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

Optimal. Leaf size=588 \[ -\frac{(d+e x) \sqrt [4]{c f^2-g (b f-a g)} \sqrt{\frac{\left (a+b x+c x^2\right ) (e f-d g)^2}{(d+e x)^2 \left (a g^2-b f g+c f^2\right )}} \left (\frac{(f+g x) \sqrt{a e^2-b d e+c d^2}}{(d+e x) \sqrt{c f^2-g (b f-a g)}}+1\right ) \sqrt{\frac{\frac{(f+g x)^2 \left (a e^2-b d e+c d^2\right )}{(d+e x)^2 \left (c f^2-g (b f-a g)\right )}-\frac{(f+g x) (2 a e g-b (d g+e f)+2 c d f)}{(d+e x) \left (a g^2-b f g+c f^2\right )}+1}{\left (\frac{(f+g x) \sqrt{a e^2-b d e+c d^2}}{(d+e x) \sqrt{c f^2-g (b f-a g)}}+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{f+g x} \sqrt [4]{a e^2-b d e+c d^2}}{\sqrt{d+e x} \sqrt [4]{a g^2-b f g+c f^2}}\right ),\frac{1}{4} \left (\frac{2 a e g-b (d g+e f)+2 c d f}{\sqrt{c d^2-e (b d-a e)} \sqrt{c f^2-g (b f-a g)}}+2\right )\right )}{\sqrt{a+b x+c x^2} (e f-d g) \sqrt [4]{a e^2-b d e+c d^2} \sqrt{\frac{(f+g x)^2 \left (a e^2-b d e+c d^2\right )}{(d+e x)^2 \left (c f^2-g (b f-a g)\right )}-\frac{(f+g x) (2 a e g-b (d g+e f)+2 c d f)}{(d+e x) \left (a g^2-b f g+c f^2\right )}+1}} \]

[Out]

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

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Rubi [A]  time = 1.16692, antiderivative size = 588, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {935, 1103} \[ -\frac{(d+e x) \sqrt [4]{c f^2-g (b f-a g)} \sqrt{\frac{\left (a+b x+c x^2\right ) (e f-d g)^2}{(d+e x)^2 \left (a g^2-b f g+c f^2\right )}} \left (\frac{(f+g x) \sqrt{a e^2-b d e+c d^2}}{(d+e x) \sqrt{c f^2-g (b f-a g)}}+1\right ) \sqrt{\frac{\frac{(f+g x)^2 \left (a e^2-b d e+c d^2\right )}{(d+e x)^2 \left (c f^2-g (b f-a g)\right )}-\frac{(f+g x) (2 a e g-b (d g+e f)+2 c d f)}{(d+e x) \left (a g^2-b f g+c f^2\right )}+1}{\left (\frac{(f+g x) \sqrt{a e^2-b d e+c d^2}}{(d+e x) \sqrt{c f^2-g (b f-a g)}}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c d^2-b e d+a e^2} \sqrt{f+g x}}{\sqrt [4]{c f^2-b g f+a g^2} \sqrt{d+e x}}\right )|\frac{1}{4} \left (\frac{2 c d f+2 a e g-b (e f+d g)}{\sqrt{c d^2-e (b d-a e)} \sqrt{c f^2-g (b f-a g)}}+2\right )\right )}{\sqrt{a+b x+c x^2} (e f-d g) \sqrt [4]{a e^2-b d e+c d^2} \sqrt{\frac{(f+g x)^2 \left (a e^2-b d e+c d^2\right )}{(d+e x)^2 \left (c f^2-g (b f-a g)\right )}-\frac{(f+g x) (2 a e g-b (d g+e f)+2 c d f)}{(d+e x) \left (a g^2-b f g+c f^2\right )}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 935

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :
> Dist[(-2*(d + e*x)*Sqrt[((e*f - d*g)^2*(a + b*x + c*x^2))/((c*f^2 - b*f*g + a*g^2)*(d + e*x)^2)])/((e*f - d*
g)*Sqrt[a + b*x + c*x^2]), Subst[Int[1/Sqrt[1 - ((2*c*d*f - b*e*f - b*d*g + 2*a*e*g)*x^2)/(c*f^2 - b*f*g + a*g
^2) + ((c*d^2 - b*d*e + a*e^2)*x^4)/(c*f^2 - b*f*g + a*g^2)], x], x, Sqrt[f + g*x]/Sqrt[d + e*x]], x] /; FreeQ
[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1103

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)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rubi steps

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

Mathematica [A]  time = 3.51726, size = 375, normalized size = 0.64 \[ \frac{2 \sqrt{2} e \sqrt{a+x (b+c x)} \sqrt{-\frac{e (f+g x) \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (-d g \sqrt{e^2 \left (b^2-4 a c\right )}+e f \sqrt{e^2 \left (b^2-4 a c\right )}-2 a e^2 g+b e (d g+e f)-2 c d e f\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{(d+e x) \sqrt{e^2 \left (b^2-4 a c\right )}+2 a e^2+b e (e x-d)-2 c d e x}{(d+e x) \sqrt{e^2 \left (b^2-4 a c\right )}}}}{\sqrt{2}}\right ),\frac{2 \sqrt{e^2 \left (b^2-4 a c\right )} (e f-d g)}{-d g \sqrt{e^2 \left (b^2-4 a c\right )}+e f \sqrt{e^2 \left (b^2-4 a c\right )}-2 a e^2 g+b e (d g+e f)-2 c d e f}\right )}{\sqrt{d+e x} \sqrt{f+g x} \sqrt{e^2 \left (b^2-4 a c\right )} \sqrt{-\frac{(a+x (b+c x)) \left (e (a e-b d)+c d^2\right )}{\left (b^2-4 a c\right ) (d+e x)^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.471, size = 605, normalized size = 1. \begin{align*} 4\,{\frac{ \left ( \sqrt{-4\,ac+{b}^{2}}{x}^{2}{e}^{2}g+b{e}^{2}g{x}^{2}-2\,c{e}^{2}f{x}^{2}+2\,\sqrt{-4\,ac+{b}^{2}}xdeg+2\,xbdeg-4\,xcdef+\sqrt{-4\,ac+{b}^{2}}{d}^{2}g+b{d}^{2}g-2\,c{d}^{2}f \right ) \sqrt{ex+d}\sqrt{gx+f}\sqrt{c{x}^{2}+bx+a}}{ \left ( dg-ef \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) \sqrt{ceg{x}^{4}+beg{x}^{3}+cdg{x}^{3}+cef{x}^{3}+aeg{x}^{2}+bdg{x}^{2}+bef{x}^{2}+cdf{x}^{2}+adgx+aefx+bdfx+adf}}{\it EllipticF} \left ( \sqrt{{\frac{ \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) \left ( gx+f \right ) }{ \left ( g\sqrt{-4\,ac+{b}^{2}}+bg-2\,cf \right ) \left ( ex+d \right ) }}},\sqrt{{\frac{ \left ( 2\,cd-be+e\sqrt{-4\,ac+{b}^{2}} \right ) \left ( g\sqrt{-4\,ac+{b}^{2}}+bg-2\,cf \right ) }{ \left ( 2\,cf-bg+g\sqrt{-4\,ac+{b}^{2}} \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) }}} \right ) \sqrt{{\frac{ \left ( dg-ef \right ) \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) }{ \left ( g\sqrt{-4\,ac+{b}^{2}}+bg-2\,cf \right ) \left ( ex+d \right ) }}}\sqrt{{\frac{ \left ( dg-ef \right ) \left ( -b-2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) }{ \left ( 2\,cf-bg+g\sqrt{-4\,ac+{b}^{2}} \right ) \left ( ex+d \right ) }}}\sqrt{{\frac{ \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) \left ( gx+f \right ) }{ \left ( g\sqrt{-4\,ac+{b}^{2}}+bg-2\,cf \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{-{\frac{ \left ( gx+f \right ) \left ( ex+d \right ) \left ( -b-2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) }{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{e x + d} \sqrt{g x + f}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(1/2)/(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(e*x + d)*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{e x + d} \sqrt{g x + f}}{c e g x^{4} +{\left (c e f +{\left (c d + b e\right )} g\right )} x^{3} + a d f +{\left ({\left (c d + b e\right )} f +{\left (b d + a e\right )} g\right )} x^{2} +{\left (a d g +{\left (b d + a e\right )} f\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(1/2)/(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(e*x + d)*sqrt(g*x + f)/(c*e*g*x^4 + (c*e*f + (c*d + b*e)*g)*x^3 + a*d*f +
((c*d + b*e)*f + (b*d + a*e)*g)*x^2 + (a*d*g + (b*d + a*e)*f)*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d + e x} \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/(e*x+d)**(1/2)/(g*x+f)**(1/2)/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral(1/(sqrt(d + e*x)*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{e x + d} \sqrt{g x + f}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(1/2)/(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(e*x + d)*sqrt(g*x + f)), x)