\(\int \frac {1}{\sqrt {1+x^2} \sqrt {2+3 x^2}} \, dx\) [230]

Optimal result

Integrand size = 21, antiderivative size = 51 \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+3 x^2}} \, dx=\frac {\sqrt {2+3 x^2} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {1+x^2} \sqrt {\frac {2+3 x^2}{1+x^2}}} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {429} \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+3 x^2}} \, dx=\frac {\sqrt {3 x^2+2} \operatorname {EllipticF}\left (\arctan (x),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {x^2+1} \sqrt {\frac {3 x^2+2}{x^2+1}}} \]

[In]

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

[Out]

(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 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]

Rubi steps \begin{align*} \text {integral}& = \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*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.37 \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+3 x^2}} \, dx=-\frac {i \operatorname {EllipticF}\left (i \text {arcsinh}(x),\frac {3}{2}\right )}{\sqrt {2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.59 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.33

method	result	size
default	\(-\frac {i F\left (i x , \frac {\sqrt {6}}{2}\right ) \sqrt {2}}{2}\)	\(17\)
elliptic	\(-\frac {i \sqrt {\left (3 x^{2}+2\right ) \left (x^{2}+1\right )}\, \sqrt {6 x^{2}+4}\, F\left (i x , \frac {\sqrt {6}}{2}\right )}{2 \sqrt {3 x^{2}+2}\, \sqrt {3 x^{4}+5 x^{2}+2}}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.22 \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+3 x^2}} \, dx=-\frac {1}{2} i \, \sqrt {2} F(\arcsin \left (i \, x\right )\,|\,\frac {3}{2}) \]

[In]

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

[Out]

-1/2*I*sqrt(2)*elliptic_f(arcsin(I*x), 3/2)

Sympy [F]

\[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+3 x^2}} \, dx=\int \frac {1}{\sqrt {x^{2} + 1} \sqrt {3 x^{2} + 2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+3 x^2}} \, dx=\int { \frac {1}{\sqrt {3 \, x^{2} + 2} \sqrt {x^{2} + 1}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+3 x^2}} \, dx=\int { \frac {1}{\sqrt {3 \, x^{2} + 2} \sqrt {x^{2} + 1}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {1+x^2} \sqrt {2+3 x^2}} \, dx=\int \frac {1}{\sqrt {x^2+1}\,\sqrt {3\,x^2+2}} \,d x \]

[In]

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

[Out]

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