\(\int x^8 (a+b \text {arctanh}(c x^3))^2 \, dx\) [117]

Optimal result

Integrand size = 16, antiderivative size = 146 \[ \int x^8 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {b^2 x^3}{9 c^2}-\frac {b^2 \text {arctanh}\left (c x^3\right )}{9 c^3}+\frac {b x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{9 c}+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{9 c^3}+\frac {1}{9} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2 b \left (a+b \text {arctanh}\left (c x^3\right )\right ) \log \left (\frac {2}{1-c x^3}\right )}{9 c^3}-\frac {b^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right )}{9 c^3} \] Output:

1/9*b^2*x^3/c^2-1/9*b^2*arctanh(c*x^3)/c^3+1/9*b*x^6*(a+b*arctanh(c*x^3))/ 
c+1/9*(a+b*arctanh(c*x^3))^2/c^3+1/9*x^9*(a+b*arctanh(c*x^3))^2-2/9*b*(a+b 
*arctanh(c*x^3))*ln(2/(-c*x^3+1))/c^3-1/9*b^2*polylog(2,1-2/(-c*x^3+1))/c^ 
3
 

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90 \[ \int x^8 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {b^2 c x^3+a b c^2 x^6+a^2 c^3 x^9+b^2 \left (-1+c^3 x^9\right ) \text {arctanh}\left (c x^3\right )^2+b \text {arctanh}\left (c x^3\right ) \left (-b+b c^2 x^6+2 a c^3 x^9-2 b \log \left (1+e^{-2 \text {arctanh}\left (c x^3\right )}\right )\right )+a b \log \left (-1+c^2 x^6\right )+b^2 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}\left (c x^3\right )}\right )}{9 c^3} \] Input:

Integrate[x^8*(a + b*ArcTanh[c*x^3])^2,x]
 

Output:

(b^2*c*x^3 + a*b*c^2*x^6 + a^2*c^3*x^9 + b^2*(-1 + c^3*x^9)*ArcTanh[c*x^3] 
^2 + b*ArcTanh[c*x^3]*(-b + b*c^2*x^6 + 2*a*c^3*x^9 - 2*b*Log[1 + E^(-2*Ar 
cTanh[c*x^3])]) + a*b*Log[-1 + c^2*x^6] + b^2*PolyLog[2, -E^(-2*ArcTanh[c* 
x^3])])/(9*c^3)
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6454, 6452, 6542, 6452, 262, 219, 6546, 6470, 2849, 2752}

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.

\(\displaystyle \int x^8 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx\)

\(\Big \downarrow \) 6454

\(\displaystyle \frac {1}{3} \int x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2dx^3\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2}{3} b c \int \frac {x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3\right )\)

\(\Big \downarrow \) 6542

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2}{3} b c \left (\frac {\int \frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )dx^3}{c^2}\right )\right )\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2}{3} b c \left (\frac {\int \frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b c \int \frac {x^6}{1-c^2 x^6}dx^3}{c^2}\right )\right )\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2}{3} b c \left (\frac {\int \frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b c \left (\frac {\int \frac {1}{1-c^2 x^6}dx^3}{c^2}-\frac {x^3}{c^2}\right )}{c^2}\right )\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2}{3} b c \left (\frac {\int \frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}-\frac {\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c x^3\right )}{c^3}-\frac {x^3}{c^2}\right )}{c^2}\right )\right )\)

\(\Big \downarrow \) 6546

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2}{3} b c \left (\frac {\frac {\int \frac {a+b \text {arctanh}\left (c x^3\right )}{1-c x^3}dx^3}{c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b c^2}}{c^2}-\frac {\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c x^3\right )}{c^3}-\frac {x^3}{c^2}\right )}{c^2}\right )\right )\)

\(\Big \downarrow \) 6470

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2}{3} b c \left (\frac {\frac {\frac {\log \left (\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{c}-b \int \frac {\log \left (\frac {2}{1-c x^3}\right )}{1-c^2 x^6}dx^3}{c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b c^2}}{c^2}-\frac {\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c x^3\right )}{c^3}-\frac {x^3}{c^2}\right )}{c^2}\right )\right )\)

\(\Big \downarrow \) 2849

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2}{3} b c \left (\frac {\frac {\frac {b \int \frac {\log \left (\frac {2}{1-c x^3}\right )}{1-\frac {2}{1-c x^3}}d\frac {1}{1-c x^3}}{c}+\frac {\log \left (\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{c}}{c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b c^2}}{c^2}-\frac {\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c x^3\right )}{c^3}-\frac {x^3}{c^2}\right )}{c^2}\right )\right )\)

\(\Big \downarrow \) 2752

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} x^9 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-\frac {2}{3} b c \left (\frac {\frac {\frac {\log \left (\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right )}{2 c}}{c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b c^2}}{c^2}-\frac {\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )-\frac {1}{2} b c \left (\frac {\text {arctanh}\left (c x^3\right )}{c^3}-\frac {x^3}{c^2}\right )}{c^2}\right )\right )\)

Input:

Int[x^8*(a + b*ArcTanh[c*x^3])^2,x]
 

Output:

((x^9*(a + b*ArcTanh[c*x^3])^2)/3 - (2*b*c*(-(((x^6*(a + b*ArcTanh[c*x^3]) 
)/2 - (b*c*(-(x^3/c^2) + ArcTanh[c*x^3]/c^3))/2)/c^2) + (-1/2*(a + b*ArcTa 
nh[c*x^3])^2/(b*c^2) + (((a + b*ArcTanh[c*x^3])*Log[2/(1 - c*x^3)])/c + (b 
*PolyLog[2, 1 - 2/(1 - c*x^3)])/(2*c))/c)/c^2))/3)/3
 

Defintions of rubi rules used

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 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 2752

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo 
g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
 

rule 2849

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Simp 
[-e/g   Subst[Int[Log[2*d*x]/(1 - 2*d*x), x], x, 1/(d + e*x)], x] /; FreeQ[ 
{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]
 

rule 6452

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : 
> Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m 
+ 1))   Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x 
], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 
] && IntegerQ[m])) && NeQ[m, -1]
 

rule 6454

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> 
 Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x 
], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Simpl 
ify[(m + 1)/n]]
 

rule 6470

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

rule 6542

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

rule 6546

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.36 (sec) , antiderivative size = 2906, normalized size of antiderivative = 19.90

method	result	size
default	\(\text {Expression too large to display}\)	\(2906\)
parts	\(\text {Expression too large to display}\)	\(2906\)

Input:

int(x^8*(a+b*arctanh(c*x^3))^2,x,method=_RETURNVERBOSE)
 

Output:

1/9*a^2*x^9+b^2*(1/9*x^9*arctanh(c*x^3)^2-2/3*c*(-1/6*arctanh(c*x^3)/c^2*x 
^6-1/6*arctanh(c*x^3)/c^4*ln(c^2*x^6-1)-1/2*c*(1/3/c^4*x^3-1/6/c^5*ln(c*x^ 
3+1)+1/6/c^5*ln(c*x^3-1)+1/c^4*(Sum(1/6*(ln(x-_alpha)*ln(c^2*x^6-1)-6*c^2* 
(1/2/c*(1/3*ln(x-_alpha)*(ln(1/2*(x+_alpha)/_alpha)+ln((RootOf(_Z^2+_Z*_al 
pha+_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2+_Z*_alpha+_alpha^2,index=1))+l 
n((RootOf(_Z^2+_Z*_alpha+_alpha^2,index=2)-x+_alpha)/RootOf(_Z^2+_Z*_alpha 
+_alpha^2,index=2)))/c+1/3*(dilog(1/2*(x+_alpha)/_alpha)+dilog((RootOf(_Z^ 
2+_Z*_alpha+_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2+_Z*_alpha+_alpha^2,ind 
ex=1))+dilog((RootOf(_Z^2+_Z*_alpha+_alpha^2,index=2)-x+_alpha)/RootOf(_Z^ 
2+_Z*_alpha+_alpha^2,index=2)))/c)+1/2/c*(1/6/_alpha^2/c*ln(x-_alpha)^2-1/ 
3*_alpha*ln(x-_alpha)*(2*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x 
+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1))*RootOf(_Z^2+3*_Z*_al 
pha+3*_alpha^2,index=1)*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=2)+3*ln(( 
RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2+3*_Z*_al 
pha+3*_alpha^2,index=1))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)*_alph 
a+6*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2+ 
3*_Z*_alpha+3*_alpha^2,index=1))*RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index= 
2)*_alpha+9*ln((RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1)-x+_alpha)/Root 
Of(_Z^2+3*_Z*_alpha+3*_alpha^2,index=1))*_alpha^2+2*ln((RootOf(_Z^2+3*_Z*_ 
alpha+3*_alpha^2,index=2)-x+_alpha)/RootOf(_Z^2+3*_Z*_alpha+3*_alpha^2,...
 

Fricas [F]

\[ \int x^8 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2} x^{8} \,d x } \] Input:

integrate(x^8*(a+b*arctanh(c*x^3))^2,x, algorithm="fricas")
 

Output:

integral(b^2*x^8*arctanh(c*x^3)^2 + 2*a*b*x^8*arctanh(c*x^3) + a^2*x^8, x)
 

Sympy [F(-1)]

Timed out. \[ \int x^8 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\text {Timed out} \] Input:

integrate(x**8*(a+b*atanh(c*x**3))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int x^8 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2} x^{8} \,d x } \] Input:

integrate(x^8*(a+b*arctanh(c*x^3))^2,x, algorithm="maxima")
 

Output:

1/9*a^2*x^9 + 1/9*(2*x^9*arctanh(c*x^3) + (x^6/c^2 + log(c^2*x^6 - 1)/c^4) 
*c)*a*b + 1/648*(18*x^9*log(-c*x^3 + 1)^2 - 2*c^4*(2*(c^2*x^9 + 3*x^3)/c^6 
 - 3*log(c*x^3 + 1)/c^7 + 3*log(c*x^3 - 1)/c^7) + 3*(x^6/c^4 + log(c^2*x^6 
 - 1)/c^6)*c^3 + 1944*c^3*integrate(1/9*x^11*log(c*x^3 + 1)/(c^4*x^6 - c^2 
), x) - 9*c^2*(2*x^3/c^4 - log(c*x^3 + 1)/c^5 + log(c*x^3 - 1)/c^5) - 6*c* 
((2*c^2*x^9 + 3*c*x^6 + 6*x^3)/c^3 + 6*log(c*x^3 - 1)/c^4)*log(-c*x^3 + 1) 
 + 972*c*integrate(1/9*x^5*log(c*x^3 + 1)/(c^4*x^6 - c^2), x) + 6*(3*c^3*x 
^9*log(c*x^3 + 1)^2 + (2*c^3*x^9 - 3*c^2*x^6 + 6*c*x^3 - 6*(c^3*x^9 + 1)*l 
og(c*x^3 + 1))*log(-c*x^3 + 1))/c^3 + (4*c^3*x^9 + 15*c^2*x^6 + 66*c*x^3 + 
 18*log(c*x^3 - 1)^2 + 66*log(c*x^3 - 1))/c^3 - 18*log(9*c^4*x^6 - 9*c^2)/ 
c^3 + 972*integrate(1/9*x^2*log(c*x^3 + 1)/(c^4*x^6 - c^2), x))*b^2
 

Giac [F]

\[ \int x^8 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{2} x^{8} \,d x } \] Input:

integrate(x^8*(a+b*arctanh(c*x^3))^2,x, algorithm="giac")
 

Output:

integrate((b*arctanh(c*x^3) + a)^2*x^8, x)
 

Mupad [F(-1)]

Timed out. \[ \int x^8 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\int x^8\,{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^2 \,d x \] Input:

int(x^8*(a + b*atanh(c*x^3))^2,x)
 

Output:

int(x^8*(a + b*atanh(c*x^3))^2, x)
 

Reduce [F]

\[ \int x^8 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \, dx=\frac {\mathit {atanh} \left (c \,x^{3}\right )^{2} b^{2} c^{3} x^{9}-\mathit {atanh} \left (c \,x^{3}\right )^{2} b^{2} c \,x^{3}+2 \mathit {atanh} \left (c \,x^{3}\right ) a b \,c^{3} x^{9}-2 \mathit {atanh} \left (c \,x^{3}\right ) a b +\mathit {atanh} \left (c \,x^{3}\right ) b^{2} c^{2} x^{6}-\mathit {atanh} \left (c \,x^{3}\right ) b^{2}+3 \left (\int \mathit {atanh} \left (c \,x^{3}\right )^{2} x^{2}d x \right ) b^{2} c +2 \,\mathrm {log}\left (c^{\frac {2}{3}} x^{2}-c^{\frac {1}{3}} x +1\right ) a b +2 \,\mathrm {log}\left (c^{\frac {2}{3}} x +c^{\frac {1}{3}}\right ) a b +a^{2} c^{3} x^{9}+a b \,c^{2} x^{6}+b^{2} c \,x^{3}}{9 c^{3}} \] Input:

int(x^8*(a+b*atanh(c*x^3))^2,x)
 

Output:

(atanh(c*x**3)**2*b**2*c**3*x**9 - atanh(c*x**3)**2*b**2*c*x**3 + 2*atanh( 
c*x**3)*a*b*c**3*x**9 - 2*atanh(c*x**3)*a*b + atanh(c*x**3)*b**2*c**2*x**6 
 - atanh(c*x**3)*b**2 + 3*int(atanh(c*x**3)**2*x**2,x)*b**2*c + 2*log(c**( 
2/3)*x**2 - c**(1/3)*x + 1)*a*b + 2*log(c**(2/3)*x + c**(1/3))*a*b + a**2* 
c**3*x**9 + a*b*c**2*x**6 + b**2*c*x**3)/(9*c**3)