Integrand size = 16, antiderivative size = 139 \[ \int x^5 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3 \, dx=\frac {b \left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 c^2}+\frac {b x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{6 c^2}+\frac {1}{6} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {b^2 \left (a+b \text {arctanh}\left (c x^3\right )\right ) \log \left (\frac {2}{1-c x^3}\right )}{c^2}-\frac {b^3 \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right )}{2 c^2} \] Output:
1/2*b*(a+b*arctanh(c*x^3))^2/c^2+1/2*b*x^3*(a+b*arctanh(c*x^3))^2/c-1/6*(a +b*arctanh(c*x^3))^3/c^2+1/6*x^6*(a+b*arctanh(c*x^3))^3-b^2*(a+b*arctanh(c *x^3))*ln(2/(-c*x^3+1))/c^2-1/2*b^3*polylog(2,1-2/(-c*x^3+1))/c^2
Time = 0.09 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.33 \[ \int x^5 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3 \, dx=\frac {6 b^2 \left (-1+c x^3\right ) \left (a+b+a c x^3\right ) \text {arctanh}\left (c x^3\right )^2+2 b^3 \left (-1+c^2 x^6\right ) \text {arctanh}\left (c x^3\right )^3+6 b \text {arctanh}\left (c x^3\right ) \left (a c x^3 \left (2 b+a c x^3\right )-2 b^2 \log \left (1+e^{-2 \text {arctanh}\left (c x^3\right )}\right )\right )+a \left (6 a b c x^3+2 a^2 c^2 x^6+3 a b \log \left (1-c x^3\right )-3 a b \log \left (1+c x^3\right )+6 b^2 \log \left (1-c^2 x^6\right )\right )+6 b^3 \operatorname {PolyLog}\left (2,-e^{-2 \text {arctanh}\left (c x^3\right )}\right )}{12 c^2} \] Input:
Integrate[x^5*(a + b*ArcTanh[c*x^3])^3,x]
Output:
(6*b^2*(-1 + c*x^3)*(a + b + a*c*x^3)*ArcTanh[c*x^3]^2 + 2*b^3*(-1 + c^2*x ^6)*ArcTanh[c*x^3]^3 + 6*b*ArcTanh[c*x^3]*(a*c*x^3*(2*b + a*c*x^3) - 2*b^2 *Log[1 + E^(-2*ArcTanh[c*x^3])]) + a*(6*a*b*c*x^3 + 2*a^2*c^2*x^6 + 3*a*b* Log[1 - c*x^3] - 3*a*b*Log[1 + c*x^3] + 6*b^2*Log[1 - c^2*x^6]) + 6*b^3*Po lyLog[2, -E^(-2*ArcTanh[c*x^3])])/(12*c^2)
Time = 1.06 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {6454, 6452, 6542, 6436, 6510, 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.
\(\displaystyle \int x^5 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3 \, dx\) |
\(\Big \downarrow \) 6454 |
\(\displaystyle \frac {1}{3} \int x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3dx^3\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {3}{2} b c \int \frac {x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{1-c^2 x^6}dx^3\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {\int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{1-c^2 x^6}dx^3}{c^2}-\frac {\int \left (a+b \text {arctanh}\left (c x^3\right )\right )^2dx^3}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6436 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {\int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{1-c^2 x^6}dx^3}{c^2}-\frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-2 b c \int \frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6510 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b c^3}-\frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-2 b c \int \frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )}{1-c^2 x^6}dx^3}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6546 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b c^3}-\frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-2 b c \left (\frac {\int \frac {a+b \text {arctanh}\left (c x^3\right )}{1-c x^3}dx^3}{c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6470 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b c^3}-\frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{c}-b \int \frac {\log \left (\frac {2}{1-c x^3}\right )}{1-c^2 x^6}dx^3}{c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2849 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b c^3}-\frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-2 b c \left (\frac {\frac {b \int \frac {\log \left (\frac {2}{1-c x^3}\right )}{1-\frac {2}{1-c x^3}}d\frac {1}{1-c x^3}}{c}+\frac {\log \left (\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{c}}{c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} x^6 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {3}{2} b c \left (\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b c^3}-\frac {x^3 \left (a+b \text {arctanh}\left (c x^3\right )\right )^2-2 b c \left (\frac {\frac {\log \left (\frac {2}{1-c x^3}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{c}+\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1-c x^3}\right )}{2 c}}{c}-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{2 b c^2}\right )}{c^2}\right )\right )\) |
Input:
Int[x^5*(a + b*ArcTanh[c*x^3])^3,x]
Output:
((x^6*(a + b*ArcTanh[c*x^3])^3)/2 - (3*b*c*((a + b*ArcTanh[c*x^3])^3/(3*b* c^3) - (x^3*(a + b*ArcTanh[c*x^3])^2 - 2*b*c*(-1/2*(a + b*ArcTanh[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))/2)/3
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]
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]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Simp[b*c*n*p Int[x^n*((a + b*ArcTanh[c*x^n]) ^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0] && (EqQ[n, 1] || EqQ[p, 1])
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]
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]]
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]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symb ol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b , c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
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]
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.61 (sec) , antiderivative size = 800, normalized size of antiderivative = 5.76
Input:
int(x^5*(a+b*arctanh(c*x^3))^3,x,method=_RETURNVERBOSE)
Output:
1/48*b^3*(c^2*x^6-1)/c^2*ln(c*x^3+1)^3+1/16*b^2*(-b*c^2*ln(-c*x^3+1)*x^6+2 *a*c^2*x^6+2*b*c*x^3+b*ln(-c*x^3+1)-2*a+2*b)/c^2*ln(c*x^3+1)^2+(1/16*b^3*( c^2*x^6-1)/c^2*ln(-c*x^3+1)^2-1/16*b^2*(2*a*c*x^3+b)^2/c^2/a*ln(-c*x^3+1)- 1/16*b*(-4*a^3*c^2*x^6-8*a^2*b*c*x^3-4*ln(-c*x^3+1)*a^2*b-4*ln(-c*x^3+1)*a *b^2-ln(-c*x^3+1)*b^3-4*a*b^2)/a/c^2)*ln(c*x^3+1)-1/48*b^3*x^6*ln(-c*x^3+1 )^3-1/8*b^3/c^2*ln(-c*x^3+1)^2+1/48*b^3/c^2*ln(-c*x^3+1)^3+1/4/c^2*b^3*ln( -c*x^3+1)+1/6*a^3*x^6-1/4/c^2*b^3*ln(c*x^3+1)-1/4/c^2*b^3*ln(c*x^3-1)+1/8* b^3/c*x^3*ln(-c*x^3+1)^2-1/4*a^2*b*x^6*ln(-c*x^3+1)+1/4*a^2*b/c^2*ln(c*x^3 -1)+1/8*a*b^2*x^6*ln(-c*x^3+1)^2+3/8/c^2*a*b^2*ln(-c*x^3+1)-1/8/c^2*a*b^2* ln(-c*x^3+1)^2-1/2*a*b^2/c*x^3*ln(-c*x^3+1)-1/4*b/c^2*ln(c*x^3+1)*a^2+1/2* b^2/c^2*ln(c*x^3+1)*a+1/8/c^2*b^2*a*ln(c*x^3-1)+3/4/c*b^2*Sum(-2/3*(ln(x-_ alpha)*ln(-c*x^3+1)+3*c*(-1/3*ln(x-_alpha)*(ln(1/2*(x+_alpha)/_alpha)+ln(( RootOf(_Z^2+_Z*_alpha+_alpha^2,index=1)-x+_alpha)/RootOf(_Z^2+_Z*_alpha+_a lpha^2,index=1))+ln((RootOf(_Z^2+_Z*_alpha+_alpha^2,index=2)-x+_alpha)/Roo tOf(_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,index=1))+dilog((RootOf(_Z^2+_Z*_alpha+_alpha^2,index=2)-x+ _alpha)/RootOf(_Z^2+_Z*_alpha+_alpha^2,index=2)))/c))*b/c,_alpha=RootOf(_Z ^3*c+1))-1/8*b^3/c^2+1/2/c*a^2*b*x^3
\[ \int x^5 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{3} x^{5} \,d x } \] Input:
integrate(x^5*(a+b*arctanh(c*x^3))^3,x, algorithm="fricas")
Output:
integral(b^3*x^5*arctanh(c*x^3)^3 + 3*a*b^2*x^5*arctanh(c*x^3)^2 + 3*a^2*b *x^5*arctanh(c*x^3) + a^3*x^5, x)
Timed out. \[ \int x^5 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3 \, dx=\text {Timed out} \] Input:
integrate(x**5*(a+b*atanh(c*x**3))**3,x)
Output:
Timed out
\[ \int x^5 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{3} x^{5} \,d x } \] Input:
integrate(x^5*(a+b*arctanh(c*x^3))^3,x, algorithm="maxima")
Output:
1/2*a*b^2*x^6*arctanh(c*x^3)^2 + 1/6*a^3*x^6 + 1/4*(2*x^6*arctanh(c*x^3) + c*(2*x^3/c^2 - log(c*x^3 + 1)/c^3 + log(c*x^3 - 1)/c^3))*a^2*b + 1/8*(4*c *(2*x^3/c^2 - log(c*x^3 + 1)/c^3 + log(c*x^3 - 1)/c^3)*arctanh(c*x^3) - (2 *(log(c*x^3 - 1) - 2)*log(c*x^3 + 1) - log(c*x^3 + 1)^2 - log(c*x^3 - 1)^2 - 4*log(c*x^3 - 1))/c^2)*a*b^2 - 1/192*(4*x^6*log(-c*x^3 + 1)^3 + 3*(x^6/ c^3 + log(c^2*x^6 - 1)/c^5)*c^3 - 6*c*((c*x^6 + 2*x^3)/c^2 + 2*log(c*x^3 - 1)/c^3)*log(-c*x^3 + 1)^2 + 21*c^2*(2*x^3/c^3 - log(c*x^3 + 1)/c^4 + log( c*x^3 - 1)/c^4) + c*(6*(c^2*x^6 + 6*c*x^3 + 2*log(c*x^3 - 1)^2 + 6*log(c*x ^3 - 1))*log(-c*x^3 + 1)/c^3 - (3*c^2*x^6 + 42*c*x^3 + 4*log(c*x^3 - 1)^3 + 18*log(c*x^3 - 1)^2 + 42*log(c*x^3 - 1))/c^3) - 1728*c*integrate(1/4*x^5 *log(c*x^3 + 1)/(c^3*x^6 - c), x) - 2*(12*c*x^3*log(c*x^3 + 1)^2 + 2*(c^2* x^6 - 1)*log(c*x^3 + 1)^3 - 3*(c^2*x^6 - 2*c*x^3 - 2*(c^2*x^6 - 1)*log(c*x ^3 + 1) + 1)*log(-c*x^3 + 1)^2 + 3*(c^2*x^6 + 6*c*x^3 - 2*(c^2*x^6 - 1)*lo g(c*x^3 + 1)^2 - 8*(c*x^3 + 1)*log(c*x^3 + 1))*log(-c*x^3 + 1))/c^2 + 18*l og(4*c^3*x^6 - 4*c)/c^2 - 576*integrate(1/4*x^2*log(c*x^3 + 1)/(c^3*x^6 - c), x))*b^3
\[ \int x^5 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3 \, dx=\int { {\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{3} x^{5} \,d x } \] Input:
integrate(x^5*(a+b*arctanh(c*x^3))^3,x, algorithm="giac")
Output:
integrate((b*arctanh(c*x^3) + a)^3*x^5, x)
Timed out. \[ \int x^5 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3 \, dx=\int x^5\,{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^3 \,d x \] Input:
int(x^5*(a + b*atanh(c*x^3))^3,x)
Output:
int(x^5*(a + b*atanh(c*x^3))^3, x)
\[ \int x^5 \left (a+b \text {arctanh}\left (c x^3\right )\right )^3 \, dx=\frac {\mathit {atanh} \left (c \,x^{3}\right )^{3} b^{3} c^{2} x^{6}-\mathit {atanh} \left (c \,x^{3}\right )^{3} b^{3}+3 \mathit {atanh} \left (c \,x^{3}\right )^{2} a \,b^{2} c^{2} x^{6}-3 \mathit {atanh} \left (c \,x^{3}\right )^{2} a \,b^{2}+3 \mathit {atanh} \left (c \,x^{3}\right )^{2} b^{3} c \,x^{3}+3 \mathit {atanh} \left (c \,x^{3}\right ) a^{2} b \,c^{2} x^{6}-3 \mathit {atanh} \left (c \,x^{3}\right ) a^{2} b +6 \mathit {atanh} \left (c \,x^{3}\right ) a \,b^{2} c \,x^{3}-6 \mathit {atanh} \left (c \,x^{3}\right ) a \,b^{2}+18 \left (\int \frac {\mathit {atanh} \left (c \,x^{3}\right ) x^{5}}{c^{2} x^{6}-1}d x \right ) b^{3} c^{2}+6 \,\mathrm {log}\left (c^{\frac {2}{3}} x^{2}-c^{\frac {1}{3}} x +1\right ) a \,b^{2}+6 \,\mathrm {log}\left (c^{\frac {2}{3}} x +c^{\frac {1}{3}}\right ) a \,b^{2}+a^{3} c^{2} x^{6}+3 a^{2} b c \,x^{3}}{6 c^{2}} \] Input:
int(x^5*(a+b*atanh(c*x^3))^3,x)
Output:
(atanh(c*x**3)**3*b**3*c**2*x**6 - atanh(c*x**3)**3*b**3 + 3*atanh(c*x**3) **2*a*b**2*c**2*x**6 - 3*atanh(c*x**3)**2*a*b**2 + 3*atanh(c*x**3)**2*b**3 *c*x**3 + 3*atanh(c*x**3)*a**2*b*c**2*x**6 - 3*atanh(c*x**3)*a**2*b + 6*at anh(c*x**3)*a*b**2*c*x**3 - 6*atanh(c*x**3)*a*b**2 + 18*int((atanh(c*x**3) *x**5)/(c**2*x**6 - 1),x)*b**3*c**2 + 6*log(c**(2/3)*x**2 - c**(1/3)*x + 1 )*a*b**2 + 6*log(c**(2/3)*x + c**(1/3))*a*b**2 + a**3*c**2*x**6 + 3*a**2*b *c*x**3)/(6*c**2)