3.299 \(\int \frac{\sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}}}{\sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}} \, dx\)

Optimal. Leaf size=478 \[ \frac{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} \text{EllipticF}\left (\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{\sqrt{b^2-4 a c}+b}}\right ),-\frac{2 \sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{2} \sqrt{c} \sqrt{\frac{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1}{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}}+\frac{x \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1}}{\sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}}-\frac{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} E\left (\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )|-\frac{2 \sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{2} \sqrt{c} \sqrt{\frac{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1}{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

[Out]

(x*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])])/Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])] - (Sqrt[b + Sqrt[b
^2 - 4*a*c]]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticE[ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^
2 - 4*a*c]]], (-2*Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[2]*Sqrt[c]*Sqrt[(1 + (2*c*x^2)/(b - Sqrt[
b^2 - 4*a*c]))/(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]) + (Sqrt[b
+ Sqrt[b^2 - 4*a*c]]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b
+ Sqrt[b^2 - 4*a*c]]], (-2*Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[2]*Sqrt[c]*Sqrt[(1 + (2*c*x^2)/(
b - Sqrt[b^2 - 4*a*c]))/(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])

________________________________________________________________________________________

Rubi [A]  time = 0.36674, antiderivative size = 478, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 59, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.068, Rules used = {422, 418, 492, 411} \[ \frac{x \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1}}{\sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}}+\frac{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} F\left (\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )|-\frac{2 \sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{2} \sqrt{c} \sqrt{\frac{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1}{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}}-\frac{\sqrt{\sqrt{b^2-4 a c}+b} \sqrt{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1} E\left (\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )|-\frac{2 \sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{2} \sqrt{c} \sqrt{\frac{\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}+1}{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \sqrt{\frac{2 c x^2}{\sqrt{b^2-4 a c}+b}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]/Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])],x]

[Out]

(x*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])])/Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])] - (Sqrt[b + Sqrt[b
^2 - 4*a*c]]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticE[ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^
2 - 4*a*c]]], (-2*Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[2]*Sqrt[c]*Sqrt[(1 + (2*c*x^2)/(b - Sqrt[
b^2 - 4*a*c]))/(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]) + (Sqrt[b
+ Sqrt[b^2 - 4*a*c]]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*EllipticF[ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b
+ Sqrt[b^2 - 4*a*c]]], (-2*Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[2]*Sqrt[c]*Sqrt[(1 + (2*c*x^2)/(
b - Sqrt[b^2 - 4*a*c]))/(1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]))]*Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[a, Int[1/(Sqrt[a + b*x^2]*Sqrt[c +
d*x^2]), x], x] + Dist[b, Int[x^2/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[
d/c] && PosQ[b/a]

Rule 418

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{\sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}}}{\sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}} \, dx &=\frac{(2 c) \int \frac{x^2}{\sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}} \, dx}{b-\sqrt{b^2-4 a c}}+\int \frac{1}{\sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}} \, dx\\ &=\frac{x \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}}}{\sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}}+\frac{\sqrt{b+\sqrt{b^2-4 a c}} \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} F\left (\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )|-\frac{2 \sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{2} \sqrt{c} \sqrt{\frac{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}}{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}}-\int \frac{\sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}}}{\left (1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}\right )^{3/2}} \, dx\\ &=\frac{x \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}}}{\sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}}-\frac{\sqrt{b+\sqrt{b^2-4 a c}} \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} E\left (\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )|-\frac{2 \sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{2} \sqrt{c} \sqrt{\frac{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}}{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}}+\frac{\sqrt{b+\sqrt{b^2-4 a c}} \sqrt{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}} F\left (\tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{b+\sqrt{b^2-4 a c}}}\right )|-\frac{2 \sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{2} \sqrt{c} \sqrt{\frac{1+\frac{2 c x^2}{b-\sqrt{b^2-4 a c}}}{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}} \sqrt{1+\frac{2 c x^2}{b+\sqrt{b^2-4 a c}}}}\\ \end{align*}

Mathematica [A]  time = 0.116894, size = 102, normalized size = 0.21 \[ \frac{\sqrt{-\sqrt{b^2-4 a c}-b} E\left (\sin ^{-1}\left (\frac{\sqrt{2} \sqrt{c} x}{\sqrt{-b-\sqrt{b^2-4 a c}}}\right )|\frac{b+\sqrt{b^2-4 a c}}{b-\sqrt{b^2-4 a c}}\right )}{\sqrt{2} \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]/Sqrt[1 + (2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])],x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.159, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{1+2\,{\frac{c{x}^{2}}{b-\sqrt{-4\,ac+{b}^{2}}}}}{\frac{1}{\sqrt{1+2\,{\frac{c{x}^{2}}{b+\sqrt{-4\,ac+{b}^{2}}}}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{2 \, c x^{2}}{b - \sqrt{b^{2} - 4 \, a c}} + 1}}{\sqrt{\frac{2 \, c x^{2}}{b + \sqrt{b^{2} - 4 \, a c}} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2),x, algorithm="maxi
ma")

[Out]

integrate(sqrt(2*c*x^2/(b - sqrt(b^2 - 4*a*c)) + 1)/sqrt(2*c*x^2/(b + sqrt(b^2 - 4*a*c)) + 1), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + \sqrt{b^{2} - 4 \, a c} x^{2} + 2 \, a\right )} \sqrt{\frac{b x^{2} + \sqrt{b^{2} - 4 \, a c} x^{2} + 2 \, a}{a}} \sqrt{\frac{b x^{2} - \sqrt{b^{2} - 4 \, a c} x^{2} + 2 \, a}{a}}}{4 \,{\left (c x^{4} + b x^{2} + a\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2),x, algorithm="fric
as")

[Out]

integral(1/4*(b*x^2 + sqrt(b^2 - 4*a*c)*x^2 + 2*a)*sqrt((b*x^2 + sqrt(b^2 - 4*a*c)*x^2 + 2*a)/a)*sqrt((b*x^2 -
sqrt(b^2 - 4*a*c)*x^2 + 2*a)/a)/(c*x^4 + b*x^2 + a), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{b + 2 c x^{2} - \sqrt{- 4 a c + b^{2}}}{b - \sqrt{- 4 a c + b^{2}}}}}{\sqrt{\frac{b + 2 c x^{2} + \sqrt{- 4 a c + b^{2}}}{b + \sqrt{- 4 a c + b^{2}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*c*x**2/(b-(-4*a*c+b**2)**(1/2)))**(1/2)/(1+2*c*x**2/(b+(-4*a*c+b**2)**(1/2)))**(1/2),x)

[Out]

Integral(sqrt((b + 2*c*x**2 - sqrt(-4*a*c + b**2))/(b - sqrt(-4*a*c + b**2)))/sqrt((b + 2*c*x**2 + sqrt(-4*a*c
+ b**2))/(b + sqrt(-4*a*c + b**2))), x)

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))^(1/2)/(1+2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))^(1/2),x, algorithm="giac
")

[Out]

Timed out