3.10.19 \(\int \frac {1}{\sqrt {d+e x} \sqrt {f+g x} \sqrt {a+b x+c x^2}} \, dx\) [919]

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

[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 = 0.68, 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 = {949, 1117} \begin {gather*} -\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 \text {ArcTan}\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}} \end {gather*}

Warning: Unable to verify antiderivative.

[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 949

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 1117

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)]/(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(
4*c))], 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 ) \text {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 = 24.77, size = 375, normalized size = 0.64 \begin {gather*} \frac {2 \sqrt {2} e \sqrt {-\frac {e \left (c d^2+e (-b d+a e)\right ) (f+g x)}{\left (-2 c d e f+e \sqrt {\left (b^2-4 a c\right ) e^2} f-2 a e^2 g-d \sqrt {\left (b^2-4 a c\right ) e^2} g+b e (e f+d g)\right ) (d+e x)}} \sqrt {a+x (b+c x)} F\left (\sin ^{-1}\left (\frac {\sqrt {\frac {2 a e^2-2 c d e x+b e (-d+e x)+\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 c d e f+e \sqrt {\left (b^2-4 a c\right ) e^2} f-2 a e^2 g-d \sqrt {\left (b^2-4 a c\right ) e^2} g+b e (e f+d g)}\right )}{\sqrt {\left (b^2-4 a c\right ) e^2} \sqrt {d+e x} \sqrt {f+g x} \sqrt {-\frac {\left (c d^2+e (-b d+a e)\right ) (a+x (b+c x))}{\left (b^2-4 a c\right ) (d+e x)^2}}} \end {gather*}

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 = 0.14, size = 550, normalized size = 0.94

method	result	size
default	\(\frac {4 \left (\sqrt {-4 a c +b^{2}}\, e^{2} g \,x^{2}+b \,e^{2} g \,x^{2}-2 c \,e^{2} f \,x^{2}+2 \sqrt {-4 a c +b^{2}}\, d e g x +2 b d e g x -4 c d e f x +\sqrt {-4 a c +b^{2}}\, d^{2} g +b \,d^{2} g -2 c \,d^{2} f \right ) \EllipticF \left (\sqrt {\frac {\left (e \sqrt {-4 a c +b^{2}}+e b -2 c d \right ) \left (g x +f \right )}{\left (g \sqrt {-4 a c +b^{2}}+b g -2 c f \right ) \left (e x +d \right )}}, \sqrt {\frac {\left (e \sqrt {-4 a c +b^{2}}-e b +2 c d \right ) \left (g \sqrt {-4 a c +b^{2}}+b g -2 c f \right )}{\left (2 c f -b g +g \sqrt {-4 a c +b^{2}}\right ) \left (e \sqrt {-4 a c +b^{2}}+e b -2 c d \right )}}\right ) \sqrt {\frac {\left (d g -e f \right ) \left (b +2 c x +\sqrt {-4 a c +b^{2}}\right )}{\left (g \sqrt {-4 a c +b^{2}}+b g -2 c f \right ) \left (e x +d \right )}}\, \sqrt {\frac {\left (d g -e f \right ) \left (-b -2 c x +\sqrt {-4 a c +b^{2}}\right )}{\left (2 c f -b g +g \sqrt {-4 a c +b^{2}}\right ) \left (e x +d \right )}}\, \sqrt {\frac {\left (e \sqrt {-4 a c +b^{2}}+e b -2 c d \right ) \left (g x +f \right )}{\left (g \sqrt {-4 a c +b^{2}}+b g -2 c f \right ) \left (e x +d \right )}}\, \sqrt {e x +d}\, \sqrt {g x +f}\, \sqrt {c \,x^{2}+b x +a}}{\sqrt {-\frac {\left (g x +f \right ) \left (e x +d \right ) \left (-b -2 c x +\sqrt {-4 a c +b^{2}}\right ) \left (b +2 c x +\sqrt {-4 a c +b^{2}}\right )}{c}}\, \left (e \sqrt {-4 a c +b^{2}}+e b -2 c d \right ) \left (d g -e f \right ) \sqrt {\left (g x +f \right ) \left (e x +d \right ) \left (c \,x^{2}+b x +a \right )}}\)	\(550\)
elliptic	\(\frac {2 \sqrt {\left (g x +f \right ) \left (e x +d \right ) \left (c \,x^{2}+b x +a \right )}\, \left (-\frac {f}{g}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \sqrt {\frac {\left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {d}{e}\right ) \left (x +\frac {f}{g}\right )}{\left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {f}{g}\right ) \left (x +\frac {d}{e}\right )}}\, \left (x +\frac {d}{e}\right )^{2} \sqrt {\frac {\left (-\frac {d}{e}+\frac {f}{g}\right ) \left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}{\left (\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {f}{g}\right ) \left (x +\frac {d}{e}\right )}}\, \sqrt {\frac {\left (-\frac {d}{e}+\frac {f}{g}\right ) \left (x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}{\left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {f}{g}\right ) \left (x +\frac {d}{e}\right )}}\, \EllipticF \left (\sqrt {\frac {\left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {d}{e}\right ) \left (x +\frac {f}{g}\right )}{\left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {f}{g}\right ) \left (x +\frac {d}{e}\right )}}, \sqrt {\frac {\left (-\frac {d}{e}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \left (-\frac {f}{g}+\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}{\left (-\frac {f}{g}-\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \left (\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}-\frac {d}{e}\right )}}\right )}{\sqrt {g x +f}\, \sqrt {e x +d}\, \sqrt {c \,x^{2}+b x +a}\, \left (-\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}+\frac {d}{e}\right ) \left (-\frac {d}{e}+\frac {f}{g}\right ) \sqrt {c e g \left (x +\frac {f}{g}\right ) \left (x +\frac {d}{e}\right ) \left (x -\frac {-b +\sqrt {-4 a c +b^{2}}}{2 c}\right ) \left (x +\frac {b +\sqrt {-4 a c +b^{2}}}{2 c}\right )}}\)	\(566\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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(g*x + f)*sqrt(x*e + d)), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(g*x + f)*sqrt(x*e + d)), x)

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

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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