\(\int \frac {\sqrt {2+x^2}}{(1+x^2)^{3/2} (a+b x^2)} \, dx\) [133]

Optimal result

Integrand size = 28, antiderivative size = 86 \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\frac {\sqrt {2} \sqrt {2+x^2} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{(a-b) \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}-\frac {2 b \operatorname {EllipticPi}\left (1-\frac {2 b}{a},\arctan \left (\frac {x}{\sqrt {2}}\right ),-1\right )}{a (a-b)} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.83 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\frac {\frac {x}{\sqrt {\frac {1+x^2}{2+x^2}}}+i E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-\frac {i (a-2 b) \operatorname {EllipticPi}\left (\frac {2 b}{a},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )}{a}}{a-b} \] Input:

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

Output:

(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*ArcSinh[x/Sqrt[2]], 2])/a)/(a - b)
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.41, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {416, 313, 414}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 416

\(\displaystyle \frac {\int \frac {\sqrt {x^2+2}}{\left (x^2+1\right )^{3/2}}dx}{a-b}-\frac {b \int \frac {\sqrt {x^2+2}}{\sqrt {x^2+1} \left (b x^2+a\right )}dx}{a-b}\)

\(\Big \downarrow \) 313

\(\displaystyle \frac {\sqrt {2} \sqrt {x^2+2} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}} (a-b)}-\frac {b \int \frac {\sqrt {x^2+2}}{\sqrt {x^2+1} \left (b x^2+a\right )}dx}{a-b}\)

\(\Big \downarrow \) 414

\(\displaystyle \frac {\sqrt {2} \sqrt {x^2+2} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}} (a-b)}-\frac {2 b \sqrt {x^2+1} \operatorname {EllipticPi}\left (1-\frac {2 b}{a},\arctan \left (\frac {x}{\sqrt {2}}\right ),-1\right )}{a \sqrt {\frac {x^2+1}{x^2+2}} \sqrt {x^2+2} (a-b)}\)

Input:

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

Output:

(Sqrt[2]*Sqrt[2 + x^2]*EllipticE[ArcTan[x], 1/2])/((a - b)*Sqrt[1 + x^2]*S 
qrt[(2 + x^2)/(1 + x^2)]) - (2*b*Sqrt[1 + x^2]*EllipticPi[1 - (2*b)/a, Arc 
Tan[x/Sqrt[2]], -1])/(a*(a - b)*Sqrt[(1 + x^2)/(2 + x^2)]*Sqrt[2 + x^2])
 

Defintions of rubi rules used

rule 313

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Sim 
p[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sqrt[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 414

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)), ArcTan[ 
Rt[d/c, 2]*x], 1 - c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ 
[d/c]
 

rule 416

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

Maple [A] (verified)

Time = 4.92 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.71

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

Input:

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

Output:

(I*EllipticE(1/2*I*x*2^(1/2),2^(1/2))*a*(x^2+1)^(1/2)*(x^2+2)^(1/2)-I*Elli 
pticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*a*(x^2+1)^(1/2)*(x^2+2)^(1/2)+2*I*El 
lipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))*b*(x^2+1)^(1/2)*(x^2+2)^(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]

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

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

Output:

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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{3/2} \left (a+b x^2\right )} \, dx=\int \frac {\sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}{b \,x^{6}+a \,x^{4}+2 b \,x^{4}+2 a \,x^{2}+b \,x^{2}+a}d x \] Input:

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

Output:

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