\(\int \frac {1}{(a+b \sqrt [3]{x}+c x^{2/3})^3} \, dx\) [55]

Optimal result

Integrand size = 18, antiderivative size = 153 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^3} \, dx=-\frac {3 \left (a b+\left (b^2-2 a c\right ) \sqrt [3]{x}\right )}{2 c \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^2}+\frac {3 \left (b^2+2 a c\right ) \left (b+2 c \sqrt [3]{x}\right )}{2 c \left (b^2-4 a c\right )^2 \left (a+b \sqrt [3]{x}+c x^{2/3}\right )}-\frac {6 \left (b^2+2 a c\right ) \text {arctanh}\left (\frac {b+2 c \sqrt [3]{x}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^3} \, dx=\frac {3 \left (a^2 \left (6 b-4 c \sqrt [3]{x}\right )+3 b^3 x^{2/3}+2 b^2 c x+2 a \left (5 b^2 \sqrt [3]{x}+3 b c x^{2/3}+2 c^2 x\right )\right )}{2 \left (b^2-4 a c\right )^2 \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^2}+\frac {6 \left (b^2+2 a c\right ) \arctan \left (\frac {b+2 c \sqrt [3]{x}}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{5/2}} \] Input:

Integrate[(a + b*x^(1/3) + c*x^(2/3))^(-3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1680, 1164, 27, 1159, 1083, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^3} \, dx\)

\(\Big \downarrow \) 1680

\(\displaystyle 3 \int \frac {x^{2/3}}{\left (a+c x^{2/3}+b \sqrt [3]{x}\right )^3}d\sqrt [3]{x}\)

\(\Big \downarrow \) 1164

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (2 a+b \sqrt [3]{x}\right )}{2 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^2}-\frac {\int \frac {2 \left (a-b \sqrt [3]{x}\right )}{\left (a+c x^{2/3}+b \sqrt [3]{x}\right )^2}d\sqrt [3]{x}}{2 \left (b^2-4 a c\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (2 a+b \sqrt [3]{x}\right )}{2 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^2}-\frac {\int \frac {a-b \sqrt [3]{x}}{\left (a+c x^{2/3}+b \sqrt [3]{x}\right )^2}d\sqrt [3]{x}}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 1159

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (2 a+b \sqrt [3]{x}\right )}{2 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^2}-\frac {-\frac {\left (2 a c+b^2\right ) \int \frac {1}{a+c x^{2/3}+b \sqrt [3]{x}}d\sqrt [3]{x}}{b^2-4 a c}-\frac {\sqrt [3]{x} \left (2 a c+b^2\right )+3 a b}{\left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )}}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (2 a+b \sqrt [3]{x}\right )}{2 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^2}-\frac {\frac {2 \left (2 a c+b^2\right ) \int \frac {1}{b^2-4 a c-x^{2/3}}d\left (b+2 c \sqrt [3]{x}\right )}{b^2-4 a c}-\frac {\sqrt [3]{x} \left (2 a c+b^2\right )+3 a b}{\left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )}}{b^2-4 a c}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle 3 \left (\frac {\sqrt [3]{x} \left (2 a+b \sqrt [3]{x}\right )}{2 \left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )^2}-\frac {\frac {2 \left (2 a c+b^2\right ) \text {arctanh}\left (\frac {b+2 c \sqrt [3]{x}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {\sqrt [3]{x} \left (2 a c+b^2\right )+3 a b}{\left (b^2-4 a c\right ) \left (a+b \sqrt [3]{x}+c x^{2/3}\right )}}{b^2-4 a c}\right )\)

Input:

Int[(a + b*x^(1/3) + c*x^(2/3))^(-3),x]
 

Output:

3*(((2*a + b*x^(1/3))*x^(1/3))/(2*(b^2 - 4*a*c)*(a + b*x^(1/3) + c*x^(2/3) 
)^2) - (-((3*a*b + (b^2 + 2*a*c)*x^(1/3))/((b^2 - 4*a*c)*(a + b*x^(1/3) + 
c*x^(2/3)))) + (2*(b^2 + 2*a*c)*ArcTanh[(b + 2*c*x^(1/3))/Sqrt[b^2 - 4*a*c 
]])/(b^2 - 4*a*c)^(3/2))/(b^2 - 4*a*c))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

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

rule 1083

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

rule 1159

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)/((p + 1)*(b^2 - 4*a*c)))*(a + b* 
x + c*x^2)^(p + 1), x] - Simp[(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] & 
& LtQ[p, -1] && NeQ[p, -3/2]
 

rule 1164

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> 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] + Simp[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] && LtQ[p, -1] && GtQ[m, 1] && Int 
QuadraticQ[a, b, c, d, e, m, p, x]
 

rule 1680

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k 
 = Denominator[n]}, Simp[k   Subst[Int[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k* 
n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && Fr 
actionQ[n]
 

Maple [A] (verified)

Time = 1.32 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {\frac {3 c \left (2 a c +b^{2}\right ) x}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}+\frac {9 b \left (2 a c +b^{2}\right ) x^{\frac {2}{3}}}{2 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right )}-\frac {3 a \left (2 a c -5 b^{2}\right ) x^{\frac {1}{3}}}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}+\frac {9 a^{2} b}{16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}}}{\left (a +b \,x^{\frac {1}{3}}+c \,x^{\frac {2}{3}}\right )^{2}}+\frac {6 \left (2 a c +b^{2}\right ) \arctan \left (\frac {b +2 c \,x^{\frac {1}{3}}}{\sqrt {4 a c -b^{2}}}\right )}{\left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \sqrt {4 a c -b^{2}}}\)	\(215\)
default	\(\text {Expression too large to display}\)	\(32464\)

Input:

int(1/(a+b*x^(1/3)+c*x^(2/3))^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 887 vs. \(2 (131) = 262\).

Time = 0.25 (sec) , antiderivative size = 1937, normalized size of antiderivative = 12.66 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*x^(1/3)+c*x^(2/3))^3,x, algorithm="fricas")
 

Output:

[3/2*(6*a^6*b^3 - 24*a^7*b*c + 6*(a*b^4*c^4 - 6*a^2*b^2*c^5 + 8*a^3*c^6)*x 
^3 + 3*(b^9 - 10*a*b^7*c + 33*a^2*b^5*c^2 - 34*a^3*b^3*c^3 - 8*a^4*b*c^4)* 
x^2 + 2*(a^6*b^2 + 2*a^7*c + (b^2*c^6 + 2*a*c^7)*x^4 + 2*(b^5*c^3 - a*b^3* 
c^4 - 6*a^2*b*c^5)*x^3 + (b^8 - 4*a*b^6*c - 3*a^2*b^4*c^2 + 20*a^3*b^2*c^3 
 + 4*a^4*c^4)*x^2 + 2*(a^3*b^5 - a^4*b^3*c - 6*a^5*b*c^2)*x)*sqrt(b^2 - 4* 
a*c)*log((2*c^4*x^2 + a^2*b^2 - 2*a^3*c + (b^3*c - 2*a*b*c^2)*x + (b^4 - 5 
*a*b^2*c + 4*a^2*c^2 - (2*c^3*x + b^3 - 3*a*b*c)*sqrt(b^2 - 4*a*c))*x^(2/3 
) - (a^2*b + (b^2*c - 2*a*c^2)*x)*sqrt(b^2 - 4*a*c) - (a*b^3 - 4*a^2*b*c - 
 (b^2*c^2 - 4*a*c^3)*x - (b*c^2*x + a*b^2 - 2*a^2*c)*sqrt(b^2 - 4*a*c))*x^ 
(1/3))/(c^3*x^2 + a^3 + (b^3 - 3*a*b*c)*x)) + 6*(2*a^3*b^6 - 15*a^4*b^4*c 
+ 30*a^5*b^2*c^2 - 8*a^6*c^3)*x + (a^4*b^5 - 2*a^5*b^3*c - 8*a^6*b*c^2 + 2 
*(b^4*c^5 - 2*a*b^2*c^6 - 8*a^2*c^7)*x^3 + (5*b^7*c^2 - 36*a*b^5*c^3 + 66* 
a^2*b^3*c^4 - 8*a^3*b*c^5)*x^2 + 2*(2*a*b^8 - 19*a^2*b^6*c + 63*a^3*b^4*c^ 
2 - 86*a^4*b^2*c^3 + 40*a^5*c^4)*x)*x^(2/3) - (2*a^5*b^4 - 4*a^6*b^2*c - 1 
6*a^7*c^2 + (b^5*c^4 - 2*a*b^3*c^5 - 8*a^2*b*c^6)*x^3 + 2*(2*b^8*c - 19*a* 
b^6*c^2 + 63*a^2*b^4*c^3 - 86*a^3*b^2*c^4 + 40*a^4*c^5)*x^2 + (5*a^2*b^7 - 
 36*a^3*b^5*c + 66*a^4*b^3*c^2 - 8*a^5*b*c^3)*x)*x^(1/3))/(a^6*b^6 - 12*a^ 
7*b^4*c + 48*a^8*b^2*c^2 - 64*a^9*c^3 + (b^6*c^6 - 12*a*b^4*c^7 + 48*a^2*b 
^2*c^8 - 64*a^3*c^9)*x^4 + 2*(b^9*c^3 - 15*a*b^7*c^4 + 84*a^2*b^5*c^5 - 20 
8*a^3*b^3*c^6 + 192*a^4*b*c^7)*x^3 + (b^12 - 18*a*b^10*c + 129*a^2*b^8*...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^3} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*x**(1/3)+c*x**(2/3))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*x^(1/3)+c*x^(2/3))^3,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^3} \, dx=\frac {6 \, {\left (b^{2} + 2 \, a c\right )} \arctan \left (\frac {2 \, c x^{\frac {1}{3}} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} + \frac {3 \, {\left (2 \, b^{2} c x + 4 \, a c^{2} x + 3 \, b^{3} x^{\frac {2}{3}} + 6 \, a b c x^{\frac {2}{3}} + 10 \, a b^{2} x^{\frac {1}{3}} - 4 \, a^{2} c x^{\frac {1}{3}} + 6 \, a^{2} b\right )}}{2 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} {\left (c x^{\frac {2}{3}} + b x^{\frac {1}{3}} + a\right )}^{2}} \] Input:

integrate(1/(a+b*x^(1/3)+c*x^(2/3))^3,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 18.97 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.08 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^3} \, dx=\frac {\frac {9\,a^2\,b}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {9\,b\,x^{2/3}\,\left (b^2+2\,a\,c\right )}{2\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}-\frac {3\,a\,x^{1/3}\,\left (2\,a\,c-5\,b^2\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}+\frac {3\,c\,x\,\left (b^2+2\,a\,c\right )}{16\,a^2\,c^2-8\,a\,b^2\,c+b^4}}{x^{2/3}\,\left (b^2+2\,a\,c\right )+a^2+c^2\,x^{4/3}+2\,b\,c\,x+2\,a\,b\,x^{1/3}}+\frac {6\,\mathrm {atan}\left (\frac {\left (\frac {3\,\left (b^2+2\,a\,c\right )\,\left (16\,a^2\,b\,c^2-8\,a\,b^3\,c+b^5\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}+\frac {6\,c\,x^{1/3}\,\left (b^2+2\,a\,c\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}}\right )\,\left (16\,a^2\,c^2-8\,a\,b^2\,c+b^4\right )}{3\,b^2+6\,a\,c}\right )\,\left (b^2+2\,a\,c\right )}{{\left (4\,a\,c-b^2\right )}^{5/2}} \] Input:

int(1/(a + b*x^(1/3) + c*x^(2/3))^3,x)
 

Output:

((9*a^2*b)/(b^4 + 16*a^2*c^2 - 8*a*b^2*c) + (9*b*x^(2/3)*(2*a*c + b^2))/(2 
*(b^4 + 16*a^2*c^2 - 8*a*b^2*c)) - (3*a*x^(1/3)*(2*a*c - 5*b^2))/(b^4 + 16 
*a^2*c^2 - 8*a*b^2*c) + (3*c*x*(2*a*c + b^2))/(b^4 + 16*a^2*c^2 - 8*a*b^2* 
c))/(x^(2/3)*(2*a*c + b^2) + a^2 + c^2*x^(4/3) + 2*b*c*x + 2*a*b*x^(1/3)) 
+ (6*atan((((3*(2*a*c + b^2)*(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)) + (6*c*x^(1/3)*(2*a*c + b^2))/(4 
*a*c - b^2)^(5/2))*(b^4 + 16*a^2*c^2 - 8*a*b^2*c))/(6*a*c + 3*b^2))*(2*a*c 
 + b^2))/(4*a*c - b^2)^(5/2)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 848, normalized size of antiderivative = 5.54 \[ \int \frac {1}{\left (a+b \sqrt [3]{x}+c x^{2/3}\right )^3} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*x^(1/3)+c*x^(2/3))^3,x)
 

Output:

(3*(16*x**(2/3)*sqrt(4*a*c - b**2)*atan((2*x**(1/3)*c + b)/sqrt(4*a*c - b* 
*2))*a**2*b*c**2 + 16*x**(2/3)*sqrt(4*a*c - b**2)*atan((2*x**(1/3)*c + b)/ 
sqrt(4*a*c - b**2))*a*b**3*c + 4*x**(2/3)*sqrt(4*a*c - b**2)*atan((2*x**(1 
/3)*c + b)/sqrt(4*a*c - b**2))*b**5 + 16*x**(1/3)*sqrt(4*a*c - b**2)*atan( 
(2*x**(1/3)*c + b)/sqrt(4*a*c - b**2))*a**2*b**2*c + 8*x**(1/3)*sqrt(4*a*c 
 - b**2)*atan((2*x**(1/3)*c + b)/sqrt(4*a*c - b**2))*a*b**4 + 8*x**(1/3)*s 
qrt(4*a*c - b**2)*atan((2*x**(1/3)*c + b)/sqrt(4*a*c - b**2))*a*b*c**3*x + 
 4*x**(1/3)*sqrt(4*a*c - b**2)*atan((2*x**(1/3)*c + b)/sqrt(4*a*c - b**2)) 
*b**3*c**2*x + 8*sqrt(4*a*c - b**2)*atan((2*x**(1/3)*c + b)/sqrt(4*a*c - b 
**2))*a**3*b*c + 4*sqrt(4*a*c - b**2)*atan((2*x**(1/3)*c + b)/sqrt(4*a*c - 
 b**2))*a**2*b**3 + 16*sqrt(4*a*c - b**2)*atan((2*x**(1/3)*c + b)/sqrt(4*a 
*c - b**2))*a*b**2*c**2*x + 8*sqrt(4*a*c - b**2)*atan((2*x**(1/3)*c + b)/s 
qrt(4*a*c - b**2))*b**4*c*x - 16*x**(2/3)*a**3*c**3 + 12*x**(2/3)*a**2*b** 
2*c**2 + 6*x**(2/3)*a*b**4*c - 2*x**(2/3)*b**6 - 32*x**(1/3)*a**3*b*c**2 + 
 40*x**(1/3)*a**2*b**3*c - 8*x**(1/3)*a**2*c**4*x - 8*x**(1/3)*a*b**5 - 2* 
x**(1/3)*a*b**2*c**3*x + x**(1/3)*b**4*c**2*x - 8*a**4*c**2 + 22*a**3*b**2 
*c - 5*a**2*b**4))/(2*b*(128*x**(2/3)*a**4*c**4 - 32*x**(2/3)*a**3*b**2*c* 
*3 - 24*x**(2/3)*a**2*b**4*c**2 + 10*x**(2/3)*a*b**6*c - x**(2/3)*b**8 + 1 
28*x**(1/3)*a**4*b*c**3 - 96*x**(1/3)*a**3*b**3*c**2 + 64*x**(1/3)*a**3*c* 
*5*x + 24*x**(1/3)*a**2*b**5*c - 48*x**(1/3)*a**2*b**2*c**4*x - 2*x**(1...