3.195 \(\int \frac {\sqrt {1+x^2}}{\sqrt {2+3 x^2}} \, dx\)

Optimal. Leaf size=131 \[ \frac {\sqrt {3 x^2+2} \operatorname {EllipticF}\left (\tan ^{-1}(x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}}+\frac {\sqrt {3 x^2+2} x}{3 \sqrt {x^2+1}}-\frac {\sqrt {2} \sqrt {3 x^2+2} E\left (\tan ^{-1}(x)|-\frac {1}{2}\right )}{3 \sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 131, 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 = {422, 418, 492, 411} \[ \frac {\sqrt {3 x^2+2} x}{3 \sqrt {x^2+1}}+\frac {\sqrt {3 x^2+2} F\left (\tan ^{-1}(x)|-\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}}-\frac {\sqrt {2} \sqrt {3 x^2+2} E\left (\tan ^{-1}(x)|-\frac {1}{2}\right )}{3 \sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^2]/Sqrt[2 + 3*x^2],x]

[Out]

(x*Sqrt[2 + 3*x^2])/(3*Sqrt[1 + x^2]) - (Sqrt[2]*Sqrt[2 + 3*x^2]*EllipticE[ArcTan[x], -1/2])/(3*Sqrt[1 + x^2]*
Sqrt[(2 + 3*x^2)/(1 + x^2)]) + (Sqrt[2 + 3*x^2]*EllipticF[ArcTan[x], -1/2])/(Sqrt[2]*Sqrt[1 + x^2]*Sqrt[(2 + 3
*x^2)/(1 + x^2)])

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]

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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.00, size = 27, normalized size = 0.21 \[ -\frac {i E\left (i \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right )|\frac {2}{3}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {x^{2} + 1}}{\sqrt {3 \, x^{2} + 2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)^(1/2)/(3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} + 1}}{\sqrt {3 \, x^{2} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)^(1/2)/(3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 30, normalized size = 0.23 \[ -\frac {i \left (2 \EllipticE \left (i x , \frac {\sqrt {6}}{2}\right )+\EllipticF \left (i x , \frac {\sqrt {6}}{2}\right )\right ) \sqrt {2}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*I*(EllipticF(I*x,1/2*6^(1/2))+2*EllipticE(I*x,1/2*6^(1/2)))*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} + 1}}{\sqrt {3 \, x^{2} + 2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)^(1/2)/(3*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {x^2+1}}{\sqrt {3\,x^2+2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {x^{2} + 1}}{\sqrt {3 x^{2} + 2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**2 + 1)/sqrt(3*x**2 + 2), x)

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