3.183 \(\int \frac{\sqrt{4+x^2}}{\sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=150 \[ \frac{4 \sqrt{c+d x^2} \text{EllipticF}\left (\tan ^{-1}\left (\frac{x}{2}\right ),1-\frac{4 d}{c}\right )}{c \sqrt{x^2+4} \sqrt{\frac{c+d x^2}{c \left (x^2+4\right )}}}+\frac{x \sqrt{c+d x^2}}{d \sqrt{x^2+4}}-\frac{\sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{x}{2}\right )|1-\frac{4 d}{c}\right )}{d \sqrt{x^2+4} \sqrt{\frac{c+d x^2}{c \left (x^2+4\right )}}} \]

[Out]

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

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Rubi [A]  time = 0.0591805, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {422, 418, 492, 411} \[ \frac{x \sqrt{c+d x^2}}{d \sqrt{x^2+4}}+\frac{4 \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{x}{2}\right )|1-\frac{4 d}{c}\right )}{c \sqrt{x^2+4} \sqrt{\frac{c+d x^2}{c \left (x^2+4\right )}}}-\frac{\sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{x}{2}\right )|1-\frac{4 d}{c}\right )}{d \sqrt{x^2+4} \sqrt{\frac{c+d x^2}{c \left (x^2+4\right )}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[4 + x^2]/Sqrt[c + d*x^2],x]

[Out]

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

Rule 422

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[a, Int[1/(Sqrt[a + b*x^2]*Sqrt[c +
d*x^2]), x], x] + Dist[b, Int[x^2/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[
d/c] && PosQ[b/a]

Rule 418

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

Rule 492

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(x*Sqrt[a + b*x^2])/(b*Sqr
t[c + d*x^2]), x] - Dist[c/b, Int[Sqrt[a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]

Rule 411

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(Sqrt[a + b*x^2]*EllipticE[ArcTan
[Rt[d/c, 2]*x], 1 - (b*c)/(a*d)])/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]), x] /;
FreeQ[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]

Rubi steps

\begin{align*} \int \frac{\sqrt{4+x^2}}{\sqrt{c+d x^2}} \, dx &=4 \int \frac{1}{\sqrt{4+x^2} \sqrt{c+d x^2}} \, dx+\int \frac{x^2}{\sqrt{4+x^2} \sqrt{c+d x^2}} \, dx\\ &=\frac{x \sqrt{c+d x^2}}{d \sqrt{4+x^2}}+\frac{4 \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{x}{2}\right )|1-\frac{4 d}{c}\right )}{c \sqrt{4+x^2} \sqrt{\frac{c+d x^2}{c \left (4+x^2\right )}}}-\frac{4 \int \frac{\sqrt{c+d x^2}}{\left (4+x^2\right )^{3/2}} \, dx}{d}\\ &=\frac{x \sqrt{c+d x^2}}{d \sqrt{4+x^2}}-\frac{\sqrt{c+d x^2} E\left (\tan ^{-1}\left (\frac{x}{2}\right )|1-\frac{4 d}{c}\right )}{d \sqrt{4+x^2} \sqrt{\frac{c+d x^2}{c \left (4+x^2\right )}}}+\frac{4 \sqrt{c+d x^2} F\left (\tan ^{-1}\left (\frac{x}{2}\right )|1-\frac{4 d}{c}\right )}{c \sqrt{4+x^2} \sqrt{\frac{c+d x^2}{c \left (4+x^2\right )}}}\\ \end{align*}

Mathematica [A]  time = 0.0311412, size = 60, normalized size = 0.4 \[ \frac{2 \sqrt{\frac{c+d x^2}{c}} E\left (\sin ^{-1}\left (\sqrt{-\frac{d}{c}} x\right )|\frac{c}{4 d}\right )}{\sqrt{-\frac{d}{c}} \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[4 + x^2]/Sqrt[c + d*x^2],x]

[Out]

(2*Sqrt[(c + d*x^2)/c]*EllipticE[ArcSin[Sqrt[-(d/c)]*x], c/(4*d)])/(Sqrt[-(d/c)]*Sqrt[c + d*x^2])

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Maple [A]  time = 0.025, size = 53, normalized size = 0.4 \begin{align*} 2\,{\frac{1}{\sqrt{d{x}^{2}+c}}{\it EllipticE} \left ( x\sqrt{-{\frac{d}{c}}},1/2\,\sqrt{{\frac{c}{d}}} \right ) \sqrt{{\frac{d{x}^{2}+c}{c}}}{\frac{1}{\sqrt{-{\frac{d}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+4)^(1/2)/(d*x^2+c)^(1/2),x)

[Out]

2*EllipticE(x*(-d/c)^(1/2),1/2*(c/d)^(1/2))*((d*x^2+c)/c)^(1/2)/(d*x^2+c)^(1/2)/(-d/c)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + 4}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+4)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(x^2 + 4)/sqrt(d*x^2 + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{2} + 4}}{\sqrt{d x^{2} + c}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+4)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^2 + 4)/sqrt(d*x^2 + c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + 4}}{\sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+4)**(1/2)/(d*x**2+c)**(1/2),x)

[Out]

Integral(sqrt(x**2 + 4)/sqrt(c + d*x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} + 4}}{\sqrt{d x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+4)^(1/2)/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(x^2 + 4)/sqrt(d*x^2 + c), x)