a^{3} b c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 144 a^{2} b^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 36 a b^{5} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{7} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )}}{2} - \frac {3 b c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \log {\left (x^{2} + \frac {192 a^{3} b c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 144 a^{2} b^{3} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 36 a b^{5} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} - 3 b^{7} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} + 3 b^{2} c}{6 b c^{2}} \right )}}{2} + \frac {- 8 a^{2} c - a b^{2} - 9 b^{2} c x^{4} - 6 b c^{2} x^{6} + x^{2} \left (- 10 a b c - 2 b^{3}\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x**2 + (-192*a**3*b*c**4*sqrt(-1/(4*a*c - b**2)**5) + 144*a**2*b**3*c**3*
sqrt(-1/(4*a*c - b**2)**5) - 36*a*b**5*c**2*sqrt(-1/(4*a*c - b**2)**5) + 3*b**7*c*sqrt(-1/(4*a*c - b**2)**5) +
3*b**2*c)/(6*b*c**2))/2 - 3*b*c*sqrt(-1/(4*a*c - b**2)**5)*log(x**2 + (192*a**3*b*c**4*sqrt(-1/(4*a*c - b**2)
**5) - 144*a**2*b**3*c**3*sqrt(-1/(4*a*c - b**2)**5) + 36*a*b**5*c**2*sqrt(-1/(4*a*c - b**2)**5) - 3*b**7*c*sq
rt(-1/(4*a*c - b**2)**5) + 3*b**2*c)/(6*b*c**2))/2 + (-8*a**2*c - a*b**2 - 9*b**2*c*x**4 - 6*b*c**2*x**6 + x**
2*(-10*a*b*c - 2*b**3))/(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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