\(\int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{3/2}} \, dx\) [231]

Optimal result

Integrand size = 28, antiderivative size = 229 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{3/2}} \, dx=-\frac {\sqrt {a+b x+c x^2}}{c d \sqrt {b d+2 c d x}}+\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{c^2 d^{3/2} \sqrt {a+b x+c x^2}}-\frac {\left (b^2-4 a c\right )^{3/4} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{c^2 d^{3/2} \sqrt {a+b x+c x^2}} \] Output:

-(c*x^2+b*x+a)^(1/2)/c/d/(2*c*d*x+b*d)^(1/2)+(-4*a*c+b^2)^(3/4)*(-c*(c*x^2 
+b*x+a)/(-4*a*c+b^2))^(1/2)*EllipticE((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/ 
4)/d^(1/2),I)/c^2/d^(3/2)/(c*x^2+b*x+a)^(1/2)-(-4*a*c+b^2)^(3/4)*(-c*(c*x^ 
2+b*x+a)/(-4*a*c+b^2))^(1/2)*EllipticF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1 
/4)/d^(1/2),I)/c^2/d^(3/2)/(c*x^2+b*x+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.40 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{3/2}} \, dx=-\frac {\sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{4},\frac {3}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2 c d \sqrt {d (b+2 c x)} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

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

Output:

-1/2*(Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-1/2, -1/4, 3/4, (b + 2*c*x) 
^2/(b^2 - 4*a*c)])/(c*d*Sqrt[d*(b + 2*c*x)]*Sqrt[(c*(a + x*(b + c*x)))/(-b 
^2 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.83, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1108, 1115, 1114, 836, 27, 762, 1389, 327}

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.

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

\(\Big \downarrow \) 1108

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

\(\Big \downarrow \) 1115

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

\(\Big \downarrow \) 1114

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

\(\Big \downarrow \) 836

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 762

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

\(\Big \downarrow \) 1389

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

\(\Big \downarrow \) 327

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

Input:

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

Output:

-(Sqrt[a + b*x + c*x^2]/(c*d*Sqrt[b*d + 2*c*d*x])) + (Sqrt[-((c*(a + b*x + 
 c*x^2))/(b^2 - 4*a*c))]*((b^2 - 4*a*c)^(3/4)*d^(3/2)*EllipticE[ArcSin[Sqr 
t[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d])], -1] - (b^2 - 4*a*c)^(3/4) 
*d^(3/2)*EllipticF[ArcSin[Sqrt[b*d + 2*c*d*x]/((b^2 - 4*a*c)^(1/4)*Sqrt[d] 
)], -1]))/(c^2*d^3*Sqrt[a + b*x + c*x^2])
 

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 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 836

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

rule 1108

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 1))), x] - Si 
mp[b*(p/(d*e*(m + 1)))   Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x 
], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 
3, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !(IntegerQ[m/2] && LtQ[m + 2*p + 3, 0] 
) && IntegerQ[2*p]
 

rule 1114

Int[Sqrt[(d_) + (e_.)*(x_)]/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symb 
ol] :> Simp[(4/e)*Sqrt[-c/(b^2 - 4*a*c)]   Subst[Int[x^2/Sqrt[Simp[1 - b^2* 
(x^4/(d^2*(b^2 - 4*a*c))), x]], x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c 
, d, e}, x] && EqQ[2*c*d - b*e, 0] && LtQ[c/(b^2 - 4*a*c), 0]
 

rule 1115

Int[((d_) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sym 
bol] :> Simp[Sqrt[(-c)*((a + b*x + c*x^2)/(b^2 - 4*a*c))]/Sqrt[a + b*x + c* 
x^2]   Int[(d + e*x)^m/Sqrt[(-a)*(c/(b^2 - 4*a*c)) - b*c*(x/(b^2 - 4*a*c)) 
- c^2*(x^2/(b^2 - 4*a*c))], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c* 
d - b*e, 0] && EqQ[m^2, 1/4]
 

rule 1389

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Simp[d/Sq 
rt[a]   Int[Sqrt[1 + e*(x^2/d)]/Sqrt[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, 
 d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && NegQ[c/a] && GtQ[a, 0]
 

Maple [A] (verified)

Time = 1.34 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.42

method	result	size
default	\(\frac {\sqrt {c \,x^{2}+b x +a}\, \sqrt {d \left (2 c x +b \right )}\, \left (4 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticE}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) a c -\sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticE}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b^{2}-2 c^{2} x^{2}-2 c b x -2 a c \right )}{2 d^{2} \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) c^{2}}\)	\(325\)
elliptic	\(\text {Expression too large to display}\)	\(1003\)

Input:

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

Output:

1/2*(c*x^2+b*x+a)^(1/2)*(d*(2*c*x+b))^(1/2)*(4*(1/(-4*a*c+b^2)^(1/2)*(2*c* 
x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a*c+b^2)^(1/2))^(1/2)*((-2* 
c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2)*EllipticE(1/2*2^(1/2)* 
(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2),2^(1/2))*a*c-(1/ 
(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b))^(1/2)*(-(2*c*x+b)/(-4*a*c 
+b^2)^(1/2))^(1/2)*((-2*c*x+(-4*a*c+b^2)^(1/2)-b)/(-4*a*c+b^2)^(1/2))^(1/2 
)*EllipticE(1/2*2^(1/2)*(1/(-4*a*c+b^2)^(1/2)*(2*c*x+(-4*a*c+b^2)^(1/2)+b) 
)^(1/2),2^(1/2))*b^2-2*c^2*x^2-2*c*b*x-2*a*c)/d^2/(2*c^2*x^3+3*b*c*x^2+2*a 
*c*x+b^2*x+a*b)/c^2
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{3/2}} \, dx=-\frac {\sqrt {2} \sqrt {c^{2} d} {\left (2 \, c x + b\right )} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a} c}{2 \, c^{3} d^{2} x + b c^{2} d^{2}} \] Input:

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

Output:

-(sqrt(2)*sqrt(c^2*d)*(2*c*x + b)*weierstrassZeta((b^2 - 4*a*c)/c^2, 0, we 
ierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c)) + sqrt(2*c*d*x 
 + b*d)*sqrt(c*x^2 + b*x + a)*c)/(2*c^3*d^2*x + b*c^2*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {a+b x+c x^2}}{(b d+2 c d x)^{3/2}} \, dx=\text {too large to display} \] Input:

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

Output:

(sqrt(d)*( - 2*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a + 8*int((sqrt(b + 
2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**2*b**2*c + 8*a**2*b*c**2*x + 8*a 
**2*c**3*x**2 - a*b**4 - 2*a*b**3*c*x + 6*a*b**2*c**2*x**2 + 16*a*b*c**3*x 
**3 + 8*a*c**4*x**4 - b**5*x - 5*b**4*c*x**2 - 8*b**3*c**2*x**3 - 4*b**2*c 
**3*x**4),x)*a**2*b*c**3 + 16*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)* 
x**2)/(2*a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a*b**4 - 2*a*b 
**3*c*x + 6*a*b**2*c**2*x**2 + 16*a*b*c**3*x**3 + 8*a*c**4*x**4 - b**5*x - 
 5*b**4*c*x**2 - 8*b**3*c**2*x**3 - 4*b**2*c**3*x**4),x)*a**2*c**4*x - 6*i 
nt((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**2*b**2*c + 8*a**2*b 
*c**2*x + 8*a**2*c**3*x**2 - a*b**4 - 2*a*b**3*c*x + 6*a*b**2*c**2*x**2 + 
16*a*b*c**3*x**3 + 8*a*c**4*x**4 - b**5*x - 5*b**4*c*x**2 - 8*b**3*c**2*x* 
*3 - 4*b**2*c**3*x**4),x)*a*b**3*c**2 - 12*int((sqrt(b + 2*c*x)*sqrt(a + b 
*x + c*x**2)*x**2)/(2*a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a 
*b**4 - 2*a*b**3*c*x + 6*a*b**2*c**2*x**2 + 16*a*b*c**3*x**3 + 8*a*c**4*x* 
*4 - b**5*x - 5*b**4*c*x**2 - 8*b**3*c**2*x**3 - 4*b**2*c**3*x**4),x)*a*b* 
*2*c**3*x + int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(2*a**2*b**2 
*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a*b**4 - 2*a*b**3*c*x + 6*a*b**2 
*c**2*x**2 + 16*a*b*c**3*x**3 + 8*a*c**4*x**4 - b**5*x - 5*b**4*c*x**2 - 8 
*b**3*c**2*x**3 - 4*b**2*c**3*x**4),x)*b**5*c + 2*int((sqrt(b + 2*c*x)*sqr 
t(a + b*x + c*x**2)*x**2)/(2*a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**...