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 \[ -\frac {(d+e x) \sqrt [4]{c f^2-g (b f-a g)} \sqrt {\frac {\left (a+b x+c x^2\right ) (e f-d g)^2}{(d+e x)^2 \left (a g^2-b f g+c f^2\right )}} \left (\frac {(f+g x) \sqrt {a e^2-b d e+c d^2}}{(d+e x) \sqrt {c f^2-g (b f-a g)}}+1\right ) \sqrt {\frac {\frac {(f+g x)^2 \left (a e^2-b d e+c d^2\right )}{(d+e x)^2 \left (c f^2-g (b f-a g)\right )}-\frac {(f+g x) (2 a e g-b (d g+e f)+2 c d f)}{(d+e x) \left (a g^2-b f g+c f^2\right )}+1}{\left (\frac {(f+g x) \sqrt {a e^2-b d e+c d^2}}{(d+e x) \sqrt {c f^2-g (b f-a g)}}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c d^2-b e d+a e^2} \sqrt {f+g x}}{\sqrt [4]{c f^2-b g f+a g^2} \sqrt {d+e x}}\right )|\frac {1}{4} \left (\frac {2 c d f+2 a e g-b (e f+d g)}{\sqrt {c d^2-e (b d-a e)} \sqrt {c f^2-g (b f-a g)}}+2\right )\right )}{\sqrt {a+b x+c x^2} (e f-d g) \sqrt [4]{a e^2-b d e+c d^2} \sqrt {\frac {(f+g x)^2 \left (a e^2-b d e+c d^2\right )}{(d+e x)^2 \left (c f^2-g (b f-a g)\right )}-\frac {(f+g x) (2 a e g-b (d g+e f)+2 c d f)}{(d+e x) \left (a g^2-b f g+c f^2\right )}+1}} \]

[Out]

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

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Rubi [A]  time = 1.17, antiderivative size = 588, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 3.52, size = 375, normalized size = 0.64 \[ \frac {2 \sqrt {2} e \sqrt {a+x (b+c x)} \sqrt {-\frac {e (f+g x) \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (-d g \sqrt {e^2 \left (b^2-4 a c\right )}+e f \sqrt {e^2 \left (b^2-4 a c\right )}-2 a e^2 g+b e (d g+e f)-2 c d e f\right )}} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {2 a e^2-2 c d x e+b (e x-d) e+\sqrt {\left (b^2-4 a c\right ) e^2} (d+e x)}{\sqrt {\left (b^2-4 a c\right ) e^2} (d+e x)}}}{\sqrt {2}}\right )|\frac {2 \sqrt {\left (b^2-4 a c\right ) e^2} (e f-d g)}{-2 a g e^2-2 c d f e+\sqrt {\left (b^2-4 a c\right ) e^2} f e+b (e f+d g) e-d \sqrt {\left (b^2-4 a c\right ) e^2} g}\right )}{\sqrt {d+e x} \sqrt {f+g x} \sqrt {e^2 \left (b^2-4 a c\right )} \sqrt {-\frac {(a+x (b+c x)) \left (e (a e-b d)+c d^2\right )}{\left (b^2-4 a c\right ) (d+e x)^2}}} \]

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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fricas [F]  time = 2.23, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c x^{2} + b x + a} \sqrt {e x + d} \sqrt {g x + f}}{c e g x^{4} + {\left (c e f + {\left (c d + b e\right )} g\right )} x^{3} + a d f + {\left ({\left (c d + b e\right )} f + {\left (b d + a e\right )} g\right )} x^{2} + {\left (a d g + {\left (b d + a e\right )} f\right )} x}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{2} + b x + a} \sqrt {e x + d} \sqrt {g x + f}}\,{d x} \]

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)

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maple [A]  time = 0.10, size = 605, normalized size = 1.03 \[ \frac {4 \left (b \,e^{2} g \,x^{2}-2 c \,e^{2} f \,x^{2}+2 b d e g x -4 c d e f x +\sqrt {-4 a c +b^{2}}\, e^{2} g \,x^{2}+b \,d^{2} g -2 c \,d^{2} f +2 \sqrt {-4 a c +b^{2}}\, d e g x +\sqrt {-4 a c +b^{2}}\, d^{2} g \right ) \sqrt {\frac {\left (d g -e f \right ) \left (2 c x +b +\sqrt {-4 a c +b^{2}}\right )}{\left (b g -2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) \left (e x +d \right )}}\, \sqrt {\frac {\left (d g -e f \right ) \left (-2 c x -b +\sqrt {-4 a c +b^{2}}\right )}{\left (-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) \left (e x +d \right )}}\, \sqrt {\frac {\left (b e -2 c d +\sqrt {-4 a c +b^{2}}\, e \right ) \left (g x +f \right )}{\left (b g -2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) \left (e x +d \right )}}\, \sqrt {e x +d}\, \sqrt {g x +f}\, \sqrt {c \,x^{2}+b x +a}\, \EllipticF \left (\sqrt {\frac {\left (b e -2 c d +\sqrt {-4 a c +b^{2}}\, e \right ) \left (g x +f \right )}{\left (b g -2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) \left (e x +d \right )}}, \sqrt {\frac {\left (-b e +2 c d +\sqrt {-4 a c +b^{2}}\, e \right ) \left (b g -2 c f +\sqrt {-4 a c +b^{2}}\, g \right )}{\left (-b g +2 c f +\sqrt {-4 a c +b^{2}}\, g \right ) \left (b e -2 c d +\sqrt {-4 a c +b^{2}}\, e \right )}}\right )}{\sqrt {-\frac {\left (g x +f \right ) \left (e x +d \right ) \left (-2 c x -b +\sqrt {-4 a c +b^{2}}\right ) \left (2 c x +b +\sqrt {-4 a c +b^{2}}\right )}{c}}\, \left (d g -e f \right ) \left (b e -2 c d +\sqrt {-4 a c +b^{2}}\, e \right ) \sqrt {c e g \,x^{4}+b e g \,x^{3}+c d g \,x^{3}+c e f \,x^{3}+a e g \,x^{2}+b d g \,x^{2}+b e f \,x^{2}+c d f \,x^{2}+a d g x +a e f x +b d f x +a d f}} \]

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*(b*e^2*g*x^2-2*c*e^2*f*x^2+2*b*d*e*g*x-4*c*d*e*f*x+(-4*a*c+b^2)^(1/2)*e^2*g*x^2+b*d^2*g-2*c*d^2*f+2*(-4*a*c+
b^2)^(1/2)*d*e*g*x+(-4*a*c+b^2)^(1/2)*d^2*g)*EllipticF(((b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*(g*x+f)/(b*g-2*c*f+(-
4*a*c+b^2)^(1/2)*g)/(e*x+d))^(1/2),((-b*e+2*c*d+(-4*a*c+b^2)^(1/2)*e)*(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)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e))^(1/2))*((d*g-e*f)*(2*c*x+b+(-4*a*c+b^2)^(1/2))/(b
*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)/(e*x+d))^(1/2)*((d*g-e*f)*(-2*c*x-b+(-4*a*c+b^2)^(1/2))/(-b*g+2*c*f+(-4*a*c+b^2
)^(1/2)*g)/(e*x+d))^(1/2)*((b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)*(g*x+f)/(b*g-2*c*f+(-4*a*c+b^2)^(1/2)*g)/(e*x+d))^
(1/2)*(e*x+d)^(1/2)*(g*x+f)^(1/2)*(c*x^2+b*x+a)^(1/2)/(-(g*x+f)*(e*x+d)*(-2*c*x-b+(-4*a*c+b^2)^(1/2))*(2*c*x+b
+(-4*a*c+b^2)^(1/2))/c)^(1/2)/(d*g-e*f)/(b*e-2*c*d+(-4*a*c+b^2)^(1/2)*e)/(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {c x^{2} + b x + a} \sqrt {e x + d} \sqrt {g x + f}}\,{d x} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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