### 3.2465 $$\int \frac{1}{\sqrt{d+e x} \sqrt{a+b x+c x^2}} \, dx$$

Optimal. Leaf size=189 $\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right ),-\frac{2 e \sqrt{b^2-4 a c}}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )}{c \sqrt{d+e x} \sqrt{a+b x+c x^2}}$

[Out]

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

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Rubi [A]  time = 0.0711155, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.083, Rules used = {718, 419} $\frac{2 \sqrt{2} \sqrt{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \sqrt{\frac{c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F\left (\sin ^{-1}\left (\frac{\sqrt{\frac{b+2 c x+\sqrt{b^2-4 a c}}{\sqrt{b^2-4 a c}}}}{\sqrt{2}}\right )|-\frac{2 \sqrt{b^2-4 a c} e}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{c \sqrt{d+e x} \sqrt{a+b x+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 718

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

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rubi steps

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

Mathematica [C]  time = 0.694371, size = 308, normalized size = 1.63 $\frac{i (d+e x) \sqrt{2-\frac{4 \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (\sqrt{e^2 \left (b^2-4 a c\right )}-b e+2 c d\right )}} \sqrt{\frac{2 \left (e (a e-b d)+c d^2\right )}{(d+e x) \left (\sqrt{e^2 \left (b^2-4 a c\right )}+b e-2 c d\right )}+1} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{2} \sqrt{\frac{a e^2-b d e+c d^2}{\sqrt{e^2 \left (b^2-4 a c\right )}+b e-2 c d}}}{\sqrt{d+e x}}\right ),-\frac{\sqrt{e^2 \left (b^2-4 a c\right )}+b e-2 c d}{\sqrt{e^2 \left (b^2-4 a c\right )}-b e+2 c d}\right )}{e \sqrt{a+x (b+c x)} \sqrt{\frac{e (a e-b d)+c d^2}{\sqrt{e^2 \left (b^2-4 a c\right )}+b e-2 c d}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.306, size = 287, normalized size = 1.5 \begin{align*}{\frac{\sqrt{2}}{ce \left ( ce{x}^{3}+be{x}^{2}+cd{x}^{2}+aex+bdx+ad \right ) } \left ( -e\sqrt{-4\,ac+{b}^{2}}-be+2\,cd \right ){\it EllipticF} \left ( \sqrt{2}\sqrt{-{c \left ( ex+d \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) ^{-1}}},\sqrt{-{ \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) \left ( 2\,cd-be+e\sqrt{-4\,ac+{b}^{2}} \right ) ^{-1}}} \right ) \sqrt{{e \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) ^{-1}}}\sqrt{{e \left ( -b-2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ) \left ( 2\,cd-be+e\sqrt{-4\,ac+{b}^{2}} \right ) ^{-1}}}\sqrt{-{c \left ( ex+d \right ) \left ( e\sqrt{-4\,ac+{b}^{2}}+be-2\,cd \right ) ^{-1}}}\sqrt{ex+d}\sqrt{c{x}^{2}+bx+a}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(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)/(c*e*x^3 + (c*d + b*e)*x^2 + a*d + (b*d + a*e)*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{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)/(c*x**2+b*x+a)**(1/2),x)

[Out]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^(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)), x)