3.22.34 \(\int \frac {f+g x}{(a+b x+c x^2)^3} \, dx\)

Optimal. Leaf size=131 \[ \frac {3 (b+2 c x) (2 c f-b g)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {-2 a g+x (2 c f-b g)+b f}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {6 c (2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {638, 614, 618, 206} \begin {gather*} \frac {3 (b+2 c x) (2 c f-b g)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {-2 a g+x (2 c f-b g)+b f}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {6 c (2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f + g*x)/(a + b*x + c*x^2)^3,x]

[Out]

-(b*f - 2*a*g + (2*c*f - b*g)*x)/(2*(b^2 - 4*a*c)*(a + b*x + c*x^2)^2) + (3*(2*c*f - b*g)*(b + 2*c*x))/(2*(b^2
- 4*a*c)^2*(a + b*x + c*x^2)) - (6*c*(2*c*f - b*g)*ArcTanh[(b + 2*c*x)/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]

Rubi steps

\begin {align*} \int \frac {f+g x}{\left (a+b x+c x^2\right )^3} \, dx &=-\frac {b f-2 a g+(2 c f-b g) x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}-\frac {(3 (2 c f-b g)) \int \frac {1}{\left (a+b x+c x^2\right )^2} \, dx}{2 \left (b^2-4 a c\right )}\\ &=-\frac {b f-2 a g+(2 c f-b g) x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 (2 c f-b g) (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}+\frac {(3 c (2 c f-b g)) \int \frac {1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=-\frac {b f-2 a g+(2 c f-b g) x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 (2 c f-b g) (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {(6 c (2 c f-b g)) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{\left (b^2-4 a c\right )^2}\\ &=-\frac {b f-2 a g+(2 c f-b g) x}{2 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^2}+\frac {3 (2 c f-b g) (b+2 c x)}{2 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )}-\frac {6 c (2 c f-b g) \tanh ^{-1}\left (\frac {b+2 c x}{\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 = 128, normalized size = 0.98 \begin {gather*} \frac {\frac {\left (b^2-4 a c\right ) (2 a g-b f+b g x-2 c f x)}{(a+x (b+c x))^2}-\frac {12 c (b g-2 c f) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {3 (b+2 c x) (2 c f-b g)}{a+x (b+c x)}}{2 \left (b^2-4 a c\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((b^2 - 4*a*c)*(-(b*f) + 2*a*g - 2*c*f*x + b*g*x))/(a + x*(b + c*x))^2 + (3*(2*c*f - b*g)*(b + 2*c*x))/(a + x
*(b + c*x)) - (12*c*(-2*c*f + b*g)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c])/(2*(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 {f+g x}{\left (a+b x+c x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 1116, normalized size = 8.52

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(6*(2*(b^2*c^3 - 4*a*c^4)*f - (b^3*c^2 - 4*a*b*c^3)*g)*x^3 + 9*(2*(b^3*c^2 - 4*a*b*c^3)*f - (b^4*c - 4*a*
b^2*c^2)*g)*x^2 - 6*(2*a^2*c^2*f - a^2*b*c*g + (2*c^4*f - b*c^3*g)*x^4 + 2*(2*b*c^3*f - b^2*c^2*g)*x^3 + (2*(b
^2*c^2 + 2*a*c^3)*f - (b^3*c + 2*a*b*c^2)*g)*x^2 + 2*(2*a*b*c^2*f - a*b^2*c*g)*x)*sqrt(b^2 - 4*a*c)*log((2*c^2
*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - (b^5 - 14*a*b^3*c + 40*a^2*
b*c^2)*f - (a*b^4 + 4*a^2*b^2*c - 32*a^3*c^2)*g + 2*(2*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*f - (b^5 + a*b^3*c - 2
0*a^2*b*c^2)*g)*x)/(a^2*b^6 - 12*a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^
2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (b^8 - 10*a*b^6*c + 2
4*a^2*b^4*c^2 + 32*a^3*b^2*c^3 - 128*a^4*c^4)*x^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x
), 1/2*(6*(2*(b^2*c^3 - 4*a*c^4)*f - (b^3*c^2 - 4*a*b*c^3)*g)*x^3 + 9*(2*(b^3*c^2 - 4*a*b*c^3)*f - (b^4*c - 4*
a*b^2*c^2)*g)*x^2 - 12*(2*a^2*c^2*f - a^2*b*c*g + (2*c^4*f - b*c^3*g)*x^4 + 2*(2*b*c^3*f - b^2*c^2*g)*x^3 + (2
*(b^2*c^2 + 2*a*c^3)*f - (b^3*c + 2*a*b*c^2)*g)*x^2 + 2*(2*a*b*c^2*f - a*b^2*c*g)*x)*sqrt(-b^2 + 4*a*c)*arctan
(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - (b^5 - 14*a*b^3*c + 40*a^2*b*c^2)*f - (a*b^4 + 4*a^2*b^2*c -
32*a^3*c^2)*g + 2*(2*(b^4*c + a*b^2*c^2 - 20*a^2*c^3)*f - (b^5 + a*b^3*c - 20*a^2*b*c^2)*g)*x)/(a^2*b^6 - 12*
a^3*b^4*c + 48*a^4*b^2*c^2 - 64*a^5*c^3 + (b^6*c^2 - 12*a*b^4*c^3 + 48*a^2*b^2*c^4 - 64*a^3*c^5)*x^4 + 2*(b^7*
c - 12*a*b^5*c^2 + 48*a^2*b^3*c^3 - 64*a^3*b*c^4)*x^3 + (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^2 + 2*(a*b^7 - 12*a^2*b^5*c + 48*a^3*b^3*c^2 - 64*a^4*b*c^3)*x)]

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giac [A]  time = 0.17, size = 199, normalized size = 1.52 \begin {gather*} \frac {6 \, {\left (2 \, c^{2} f - b c g\right )} \arctan \left (\frac {2 \, c x + 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 {12 \, c^{3} f x^{3} - 6 \, b c^{2} g x^{3} + 18 \, b c^{2} f x^{2} - 9 \, b^{2} c g x^{2} + 4 \, b^{2} c f x + 20 \, a c^{2} f x - 2 \, b^{3} g x - 10 \, a b c g x - b^{3} f + 10 \, a b c f - a b^{2} g - 8 \, a^{2} c g}{2 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (c x^{2} + b x + a\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*(2*c^2*f - b*c*g)*arctan((2*c*x + 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))
+ 1/2*(12*c^3*f*x^3 - 6*b*c^2*g*x^3 + 18*b*c^2*f*x^2 - 9*b^2*c*g*x^2 + 4*b^2*c*f*x + 20*a*c^2*f*x - 2*b^3*g*x
- 10*a*b*c*g*x - b^3*f + 10*a*b*c*f - a*b^2*g - 8*a^2*c*g)/((b^4 - 8*a*b^2*c + 16*a^2*c^2)*(c*x^2 + b*x + a)^
2)

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maple [A]  time = 0.05, size = 242, normalized size = 1.85 \begin {gather*} -\frac {3 b c g x}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )}-\frac {6 b c g \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {5}{2}}}+\frac {6 c^{2} f x}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )}+\frac {12 c^{2} f \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {5}{2}}}-\frac {3 b^{2} g}{2 \left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )}+\frac {3 b c f}{\left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )}+\frac {-2 a g +b f +\left (-b g +2 c f \right ) x}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*f-2*a*g+(-b*g+2*c*f)*x)/(4*a*c-b^2)/(c*x^2+b*x+a)^2-3/(4*a*c-b^2)^2/(c*x^2+b*x+a)*c*x*b*g+6/(4*a*c-b^2)
^2/(c*x^2+b*x+a)*c^2*x*f-3/2/(4*a*c-b^2)^2/(c*x^2+b*x+a)*b^2*g+3/(4*a*c-b^2)^2/(c*x^2+b*x+a)*b*c*f-6/(4*a*c-b^
2)^(5/2)*c*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b*g+12/(4*a*c-b^2)^(5/2)*c^2*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)
)*f

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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((g*x+f)/(c*x^2+b*x+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 = 2.57, size = 353, normalized size = 2.69 \begin {gather*} \frac {6\,c\,\mathrm {atan}\left (\frac {\left (\frac {6\,c^2\,x\,\left (b\,g-2\,c\,f\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}+\frac {3\,c\,\left (b\,g-2\,c\,f\right )\,\left (16\,a^2\,b\,c^2-8\,a\,b^3\,c+b^5\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{6\,c^2\,f-3\,b\,c\,g}\right )\,\left (b\,g-2\,c\,f\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}-\frac {\frac {8\,c\,g\,a^2+g\,a\,b^2-10\,c\,f\,a\,b+f\,b^3}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {x\,\left (b^2+5\,a\,c\right )\,\left (b\,g-2\,c\,f\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {3\,c^2\,x^3\,\left (b\,g-2\,c\,f\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {9\,b\,c\,x^2\,\left (b\,g-2\,c\,f\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}}{x^2\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^4+2\,a\,b\,x+2\,b\,c\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f + g*x)/(a + b*x + c*x^2)^3,x)

[Out]

(6*c*atan((((6*c^2*x*(b*g - 2*c*f))/(4*a*c - b^2)^(5/2) + (3*c*(b*g - 2*c*f)*(b^5 + 16*a^2*b*c^2 - 8*a*b^3*c))
/((4*a*c - b^2)^(5/2)*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(6*c^2*f - 3*b*c*g))*(b
*g - 2*c*f))/(4*a*c - b^2)^(5/2) - ((b^3*f + a*b^2*g + 8*a^2*c*g - 10*a*b*c*f)/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*
c)) + (x*(5*a*c + b^2)*(b*g - 2*c*f))/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (3*c^2*x^3*(b*g - 2*c*f))/(b^4 + 16*a^2
*c^2 - 8*a*b^2*c) + (9*b*c*x^2*(b*g - 2*c*f))/(2*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)))/(x^2*(2*a*c + b^2) + a^2 + c
^2*x^4 + 2*a*b*x + 2*b*c*x^3)

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sympy [B]  time = 1.68, size = 651, normalized size = 4.97 \begin {gather*} 3 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) \log {\left (x + \frac {- 192 a^{3} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) + 144 a^{2} b^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) - 36 a b^{4} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) + 3 b^{6} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) + 3 b^{2} c g - 6 b c^{2} f}{6 b c^{2} g - 12 c^{3} f} \right )} - 3 c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) \log {\left (x + \frac {192 a^{3} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) - 144 a^{2} b^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) + 36 a b^{4} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) - 3 b^{6} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{5}}} \left (b g - 2 c f\right ) + 3 b^{2} c g - 6 b c^{2} f}{6 b c^{2} g - 12 c^{3} f} \right )} + \frac {- 8 a^{2} c g - a b^{2} g + 10 a b c f - b^{3} f + x^{3} \left (- 6 b c^{2} g + 12 c^{3} f\right ) + x^{2} \left (- 9 b^{2} c g + 18 b c^{2} f\right ) + x \left (- 10 a b c g + 20 a c^{2} f - 2 b^{3} g + 4 b^{2} c f\right )}{32 a^{4} c^{2} - 16 a^{3} b^{2} c + 2 a^{2} b^{4} + x^{4} \left (32 a^{2} c^{4} - 16 a b^{2} c^{3} + 2 b^{4} c^{2}\right ) + x^{3} \left (64 a^{2} b c^{3} - 32 a b^{3} c^{2} + 4 b^{5} c\right ) + x^{2} \left (64 a^{3} c^{3} - 12 a b^{4} c + 2 b^{6}\right ) + x \left (64 a^{3} b c^{2} - 32 a^{2} b^{3} c + 4 a b^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*c*sqrt(-1/(4*a*c - b**2)**5)*(b*g - 2*c*f)*log(x + (-192*a**3*c**4*sqrt(-1/(4*a*c - b**2)**5)*(b*g - 2*c*f)
+ 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(b*g - 2*c*f) - 36*a*b**4*c**2*sqrt(-1/(4*a*c - b**2)**5)*(b*g
- 2*c*f) + 3*b**6*c*sqrt(-1/(4*a*c - b**2)**5)*(b*g - 2*c*f) + 3*b**2*c*g - 6*b*c**2*f)/(6*b*c**2*g - 12*c**3
*f)) - 3*c*sqrt(-1/(4*a*c - b**2)**5)*(b*g - 2*c*f)*log(x + (192*a**3*c**4*sqrt(-1/(4*a*c - b**2)**5)*(b*g - 2
*c*f) - 144*a**2*b**2*c**3*sqrt(-1/(4*a*c - b**2)**5)*(b*g - 2*c*f) + 36*a*b**4*c**2*sqrt(-1/(4*a*c - b**2)**5
)*(b*g - 2*c*f) - 3*b**6*c*sqrt(-1/(4*a*c - b**2)**5)*(b*g - 2*c*f) + 3*b**2*c*g - 6*b*c**2*f)/(6*b*c**2*g - 1
2*c**3*f)) + (-8*a**2*c*g - a*b**2*g + 10*a*b*c*f - b**3*f + x**3*(-6*b*c**2*g + 12*c**3*f) + x**2*(-9*b**2*c*
g + 18*b*c**2*f) + x*(-10*a*b*c*g + 20*a*c**2*f - 2*b**3*g + 4*b**2*c*f))/(32*a**4*c**2 - 16*a**3*b**2*c + 2*a
**2*b**4 + x**4*(32*a**2*c**4 - 16*a*b**2*c**3 + 2*b**4*c**2) + x**3*(64*a**2*b*c**3 - 32*a*b**3*c**2 + 4*b**5
*c) + x**2*(64*a**3*c**3 - 12*a*b**4*c + 2*b**6) + x*(64*a**3*b*c**2 - 32*a**2*b**3*c + 4*a*b**5))

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