3.903 \(\int \frac {\sqrt {f+g x}}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=188 \[ \frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{c \sqrt {a+b x+c x^2} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {718, 424} \[ \frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+2 c x+\sqrt {b^2-4 a c}}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{c \sqrt {a+b x+c x^2} \sqrt {\frac {c (f+g x)}{2 c f-g \left (\sqrt {b^2-4 a c}+b\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[f + g*x]*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))]*EllipticE[ArcSin[Sqrt[(b
+ Sqrt[b^2 - 4*a*c] + 2*c*x)/Sqrt[b^2 - 4*a*c]]/Sqrt[2]], (-2*Sqrt[b^2 - 4*a*c]*g)/(2*c*f - (b + Sqrt[b^2 - 4
*a*c])*g)])/(c*Sqrt[(c*(f + g*x))/(2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[a + b*x + c*x^2])

Rule 424

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

Rule 718

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*Rt[b^2 - 4*a*c, 2]
*(d + e*x)^m*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))])/(c*Sqrt[a + b*x + c*x^2]*((2*c*(d + e*x))/(2*c*d -
b*e - e*Rt[b^2 - 4*a*c, 2]))^m), Subst[Int[(1 + (2*e*Rt[b^2 - 4*a*c, 2]*x^2)/(2*c*d - b*e - e*Rt[b^2 - 4*a*c,
2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], 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] && NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4]

Rubi steps

\begin {align*} \int \frac {\sqrt {f+g x}}{\sqrt {a+b x+c x^2}} \, dx &=\frac {\left (\sqrt {2} \sqrt {b^2-4 a c} \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {2 \sqrt {b^2-4 a c} g x^2}{2 c f-b g-\sqrt {b^2-4 a c} g}}}{\sqrt {1-x^2}} \, dx,x,\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )}{c \sqrt {\frac {c (f+g x)}{2 c f-b g-\sqrt {b^2-4 a c} g}} \sqrt {a+b x+c x^2}}\\ &=\frac {\sqrt {2} \sqrt {b^2-4 a c} \sqrt {f+g x} \sqrt {-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\sin ^{-1}\left (\frac {\sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x}{\sqrt {b^2-4 a c}}}}{\sqrt {2}}\right )|-\frac {2 \sqrt {b^2-4 a c} g}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}\right )}{c \sqrt {\frac {c (f+g x)}{2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}} \sqrt {a+b x+c x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.78, size = 365, normalized size = 1.94 \[ \frac {i \left (g \left (\sqrt {b^2-4 a c}-b\right )+2 c f\right ) \sqrt {\frac {g \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{g \left (\sqrt {b^2-4 a c}+b\right )-2 c f}} \sqrt {1-\frac {2 c (f+g x)}{g \left (\sqrt {b^2-4 a c}-b\right )+2 c f}} \left (E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{\left (b+\sqrt {b^2-4 a c}\right ) g-2 c f}} \sqrt {f+g x}\right )|\frac {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}{2 c f+\left (\sqrt {b^2-4 a c}-b\right ) g}\right )-F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{\left (b+\sqrt {b^2-4 a c}\right ) g-2 c f}} \sqrt {f+g x}\right )|\frac {2 c f-\left (b+\sqrt {b^2-4 a c}\right ) g}{2 c f+\left (\sqrt {b^2-4 a c}-b\right ) g}\right )\right )}{\sqrt {2} c g \sqrt {a+x (b+c x)} \sqrt {\frac {c}{g \left (\sqrt {b^2-4 a c}+b\right )-2 c f}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(2*c*f + (-b + Sqrt[b^2 - 4*a*c])*g)*Sqrt[(g*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(-2*c*f + (b + Sqrt[b^2 - 4*a
*c])*g)]*Sqrt[1 - (2*c*(f + g*x))/(2*c*f + (-b + Sqrt[b^2 - 4*a*c])*g)]*(EllipticE[I*ArcSinh[Sqrt[2]*Sqrt[c/(-
2*c*f + (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[f + g*x]], (2*c*f - (b + Sqrt[b^2 - 4*a*c])*g)/(2*c*f + (-b + Sqrt[b^
2 - 4*a*c])*g)] - EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(-2*c*f + (b + Sqrt[b^2 - 4*a*c])*g)]*Sqrt[f + g*x]], (2*
c*f - (b + Sqrt[b^2 - 4*a*c])*g)/(2*c*f + (-b + Sqrt[b^2 - 4*a*c])*g)]))/(Sqrt[2]*c*g*Sqrt[c/(-2*c*f + (b + Sq
rt[b^2 - 4*a*c])*g)]*Sqrt[a + x*(b + c*x)])

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {g x + f}}{\sqrt {c x^{2} + b x + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(g*x + f)/sqrt(c*x^2 + b*x + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {g x + f}}{\sqrt {c x^{2} + b x + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 747, normalized size = 3.97 \[ \frac {\sqrt {g x +f}\, \sqrt {c \,x^{2}+b x +a}\, \left (b g -2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) \sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}\, \sqrt {\frac {\left (-2 c x -b +\sqrt {-4 a c +b^{2}}\right ) g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\, \sqrt {\frac {\left (2 c x +b +\sqrt {-4 a c +b^{2}}\right ) g}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}\, \left (-b g \EllipticE \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )+b g \EllipticF \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )+2 c f \EllipticE \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )-2 c f \EllipticF \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )+\sqrt {-4 a c +b^{2}}\, g \EllipticE \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )-\sqrt {-4 a c +b^{2}}\, g \EllipticF \left (\sqrt {2}\, \sqrt {-\frac {\left (g x +f \right ) c}{b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}}, \sqrt {-\frac {b g -2 c f +\sqrt {-4 a c +b^{2}}\, g}{-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g}}\right )\right )}{2 \left (c g \,x^{3}+b g \,x^{2}+c f \,x^{2}+a g x +b f x +a f \right ) c^{2} g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(g*x+f)^(1/2)*(c*x^2+b*x+a)^(1/2)*(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)*2^(1/2)*(-(g*x+f)/(b*g-2*c*f+(-4*a*c+b^
2)^(1/2)*g)*c)^(1/2)*((-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*g+2*c*f+(-4*a*c+b^2)^(1/2)*g)*g)^(1/2)*((2*c*x+b+(-4*a
*c+b^2)^(1/2))/(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)*g)^(1/2)*(EllipticF(2^(1/2)*(-(g*x+f)/(b*g-2*c*f+(-4*a*c+b^2)^
(1/2)*g)*c)^(1/2),(-(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)/(-b*g+2*c*f+(-4*a*c+b^2)^(1/2)*g))^(1/2))*g*b-2*f*Ellipti
cF(2^(1/2)*(-(g*x+f)/(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)*c)^(1/2),(-(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)/(-b*g+2*c*f+
(-4*a*c+b^2)^(1/2)*g))^(1/2))*c-EllipticF(2^(1/2)*(-(g*x+f)/(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)*c)^(1/2),(-(b*g-2
*c*f+(-4*a*c+b^2)^(1/2)*g)/(-b*g+2*c*f+(-4*a*c+b^2)^(1/2)*g))^(1/2))*g*(-4*a*c+b^2)^(1/2)-EllipticE(2^(1/2)*(-
(g*x+f)/(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)*c)^(1/2),(-(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)/(-b*g+2*c*f+(-4*a*c+b^2)^
(1/2)*g))^(1/2))*b*g+2*EllipticE(2^(1/2)*(-(g*x+f)/(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)*c)^(1/2),(-(b*g-2*c*f+(-4*
a*c+b^2)^(1/2)*g)/(-b*g+2*c*f+(-4*a*c+b^2)^(1/2)*g))^(1/2))*c*f+EllipticE(2^(1/2)*(-(g*x+f)/(b*g-2*c*f+(-4*a*c
+b^2)^(1/2)*g)*c)^(1/2),(-(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)/(-b*g+2*c*f+(-4*a*c+b^2)^(1/2)*g))^(1/2))*(-4*a*c+b
^2)^(1/2)*g)/g/(c*g*x^3+b*g*x^2+c*f*x^2+a*g*x+b*f*x+a*f)/c^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {g x + f}}{\sqrt {c x^{2} + b x + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {f+g\,x}}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {f + g x}}{\sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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