3.2218 \(\int \frac {1}{(a+b x+c x^2)^4} \, dx\)

Optimal. Leaf size=136 \[ \frac {40 c^3 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2}}-\frac {10 c^2 (b+2 c x)}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}+\frac {5 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}-\frac {b+2 c x}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3} \]

[Out]

1/3*(-2*c*x-b)/(-4*a*c+b^2)/(c*x^2+b*x+a)^3+5/3*c*(2*c*x+b)/(-4*a*c+b^2)^2/(c*x^2+b*x+a)^2-10*c^2*(2*c*x+b)/(-
4*a*c+b^2)^3/(c*x^2+b*x+a)+40*c^3*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(7/2)

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Rubi [A]  time = 0.05, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {614, 618, 206} \[ -\frac {10 c^2 (b+2 c x)}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}+\frac {40 c^3 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2}}+\frac {5 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}-\frac {b+2 c x}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(-4),x]

[Out]

-(b + 2*c*x)/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^3) + (5*c*(b + 2*c*x))/(3*(b^2 - 4*a*c)^2*(a + b*x + c*x^2)^2)
- (10*c^2*(b + 2*c*x))/((b^2 - 4*a*c)^3*(a + b*x + c*x^2)) + (40*c^3*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/
(b^2 - 4*a*c)^(7/2)

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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b x+c x^2\right )^4} \, dx &=-\frac {b+2 c x}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}-\frac {(10 c) \int \frac {1}{\left (a+b x+c x^2\right )^3} \, dx}{3 \left (b^2-4 a c\right )}\\ &=-\frac {b+2 c x}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {5 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}+\frac {\left (10 c^2\right ) \int \frac {1}{\left (a+b x+c x^2\right )^2} \, dx}{\left (b^2-4 a c\right )^2}\\ &=-\frac {b+2 c x}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {5 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}-\frac {10 c^2 (b+2 c x)}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}-\frac {\left (20 c^3\right ) \int \frac {1}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right )^3}\\ &=-\frac {b+2 c x}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {5 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}-\frac {10 c^2 (b+2 c x)}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}+\frac {\left (40 c^3\right ) \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 )^3}\\ &=-\frac {b+2 c x}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^3}+\frac {5 c (b+2 c x)}{3 \left (b^2-4 a c\right )^2 \left (a+b x+c x^2\right )^2}-\frac {10 c^2 (b+2 c x)}{\left (b^2-4 a c\right )^3 \left (a+b x+c x^2\right )}+\frac {40 c^3 \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 134, normalized size = 0.99 \[ -\frac {\frac {120 c^3 \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}-\frac {5 c \left (b^2-4 a c\right ) (b+2 c x)}{(a+x (b+c x))^2}+\frac {\left (b^2-4 a c\right )^2 (b+2 c x)}{(a+x (b+c x))^3}+\frac {30 c^2 (b+2 c x)}{a+x (b+c x)}}{3 \left (b^2-4 a c\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(-4),x]

[Out]

-1/3*(((b^2 - 4*a*c)^2*(b + 2*c*x))/(a + x*(b + c*x))^3 - (5*c*(b^2 - 4*a*c)*(b + 2*c*x))/(a + x*(b + c*x))^2
+ (30*c^2*(b + 2*c*x))/(a + x*(b + c*x)) + (120*c^3*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c]
)/(b^2 - 4*a*c)^3

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fricas [B]  time = 0.96, size = 1337, normalized size = 9.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x+a)^4,x, algorithm="fricas")

[Out]

[-1/3*(b^7 - 17*a*b^5*c + 118*a^2*b^3*c^2 - 264*a^3*b*c^3 + 60*(b^2*c^5 - 4*a*c^6)*x^5 + 150*(b^3*c^4 - 4*a*b*
c^5)*x^4 + 10*(11*b^4*c^3 - 28*a*b^2*c^4 - 64*a^2*c^5)*x^3 + 15*(b^5*c^2 + 12*a*b^3*c^3 - 64*a^2*b*c^4)*x^2 +
60*(c^6*x^6 + 3*b*c^5*x^5 + 3*a^2*b*c^3*x + a^3*c^3 + 3*(b^2*c^4 + a*c^5)*x^4 + (b^3*c^3 + 6*a*b*c^4)*x^3 + 3*
(a*b^2*c^3 + a^2*c^4)*x^2)*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)) - 3*(b^6*c - 22*a*b^4*c^2 + 28*a^2*b^2*c^3 + 176*a^3*c^4)*x)/(a^3*b^8 - 16*a^4*b^6*c
+ 96*a^5*b^4*c^2 - 256*a^6*b^2*c^3 + 256*a^7*c^4 + (b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^3*b^2*c^6
+ 256*a^4*c^7)*x^6 + 3*(b^9*c^2 - 16*a*b^7*c^3 + 96*a^2*b^5*c^4 - 256*a^3*b^3*c^5 + 256*a^4*b*c^6)*x^5 + 3*(b
^10*c - 15*a*b^8*c^2 + 80*a^2*b^6*c^3 - 160*a^3*b^4*c^4 + 256*a^5*c^6)*x^4 + (b^11 - 10*a*b^9*c + 320*a^3*b^5*
c^3 - 1280*a^4*b^3*c^4 + 1536*a^5*b*c^5)*x^3 + 3*(a*b^10 - 15*a^2*b^8*c + 80*a^3*b^6*c^2 - 160*a^4*b^4*c^3 + 2
56*a^6*c^5)*x^2 + 3*(a^2*b^9 - 16*a^3*b^7*c + 96*a^4*b^5*c^2 - 256*a^5*b^3*c^3 + 256*a^6*b*c^4)*x), -1/3*(b^7
- 17*a*b^5*c + 118*a^2*b^3*c^2 - 264*a^3*b*c^3 + 60*(b^2*c^5 - 4*a*c^6)*x^5 + 150*(b^3*c^4 - 4*a*b*c^5)*x^4 +
10*(11*b^4*c^3 - 28*a*b^2*c^4 - 64*a^2*c^5)*x^3 + 15*(b^5*c^2 + 12*a*b^3*c^3 - 64*a^2*b*c^4)*x^2 - 120*(c^6*x^
6 + 3*b*c^5*x^5 + 3*a^2*b*c^3*x + a^3*c^3 + 3*(b^2*c^4 + a*c^5)*x^4 + (b^3*c^3 + 6*a*b*c^4)*x^3 + 3*(a*b^2*c^3
+ a^2*c^4)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - 3*(b^6*c - 22*a*b^
4*c^2 + 28*a^2*b^2*c^3 + 176*a^3*c^4)*x)/(a^3*b^8 - 16*a^4*b^6*c + 96*a^5*b^4*c^2 - 256*a^6*b^2*c^3 + 256*a^7*
c^4 + (b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^3*b^2*c^6 + 256*a^4*c^7)*x^6 + 3*(b^9*c^2 - 16*a*b^7*c^
3 + 96*a^2*b^5*c^4 - 256*a^3*b^3*c^5 + 256*a^4*b*c^6)*x^5 + 3*(b^10*c - 15*a*b^8*c^2 + 80*a^2*b^6*c^3 - 160*a^
3*b^4*c^4 + 256*a^5*c^6)*x^4 + (b^11 - 10*a*b^9*c + 320*a^3*b^5*c^3 - 1280*a^4*b^3*c^4 + 1536*a^5*b*c^5)*x^3 +
3*(a*b^10 - 15*a^2*b^8*c + 80*a^3*b^6*c^2 - 160*a^4*b^4*c^3 + 256*a^6*c^5)*x^2 + 3*(a^2*b^9 - 16*a^3*b^7*c +
96*a^4*b^5*c^2 - 256*a^5*b^3*c^3 + 256*a^6*b*c^4)*x)]

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giac [A]  time = 0.22, size = 220, normalized size = 1.62 \[ -\frac {40 \, c^{3} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {60 \, c^{5} x^{5} + 150 \, b c^{4} x^{4} + 110 \, b^{2} c^{3} x^{3} + 160 \, a c^{4} x^{3} + 15 \, b^{3} c^{2} x^{2} + 240 \, a b c^{3} x^{2} - 3 \, b^{4} c x + 54 \, a b^{2} c^{2} x + 132 \, a^{2} c^{3} x + b^{5} - 13 \, a b^{3} c + 66 \, a^{2} b c^{2}}{3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} {\left (c x^{2} + b x + a\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x^2+b*x+a)^4,x, algorithm="giac")

[Out]

-40*c^3*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(-b^2 + 4
*a*c)) - 1/3*(60*c^5*x^5 + 150*b*c^4*x^4 + 110*b^2*c^3*x^3 + 160*a*c^4*x^3 + 15*b^3*c^2*x^2 + 240*a*b*c^3*x^2
- 3*b^4*c*x + 54*a*b^2*c^2*x + 132*a^2*c^3*x + b^5 - 13*a*b^3*c + 66*a^2*b*c^2)/((b^6 - 12*a*b^4*c + 48*a^2*b^
2*c^2 - 64*a^3*c^3)*(c*x^2 + b*x + a)^3)

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maple [A]  time = 0.06, size = 189, normalized size = 1.39 \[ \frac {20 c^{3} x}{\left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )}+\frac {40 c^{3} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {7}{2}}}+\frac {10 b \,c^{2}}{\left (4 a c -b^{2}\right )^{3} \left (c \,x^{2}+b x +a \right )}+\frac {10 c^{2} x}{3 \left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )^{2}}+\frac {5 b c}{3 \left (4 a c -b^{2}\right )^{2} \left (c \,x^{2}+b x +a \right )^{2}}+\frac {2 c x +b}{3 \left (4 a c -b^{2}\right ) \left (c \,x^{2}+b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x^2+b*x+a)^4,x)

[Out]

1/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^3+10/3*c^2/(4*a*c-b^2)^2/(c*x^2+b*x+a)^2*x+5/3*c/(4*a*c-b^2)^2/(c*x^2+
b*x+a)^2*b+20*c^3/(4*a*c-b^2)^3/(c*x^2+b*x+a)*x+10*c^2/(4*a*c-b^2)^3/(c*x^2+b*x+a)*b+40*c^3/(4*a*c-b^2)^(7/2)*
arctan((2*c*x+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(1/(c*x^2+b*x+a)^4,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 [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \left \{\begin {array}{cl} \frac {20\,\left (\frac {b}{2}+c\,x\right )\,\left (\frac {c^2}{6\,{\left (4\,a\,c-b^2\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^2}+\frac {c^3}{{\left (4\,a\,c-b^2\right )}^3\,\left (c\,x^2+b\,x+a\right )}+\frac {c}{30\,\left (4\,a\,c-b^2\right )\,{\left (c\,x^2+b\,x+a\right )}^3}\right )}{c}-\frac {20\,c^3\,\ln \left (\frac {\frac {b}{2}-\sqrt {\frac {b^2}{4}-a\,c}+c\,x}{\frac {b}{2}+\sqrt {\frac {b^2}{4}-a\,c}+c\,x}\right )}{{\left (b^2-4\,a\,c\right )}^{7/2}} & \text {\ if\ \ }0<b^2-4\,a\,c\\ \frac {20\,\left (\frac {b}{2}+c\,x\right )\,\left (\frac {c^2}{6\,{\left (4\,a\,c-b^2\right )}^2\,{\left (c\,x^2+b\,x+a\right )}^2}+\frac {c^3}{{\left (4\,a\,c-b^2\right )}^3\,\left (c\,x^2+b\,x+a\right )}+\frac {c}{30\,\left (4\,a\,c-b^2\right )\,{\left (c\,x^2+b\,x+a\right )}^3}\right )}{c}+\frac {20\,c^3\,\mathrm {atan}\left (\frac {\frac {b}{2}+c\,x}{\sqrt {a\,c-\frac {b^2}{4}}}\right )}{\sqrt {a\,c-\frac {b^2}{4}}\,{\left (4\,a\,c-b^2\right )}^3} & \text {\ if\ \ }b^2-4\,a\,c<0\\ \int \frac {1}{{\left (c\,x^2+b\,x+a\right )}^4} \,d x & \text {\ if\ \ }b^2-4\,a\,c\notin \mathbb {R}\vee b^2=4\,a\,c \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x + c*x^2)^4,x)

[Out]

piecewise(0 < - 4*a*c + b^2, - (20*c^3*log((b/2 - (- a*c + b^2/4)^(1/2) + c*x)/(b/2 + (- a*c + b^2/4)^(1/2) +
c*x)))/(- 4*a*c + b^2)^(7/2) + (20*(b/2 + c*x)*(c^2/(6*(4*a*c - b^2)^2*(a + b*x + c*x^2)^2) + c^3/((4*a*c - b^
2)^3*(a + b*x + c*x^2)) + c/(30*(4*a*c - b^2)*(a + b*x + c*x^2)^3)))/c, - 4*a*c + b^2 < 0, (20*(b/2 + c*x)*(c^
2/(6*(4*a*c - b^2)^2*(a + b*x + c*x^2)^2) + c^3/((4*a*c - b^2)^3*(a + b*x + c*x^2)) + c/(30*(4*a*c - b^2)*(a +
b*x + c*x^2)^3)))/c + (20*c^3*atan((b/2 + c*x)/(a*c - b^2/4)^(1/2)))/((a*c - b^2/4)^(1/2)*(4*a*c - b^2)^3), ~
in(- 4*a*c + b^2, 'real') | b^2 == 4*a*c, int(1/(a + b*x + c*x^2)^4, x))

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sympy [B]  time = 1.86, size = 777, normalized size = 5.71 \[ - 20 c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \log {\left (x + \frac {- 5120 a^{4} c^{7} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 5120 a^{3} b^{2} c^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} - 1920 a^{2} b^{4} c^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 320 a b^{6} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} - 20 b^{8} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 20 b c^{3}}{40 c^{4}} \right )} + 20 c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} \log {\left (x + \frac {5120 a^{4} c^{7} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} - 5120 a^{3} b^{2} c^{6} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 1920 a^{2} b^{4} c^{5} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} - 320 a b^{6} c^{4} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 20 b^{8} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{7}}} + 20 b c^{3}}{40 c^{4}} \right )} + \frac {66 a^{2} b c^{2} - 13 a b^{3} c + b^{5} + 150 b c^{4} x^{4} + 60 c^{5} x^{5} + x^{3} \left (160 a c^{4} + 110 b^{2} c^{3}\right ) + x^{2} \left (240 a b c^{3} + 15 b^{3} c^{2}\right ) + x \left (132 a^{2} c^{3} + 54 a b^{2} c^{2} - 3 b^{4} c\right )}{192 a^{6} c^{3} - 144 a^{5} b^{2} c^{2} + 36 a^{4} b^{4} c - 3 a^{3} b^{6} + x^{6} \left (192 a^{3} c^{6} - 144 a^{2} b^{2} c^{5} + 36 a b^{4} c^{4} - 3 b^{6} c^{3}\right ) + x^{5} \left (576 a^{3} b c^{5} - 432 a^{2} b^{3} c^{4} + 108 a b^{5} c^{3} - 9 b^{7} c^{2}\right ) + x^{4} \left (576 a^{4} c^{5} + 144 a^{3} b^{2} c^{4} - 324 a^{2} b^{4} c^{3} + 99 a b^{6} c^{2} - 9 b^{8} c\right ) + x^{3} \left (1152 a^{4} b c^{4} - 672 a^{3} b^{3} c^{3} + 72 a^{2} b^{5} c^{2} + 18 a b^{7} c - 3 b^{9}\right ) + x^{2} \left (576 a^{5} c^{4} + 144 a^{4} b^{2} c^{3} - 324 a^{3} b^{4} c^{2} + 99 a^{2} b^{6} c - 9 a b^{8}\right ) + x \left (576 a^{5} b c^{3} - 432 a^{4} b^{3} c^{2} + 108 a^{3} b^{5} c - 9 a^{2} b^{7}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x**2+b*x+a)**4,x)

[Out]

-20*c**3*sqrt(-1/(4*a*c - b**2)**7)*log(x + (-5120*a**4*c**7*sqrt(-1/(4*a*c - b**2)**7) + 5120*a**3*b**2*c**6*
sqrt(-1/(4*a*c - b**2)**7) - 1920*a**2*b**4*c**5*sqrt(-1/(4*a*c - b**2)**7) + 320*a*b**6*c**4*sqrt(-1/(4*a*c -
b**2)**7) - 20*b**8*c**3*sqrt(-1/(4*a*c - b**2)**7) + 20*b*c**3)/(40*c**4)) + 20*c**3*sqrt(-1/(4*a*c - b**2)*
*7)*log(x + (5120*a**4*c**7*sqrt(-1/(4*a*c - b**2)**7) - 5120*a**3*b**2*c**6*sqrt(-1/(4*a*c - b**2)**7) + 1920
*a**2*b**4*c**5*sqrt(-1/(4*a*c - b**2)**7) - 320*a*b**6*c**4*sqrt(-1/(4*a*c - b**2)**7) + 20*b**8*c**3*sqrt(-1
/(4*a*c - b**2)**7) + 20*b*c**3)/(40*c**4)) + (66*a**2*b*c**2 - 13*a*b**3*c + b**5 + 150*b*c**4*x**4 + 60*c**5
*x**5 + x**3*(160*a*c**4 + 110*b**2*c**3) + x**2*(240*a*b*c**3 + 15*b**3*c**2) + x*(132*a**2*c**3 + 54*a*b**2*
c**2 - 3*b**4*c))/(192*a**6*c**3 - 144*a**5*b**2*c**2 + 36*a**4*b**4*c - 3*a**3*b**6 + x**6*(192*a**3*c**6 - 1
44*a**2*b**2*c**5 + 36*a*b**4*c**4 - 3*b**6*c**3) + x**5*(576*a**3*b*c**5 - 432*a**2*b**3*c**4 + 108*a*b**5*c*
*3 - 9*b**7*c**2) + x**4*(576*a**4*c**5 + 144*a**3*b**2*c**4 - 324*a**2*b**4*c**3 + 99*a*b**6*c**2 - 9*b**8*c)
+ x**3*(1152*a**4*b*c**4 - 672*a**3*b**3*c**3 + 72*a**2*b**5*c**2 + 18*a*b**7*c - 3*b**9) + x**2*(576*a**5*c*
*4 + 144*a**4*b**2*c**3 - 324*a**3*b**4*c**2 + 99*a**2*b**6*c - 9*a*b**8) + x*(576*a**5*b*c**3 - 432*a**4*b**3
*c**2 + 108*a**3*b**5*c - 9*a**2*b**7))

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