\(\int \frac {\sqrt {d x}}{\sqrt {a+b x+c x^2}} \, dx\) [328]

Optimal result

Integrand size = 22, antiderivative size = 156 \[ \int \frac {\sqrt {d x}}{\sqrt {a+b x+c x^2}} \, dx=\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {d x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\arcsin \left (\frac {\sqrt {1+\frac {b+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|\frac {2 \sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}}\right )}{c \sqrt {-\frac {c x}{b+\sqrt {b^2-4 a c}}} \sqrt {a+b x+c x^2}} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.22 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {d x}}{\sqrt {a+b x+c x^2}} \, dx=\frac {i \left (-b+\sqrt {b^2-4 a c}\right ) \sqrt {d x} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}} \left (E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {x}\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {x}\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{\sqrt {2} c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {x} \sqrt {a+x (b+c x)}} \] Input:

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

Output:

(I*(-b + Sqrt[b^2 - 4*a*c])*Sqrt[d*x]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x) 
/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c])]*(Ellip 
ticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[x]], (b + Sqrt 
[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])] - EllipticF[I*ArcSinh[Sqrt[2]*Sqrt 
[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[x]], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^ 
2 - 4*a*c])]))/(Sqrt[2]*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[x]*Sqrt[a + 
 x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.77, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1171, 1170, 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.

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

\(\Big \downarrow \) 1171

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

\(\Big \downarrow \) 1170

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

\(\Big \downarrow \) 1459

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1416

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

\(\Big \downarrow \) 1509

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

Input:

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

Output:

(2*Sqrt[d*x]*(-((-((Sqrt[x]*Sqrt[a + b*x + c*x^2])/(Sqrt[a] + Sqrt[c]*x)) 
+ (a^(1/4)*(Sqrt[a] + Sqrt[c]*x)*Sqrt[(a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c] 
*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt 
[c]))/4])/(c^(1/4)*Sqrt[a + b*x + c*x^2]))/Sqrt[c]) + (a^(1/4)*(Sqrt[a] + 
Sqrt[c]*x)*Sqrt[(a + b*x + c*x^2)/(Sqrt[a] + Sqrt[c]*x)^2]*EllipticF[2*Arc 
Tan[(c^(1/4)*Sqrt[x])/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*c^(3/4)*S 
qrt[a + b*x + c*x^2])))/Sqrt[x]
 

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 1170

Int[(x_)^(m_)/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 
 Subst[Int[x^(2*m + 1)/Sqrt[a + b*x^2 + c*x^4], x], x, Sqrt[x]], x] /; Free 
Q[{a, b, c}, x] && EqQ[m^2, 1/4]
 

rule 1171

Int[((e_)*(x_))^(m_)/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> S 
imp[(e*x)^m/x^m   Int[x^m/Sqrt[a + b*x + c*x^2], x], x] /; FreeQ[{a, b, c, 
e}, x] && EqQ[m^2, 1/4]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(340\) vs. \(2(137)=274\).

Time = 1.16 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.19

method	result	size
default	\(-\frac {\sqrt {d x}\, \left (b +\sqrt {-4 a c +b^{2}}\right ) \sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{b +\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {-2 c x +\sqrt {-4 a c +b^{2}}-b}{\sqrt {-4 a c +b^{2}}}}\, \sqrt {-\frac {c x}{b +\sqrt {-4 a c +b^{2}}}}\, \left (2 \sqrt {-4 a c +b^{2}}\, \operatorname {EllipticE}\left (\sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}}{2}\right )-\operatorname {EllipticF}\left (\sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}}{2}\right ) \sqrt {-4 a c +b^{2}}+\operatorname {EllipticF}\left (\sqrt {\frac {2 c x +\sqrt {-4 a c +b^{2}}+b}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {2}\, \sqrt {\frac {b +\sqrt {-4 a c +b^{2}}}{\sqrt {-4 a c +b^{2}}}}}{2}\right ) b \right )}{2 \sqrt {c \,x^{2}+b x +a}\, x \,c^{2}}\)	\(341\)
elliptic	\(\frac {\sqrt {d x}\, \sqrt {d x \left (c \,x^{2}+b x +a \right )}\, \left (b +\sqrt {-4 a c +b^{2}}\right ) \sqrt {2}\, \sqrt {\frac {\left (x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) c}{b +\sqrt {-4 a c +b^{2}}}}\, \sqrt {\frac {x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}{-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}}}\, \sqrt {-\frac {2 c x}{b +\sqrt {-4 a c +b^{2}}}}\, \left (\left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \operatorname {EllipticE}\left (\sqrt {2}\, \sqrt {\frac {\left (x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) c}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {-\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right )}{c \left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}}{2}\right )+\frac {\left (-b +\sqrt {-4 a c +b^{2}}\right ) \operatorname {EllipticF}\left (\sqrt {2}\, \sqrt {\frac {\left (x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) c}{b +\sqrt {-4 a c +b^{2}}}}, \frac {\sqrt {-\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right )}{c \left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}}{2}\right )}{2 c}\right )}{x \sqrt {c \,x^{2}+b x +a}\, c \sqrt {c d \,x^{3}+b d \,x^{2}+a d x}}\)	\(456\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {d x}}{\sqrt {a+b x+c x^2}} \, dx=-\frac {2 \, {\left (\sqrt {c d} b {\rm weierstrassPInverse}\left (\frac {4 \, {\left (b^{2} - 3 \, a c\right )}}{3 \, c^{2}}, -\frac {4 \, {\left (2 \, b^{3} - 9 \, a b c\right )}}{27 \, c^{3}}, \frac {3 \, c x + b}{3 \, c}\right ) + 3 \, \sqrt {c d} c {\rm weierstrassZeta}\left (\frac {4 \, {\left (b^{2} - 3 \, a c\right )}}{3 \, c^{2}}, -\frac {4 \, {\left (2 \, b^{3} - 9 \, a b c\right )}}{27 \, c^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (b^{2} - 3 \, a c\right )}}{3 \, c^{2}}, -\frac {4 \, {\left (2 \, b^{3} - 9 \, a b c\right )}}{27 \, c^{3}}, \frac {3 \, c x + b}{3 \, c}\right )\right )\right )}}{3 \, c^{2}} \] Input:

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

Output:

-2/3*(sqrt(c*d)*b*weierstrassPInverse(4/3*(b^2 - 3*a*c)/c^2, -4/27*(2*b^3 
- 9*a*b*c)/c^3, 1/3*(3*c*x + b)/c) + 3*sqrt(c*d)*c*weierstrassZeta(4/3*(b^ 
2 - 3*a*c)/c^2, -4/27*(2*b^3 - 9*a*b*c)/c^3, weierstrassPInverse(4/3*(b^2 
- 3*a*c)/c^2, -4/27*(2*b^3 - 9*a*b*c)/c^3, 1/3*(3*c*x + b)/c)))/c^2
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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