\(\int \frac {\sqrt {2+x^2}}{(1+x^2)^{3/2} (a+b x^2)} \, dx\) [92]

Optimal result

Integrand size = 28, antiderivative size = 121 \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\frac {\sqrt {2} \sqrt {2+x^2} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{(a-b) \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}-\frac {2 b \sqrt {1+x^2} \operatorname {EllipticPi}\left (1-\frac {2 b}{a},\arctan \left (\frac {x}{\sqrt {2}}\right ),-1\right )}{a (a-b) \sqrt {\frac {1+x^2}{2+x^2}} \sqrt {2+x^2}} \]

[Out]

-2*b*(1/(2*x^2+4))^(1/2)*(2*x^2+4)^(1/2)*EllipticPi(x*2^(1/2)/(2*x^2+4)^(1/2),1-2*b/a,I)*(x^2+1)^(1/2)/a/(a-b)
/((x^2+1)/(x^2+2))^(1/2)/(x^2+2)^(1/2)+(1/(x^2+1))^(1/2)*EllipticE(x/(x^2+1)^(1/2),1/2*2^(1/2))*2^(1/2)*(x^2+2
)^(1/2)/(a-b)/((x^2+2)/(x^2+1))^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {555, 553, 422} \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\frac {\sqrt {2} \sqrt {x^2+2} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}} (a-b)}-\frac {2 b \sqrt {x^2+1} \operatorname {EllipticPi}\left (1-\frac {2 b}{a},\arctan \left (\frac {x}{\sqrt {2}}\right ),-1\right )}{a \sqrt {\frac {x^2+1}{x^2+2}} \sqrt {x^2+2} (a-b)} \]

[In]

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

[Out]

(Sqrt[2]*Sqrt[2 + x^2]*EllipticE[ArcTan[x], 1/2])/((a - b)*Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x^2)]) - (2*b*Sqr
t[1 + x^2]*EllipticPi[1 - (2*b)/a, ArcTan[x/Sqrt[2]], -1])/(a*(a - b)*Sqrt[(1 + x^2)/(2 + x^2)]*Sqrt[2 + 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 553

Int[Sqrt[(c_) + (d_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[c*(Sqrt[e +
f*x^2]/(a*e*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((e + f*x^2)/(e*(c + d*x^2)))]))*EllipticPi[1 - b*(c/(a*d)), Ar
cTan[Rt[d/c, 2]*x], 1 - c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c]

Rule 555

Int[Sqrt[(e_) + (f_.)*(x_)^2]/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)^(3/2)), x_Symbol] :> Dist[b/(b*c -
a*d), Int[Sqrt[e + f*x^2]/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] - Dist[d/(b*c - a*d), Int[Sqrt[e + f*x^2]/(c +
d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[d/c] && PosQ[f/e]

Rubi steps \begin{align*} \text {integral}& = -\frac {b \int \frac {\sqrt {2+x^2}}{\sqrt {1+x^2} \left (a+b x^2\right )} \, dx}{a-b}-\frac {\int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2}} \, dx}{-a+b} \\ & = \frac {\sqrt {2} \sqrt {2+x^2} E\left (\tan ^{-1}(x)|\frac {1}{2}\right )}{(a-b) \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}-\frac {2 b \sqrt {1+x^2} \Pi \left (1-\frac {2 b}{a};\left .\tan ^{-1}\left (\frac {x}{\sqrt {2}}\right )\right |-1\right )}{a (a-b) \sqrt {\frac {1+x^2}{2+x^2}} \sqrt {2+x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.31 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\frac {\frac {x}{\sqrt {\frac {1+x^2}{2+x^2}}}+i E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-\frac {i (a-2 b) \operatorname {EllipticPi}\left (\frac {2 b}{a},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )}{a}}{a-b} \]

[In]

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

[Out]

(x/Sqrt[(1 + x^2)/(2 + x^2)] + I*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] - (I*(a - 2*b)*EllipticPi[(2*b)/a, I*ArcSi
nh[x/Sqrt[2]], 2])/a)/(a - b)

Maple [A] (verified)

Time = 3.35 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.21

method	result	size
default	\(\frac {\left (i E\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a \sqrt {x^{2}+2}\, \sqrt {x^{2}+1}-i \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) a \sqrt {x^{2}+2}\, \sqrt {x^{2}+1}+2 i \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) b \sqrt {x^{2}+2}\, \sqrt {x^{2}+1}+a \,x^{3}+2 a x \right ) \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}{a \left (x^{4}+3 x^{2}+2\right ) \left (a -b \right )}\)	\(147\)
elliptic	\(\frac {\sqrt {\left (x^{2}+1\right ) \left (x^{2}+2\right )}\, \left (\frac {\left (x^{2}+2\right ) x}{\left (a -b \right ) \sqrt {\left (x^{2}+1\right ) \left (x^{2}+2\right )}}+\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, E\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{2 \left (a -b \right ) \sqrt {x^{4}+3 x^{2}+2}}-\frac {i \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{\left (a -b \right ) \sqrt {x^{4}+3 x^{2}+2}}+\frac {2 i b \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \Pi \left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{\left (a -b \right ) a \sqrt {x^{4}+3 x^{2}+2}}\right )}{\sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}\)	\(229\)

[In]

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

[Out]

(I*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*a*(x^2+2)^(1/2)*(x^2+1)^(1/2)-I*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2)
)*a*(x^2+2)^(1/2)*(x^2+1)^(1/2)+2*I*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*b*(x^2+2)^(1/2)*(x^2+1)^(1/2)+a*
x^3+2*a*x)*(x^2+1)^(1/2)*(x^2+2)^(1/2)/a/(x^4+3*x^2+2)/(a-b)

Fricas [F]

\[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\int { \frac {\sqrt {x^{2} + 2}}{{\left (b x^{2} + a\right )} {\left (x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integral(sqrt(x^2 + 2)*sqrt(x^2 + 1)/(b*x^6 + (a + 2*b)*x^4 + (2*a + b)*x^2 + a), x)

Sympy [F]

\[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\int \frac {\sqrt {x^{2} + 2}}{\left (a + b x^{2}\right ) \left (x^{2} + 1\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\int { \frac {\sqrt {x^{2} + 2}}{{\left (b x^{2} + a\right )} {\left (x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\int { \frac {\sqrt {x^{2} + 2}}{{\left (b x^{2} + a\right )} {\left (x^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\int \frac {\sqrt {x^2+2}}{{\left (x^2+1\right )}^{3/2}\,\left (b\,x^2+a\right )} \,d x \]

[In]

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

[Out]

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