\(\int \frac {A+B x^2+C x^4}{(a+b x^2)^{7/2}} \, dx\) [120]

Optimal result

Integrand size = 24, antiderivative size = 115 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {\left (\frac {A}{a}-\frac {b B-a C}{b^2}\right ) x}{5 \left (a+b x^2\right )^{5/2}}+\frac {\left (4 A b^2+a (b B-6 a C)\right ) x}{15 a^2 b^2 \left (a+b x^2\right )^{3/2}}+\frac {\left (8 A b^2+a (2 b B+3 a C)\right ) x}{15 a^3 b^2 \sqrt {a+b x^2}} \] Output:

1/5*(A/a-(B*b-C*a)/b^2)*x/(b*x^2+a)^(5/2)+1/15*(4*A*b^2+a*(B*b-6*C*a))*x/a 
^2/b^2/(b*x^2+a)^(3/2)+1/15*(8*A*b^2+a*(2*B*b+3*C*a))*x/a^3/b^2/(b*x^2+a)^ 
(1/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.57 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {8 A b^2 x^5+2 a b x^3 \left (10 A+B x^2\right )+a^2 \left (15 A x+5 B x^3+3 C x^5\right )}{15 a^3 \left (a+b x^2\right )^{5/2}} \] Input:

Integrate[(A + B*x^2 + C*x^4)/(a + b*x^2)^(7/2),x]
 

Output:

(8*A*b^2*x^5 + 2*a*b*x^3*(10*A + B*x^2) + a^2*(15*A*x + 5*B*x^3 + 3*C*x^5) 
)/(15*a^3*(a + b*x^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1469, 2075, 362, 242}

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.

\(\displaystyle \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 1469

\(\displaystyle \frac {\int \frac {x^2 \left (4 A b+a \left (C x^2+B\right )\right )}{\left (b x^2+a\right )^{7/2}}dx}{a}+\frac {A x}{a \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2075

\(\displaystyle \frac {\int \frac {x^2 \left (a C x^2+4 A b+a B\right )}{\left (b x^2+a\right )^{7/2}}dx}{a}+\frac {A x}{a \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 362

\(\displaystyle \frac {\frac {1}{5} \left (\frac {8 A b}{a}+\frac {3 a C}{b}+2 B\right ) \int \frac {x^2}{\left (b x^2+a\right )^{5/2}}dx+\frac {x^3 \left (a \left (B-\frac {a C}{b}\right )+4 A b\right )}{5 a \left (a+b x^2\right )^{5/2}}}{a}+\frac {A x}{a \left (a+b x^2\right )^{5/2}}\)

\(\Big \downarrow \) 242

\(\displaystyle \frac {\frac {x^3 \left (\frac {8 A b}{a}+\frac {3 a C}{b}+2 B\right )}{15 a \left (a+b x^2\right )^{3/2}}+\frac {x^3 \left (a \left (B-\frac {a C}{b}\right )+4 A b\right )}{5 a \left (a+b x^2\right )^{5/2}}}{a}+\frac {A x}{a \left (a+b x^2\right )^{5/2}}\)

Input:

Int[(A + B*x^2 + C*x^4)/(a + b*x^2)^(7/2),x]
 

Output:

(A*x)/(a*(a + b*x^2)^(5/2)) + (((4*A*b + a*(B - (a*C)/b))*x^3)/(5*a*(a + b 
*x^2)^(5/2)) + (((8*A*b)/a + 2*B + (3*a*C)/b)*x^3)/(15*a*(a + b*x^2)^(3/2) 
))/a
 

Defintions of rubi rules used

rule 242

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x 
] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

rule 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

rule 1469

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), 
x_Symbol] :> Simp[a^p*x*((d + e*x^2)^(q + 1)/d), x] + Simp[1/d   Int[x^2*(d 
 + e*x^2)^q*(d*PolynomialQuotient[(a + b*x^2 + c*x^4)^p - a^p, x^2, x] - e* 
a^p*(2*q + 3)), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] 
&& NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && ILtQ[q + 1/2, 0] && LtQ[4 
*p + 2*q + 1, 0]
 

rule 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.50

method	result	size
pseudoelliptic	\(\frac {x \left (\left (\frac {1}{5} C \,x^{4}+\frac {1}{3} x^{2} B +A \right ) a^{2}+\frac {4 b \left (\frac {x^{2} B}{10}+A \right ) x^{2} a}{3}+\frac {8 A \,b^{2} x^{4}}{15}\right )}{\left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\)	\(58\)
gosper	\(\frac {x \left (8 A \,b^{2} x^{4}+2 B a b \,x^{4}+3 C \,a^{2} x^{4}+20 a A b \,x^{2}+5 B \,a^{2} x^{2}+15 a^{2} A \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\)	\(66\)
trager	\(\frac {x \left (8 A \,b^{2} x^{4}+2 B a b \,x^{4}+3 C \,a^{2} x^{4}+20 a A b \,x^{2}+5 B \,a^{2} x^{2}+15 a^{2} A \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\)	\(66\)
orering	\(\frac {x \left (8 A \,b^{2} x^{4}+2 B a b \,x^{4}+3 C \,a^{2} x^{4}+20 a A b \,x^{2}+5 B \,a^{2} x^{2}+15 a^{2} A \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{2}} a^{3}}\)	\(66\)
default	\(A \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )+C \left (-\frac {x^{3}}{2 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {3 a \left (-\frac {x}{4 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {a \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )}{4 b}\right )}{2 b}\right )+B \left (-\frac {x}{4 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {a \left (\frac {x}{5 a \left (b \,x^{2}+a \right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {8 x}{15 a^{2} \sqrt {b \,x^{2}+a}}}{a}\right )}{4 b}\right )\)	\(232\)

Input:

int((C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/(b*x^2+a)^(5/2)*x*((1/5*C*x^4+1/3*x^2*B+A)*a^2+4/3*b*(1/10*x^2*B+A)*x^2* 
a+8/15*A*b^2*x^4)/a^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {{\left ({\left (3 \, C a^{2} + 2 \, B a b + 8 \, A b^{2}\right )} x^{5} + 15 \, A a^{2} x + 5 \, {\left (B a^{2} + 4 \, A a b\right )} x^{3}\right )} \sqrt {b x^{2} + a}}{15 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} \] Input:

integrate((C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="fricas")
 

Output:

1/15*((3*C*a^2 + 2*B*a*b + 8*A*b^2)*x^5 + 15*A*a^2*x + 5*(B*a^2 + 4*A*a*b) 
*x^3)*sqrt(b*x^2 + a)/(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^5*b*x^2 + a^6)
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 639 vs. \(2 (107) = 214\).

Time = 14.88 (sec) , antiderivative size = 639, normalized size of antiderivative = 5.56 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{7/2}} \, dx=A \left (\frac {15 a^{5} x}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {15}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {13}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {11}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {35 a^{4} b x^{3}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {15}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {13}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {11}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {28 a^{3} b^{2} x^{5}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {15}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {13}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {11}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {8 a^{2} b^{3} x^{7}}{15 a^{\frac {17}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {15}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 45 a^{\frac {13}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {11}{2}} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B \left (\frac {5 a x^{3}}{15 a^{\frac {9}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 30 a^{\frac {7}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {5}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {2 b x^{5}}{15 a^{\frac {9}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 30 a^{\frac {7}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 15 a^{\frac {5}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + \frac {C x^{5}}{5 a^{\frac {7}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 10 a^{\frac {5}{2}} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}} + 5 a^{\frac {3}{2}} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}} \] Input:

integrate((C*x**4+B*x**2+A)/(b*x**2+a)**(7/2),x)
 

Output:

A*(15*a**5*x/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 
 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b* 
*3*x**6*sqrt(1 + b*x**2/a)) + 35*a**4*b*x**3/(15*a**(17/2)*sqrt(1 + b*x**2 
/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt 
(1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6*sqrt(1 + b*x**2/a)) + 28*a**3*b**2 
*x**5/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 45*a**(15/2)*b*x**2*sqrt(1 + b*x* 
*2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + b*x**2/a) + 15*a**(11/2)*b**3*x**6 
*sqrt(1 + b*x**2/a)) + 8*a**2*b**3*x**7/(15*a**(17/2)*sqrt(1 + b*x**2/a) + 
 45*a**(15/2)*b*x**2*sqrt(1 + b*x**2/a) + 45*a**(13/2)*b**2*x**4*sqrt(1 + 
b*x**2/a) + 15*a**(11/2)*b**3*x**6*sqrt(1 + b*x**2/a))) + B*(5*a*x**3/(15* 
a**(9/2)*sqrt(1 + b*x**2/a) + 30*a**(7/2)*b*x**2*sqrt(1 + b*x**2/a) + 15*a 
**(5/2)*b**2*x**4*sqrt(1 + b*x**2/a)) + 2*b*x**5/(15*a**(9/2)*sqrt(1 + b*x 
**2/a) + 30*a**(7/2)*b*x**2*sqrt(1 + b*x**2/a) + 15*a**(5/2)*b**2*x**4*sqr 
t(1 + b*x**2/a))) + C*x**5/(5*a**(7/2)*sqrt(1 + b*x**2/a) + 10*a**(5/2)*b* 
x**2*sqrt(1 + b*x**2/a) + 5*a**(3/2)*b**2*x**4*sqrt(1 + b*x**2/a))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.50 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{7/2}} \, dx=-\frac {C x^{3}}{2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {8 \, A x}{15 \, \sqrt {b x^{2} + a} a^{3}} + \frac {4 \, A x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {A x}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a} + \frac {C x}{10 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} + \frac {C x}{5 \, \sqrt {b x^{2} + a} a b^{2}} - \frac {3 \, C a x}{10 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{2}} - \frac {B x}{5 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b} + \frac {2 \, B x}{15 \, \sqrt {b x^{2} + a} a^{2} b} + \frac {B x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b} \] Input:

integrate((C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="maxima")
 

Output:

-1/2*C*x^3/((b*x^2 + a)^(5/2)*b) + 8/15*A*x/(sqrt(b*x^2 + a)*a^3) + 4/15*A 
*x/((b*x^2 + a)^(3/2)*a^2) + 1/5*A*x/((b*x^2 + a)^(5/2)*a) + 1/10*C*x/((b* 
x^2 + a)^(3/2)*b^2) + 1/5*C*x/(sqrt(b*x^2 + a)*a*b^2) - 3/10*C*a*x/((b*x^2 
 + a)^(5/2)*b^2) - 1/5*B*x/((b*x^2 + a)^(5/2)*b) + 2/15*B*x/(sqrt(b*x^2 + 
a)*a^2*b) + 1/15*B*x/((b*x^2 + a)^(3/2)*a*b)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.70 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {{\left (x^{2} {\left (\frac {{\left (3 \, C a^{2} b^{2} + 2 \, B a b^{3} + 8 \, A b^{4}\right )} x^{2}}{a^{3} b^{2}} + \frac {5 \, {\left (B a^{2} b^{2} + 4 \, A a b^{3}\right )}}{a^{3} b^{2}}\right )} + \frac {15 \, A}{a}\right )} x}{15 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}}} \] Input:

integrate((C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x, algorithm="giac")
 

Output:

1/15*(x^2*((3*C*a^2*b^2 + 2*B*a*b^3 + 8*A*b^4)*x^2/(a^3*b^2) + 5*(B*a^2*b^ 
2 + 4*A*a*b^3)/(a^3*b^2)) + 15*A/a)*x/(b*x^2 + a)^(5/2)
 

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.16 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {3\,C\,a^4\,x-6\,C\,a^3\,x\,\left (b\,x^2+a\right )-3\,B\,a^3\,b\,x+8\,A\,b^2\,x\,{\left (b\,x^2+a\right )}^2+3\,C\,a^2\,x\,{\left (b\,x^2+a\right )}^2+3\,A\,a^2\,b^2\,x+4\,A\,a\,b^2\,x\,\left (b\,x^2+a\right )+2\,B\,a\,b\,x\,{\left (b\,x^2+a\right )}^2+B\,a^2\,b\,x\,\left (b\,x^2+a\right )}{15\,a^3\,b^2\,{\left (b\,x^2+a\right )}^{5/2}} \] Input:

int((A + B*x^2 + C*x^4)/(a + b*x^2)^(7/2),x)
 

Output:

(3*C*a^4*x - 6*C*a^3*x*(a + b*x^2) - 3*B*a^3*b*x + 8*A*b^2*x*(a + b*x^2)^2 
 + 3*C*a^2*x*(a + b*x^2)^2 + 3*A*a^2*b^2*x + 4*A*a*b^2*x*(a + b*x^2) + 2*B 
*a*b*x*(a + b*x^2)^2 + B*a^2*b*x*(a + b*x^2))/(15*a^3*b^2*(a + b*x^2)^(5/2 
))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.72 \[ \int \frac {A+B x^2+C x^4}{\left (a+b x^2\right )^{7/2}} \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} x +25 \sqrt {b \,x^{2}+a}\, a \,b^{4} x^{3}+3 \sqrt {b \,x^{2}+a}\, a \,b^{3} c \,x^{5}+10 \sqrt {b \,x^{2}+a}\, b^{5} x^{5}+3 \sqrt {b}\, a^{4} c -10 \sqrt {b}\, a^{3} b^{2}+9 \sqrt {b}\, a^{3} b c \,x^{2}-30 \sqrt {b}\, a^{2} b^{3} x^{2}+9 \sqrt {b}\, a^{2} b^{2} c \,x^{4}-30 \sqrt {b}\, a \,b^{4} x^{4}+3 \sqrt {b}\, a \,b^{3} c \,x^{6}-10 \sqrt {b}\, b^{5} x^{6}}{15 a^{2} b^{3} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:

int((C*x^4+B*x^2+A)/(b*x^2+a)^(7/2),x)
 

Output:

(15*sqrt(a + b*x**2)*a**2*b**3*x + 25*sqrt(a + b*x**2)*a*b**4*x**3 + 3*sqr 
t(a + b*x**2)*a*b**3*c*x**5 + 10*sqrt(a + b*x**2)*b**5*x**5 + 3*sqrt(b)*a* 
*4*c - 10*sqrt(b)*a**3*b**2 + 9*sqrt(b)*a**3*b*c*x**2 - 30*sqrt(b)*a**2*b* 
*3*x**2 + 9*sqrt(b)*a**2*b**2*c*x**4 - 30*sqrt(b)*a*b**4*x**4 + 3*sqrt(b)* 
a*b**3*c*x**6 - 10*sqrt(b)*b**5*x**6)/(15*a**2*b**3*(a**3 + 3*a**2*b*x**2 
+ 3*a*b**2*x**4 + b**3*x**6))