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

Optimal. Leaf size=1077 \[ \text{result too large to display} \]

[Out]

-(((b^2*d + b*Sqrt[b^2 - 4*a*c]*d - 2*a*(c*d - a*f))^(1/4)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)^(3/2)*Sqrt[2*a + (b
+ Sqrt[b^2 - 4*a*c])*x]*Sqrt[((4*a*c - (b + Sqrt[b^2 - 4*a*c])^2)^2*(d + f*x^2))/(((b + Sqrt[b^2 - 4*a*c])^2*
d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)^2)]*(1 + (Sqrt[2*c^2*d - 2*a*c*f + b*(b + Sqrt[b^2 - 4*a*c])*f]*(
2*a + (b + Sqrt[b^2 - 4*a*c])*x))/(Sqrt[b^2*d + b*Sqrt[b^2 - 4*a*c]*d - 2*a*(c*d - a*f)]*(b + Sqrt[b^2 - 4*a*c
] + 2*c*x)))*Sqrt[(1 - (4*(b + Sqrt[b^2 - 4*a*c])*(c*d + a*f)*(2*a + (b + Sqrt[b^2 - 4*a*c])*x))/(((b + Sqrt[b
^2 - 4*a*c])^2*d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)) + ((4*c^2*d + (b + Sqrt[b^2 - 4*a*c])^2*f)*(2*a +
(b + Sqrt[b^2 - 4*a*c])*x)^2)/(((b + Sqrt[b^2 - 4*a*c])^2*d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)^2))/(1
+ (Sqrt[2*c^2*d - 2*a*c*f + b*(b + Sqrt[b^2 - 4*a*c])*f]*(2*a + (b + Sqrt[b^2 - 4*a*c])*x))/(Sqrt[b^2*d + b*S
qrt[b^2 - 4*a*c]*d - 2*a*(c*d - a*f)]*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)))^2]*EllipticF[2*ArcTan[((2*c^2*d - 2*a*
c*f + b*(b + Sqrt[b^2 - 4*a*c])*f)^(1/4)*Sqrt[2*a + (b + Sqrt[b^2 - 4*a*c])*x])/((b^2*d + b*Sqrt[b^2 - 4*a*c]*
d - 2*a*(c*d - a*f))^(1/4)*Sqrt[b + Sqrt[b^2 - 4*a*c] + 2*c*x])], (1 + ((b + Sqrt[b^2 - 4*a*c])*(c*d + a*f))/(
Sqrt[2*c^2*d - 2*a*c*f + b*(b + Sqrt[b^2 - 4*a*c])*f]*Sqrt[b^2*d + b*Sqrt[b^2 - 4*a*c]*d - 2*a*(c*d - a*f)]))/
2])/((4*a*c - (b + Sqrt[b^2 - 4*a*c])^2)*(2*c^2*d - 2*a*c*f + b*(b + Sqrt[b^2 - 4*a*c])*f)^(1/4)*Sqrt[a + b*x
+ c*x^2]*Sqrt[d + f*x^2]*Sqrt[1 - (4*(b + Sqrt[b^2 - 4*a*c])*(c*d + a*f)*(2*a + (b + Sqrt[b^2 - 4*a*c])*x))/((
(b + Sqrt[b^2 - 4*a*c])^2*d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)) + ((4*c^2*d + (b + Sqrt[b^2 - 4*a*c])^
2*f)*(2*a + (b + Sqrt[b^2 - 4*a*c])*x)^2)/(((b + Sqrt[b^2 - 4*a*c])^2*d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*
c*x)^2)]))

________________________________________________________________________________________

Rubi [A]  time = 3.09899, antiderivative size = 1077, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {993, 936, 1103} \[ -\frac{\sqrt [4]{d b^2+\sqrt{b^2-4 a c} d b-2 a (c d-a f)} \left (b+2 c x+\sqrt{b^2-4 a c}\right )^{3/2} \sqrt{2 a+\left (b+\sqrt{b^2-4 a c}\right ) x} \sqrt{\frac{\left (4 a c-\left (b+\sqrt{b^2-4 a c}\right )^2\right )^2 \left (f x^2+d\right )}{\left (4 f a^2+\left (b+\sqrt{b^2-4 a c}\right )^2 d\right ) \left (b+2 c x+\sqrt{b^2-4 a c}\right )^2}} \left (\frac{\sqrt{2 d c^2-2 a f c+b \left (b+\sqrt{b^2-4 a c}\right ) f} \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )}{\sqrt{d b^2+\sqrt{b^2-4 a c} d b-2 a (c d-a f)} \left (b+2 c x+\sqrt{b^2-4 a c}\right )}+1\right ) \sqrt{\frac{\frac{\left (4 d c^2+\left (b+\sqrt{b^2-4 a c}\right )^2 f\right ) \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )^2}{\left (4 f a^2+\left (b+\sqrt{b^2-4 a c}\right )^2 d\right ) \left (b+2 c x+\sqrt{b^2-4 a c}\right )^2}-\frac{4 \left (b+\sqrt{b^2-4 a c}\right ) (c d+a f) \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )}{\left (4 f a^2+\left (b+\sqrt{b^2-4 a c}\right )^2 d\right ) \left (b+2 c x+\sqrt{b^2-4 a c}\right )}+1}{\left (\frac{\sqrt{2 d c^2-2 a f c+b \left (b+\sqrt{b^2-4 a c}\right ) f} \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )}{\sqrt{d b^2+\sqrt{b^2-4 a c} d b-2 a (c d-a f)} \left (b+2 c x+\sqrt{b^2-4 a c}\right )}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{2 d c^2-2 a f c+b \left (b+\sqrt{b^2-4 a c}\right ) f} \sqrt{2 a+\left (b+\sqrt{b^2-4 a c}\right ) x}}{\sqrt [4]{d b^2+\sqrt{b^2-4 a c} d b-2 a (c d-a f)} \sqrt{b+2 c x+\sqrt{b^2-4 a c}}}\right )|\frac{1}{2} \left (\frac{\left (b+\sqrt{b^2-4 a c}\right ) (c d+a f)}{\sqrt{2 d c^2-2 a f c+b \left (b+\sqrt{b^2-4 a c}\right ) f} \sqrt{d b^2+\sqrt{b^2-4 a c} d b-2 a (c d-a f)}}+1\right )\right )}{\left (4 a c-\left (b+\sqrt{b^2-4 a c}\right )^2\right ) \sqrt [4]{2 d c^2-2 a f c+b \left (b+\sqrt{b^2-4 a c}\right ) f} \sqrt{c x^2+b x+a} \sqrt{f x^2+d} \sqrt{\frac{\left (4 d c^2+\left (b+\sqrt{b^2-4 a c}\right )^2 f\right ) \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )^2}{\left (4 f a^2+\left (b+\sqrt{b^2-4 a c}\right )^2 d\right ) \left (b+2 c x+\sqrt{b^2-4 a c}\right )^2}-\frac{4 \left (b+\sqrt{b^2-4 a c}\right ) (c d+a f) \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )}{\left (4 f a^2+\left (b+\sqrt{b^2-4 a c}\right )^2 d\right ) \left (b+2 c x+\sqrt{b^2-4 a c}\right )}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b^2*d + b*Sqrt[b^2 - 4*a*c]*d - 2*a*(c*d - a*f))^(1/4)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)^(3/2)*Sqrt[2*a + (b
+ Sqrt[b^2 - 4*a*c])*x]*Sqrt[((4*a*c - (b + Sqrt[b^2 - 4*a*c])^2)^2*(d + f*x^2))/(((b + Sqrt[b^2 - 4*a*c])^2*
d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)^2)]*(1 + (Sqrt[2*c^2*d - 2*a*c*f + b*(b + Sqrt[b^2 - 4*a*c])*f]*(
2*a + (b + Sqrt[b^2 - 4*a*c])*x))/(Sqrt[b^2*d + b*Sqrt[b^2 - 4*a*c]*d - 2*a*(c*d - a*f)]*(b + Sqrt[b^2 - 4*a*c
] + 2*c*x)))*Sqrt[(1 - (4*(b + Sqrt[b^2 - 4*a*c])*(c*d + a*f)*(2*a + (b + Sqrt[b^2 - 4*a*c])*x))/(((b + Sqrt[b
^2 - 4*a*c])^2*d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)) + ((4*c^2*d + (b + Sqrt[b^2 - 4*a*c])^2*f)*(2*a +
(b + Sqrt[b^2 - 4*a*c])*x)^2)/(((b + Sqrt[b^2 - 4*a*c])^2*d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)^2))/(1
+ (Sqrt[2*c^2*d - 2*a*c*f + b*(b + Sqrt[b^2 - 4*a*c])*f]*(2*a + (b + Sqrt[b^2 - 4*a*c])*x))/(Sqrt[b^2*d + b*S
qrt[b^2 - 4*a*c]*d - 2*a*(c*d - a*f)]*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)))^2]*EllipticF[2*ArcTan[((2*c^2*d - 2*a*
c*f + b*(b + Sqrt[b^2 - 4*a*c])*f)^(1/4)*Sqrt[2*a + (b + Sqrt[b^2 - 4*a*c])*x])/((b^2*d + b*Sqrt[b^2 - 4*a*c]*
d - 2*a*(c*d - a*f))^(1/4)*Sqrt[b + Sqrt[b^2 - 4*a*c] + 2*c*x])], (1 + ((b + Sqrt[b^2 - 4*a*c])*(c*d + a*f))/(
Sqrt[2*c^2*d - 2*a*c*f + b*(b + Sqrt[b^2 - 4*a*c])*f]*Sqrt[b^2*d + b*Sqrt[b^2 - 4*a*c]*d - 2*a*(c*d - a*f)]))/
2])/((4*a*c - (b + Sqrt[b^2 - 4*a*c])^2)*(2*c^2*d - 2*a*c*f + b*(b + Sqrt[b^2 - 4*a*c])*f)^(1/4)*Sqrt[a + b*x
+ c*x^2]*Sqrt[d + f*x^2]*Sqrt[1 - (4*(b + Sqrt[b^2 - 4*a*c])*(c*d + a*f)*(2*a + (b + Sqrt[b^2 - 4*a*c])*x))/((
(b + Sqrt[b^2 - 4*a*c])^2*d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)) + ((4*c^2*d + (b + Sqrt[b^2 - 4*a*c])^
2*f)*(2*a + (b + Sqrt[b^2 - 4*a*c])*x)^2)/(((b + Sqrt[b^2 - 4*a*c])^2*d + 4*a^2*f)*(b + Sqrt[b^2 - 4*a*c] + 2*
c*x)^2)]))

Rule 993

Int[1/(Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{r = Rt[b^2 - 4*a*
c, 2]}, Dist[(Sqrt[b + r + 2*c*x]*Sqrt[2*a + (b + r)*x])/Sqrt[a + b*x + c*x^2], Int[1/(Sqrt[b + r + 2*c*x]*Sqr
t[2*a + (b + r)*x]*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 936

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(f_.) + (g_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[(-2*(d
+ e*x)*Sqrt[((e*f - d*g)^2*(a + c*x^2))/((c*f^2 + a*g^2)*(d + e*x)^2)])/((e*f - d*g)*Sqrt[a + c*x^2]), Subst[I
nt[1/Sqrt[1 - ((2*c*d*f + 2*a*e*g)*x^2)/(c*f^2 + a*g^2) + ((c*d^2 + a*e^2)*x^4)/(c*f^2 + a*g^2)], x], x, Sqrt[
f + g*x]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[c*d^2 + 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{a+b x+c x^2} \sqrt{d+f x^2}} \, dx &=\frac{\left (\sqrt{b+\sqrt{b^2-4 a c}+2 c x} \sqrt{2 a+\left (b+\sqrt{b^2-4 a c}\right ) x}\right ) \int \frac{1}{\sqrt{b+\sqrt{b^2-4 a c}+2 c x} \sqrt{2 a+\left (b+\sqrt{b^2-4 a c}\right ) x} \sqrt{d+f x^2}} \, dx}{\sqrt{a+b x+c x^2}}\\ &=-\frac{\left (2 \left (b+\sqrt{b^2-4 a c}+2 c x\right )^{3/2} \sqrt{2 a+\left (b+\sqrt{b^2-4 a c}\right ) x} \sqrt{\frac{\left (4 a c-\left (b+\sqrt{b^2-4 a c}\right )^2\right )^2 \left (d+f x^2\right )}{\left (\left (b+\sqrt{b^2-4 a c}\right )^2 d+4 a^2 f\right ) \left (b+\sqrt{b^2-4 a c}+2 c x\right )^2}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{\left (4 c \left (b+\sqrt{b^2-4 a c}\right ) d+4 a \left (b+\sqrt{b^2-4 a c}\right ) f\right ) x^2}{\left (b+\sqrt{b^2-4 a c}\right )^2 d+4 a^2 f}+\frac{\left (4 c^2 d+\left (b+\sqrt{b^2-4 a c}\right )^2 f\right ) x^4}{\left (b+\sqrt{b^2-4 a c}\right )^2 d+4 a^2 f}}} \, dx,x,\frac{\sqrt{2 a+\left (b+\sqrt{b^2-4 a c}\right ) x}}{\sqrt{b+\sqrt{b^2-4 a c}+2 c x}}\right )}{\left (4 a c-\left (b+\sqrt{b^2-4 a c}\right )^2\right ) \sqrt{a+b x+c x^2} \sqrt{d+f x^2}}\\ &=-\frac{\sqrt [4]{b^2 d+b \sqrt{b^2-4 a c} d-2 a (c d-a f)} \left (b+\sqrt{b^2-4 a c}+2 c x\right )^{3/2} \sqrt{2 a+\left (b+\sqrt{b^2-4 a c}\right ) x} \sqrt{\frac{\left (4 a c-\left (b+\sqrt{b^2-4 a c}\right )^2\right )^2 \left (d+f x^2\right )}{\left (\left (b+\sqrt{b^2-4 a c}\right )^2 d+4 a^2 f\right ) \left (b+\sqrt{b^2-4 a c}+2 c x\right )^2}} \left (1+\frac{\sqrt{2 c^2 d-2 a c f+b \left (b+\sqrt{b^2-4 a c}\right ) f} \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )}{\sqrt{b^2 d+b \sqrt{b^2-4 a c} d-2 a (c d-a f)} \left (b+\sqrt{b^2-4 a c}+2 c x\right )}\right ) \sqrt{\frac{1-\frac{4 \left (b+\sqrt{b^2-4 a c}\right ) (c d+a f) \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )}{\left (\left (b+\sqrt{b^2-4 a c}\right )^2 d+4 a^2 f\right ) \left (b+\sqrt{b^2-4 a c}+2 c x\right )}+\frac{\left (4 c^2 d+\left (b+\sqrt{b^2-4 a c}\right )^2 f\right ) \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )^2}{\left (\left (b+\sqrt{b^2-4 a c}\right )^2 d+4 a^2 f\right ) \left (b+\sqrt{b^2-4 a c}+2 c x\right )^2}}{\left (1+\frac{\sqrt{2 c^2 d-2 a c f+b \left (b+\sqrt{b^2-4 a c}\right ) f} \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )}{\sqrt{b^2 d+b \sqrt{b^2-4 a c} d-2 a (c d-a f)} \left (b+\sqrt{b^2-4 a c}+2 c x\right )}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{2 c^2 d-2 a c f+b \left (b+\sqrt{b^2-4 a c}\right ) f} \sqrt{2 a+\left (b+\sqrt{b^2-4 a c}\right ) x}}{\sqrt [4]{b^2 d+b \sqrt{b^2-4 a c} d-2 a (c d-a f)} \sqrt{b+\sqrt{b^2-4 a c}+2 c x}}\right )|\frac{1}{2} \left (1+\frac{\left (b+\sqrt{b^2-4 a c}\right ) (c d+a f)}{\sqrt{2 c^2 d-2 a c f+b \left (b+\sqrt{b^2-4 a c}\right ) f} \sqrt{b^2 d+b \sqrt{b^2-4 a c} d-2 a (c d-a f)}}\right )\right )}{\left (4 a c-\left (b+\sqrt{b^2-4 a c}\right )^2\right ) \sqrt [4]{2 c^2 d-2 a c f+b \left (b+\sqrt{b^2-4 a c}\right ) f} \sqrt{a+b x+c x^2} \sqrt{d+f x^2} \sqrt{1-\frac{4 \left (b+\sqrt{b^2-4 a c}\right ) (c d+a f) \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )}{\left (\left (b+\sqrt{b^2-4 a c}\right )^2 d+4 a^2 f\right ) \left (b+\sqrt{b^2-4 a c}+2 c x\right )}+\frac{\left (4 c^2 d+\left (b+\sqrt{b^2-4 a c}\right )^2 f\right ) \left (2 a+\left (b+\sqrt{b^2-4 a c}\right ) x\right )^2}{\left (\left (b+\sqrt{b^2-4 a c}\right )^2 d+4 a^2 f\right ) \left (b+\sqrt{b^2-4 a c}+2 c x\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 1.52761, size = 600, normalized size = 0.56 \[ -\frac{2 \sqrt{2} \left (\sqrt{f} x-i \sqrt{d}\right ) \left (\sqrt{b^2-4 a c}-b-2 c x\right ) \sqrt{-\frac{c \sqrt{b^2-4 a c} \left (\sqrt{f} x+i \sqrt{d}\right )}{\left (\sqrt{b^2-4 a c}-b-2 c x\right ) \left (\sqrt{f} \left (\sqrt{b^2-4 a c}+b\right )-2 i c \sqrt{d}\right )}} \sqrt{\frac{c \left (-i \sqrt{d} \left (\sqrt{b^2-4 a c}+2 c x\right )+\sqrt{f} \left (x \sqrt{b^2-4 a c}-2 a\right )+b \left (-\sqrt{f} x-i \sqrt{d}\right )\right )}{\left (\sqrt{b^2-4 a c}-b-2 c x\right ) \left (\sqrt{f} \left (\sqrt{b^2-4 a c}+b\right )+2 i c \sqrt{d}\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{\frac{\left (\sqrt{b^2-4 a c}+b+2 c x\right ) \left (\sqrt{f} \left (\sqrt{b^2-4 a c}-b\right )-2 i c \sqrt{d}\right )}{\left (\sqrt{b^2-4 a c}-b-2 c x\right ) \left (\sqrt{f} \left (\sqrt{b^2-4 a c}+b\right )+2 i c \sqrt{d}\right )}}\right ),\frac{-i \sqrt{d} \sqrt{f} \sqrt{b^2-4 a c}+a f+c d}{i \sqrt{d} \sqrt{f} \sqrt{b^2-4 a c}+a f+c d}\right )}{\sqrt{d+f x^2} \sqrt{a+x (b+c x)} \left (\sqrt{f} \left (\sqrt{b^2-4 a c}-b\right )-2 i c \sqrt{d}\right ) \sqrt{\frac{i c \sqrt{b^2-4 a c} \left (\sqrt{d}+i \sqrt{f} x\right )}{\left (\sqrt{b^2-4 a c}-b-2 c x\right ) \left (\sqrt{f} \left (\sqrt{b^2-4 a c}+b\right )+2 i c \sqrt{d}\right )}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)*((-I)*Sqrt[d] + Sqrt[f]*x)*Sqrt[-((c*Sqrt[b^2 - 4*a*c]*(I*Sqrt[d]
+ Sqrt[f]*x))/(((-2*I)*c*Sqrt[d] + (b + Sqrt[b^2 - 4*a*c])*Sqrt[f])*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)))]*Sqrt[
(c*((-I)*Sqrt[d]*(Sqrt[b^2 - 4*a*c] + 2*c*x) + Sqrt[f]*(-2*a + Sqrt[b^2 - 4*a*c]*x) + b*((-I)*Sqrt[d] - Sqrt[f
]*x)))/(((2*I)*c*Sqrt[d] + (b + Sqrt[b^2 - 4*a*c])*Sqrt[f])*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))]*EllipticF[ArcSi
n[Sqrt[(((-2*I)*c*Sqrt[d] + (-b + Sqrt[b^2 - 4*a*c])*Sqrt[f])*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(((2*I)*c*Sqrt[
d] + (b + Sqrt[b^2 - 4*a*c])*Sqrt[f])*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))]], (c*d - I*Sqrt[b^2 - 4*a*c]*Sqrt[d]*
Sqrt[f] + a*f)/(c*d + I*Sqrt[b^2 - 4*a*c]*Sqrt[d]*Sqrt[f] + a*f)])/(((-2*I)*c*Sqrt[d] + (-b + Sqrt[b^2 - 4*a*c
])*Sqrt[f])*Sqrt[(I*c*Sqrt[b^2 - 4*a*c]*(Sqrt[d] + I*Sqrt[f]*x))/(((2*I)*c*Sqrt[d] + (b + Sqrt[b^2 - 4*a*c])*S
qrt[f])*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))]*Sqrt[d + f*x^2]*Sqrt[a + x*(b + c*x)])

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Maple [A]  time = 0.618, size = 661, normalized size = 0.6 \begin{align*} 4\,{\frac{ \left ( b{f}^{2}{x}^{2}+2\,{x}^{2}cf\sqrt{-df}+\sqrt{-4\,ac+{b}^{2}}{f}^{2}{x}^{2}+2\,xbf\sqrt{-df}-4\,cxfd+2\,xf\sqrt{-4\,ac+{b}^{2}}\sqrt{-df}-bdf-2\,cd\sqrt{-df}-\sqrt{-4\,ac+{b}^{2}}df \right ) \sqrt{c{x}^{2}+bx+a}\sqrt{f{x}^{2}+d}}{\sqrt{-df} \left ( f\sqrt{-4\,ac+{b}^{2}}-2\,\sqrt{-df}c+bf \right ) \sqrt{cf{x}^{4}+bf{x}^{3}+af{x}^{2}+cd{x}^{2}+bdx+ad}}{\it EllipticF} \left ( \sqrt{-{\frac{ \left ( f\sqrt{-4\,ac+{b}^{2}}-2\,\sqrt{-df}c+bf \right ) \left ( -fx+\sqrt{-df} \right ) }{ \left ( f\sqrt{-4\,ac+{b}^{2}}+2\,\sqrt{-df}c+bf \right ) \left ( fx+\sqrt{-df} \right ) }}},\sqrt{{\frac{ \left ( f\sqrt{-4\,ac+{b}^{2}}+2\,\sqrt{-df}c-bf \right ) \left ( f\sqrt{-4\,ac+{b}^{2}}+2\,\sqrt{-df}c+bf \right ) }{ \left ( f\sqrt{-4\,ac+{b}^{2}}-2\,\sqrt{-df}c-bf \right ) \left ( f\sqrt{-4\,ac+{b}^{2}}-2\,\sqrt{-df}c+bf \right ) }}} \right ) \sqrt{{\frac{\sqrt{-df} \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) f}{ \left ( f\sqrt{-4\,ac+{b}^{2}}+2\,\sqrt{-df}c+bf \right ) \left ( fx+\sqrt{-df} \right ) }}}\sqrt{{\frac{\sqrt{-df} \left ( -b-2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) f}{ \left ( f\sqrt{-4\,ac+{b}^{2}}-2\,\sqrt{-df}c-bf \right ) \left ( fx+\sqrt{-df} \right ) }}}\sqrt{-{\frac{ \left ( f\sqrt{-4\,ac+{b}^{2}}-2\,\sqrt{-df}c+bf \right ) \left ( -fx+\sqrt{-df} \right ) }{ \left ( f\sqrt{-4\,ac+{b}^{2}}+2\,\sqrt{-df}c+bf \right ) \left ( fx+\sqrt{-df} \right ) }}}{\frac{1}{\sqrt{{\frac{ \left ( -fx+\sqrt{-df} \right ) \left ( fx+\sqrt{-df} \right ) \left ( -b-2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) }{cf}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(b*f^2*x^2+2*x^2*c*f*(-d*f)^(1/2)+(-4*a*c+b^2)^(1/2)*f^2*x^2+2*x*b*f*(-d*f)^(1/2)-4*c*x*f*d+2*x*f*(-4*a*c+b^
2)^(1/2)*(-d*f)^(1/2)-b*d*f-2*c*d*(-d*f)^(1/2)-(-4*a*c+b^2)^(1/2)*d*f)*EllipticF((-(f*(-4*a*c+b^2)^(1/2)-2*(-d
*f)^(1/2)*c+b*f)*(-f*x+(-d*f)^(1/2))/(f*(-4*a*c+b^2)^(1/2)+2*(-d*f)^(1/2)*c+b*f)/(f*x+(-d*f)^(1/2)))^(1/2),((f
*(-4*a*c+b^2)^(1/2)+2*(-d*f)^(1/2)*c-b*f)*(f*(-4*a*c+b^2)^(1/2)+2*(-d*f)^(1/2)*c+b*f)/(f*(-4*a*c+b^2)^(1/2)-2*
(-d*f)^(1/2)*c-b*f)/(f*(-4*a*c+b^2)^(1/2)-2*(-d*f)^(1/2)*c+b*f))^(1/2))*((-d*f)^(1/2)*(b+2*c*x+(-4*a*c+b^2)^(1
/2))*f/(f*(-4*a*c+b^2)^(1/2)+2*(-d*f)^(1/2)*c+b*f)/(f*x+(-d*f)^(1/2)))^(1/2)*((-d*f)^(1/2)*(-b-2*c*x+(-4*a*c+b
^2)^(1/2))*f/(f*(-4*a*c+b^2)^(1/2)-2*(-d*f)^(1/2)*c-b*f)/(f*x+(-d*f)^(1/2)))^(1/2)*(-(f*(-4*a*c+b^2)^(1/2)-2*(
-d*f)^(1/2)*c+b*f)*(-f*x+(-d*f)^(1/2))/(f*(-4*a*c+b^2)^(1/2)+2*(-d*f)^(1/2)*c+b*f)/(f*x+(-d*f)^(1/2)))^(1/2)*(
c*x^2+b*x+a)^(1/2)*(f*x^2+d)^(1/2)/(1/c/f*(-f*x+(-d*f)^(1/2))*(f*x+(-d*f)^(1/2))*(-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*f)^(1/2)/(f*(-4*a*c+b^2)^(1/2)-2*(-d*f)^(1/2)*c+b*f)/(c*f*x^4+b*f*x^3
+a*f*x^2+c*d*x^2+b*d*x+a*d)^(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{f x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(d + f*x**2)*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{f x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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