3.1365 \(\int \frac{1}{\sqrt{b d+2 c d x} \sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=97 \[ \frac{2 \sqrt [4]{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{b d+2 c d x}}{\sqrt{d} \sqrt [4]{b^2-4 a c}}\right ),-1\right )}{c \sqrt{d} \sqrt{a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0918113, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {691, 689, 221} \[ \frac{2 \sqrt [4]{b^2-4 a c} \sqrt{-\frac{c \left (a+b x+c x^2\right )}{b^2-4 a c}} F\left (\left .\sin ^{-1}\left (\frac{\sqrt{b d+2 c x d}}{\sqrt [4]{b^2-4 a c} \sqrt{d}}\right )\right |-1\right )}{c \sqrt{d} \sqrt{a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 691

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

Rule 689

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

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[(Rt[-b, 4]*x)/Rt[a, 4]], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rubi steps

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

Mathematica [C]  time = 0.0592199, size = 89, normalized size = 0.92 \[ \frac{2 \sqrt{\frac{c (a+x (b+c x))}{4 a c-b^2}} \sqrt{d (b+2 c x)} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};\frac{(b+2 c x)^2}{b^2-4 a c}\right )}{c d \sqrt{a+x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[d*(b + 2*c*x)]*Sqrt[(c*(a + x*(b + c*x)))/(-b^2 + 4*a*c)]*Hypergeometric2F1[1/4, 1/2, 5/4, (b + 2*c*x)
^2/(b^2 - 4*a*c)])/(c*d*Sqrt[a + x*(b + c*x)])

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Maple [B]  time = 0.212, size = 189, normalized size = 2. \begin{align*}{\frac{1}{cd \left ( 2\,{c}^{2}{x}^{3}+3\,bc{x}^{2}+2\,acx+{b}^{2}x+ab \right ) }{\it EllipticF} \left ({\frac{\sqrt{2}}{2}\sqrt{{ \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}}}},\sqrt{2} \right ) \sqrt{{ \left ( -b-2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}}}\sqrt{-{(2\,cx+b){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}}}\sqrt{{ \left ( b+2\,cx+\sqrt{-4\,ac+{b}^{2}} \right ){\frac{1}{\sqrt{-4\,ac+{b}^{2}}}}}}\sqrt{-4\,ac+{b}^{2}}\sqrt{d \left ( 2\,cx+b \right ) }\sqrt{c{x}^{2}+bx+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}}{2 \, c^{2} d x^{3} + 3 \, b c d x^{2} + a b d +{\left (b^{2} + 2 \, a c\right )} d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(d*(b + 2*c*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{2 \, c d x + b d} \sqrt{c x^{2} + b x + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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