\(\int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx\) [63]

Optimal result

Integrand size = 24, antiderivative size = 330 \[ \int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx=\frac {\sqrt {c} x^{3/2} \left (a+b x^2+c x^4\right )}{a \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt {a x+b x^3+c x^5}}{a x^{3/2}}-\frac {\sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{a^{3/4} \sqrt {a x+b x^3+c x^5}}+\frac {\sqrt [4]{c} \sqrt {x} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 a^{3/4} \sqrt {a x+b x^3+c x^5}} \] Output:

c^(1/2)*x^(3/2)*(c*x^4+b*x^2+a)/a/(a^(1/2)+c^(1/2)*x^2)/(c*x^5+b*x^3+a*x)^ 
(1/2)-(c*x^5+b*x^3+a*x)^(1/2)/a/x^(3/2)-c^(1/4)*x^(1/2)*(a^(1/2)+c^(1/2)*x 
^2)*((c*x^4+b*x^2+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*EllipticE(sin(2*arctan 
(c^(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))/a^(3/4)/(c*x^5+b*x^3 
+a*x)^(1/2)+1/2*c^(1/4)*x^(1/2)*(a^(1/2)+c^(1/2)*x^2)*((c*x^4+b*x^2+a)/(a^ 
(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4)*x/a^(1/4)),1/ 
2*(2-b/a^(1/2)/c^(1/2))^(1/2))/a^(3/4)/(c*x^5+b*x^3+a*x)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 11.28 (sec) , antiderivative size = 303, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx=\frac {-4 \left (a+b x^2+c x^4\right )+\frac {i \sqrt {2} \left (-b+\sqrt {b^2-4 a c}\right ) x \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \left (E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{\sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}}}}{4 a \sqrt {x} \sqrt {x \left (a+b x^2+c x^4\right )}} \] Input:

Integrate[1/(x^(3/2)*Sqrt[a*x + b*x^3 + c*x^5]),x]
 

Output:

(-4*(a + b*x^2 + c*x^4) + (I*Sqrt[2]*(-b + Sqrt[b^2 - 4*a*c])*x*Sqrt[(b + 
Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/( 
b - Sqrt[b^2 - 4*a*c])]*(EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 
- 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])] - Elliptic 
F[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4* 
a*c])/(b - Sqrt[b^2 - 4*a*c])]))/Sqrt[c/(b + Sqrt[b^2 - 4*a*c])])/(4*a*Sqr 
t[x]*Sqrt[x*(a + b*x^2 + c*x^4)])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1976, 27, 1961, 1459, 27, 1416, 1509}

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.

\(\displaystyle \int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx\)

\(\Big \downarrow \) 1976

\(\displaystyle \frac {\int \frac {c x^{5/2}}{\sqrt {c x^5+b x^3+a x}}dx}{a}-\frac {\sqrt {a x+b x^3+c x^5}}{a x^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {c \int \frac {x^{5/2}}{\sqrt {c x^5+b x^3+a x}}dx}{a}-\frac {\sqrt {a x+b x^3+c x^5}}{a x^{3/2}}\)

\(\Big \downarrow \) 1961

\(\displaystyle \frac {c \sqrt {x} \sqrt {a+b x^2+c x^4} \int \frac {x^2}{\sqrt {c x^4+b x^2+a}}dx}{a \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt {a x+b x^3+c x^5}}{a x^{3/2}}\)

\(\Big \downarrow \) 1459

\(\displaystyle \frac {c \sqrt {x} \sqrt {a+b x^2+c x^4} \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )}{a \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt {a x+b x^3+c x^5}}{a x^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {c \sqrt {x} \sqrt {a+b x^2+c x^4} \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )}{a \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt {a x+b x^3+c x^5}}{a x^{3/2}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {c \sqrt {x} \sqrt {a+b x^2+c x^4} \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 c^{3/4} \sqrt {a+b x^2+c x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )}{a \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt {a x+b x^3+c x^5}}{a x^{3/2}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {c \sqrt {x} \sqrt {a+b x^2+c x^4} \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 c^{3/4} \sqrt {a+b x^2+c x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{\sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {x \sqrt {a+b x^2+c x^4}}{\sqrt {a}+\sqrt {c} x^2}}{\sqrt {c}}\right )}{a \sqrt {a x+b x^3+c x^5}}-\frac {\sqrt {a x+b x^3+c x^5}}{a x^{3/2}}\)

Input:

Int[1/(x^(3/2)*Sqrt[a*x + b*x^3 + c*x^5]),x]
 

Output:

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

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 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1459

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 2]}, Simp[1/q   Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Simp[1/q 
 Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]] /; FreeQ[{a, b, c}, x] && 
 NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1509

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q 
^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* 
x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 
/(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 
- 4*a*c, 0] && PosQ[c/a]
 

rule 1961

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)] 
, x_Symbol] :> Simp[x^(q/2)*(Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]/Sqrt[a 
*x^q + b*x^n + c*x^(2*n - q)])   Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) + c*x 
^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && 
 PosQ[n - q] && ((EqQ[m, 1] && EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || 
 EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q, 1]))
 

rule 1976

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ 
), x_Symbol] :> Simp[x^(m - q + 1)*((a*x^q + b*x^n + c*x^(2*n - q))^(p + 1) 
/(a*(m + p*q + 1))), x] - Simp[1/(a*(m + p*q + 1))   Int[x^(m + n - q)*(b*( 
m + p*q + (n - q)*(p + 1) + 1) + c*(m + p*q + 2*(n - q)*(p + 1) + 1)*x^(n - 
 q))*(a*x^q + b*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c}, x] && Eq 
Q[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGt 
Q[n, 0] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && LtQ[m + p*q + 1, 0 
]
 

Maple [A] (verified)

Time = 1.01 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {c \,x^{4}+b \,x^{2}+a}{a \sqrt {x}\, \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}-\frac {c \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right ) \sqrt {x}}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \left (b +\sqrt {-4 a c +b^{2}}\right ) \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}\)	\(258\)
default	\(\frac {\left (-\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, c \,x^{4}-\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, b c \,x^{4}-c \sqrt {-\frac {2 \left (\sqrt {-4 a c +b^{2}}\, x^{2}-b \,x^{2}-2 a \right )}{a}}\, \sqrt {\frac {\sqrt {-4 a c +b^{2}}\, x^{2}+b \,x^{2}+2 a}{a}}\, x a \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {2}\, \sqrt {\frac {b \sqrt {-4 a c +b^{2}}-2 a c +b^{2}}{a c}}}{2}\right )+c \sqrt {-\frac {2 \left (\sqrt {-4 a c +b^{2}}\, x^{2}-b \,x^{2}-2 a \right )}{a}}\, \sqrt {\frac {\sqrt {-4 a c +b^{2}}\, x^{2}+b \,x^{2}+2 a}{a}}\, x a \operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {2}\, \sqrt {\frac {b \sqrt {-4 a c +b^{2}}-2 a c +b^{2}}{a c}}}{2}\right )-\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, b \,x^{2}-\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, b^{2} x^{2}-\sqrt {-4 a c +b^{2}}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, a -\sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, a b \right ) \sqrt {x \left (c \,x^{4}+b \,x^{2}+a \right )}}{x^{\frac {3}{2}} \left (c \,x^{4}+b \,x^{2}+a \right ) a \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\)	\(508\)

Input:

int(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/a*(c*x^4+b*x^2+a)/x^(1/2)/(x*(c*x^4+b*x^2+a))^(1/2)-1/2*c*2^(1/2)/((-b+ 
(-4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(-b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)*(4+ 
2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1/ 
2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^ 
(1/2))/a/c)^(1/2))-EllipticE(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/ 
2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2)))*x^(1/2)/(x*(c*x^4+b*x^2 
+a))^(1/2)
 

Fricas [F]

\[ \int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {1}{\sqrt {c x^{5} + b x^{3} + a x} x^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

integral(sqrt(c*x^5 + b*x^3 + a*x)*sqrt(x)/(c*x^7 + b*x^5 + a*x^3), x)
 

Sympy [F]

\[ \int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx=\int \frac {1}{x^{\frac {3}{2}} \sqrt {x \left (a + b x^{2} + c x^{4}\right )}}\, dx \] Input:

integrate(1/x**(3/2)/(c*x**5+b*x**3+a*x)**(1/2),x)
 

Output:

Integral(1/(x**(3/2)*sqrt(x*(a + b*x**2 + c*x**4))), x)
 

Maxima [F]

\[ \int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {1}{\sqrt {c x^{5} + b x^{3} + a x} x^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(c*x^5 + b*x^3 + a*x)*x^(3/2)), x)
 

Giac [F]

\[ \int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx=\int { \frac {1}{\sqrt {c x^{5} + b x^{3} + a x} x^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(c*x^5 + b*x^3 + a*x)*x^(3/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx=\int \frac {1}{x^{3/2}\,\sqrt {c\,x^5+b\,x^3+a\,x}} \,d x \] Input:

int(1/(x^(3/2)*(a*x + b*x^3 + c*x^5)^(1/2)),x)
 

Output:

int(1/(x^(3/2)*(a*x + b*x^3 + c*x^5)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^{3/2} \sqrt {a x+b x^3+c x^5}} \, dx=\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{6}+b \,x^{4}+a \,x^{2}}d x \] Input:

int(1/x^(3/2)/(c*x^5+b*x^3+a*x)^(1/2),x)
 

Output:

int(sqrt(a + b*x**2 + c*x**4)/(a*x**2 + b*x**4 + c*x**6),x)