3.912 \(\int \frac{1}{\sqrt{f+g x} \sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=189 \[ \frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt{\frac{c (f+g x)}{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 g \sqrt{b^2-4 a c}}{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}\right )}{c \sqrt{f+g x} \sqrt{a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0722348, antiderivative size = 189, normalized size of antiderivative = 1., 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, 419} \[ \frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt{\frac{c (f+g x)}{2 c f-g \left (\sqrt{b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} g}{2 c f-\left (b+\sqrt{b^2-4 a c}\right ) g}\right )}{c \sqrt{f+g x} \sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

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]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rubi steps

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

Mathematica [C]  time = 0.683923, size = 308, normalized size = 1.63 \[ \frac{i (f+g x) \sqrt{2-\frac{4 \left (g (a g-b f)+c f^2\right )}{(f+g x) \left (\sqrt{g^2 \left (b^2-4 a c\right )}-b g+2 c f\right )}} \sqrt{\frac{2 \left (g (a g-b f)+c f^2\right )}{(f+g x) \left (\sqrt{g^2 \left (b^2-4 a c\right )}+b g-2 c f\right )}+1} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a g^2-b f g+c f^2}{\sqrt{g^2 \left (b^2-4 a c\right )}+b g-2 c f}}}{\sqrt{f+g x}}\right ),-\frac{\sqrt{g^2 \left (b^2-4 a c\right )}+b g-2 c f}{\sqrt{g^2 \left (b^2-4 a c\right )}-b g+2 c f}\right )}{g \sqrt{a+x (b+c x)} \sqrt{\frac{g (a g-b f)+c f^2}{\sqrt{g^2 \left (b^2-4 a c\right )}+b g-2 c f}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[f + g*x]*Sqrt[a + b*x + c*x^2]),x]

[Out]

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

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Maple [A]  time = 0.311, size = 287, normalized size = 1.5 \begin{align*}{\frac{\sqrt{2}}{cg \left ( cg{x}^{3}+bg{x}^{2}+cf{x}^{2}+agx+bfx+af \right ) } \left ( -g\sqrt{-4\,ac+{b}^{2}}-bg+2\,cf \right ){\it EllipticF} \left ( \sqrt{2}\sqrt{-{c \left ( gx+f \right ) \left ( g\sqrt{-4\,ac+{b}^{2}}+bg-2\,cf \right ) ^{-1}}},\sqrt{-{ \left ( g\sqrt{-4\,ac+{b}^{2}}+bg-2\,cf \right ) \left ( 2\,cf-bg+g\sqrt{-4\,ac+{b}^{2}} \right ) ^{-1}}} \right ) \sqrt{{g \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) \left ( g\sqrt{-4\,ac+{b}^{2}}+bg-2\,cf \right ) ^{-1}}}\sqrt{{g \left ( -b-2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) \left ( 2\,cf-bg+g\sqrt{-4\,ac+{b}^{2}} \right ) ^{-1}}}\sqrt{-{c \left ( gx+f \right ) \left ( g\sqrt{-4\,ac+{b}^{2}}+bg-2\,cf \right ) ^{-1}}}\sqrt{gx+f}\sqrt{c{x}^{2}+bx+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{2} + b x + a} \sqrt{g x + f}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{2} + b x + a} \sqrt{g x + f}}{c g x^{3} +{\left (c f + b g\right )} x^{2} + a f +{\left (b f + a g\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^2 + b*x + a)*sqrt(g*x + f)/(c*g*x^3 + (c*f + b*g)*x^2 + a*f + (b*f + a*g)*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{f + g x} \sqrt{a + b x + c x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{2} + b x + a} \sqrt{g x + f}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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