\(\int \frac {\arctan (a x)}{x^2 (c+a^2 c x^2)^{5/2}} \, dx\) [247]

Optimal result

Integrand size = 22, antiderivative size = 158 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=-\frac {a}{9 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {a^2 x \arctan (a x)}{3 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {5 a^2 x \arctan (a x)}{3 c^2 \sqrt {c+a^2 c x^2}}-\frac {\sqrt {c+a^2 c x^2} \arctan (a x)}{c^3 x}-\frac {a \text {arctanh}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )}{c^{5/2}} \] Output:

-1/9*a/c/(a^2*c*x^2+c)^(3/2)-5/3*a/c^2/(a^2*c*x^2+c)^(1/2)-1/3*a^2*x*arcta 
n(a*x)/c/(a^2*c*x^2+c)^(3/2)-5/3*a^2*x*arctan(a*x)/c^2/(a^2*c*x^2+c)^(1/2) 
-(a^2*c*x^2+c)^(1/2)*arctan(a*x)/c^3/x-a*arctanh((a^2*c*x^2+c)^(1/2)/c^(1/ 
2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.96 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {-3 \sqrt {c+a^2 c x^2} \left (3+12 a^2 x^2+8 a^4 x^4\right ) \arctan (a x)+a x \left (-\left (\left (16+15 a^2 x^2\right ) \sqrt {c+a^2 c x^2}\right )+9 \sqrt {c} \left (1+a^2 x^2\right )^2 \log (x)-9 \sqrt {c} \left (1+a^2 x^2\right )^2 \log \left (c+\sqrt {c} \sqrt {c+a^2 c x^2}\right )\right )}{9 c^3 x \left (1+a^2 x^2\right )^2} \] Input:

Integrate[ArcTan[a*x]/(x^2*(c + a^2*c*x^2)^(5/2)),x]
 

Output:

(-3*Sqrt[c + a^2*c*x^2]*(3 + 12*a^2*x^2 + 8*a^4*x^4)*ArcTan[a*x] + a*x*(-( 
(16 + 15*a^2*x^2)*Sqrt[c + a^2*c*x^2]) + 9*Sqrt[c]*(1 + a^2*x^2)^2*Log[x] 
- 9*Sqrt[c]*(1 + a^2*x^2)^2*Log[c + Sqrt[c]*Sqrt[c + a^2*c*x^2]]))/(9*c^3* 
x*(1 + a^2*x^2)^2)
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.42, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {5501, 5431, 5429, 5501, 5429, 5479, 243, 73, 221}

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.

\(\displaystyle \int \frac {\arctan (a x)}{x^2 \left (a^2 c x^2+c\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 5501

\(\displaystyle \frac {\int \frac {\arctan (a x)}{x^2 \left (a^2 c x^2+c\right )^{3/2}}dx}{c}-a^2 \int \frac {\arctan (a x)}{\left (a^2 c x^2+c\right )^{5/2}}dx\)

\(\Big \downarrow \) 5431

\(\displaystyle \frac {\int \frac {\arctan (a x)}{x^2 \left (a^2 c x^2+c\right )^{3/2}}dx}{c}-a^2 \left (\frac {2 \int \frac {\arctan (a x)}{\left (a^2 c x^2+c\right )^{3/2}}dx}{3 c}+\frac {x \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {1}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 5429

\(\displaystyle \frac {\int \frac {\arctan (a x)}{x^2 \left (a^2 c x^2+c\right )^{3/2}}dx}{c}-a^2 \left (\frac {x \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{3 c}+\frac {1}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 5501

\(\displaystyle \frac {\frac {\int \frac {\arctan (a x)}{x^2 \sqrt {a^2 c x^2+c}}dx}{c}-a^2 \int \frac {\arctan (a x)}{\left (a^2 c x^2+c\right )^{3/2}}dx}{c}-a^2 \left (\frac {x \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{3 c}+\frac {1}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 5429

\(\displaystyle \frac {\frac {\int \frac {\arctan (a x)}{x^2 \sqrt {a^2 c x^2+c}}dx}{c}-a^2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{c}-a^2 \left (\frac {x \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{3 c}+\frac {1}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 5479

\(\displaystyle \frac {\frac {a \int \frac {1}{x \sqrt {a^2 c x^2+c}}dx-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{c x}}{c}-a^2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{c}-a^2 \left (\frac {x \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{3 c}+\frac {1}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\frac {\frac {1}{2} a \int \frac {1}{x^2 \sqrt {a^2 c x^2+c}}dx^2-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{c x}}{c}-a^2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{c}-a^2 \left (\frac {x \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{3 c}+\frac {1}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {\int \frac {1}{\frac {x^4}{a^2 c}-\frac {1}{a^2}}d\sqrt {a^2 c x^2+c}}{a c}-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{c x}}{c}-a^2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{c}-a^2 \left (\frac {x \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{3 c}+\frac {1}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {-\frac {\arctan (a x) \sqrt {a^2 c x^2+c}}{c x}-\frac {a \text {arctanh}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )}{\sqrt {c}}}{c}-a^2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{c}-a^2 \left (\frac {x \arctan (a x)}{3 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {2 \left (\frac {x \arctan (a x)}{c \sqrt {a^2 c x^2+c}}+\frac {1}{a c \sqrt {a^2 c x^2+c}}\right )}{3 c}+\frac {1}{9 a c \left (a^2 c x^2+c\right )^{3/2}}\right )\)

Input:

Int[ArcTan[a*x]/(x^2*(c + a^2*c*x^2)^(5/2)),x]
 

Output:

-(a^2*(1/(9*a*c*(c + a^2*c*x^2)^(3/2)) + (x*ArcTan[a*x])/(3*c*(c + a^2*c*x 
^2)^(3/2)) + (2*(1/(a*c*Sqrt[c + a^2*c*x^2]) + (x*ArcTan[a*x])/(c*Sqrt[c + 
 a^2*c*x^2])))/(3*c))) + (-(a^2*(1/(a*c*Sqrt[c + a^2*c*x^2]) + (x*ArcTan[a 
*x])/(c*Sqrt[c + a^2*c*x^2]))) + (-((Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(c*x 
)) - (a*ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/Sqrt[c])/c)/c
 

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 5429

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbo 
l] :> Simp[b/(c*d*Sqrt[d + e*x^2]), x] + Simp[x*((a + b*ArcTan[c*x])/(d*Sqr 
t[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d]
 

rule 5431

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol 
] :> Simp[b*((d + e*x^2)^(q + 1)/(4*c*d*(q + 1)^2)), x] + (-Simp[x*(d + e*x 
^2)^(q + 1)*((a + b*ArcTan[c*x])/(2*d*(q + 1))), x] + Simp[(2*q + 3)/(2*d*( 
q + 1))   Int[(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]), x], x]) /; FreeQ[{a, 
 b, c, d, e}, x] && EqQ[e, c^2*d] && LtQ[q, -1] && NeQ[q, -3/2]
 

rule 5479

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ 
.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + 
 b*ArcTan[c*x])^p/(d*f*(m + 1))), x] - Simp[b*c*(p/(f*(m + 1)))   Int[(f*x) 
^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, 
c, d, e, f, m, q}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] 
&& NeQ[m, -1]
 

rule 5501

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2 
)^(q_), x_Symbol] :> Simp[1/d   Int[x^m*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c 
*x])^p, x], x] - Simp[e/d   Int[x^(m + 2)*(d + e*x^2)^q*(a + b*ArcTan[c*x]) 
^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2* 
q] && LtQ[q, -1] && ILtQ[m, 0] && NeQ[p, -1]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.40 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.06

method	result	size
default	\(-\frac {\left (9 \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a^{5} x^{5}-9 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) a^{5} x^{5}+24 \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{4} x^{4}+15 \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}+18 \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a^{3} x^{3}-18 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) a^{3} x^{3}+36 \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a^{2} x^{2}+16 \sqrt {a^{2} x^{2}+1}\, a x +9 \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right ) a x -9 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right ) a x +9 \arctan \left (a x \right ) \sqrt {a^{2} x^{2}+1}\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{9 \sqrt {a^{2} x^{2}+1}\, x \,c^{3} \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )}\)	\(325\)

Input:

int(arctan(a*x)/x^2/(a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/9*(9*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))*a^5*x^5-9*ln((1+I*a*x)/(a^2*x^2+ 
1)^(1/2)-1)*a^5*x^5+24*arctan(a*x)*(a^2*x^2+1)^(1/2)*a^4*x^4+15*(a^2*x^2+1 
)^(1/2)*a^3*x^3+18*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))*a^3*x^3-18*ln((1+I*a* 
x)/(a^2*x^2+1)^(1/2)-1)*a^3*x^3+36*arctan(a*x)*(a^2*x^2+1)^(1/2)*a^2*x^2+1 
6*(a^2*x^2+1)^(1/2)*a*x+9*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))*a*x-9*ln((1+I* 
a*x)/(a^2*x^2+1)^(1/2)-1)*a*x+9*arctan(a*x)*(a^2*x^2+1)^(1/2))/(a^2*x^2+1) 
^(1/2)*(c*(a*x-I)*(a*x+I))^(1/2)/x/c^3/(a^4*x^4+2*a^2*x^2+1)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.90 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {9 \, {\left (a^{5} x^{5} + 2 \, a^{3} x^{3} + a x\right )} \sqrt {c} \log \left (-\frac {a^{2} c x^{2} - 2 \, \sqrt {a^{2} c x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) - 2 \, {\left (15 \, a^{3} x^{3} + 16 \, a x + 3 \, {\left (8 \, a^{4} x^{4} + 12 \, a^{2} x^{2} + 3\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{18 \, {\left (a^{4} c^{3} x^{5} + 2 \, a^{2} c^{3} x^{3} + c^{3} x\right )}} \] Input:

integrate(arctan(a*x)/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

1/18*(9*(a^5*x^5 + 2*a^3*x^3 + a*x)*sqrt(c)*log(-(a^2*c*x^2 - 2*sqrt(a^2*c 
*x^2 + c)*sqrt(c) + 2*c)/x^2) - 2*(15*a^3*x^3 + 16*a*x + 3*(8*a^4*x^4 + 12 
*a^2*x^2 + 3)*arctan(a*x))*sqrt(a^2*c*x^2 + c))/(a^4*c^3*x^5 + 2*a^2*c^3*x 
^3 + c^3*x)
 

Sympy [F]

\[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\operatorname {atan}{\left (a x \right )}}{x^{2} \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(atan(a*x)/x**2/(a**2*c*x**2+c)**(5/2),x)
 

Output:

Integral(atan(a*x)/(x**2*(c*(a**2*x**2 + 1))**(5/2)), x)
 

Maxima [F]

\[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x^{2}} \,d x } \] Input:

integrate(arctan(a*x)/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(arctan(a*x)/((a^2*c*x^2 + c)^(5/2)*x^2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(arctan(a*x)/x^2/(a^2*c*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\mathrm {atan}\left (a\,x\right )}{x^2\,{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \] Input:

int(atan(a*x)/(x^2*(c + a^2*c*x^2)^(5/2)),x)
 

Output:

int(atan(a*x)/(x^2*(c + a^2*c*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.63 \[ \int \frac {\arctan (a x)}{x^2 \left (c+a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {c}\, \left (-24 \sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right ) a^{4} x^{4}-36 \sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right ) a^{2} x^{2}-9 \sqrt {a^{2} x^{2}+1}\, \mathit {atan} \left (a x \right )-15 \sqrt {a^{2} x^{2}+1}\, a^{3} x^{3}-16 \sqrt {a^{2} x^{2}+1}\, a x +9 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x -1\right ) a^{5} x^{5}+18 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x -1\right ) a^{3} x^{3}+9 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x -1\right ) a x -9 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x +1\right ) a^{5} x^{5}-18 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x +1\right ) a^{3} x^{3}-9 \,\mathrm {log}\left (\sqrt {a^{2} x^{2}+1}+a x +1\right ) a x \right )}{9 c^{3} x \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right )} \] Input:

int(atan(a*x)/x^2/(a^2*c*x^2+c)^(5/2),x)
 

Output:

(sqrt(c)*( - 24*sqrt(a**2*x**2 + 1)*atan(a*x)*a**4*x**4 - 36*sqrt(a**2*x** 
2 + 1)*atan(a*x)*a**2*x**2 - 9*sqrt(a**2*x**2 + 1)*atan(a*x) - 15*sqrt(a** 
2*x**2 + 1)*a**3*x**3 - 16*sqrt(a**2*x**2 + 1)*a*x + 9*log(sqrt(a**2*x**2 
+ 1) + a*x - 1)*a**5*x**5 + 18*log(sqrt(a**2*x**2 + 1) + a*x - 1)*a**3*x** 
3 + 9*log(sqrt(a**2*x**2 + 1) + a*x - 1)*a*x - 9*log(sqrt(a**2*x**2 + 1) + 
 a*x + 1)*a**5*x**5 - 18*log(sqrt(a**2*x**2 + 1) + a*x + 1)*a**3*x**3 - 9* 
log(sqrt(a**2*x**2 + 1) + a*x + 1)*a*x))/(9*c**3*x*(a**4*x**4 + 2*a**2*x** 
2 + 1))