3.2.97 \(\int \frac {\sqrt {1+4 x^2}}{\sqrt {2+3 x^2}} \, dx\) [197]

Optimal. Leaf size=148 \[ \frac {4 x \sqrt {2+3 x^2}}{3 \sqrt {1+4 x^2}}-\frac {2 \sqrt {2} \sqrt {2+3 x^2} E\left (\tan ^{-1}(2 x)|\frac {5}{8}\right )}{3 \sqrt {\frac {2+3 x^2}{1+4 x^2}} \sqrt {1+4 x^2}}+\frac {\sqrt {2+3 x^2} F\left (\tan ^{-1}(2 x)|\frac {5}{8}\right )}{2 \sqrt {2} \sqrt {\frac {2+3 x^2}{1+4 x^2}} \sqrt {1+4 x^2}} \]

[Out]

4/3*x*(3*x^2+2)^(1/2)/(4*x^2+1)^(1/2)+1/4*(1/(4*x^2+1))^(1/2)*EllipticF(2*x/(4*x^2+1)^(1/2),1/4*10^(1/2))*(3*x
^2+2)^(1/2)*2^(1/2)/((3*x^2+2)/(4*x^2+1))^(1/2)-2/3*(1/(4*x^2+1))^(1/2)*EllipticE(2*x/(4*x^2+1)^(1/2),1/4*10^(
1/2))*2^(1/2)*(3*x^2+2)^(1/2)/((3*x^2+2)/(4*x^2+1))^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {433, 429, 506, 422} \begin {gather*} \frac {\sqrt {3 x^2+2} F\left (\text {ArcTan}(2 x)\left |\frac {5}{8}\right .\right )}{2 \sqrt {2} \sqrt {\frac {3 x^2+2}{4 x^2+1}} \sqrt {4 x^2+1}}-\frac {2 \sqrt {2} \sqrt {3 x^2+2} E\left (\text {ArcTan}(2 x)\left |\frac {5}{8}\right .\right )}{3 \sqrt {\frac {3 x^2+2}{4 x^2+1}} \sqrt {4 x^2+1}}+\frac {4 \sqrt {3 x^2+2} x}{3 \sqrt {4 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + 4*x^2]/Sqrt[2 + 3*x^2],x]

[Out]

(4*x*Sqrt[2 + 3*x^2])/(3*Sqrt[1 + 4*x^2]) - (2*Sqrt[2]*Sqrt[2 + 3*x^2]*EllipticE[ArcTan[2*x], 5/8])/(3*Sqrt[(2
+ 3*x^2)/(1 + 4*x^2)]*Sqrt[1 + 4*x^2]) + (Sqrt[2 + 3*x^2]*EllipticF[ArcTan[2*x], 5/8])/(2*Sqrt[2]*Sqrt[(2 + 3
*x^2)/(1 + 4*x^2)]*Sqrt[1 + 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 {1+4 x^2}}{\sqrt {2+3 x^2}} \, dx &=4 \int \frac {x^2}{\sqrt {2+3 x^2} \sqrt {1+4 x^2}} \, dx+\int \frac {1}{\sqrt {2+3 x^2} \sqrt {1+4 x^2}} \, dx\\ &=\frac {4 x \sqrt {2+3 x^2}}{3 \sqrt {1+4 x^2}}+\frac {\sqrt {2+3 x^2} F\left (\tan ^{-1}(2 x)|\frac {5}{8}\right )}{2 \sqrt {2} \sqrt {\frac {2+3 x^2}{1+4 x^2}} \sqrt {1+4 x^2}}-\frac {4}{3} \int \frac {\sqrt {2+3 x^2}}{\left (1+4 x^2\right )^{3/2}} \, dx\\ &=\frac {4 x \sqrt {2+3 x^2}}{3 \sqrt {1+4 x^2}}-\frac {2 \sqrt {2} \sqrt {2+3 x^2} E\left (\tan ^{-1}(2 x)|\frac {5}{8}\right )}{3 \sqrt {\frac {2+3 x^2}{1+4 x^2}} \sqrt {1+4 x^2}}+\frac {\sqrt {2+3 x^2} F\left (\tan ^{-1}(2 x)|\frac {5}{8}\right )}{2 \sqrt {2} \sqrt {\frac {2+3 x^2}{1+4 x^2}} \sqrt {1+4 x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.28, size = 27, normalized size = 0.18 \begin {gather*} -\frac {i E\left (i \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right )|\frac {8}{3}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + 4*x^2]/Sqrt[2 + 3*x^2],x]

[Out]

((-I)*EllipticE[I*ArcSinh[Sqrt[3/2]*x], 8/3])/Sqrt[3]

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Maple [C] Result contains complex when optimal does not.
time = 0.07, size = 20, normalized size = 0.14

method	result	size
default	\(-\frac {i \EllipticE \left (\frac {i x \sqrt {6}}{2}, \frac {2 \sqrt {6}}{3}\right ) \sqrt {3}}{3}\)	\(20\)
elliptic	\(\frac {\sqrt {\left (3 x^{2}+2\right ) \left (4 x^{2}+1\right )}\, \left (-\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {4 x^{2}+1}\, \EllipticF \left (\frac {i x \sqrt {6}}{2}, \frac {2 \sqrt {6}}{3}\right )}{6 \sqrt {12 x^{4}+11 x^{2}+2}}+\frac {i \sqrt {6}\, \sqrt {6 x^{2}+4}\, \sqrt {4 x^{2}+1}\, \left (\EllipticF \left (\frac {i x \sqrt {6}}{2}, \frac {2 \sqrt {6}}{3}\right )-\EllipticE \left (\frac {i x \sqrt {6}}{2}, \frac {2 \sqrt {6}}{3}\right )\right )}{6 \sqrt {12 x^{4}+11 x^{2}+2}}\right )}{\sqrt {3 x^{2}+2}\, \sqrt {4 x^{2}+1}}\)	\(156\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2+1)^(1/2)/(3*x^2+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*I*EllipticE(1/2*I*x*6^(1/2),2/3*6^(1/2))*3^(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((4*x^2+1)^(1/2)/(3*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(4*x^2 + 1)/sqrt(3*x^2 + 2), 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((4*x^2+1)^(1/2)/(3*x^2+2)^(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 {4 x^{2} + 1}}{\sqrt {3 x^{2} + 2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2+1)**(1/2)/(3*x**2+2)**(1/2),x)

[Out]

Integral(sqrt(4*x**2 + 1)/sqrt(3*x**2 + 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((4*x^2+1)^(1/2)/(3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(4*x^2 + 1)/sqrt(3*x^2 + 2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {4\,x^2+1}}{\sqrt {3\,x^2+2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2 + 1)^(1/2)/(3*x^2 + 2)^(1/2),x)

[Out]

int((4*x^2 + 1)^(1/2)/(3*x^2 + 2)^(1/2), x)

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