\(\int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx\) [266]

Optimal result

Integrand size = 28, antiderivative size = 195 \[ \int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 \left (b^2-4 a c\right )^{3/4} \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right )\right |-1\right )}{c \sqrt {a+b x+c x^2}}-\frac {2 \left (b^2-4 a c\right )^{3/4} \sqrt {d} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b d+2 c d x}}{\sqrt [4]{b^2-4 a c} \sqrt {d}}\right ),-1\right )}{c \sqrt {a+b x+c x^2}} \] Output:

2*(-4*a*c+b^2)^(3/4)*d^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*Ellipti 
cE((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)/c/(c*x^2+b*x+a)^(1/2) 
-2*(-4*a*c+b^2)^(3/4)*d^(1/2)*(-c*(c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*Ellipt 
icF((2*c*d*x+b*d)^(1/2)/(-4*a*c+b^2)^(1/4)/d^(1/2),I)/c/(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 = 10.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.47 \[ \int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx=\frac {2 (d (b+2 c x))^{3/2} \sqrt {\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 c d \sqrt {a+x (b+c x)}} \] Input:

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

Output:

(2*(d*(b + 2*c*x))^(3/2)*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hyperg 
eometric2F1[1/2, 3/4, 7/4, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(3*c*d*Sqrt[a + x 
*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.79, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.

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

\(\Big \downarrow \) 1115

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

\(\Big \downarrow \) 1114

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

\(\Big \downarrow \) 836

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 762

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

\(\Big \downarrow \) 1389

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

\(\Big \downarrow \) 327

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

Input:

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

Output:

(2*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[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]))/(c*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 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 = 1.06 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {\sqrt {d \left (2 c x +b \right )}\, \sqrt {c \,x^{2}+b x +a}\, \left (4 a c -b^{2}\right ) \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {2 c x +b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \operatorname {EllipticE}\left (\frac {\sqrt {2}\, \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{\sqrt {-4 a c +b^{2}}}}}{2}, \sqrt {2}\right )}{c \left (2 x^{3} c^{2}+3 b c \,x^{2}+2 a c x +b^{2} x +a b \right )}\)	\(186\)
elliptic	\(\frac {\sqrt {d \left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )}\, \sqrt {d \left (2 c x +b \right )}\, \left (\frac {2 b d \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}+\frac {4 c d \left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {\frac {x +\frac {b}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \left (\left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )-\frac {b \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}{\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}}}, \sqrt {\frac {-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {b}{2 c}}}\right )}{2 c}\right )}{\sqrt {2 c^{2} d \,x^{3}+3 b c d \,x^{2}+2 a d x c +b^{2} d x +d a b}}\right )}{\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}\, d}\)	\(948\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.28 \[ \int \frac {\sqrt {b d+2 c d x}}{\sqrt {a+b x+c x^2}} \, dx=-\frac {2 \, \sqrt {2} \sqrt {c^{2} d} {\rm weierstrassZeta}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, {\rm weierstrassPInverse}\left (\frac {b^{2} - 4 \, a c}{c^{2}}, 0, \frac {2 \, c x + b}{2 \, c}\right )\right )}{c} \] Input:

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

Output:

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

Sympy [F]

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

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

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

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

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

Output:

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