\(\int \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2} \, dx\) [230]

Optimal result

Integrand size = 28, antiderivative size = 236 \[ \int \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2} \, dx=\frac {(b d+2 c d x)^{3/2} \sqrt {a+b x+c x^2}}{5 c d}-\frac {\left (b^2-4 a c\right )^{7/4} \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{5 c^2 \sqrt {a+b x+c x^2}}+\frac {\left (b^2-4 a c\right )^{7/4} \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{5 c^2 \sqrt {a+b x+c x^2}} \] Output:

1/5*(2*c*d*x+b*d)^(3/2)*(c*x^2+b*x+a)^(1/2)/c/d-1/5*(-4*a*c+b^2)^(7/4)*d^( 
1/2)*(-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/(c*x^2+b*x+a)^(1/2)+1/5*(-4*a*c+b^2)^(7/4 
)*d^(1/2)*(-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/(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 = 10.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.39 \[ \int \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2} \, dx=\frac {(d (b+2 c x))^{3/2} \sqrt {a+x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{6 c d \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}}} \] Input:

Integrate[Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2],x]
 

Output:

((d*(b + 2*c*x))^(3/2)*Sqrt[a + x*(b + c*x)]*Hypergeometric2F1[-1/2, 3/4, 
7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(6*c*d*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 
 + 4*a*c)])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.86, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1109, 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.

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

\(\Big \downarrow \) 1109

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}{5 c d}-\frac {\left (b^2-4 a c\right ) \int \frac {\sqrt {b d+2 c x d}}{\sqrt {c x^2+b x+a}}dx}{10 c}\)

\(\Big \downarrow \) 1115

\(\displaystyle \frac {\sqrt {a+b x+c x^2} (b d+2 c d x)^{3/2}}{5 c d}-\frac {\left (b^2-4 a c\right ) \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \int \frac {\sqrt {b d+2 c x d}}{\sqrt {-\frac {c^2 x^2}{b^2-4 a c}-\frac {b c x}{b^2-4 a c}-\frac {a c}{b^2-4 a c}}}dx}{10 c \sqrt {a+b x+c x^2}}\)

\(\Big \downarrow \) 1114

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

\(\Big \downarrow \) 836

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 762

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

\(\Big \downarrow \) 1389

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

\(\Big \downarrow \) 327

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

Input:

Int[Sqrt[b*d + 2*c*d*x]*Sqrt[a + b*x + c*x^2],x]
 

Output:

((b*d + 2*c*d*x)^(3/2)*Sqrt[a + b*x + c*x^2])/(5*c*d) - ((b^2 - 4*a*c)*Sqr 
t[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*((b^2 - 4*a*c)^(3/4)*d^(3/2)*Ell 
ipticE[ArcSin[Sqrt[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]))/(5*c^2*d*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 1109

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 + 2*p + 1))), x 
] - Simp[d*p*((b^2 - 4*a*c)/(b*e*(m + 2*p + 1)))   Int[(d + e*x)^m*(a + b*x 
 + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[2*c*d - b* 
e, 0] && NeQ[m + 2*p + 3, 0] && GtQ[p, 0] &&  !LtQ[m, -1] &&  !(IGtQ[(m - 1 
)/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p])) && RationalQ[m] && 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(491\) vs. \(2(198)=396\).

Time = 1.62 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.08

method	result	size
default	\(\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \left (16 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticE}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) a^{2} c^{2}-8 \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticE}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) a \,b^{2} c +\sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticE}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right ) b^{4}+8 c^{4} x^{4}+16 b \,c^{3} x^{3}+8 a \,c^{3} x^{2}+10 b^{2} c^{2} x^{2}+8 a b \,c^{2} x +2 b^{3} c x +2 c a \,b^{2}\right )}{10 \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right ) c^{2}}\)	\(492\)
elliptic	\(\text {Expression too large to display}\)	\(1070\)
risch	\(\text {Expression too large to display}\)	\(1356\)

Input:

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

Output:

1/10*(d*(2*c*x+b))^(1/2)*(c*x^2+b*x+a)^(1/2)*(16*(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^2*c 
^2-8*(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*b^2*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^4+8*c^4*x^4+16*b*c^ 
3*x^3+8*a*c^3*x^2+10*b^2*c^2*x^2+8*a*b*c^2*x+2*b^3*c*x+2*c*a*b^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 = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.42 \[ \int \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2} \, dx=\frac {\sqrt {2} \sqrt {c^{2} d} {\left (b^{2} - 4 \, a c\right )} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right ) + {\left (2 \, c^{2} x + b c\right )} \sqrt {2 \, c d x + b d} \sqrt {c x^{2} + b x + a}}{5 \, c^{2}} \] Input:

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

Output:

1/5*(sqrt(2)*sqrt(c^2*d)*(b^2 - 4*a*c)*weierstrassZeta((b^2 - 4*a*c)/c^2, 
0, weierstrassPInverse((b^2 - 4*a*c)/c^2, 0, 1/2*(2*c*x + b)/c)) + (2*c^2* 
x + b*c)*sqrt(2*c*d*x + b*d)*sqrt(c*x^2 + b*x + a))/c^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt {b d+2 c d x} \sqrt {a+b x+c x^2} \, dx=\frac {\sqrt {d}\, \left (4 \sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, a c +2 \sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, b^{2}+6 \sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, b c x -12 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, x^{2}}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right ) a \,c^{3}+3 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}\, x^{2}}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right ) b^{2} c^{2}-4 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right ) a^{2} c^{2}+5 \left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right ) a \,b^{2} c -\left (\int \frac {\sqrt {2 c x +b}\, \sqrt {c \,x^{2}+b x +a}}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right ) b^{4}\right )}{15 b c} \] Input:

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

Output:

(sqrt(d)*(4*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*a*c + 2*sqrt(b + 2*c*x) 
*sqrt(a + b*x + c*x**2)*b**2 + 6*sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*b* 
c*x - 12*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2)*x**2)/(a*b + 2*a*c*x 
+ b**2*x + 3*b*c*x**2 + 2*c**2*x**3),x)*a*c**3 + 3*int((sqrt(b + 2*c*x)*sq 
rt(a + b*x + c*x**2)*x**2)/(a*b + 2*a*c*x + b**2*x + 3*b*c*x**2 + 2*c**2*x 
**3),x)*b**2*c**2 - 4*int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2))/(a*b + 
2*a*c*x + b**2*x + 3*b*c*x**2 + 2*c**2*x**3),x)*a**2*c**2 + 5*int((sqrt(b 
+ 2*c*x)*sqrt(a + b*x + c*x**2))/(a*b + 2*a*c*x + b**2*x + 3*b*c*x**2 + 2* 
c**2*x**3),x)*a*b**2*c - int((sqrt(b + 2*c*x)*sqrt(a + b*x + c*x**2))/(a*b 
 + 2*a*c*x + b**2*x + 3*b*c*x**2 + 2*c**2*x**3),x)*b**4))/(15*b*c)