3.2.83 \(\int \frac {\sqrt {4+x^2}}{\sqrt {c+d x^2}} \, dx\) [183]

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

[Out]

x*(d*x^2+c)^(1/2)/d/(x^2+4)^(1/2)-(1/(x^2+4))^(1/2)*EllipticE(x/(x^2+4)^(1/2),(1-4*d/c)^(1/2))*(d*x^2+c)^(1/2)
/d/((d*x^2+c)/c/(x^2+4))^(1/2)+4*(1/(x^2+4))^(1/2)*EllipticF(x/(x^2+4)^(1/2),(1-4*d/c)^(1/2))*(d*x^2+c)^(1/2)/
c/((d*x^2+c)/c/(x^2+4))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {433, 429, 506, 422} \begin {gather*} \frac {4 \sqrt {c+d x^2} F\left (\text {ArcTan}\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 (\text {ArcTan}\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 )}}}+\frac {x \sqrt {c+d x^2}}{d \sqrt {x^2+4}} \end {gather*}

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

Rule 429

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

Rule 433

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 506

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt
[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]

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*}

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Mathematica [A]
time = 0.64, size = 60, normalized size = 0.40 \begin {gather*} \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}} \end {gather*}

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 = 0.08, size = 53, normalized size = 0.35

method	result	size
default	\(\frac {2 \EllipticE \left (x \sqrt {-\frac {d}{c}}, \frac {\sqrt {\frac {c}{d}}}{2}\right ) \sqrt {\frac {d \,x^{2}+c}{c}}}{\sqrt {d \,x^{2}+c}\, \sqrt {-\frac {d}{c}}}\)	\(53\)
elliptic	\(\frac {\sqrt {\left (d \,x^{2}+c \right ) \left (x^{2}+4\right )}\, \left (\frac {2 \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {x^{2}+4}\, \EllipticF \left (x \sqrt {-\frac {d}{c}}, \frac {\sqrt {-4+\frac {c +4 d}{d}}}{2}\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d \,x^{4}+c \,x^{2}+4 d \,x^{2}+4 c}}-\frac {2 \sqrt {1+\frac {d \,x^{2}}{c}}\, \sqrt {x^{2}+4}\, \left (\EllipticF \left (x \sqrt {-\frac {d}{c}}, \frac {\sqrt {-4+\frac {c +4 d}{d}}}{2}\right )-\EllipticE \left (x \sqrt {-\frac {d}{c}}, \frac {\sqrt {-4+\frac {c +4 d}{d}}}{2}\right )\right )}{\sqrt {-\frac {d}{c}}\, \sqrt {d \,x^{4}+c \,x^{2}+4 d \,x^{2}+4 c}}\right )}{\sqrt {d \,x^{2}+c}\, \sqrt {x^{2}+4}}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, 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((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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^{2} + 4}}{\sqrt {c + d x^{2}}}\, dx \end {gather*}

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 = 0.00, 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((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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x^2+4}}{\sqrt {d\,x^2+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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