\(\int \frac {\sqrt {2+x^2}}{(1+x^2)^{5/2} (a+b x^2)} \, dx\) [134]

Optimal result

Integrand size = 28, antiderivative size = 235 \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx=\frac {x \sqrt {2+x^2}}{3 (a-b) \left (1+x^2\right )^{3/2}}+\frac {\sqrt {2} (a-2 b) \sqrt {2+x^2} E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{(a-b)^2 \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}-\frac {\left (2 a^2-4 a b-b^2\right ) \sqrt {2+x^2} \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{3 \sqrt {2} (a-b)^3 \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}}+\frac {(a-2 b) b^2 \sqrt {2+x^2} \operatorname {EllipticPi}\left (1-\frac {b}{a},\arctan (x),\frac {1}{2}\right )}{\sqrt {2} a (a-b)^3 \sqrt {1+x^2} \sqrt {\frac {2+x^2}{1+x^2}}} \] Output:

1/3*x*(x^2+2)^(1/2)/(a-b)/(x^2+1)^(3/2)+2^(1/2)*(a-2*b)*(x^2+2)^(1/2)*Elli 
pticE(x/(x^2+1)^(1/2),1/2*2^(1/2))/(a-b)^2/(x^2+1)^(1/2)/((x^2+2)/(x^2+1)) 
^(1/2)-1/6*(2*a^2-4*a*b-b^2)*(x^2+2)^(1/2)*InverseJacobiAM(arctan(x),1/2*2 
^(1/2))*2^(1/2)/(a-b)^3/(x^2+1)^(1/2)/((x^2+2)/(x^2+1))^(1/2)+1/2*(a-2*b)* 
b^2*(x^2+2)^(1/2)*EllipticPi(x/(x^2+1)^(1/2),1-b/a,1/2*2^(1/2))*2^(1/2)/a/ 
(a-b)^3/(x^2+1)^(1/2)/((x^2+2)/(x^2+1))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.52 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.50 \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx=\frac {4 a^2 x \sqrt {1+x^2} \sqrt {2+x^2}-7 a b x \sqrt {1+x^2} \sqrt {2+x^2}+3 a^2 x^3 \sqrt {1+x^2} \sqrt {2+x^2}-6 a b x^3 \sqrt {1+x^2} \sqrt {2+x^2}+3 i a (a-2 b) \left (1+x^2\right )^2 E\left (\left .i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right )\right |2\right )-i a (a-b) \left (1+x^2\right )^2 \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )+3 i a b \operatorname {EllipticPi}\left (\frac {2 b}{a},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )-6 i b^2 \operatorname {EllipticPi}\left (\frac {2 b}{a},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )+6 i a b x^2 \operatorname {EllipticPi}\left (\frac {2 b}{a},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )-12 i b^2 x^2 \operatorname {EllipticPi}\left (\frac {2 b}{a},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )+3 i a b x^4 \operatorname {EllipticPi}\left (\frac {2 b}{a},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )-6 i b^2 x^4 \operatorname {EllipticPi}\left (\frac {2 b}{a},i \text {arcsinh}\left (\frac {x}{\sqrt {2}}\right ),2\right )}{3 a (a-b)^2 \left (1+x^2\right )^2} \] Input:

Integrate[Sqrt[2 + x^2]/((1 + x^2)^(5/2)*(a + b*x^2)),x]
 

Output:

(4*a^2*x*Sqrt[1 + x^2]*Sqrt[2 + x^2] - 7*a*b*x*Sqrt[1 + x^2]*Sqrt[2 + x^2] 
 + 3*a^2*x^3*Sqrt[1 + x^2]*Sqrt[2 + x^2] - 6*a*b*x^3*Sqrt[1 + x^2]*Sqrt[2 
+ x^2] + (3*I)*a*(a - 2*b)*(1 + x^2)^2*EllipticE[I*ArcSinh[x/Sqrt[2]], 2] 
- I*a*(a - b)*(1 + x^2)^2*EllipticF[I*ArcSinh[x/Sqrt[2]], 2] + (3*I)*a*b*E 
llipticPi[(2*b)/a, I*ArcSinh[x/Sqrt[2]], 2] - (6*I)*b^2*EllipticPi[(2*b)/a 
, I*ArcSinh[x/Sqrt[2]], 2] + (6*I)*a*b*x^2*EllipticPi[(2*b)/a, I*ArcSinh[x 
/Sqrt[2]], 2] - (12*I)*b^2*x^2*EllipticPi[(2*b)/a, I*ArcSinh[x/Sqrt[2]], 2 
] + (3*I)*a*b*x^4*EllipticPi[(2*b)/a, I*ArcSinh[x/Sqrt[2]], 2] - (6*I)*b^2 
*x^4*EllipticPi[(2*b)/a, I*ArcSinh[x/Sqrt[2]], 2])/(3*a*(a - b)^2*(1 + x^2 
)^2)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {421, 25, 401, 25, 400, 313, 320, 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.

\(\displaystyle \int \frac {\sqrt {x^2+2}}{\left (x^2+1\right )^{5/2} \left (a+b x^2\right )} \, dx\)

\(\Big \downarrow \) 421

\(\displaystyle \frac {b^2 \int \frac {\sqrt {x^2+2}}{\sqrt {x^2+1} \left (b x^2+a\right )}dx}{(a-b)^2}-\frac {\int -\frac {\sqrt {x^2+2} \left (-b x^2+a-2 b\right )}{\left (x^2+1\right )^{5/2}}dx}{(a-b)^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^2 \int \frac {\sqrt {x^2+2}}{\sqrt {x^2+1} \left (b x^2+a\right )}dx}{(a-b)^2}+\frac {\int \frac {\sqrt {x^2+2} \left (-b x^2+a-2 b\right )}{\left (x^2+1\right )^{5/2}}dx}{(a-b)^2}\)

\(\Big \downarrow \) 401

\(\displaystyle \frac {b^2 \int \frac {\sqrt {x^2+2}}{\sqrt {x^2+1} \left (b x^2+a\right )}dx}{(a-b)^2}+\frac {\frac {x \sqrt {x^2+2} (a-b)}{3 \left (x^2+1\right )^{3/2}}-\frac {1}{3} \int -\frac {(a-4 b) x^2+2 (2 a-5 b)}{\left (x^2+1\right )^{3/2} \sqrt {x^2+2}}dx}{(a-b)^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b^2 \int \frac {\sqrt {x^2+2}}{\sqrt {x^2+1} \left (b x^2+a\right )}dx}{(a-b)^2}+\frac {\frac {1}{3} \int \frac {(a-4 b) x^2+2 (2 a-5 b)}{\left (x^2+1\right )^{3/2} \sqrt {x^2+2}}dx+\frac {\sqrt {x^2+2} x (a-b)}{3 \left (x^2+1\right )^{3/2}}}{(a-b)^2}\)

\(\Big \downarrow \) 400

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

\(\Big \downarrow \) 313

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

\(\Big \downarrow \) 320

\(\displaystyle \frac {b^2 \int \frac {\sqrt {x^2+2}}{\sqrt {x^2+1} \left (b x^2+a\right )}dx}{(a-b)^2}+\frac {\frac {1}{3} \left (\frac {3 \sqrt {2} \sqrt {x^2+2} (a-2 b) E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}-\frac {\sqrt {2} \sqrt {x^2+2} (a-b) \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}\right )+\frac {\sqrt {x^2+2} x (a-b)}{3 \left (x^2+1\right )^{3/2}}}{(a-b)^2}\)

\(\Big \downarrow \) 414

\(\displaystyle \frac {2 b^2 \sqrt {x^2+1} \operatorname {EllipticPi}\left (1-\frac {2 b}{a},\arctan \left (\frac {x}{\sqrt {2}}\right ),-1\right )}{a \sqrt {\frac {x^2+1}{x^2+2}} \sqrt {x^2+2} (a-b)^2}+\frac {\frac {1}{3} \left (\frac {3 \sqrt {2} \sqrt {x^2+2} (a-2 b) E\left (\arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}-\frac {\sqrt {2} \sqrt {x^2+2} (a-b) \operatorname {EllipticF}\left (\arctan (x),\frac {1}{2}\right )}{\sqrt {x^2+1} \sqrt {\frac {x^2+2}{x^2+1}}}\right )+\frac {\sqrt {x^2+2} x (a-b)}{3 \left (x^2+1\right )^{3/2}}}{(a-b)^2}\)

Input:

Int[Sqrt[2 + x^2]/((1 + x^2)^(5/2)*(a + b*x^2)),x]
 

Output:

(((a - b)*x*Sqrt[2 + x^2])/(3*(1 + x^2)^(3/2)) + ((3*Sqrt[2]*(a - 2*b)*Sqr 
t[2 + x^2]*EllipticE[ArcTan[x], 1/2])/(Sqrt[1 + x^2]*Sqrt[(2 + x^2)/(1 + x 
^2)]) - (Sqrt[2]*(a - b)*Sqrt[2 + x^2]*EllipticF[ArcTan[x], 1/2])/(Sqrt[1 
+ x^2]*Sqrt[(2 + x^2)/(1 + x^2)]))/3)/(a - b)^2 + (2*b^2*Sqrt[1 + x^2]*Ell 
ipticPi[1 - (2*b)/a, ArcTan[x/Sqrt[2]], -1])/(a*(a - b)^2*Sqrt[(1 + x^2)/( 
2 + x^2)]*Sqrt[2 + x^2])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 320

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(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] /; Fre 
eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 400

Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)^ 
(3/2)), x_Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(Sqrt[a + b*x^2]* 
Sqrt[c + d*x^2]), x], x] - Simp[(d*e - c*f)/(b*c - a*d)   Int[Sqrt[a + b*x^ 
2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && PosQ[b/a] & 
& PosQ[d/c]
 

rule 401

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
q/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1))   Int[(a + b*x^2)^(p + 1)*( 
c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + 
(b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L 
tQ[p, -1] && GtQ[q, 0]
 

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 421

Int[(((c_) + (d_.)*(x_)^2)^(q_)*((e_) + (f_.)*(x_)^2)^(r_))/((a_) + (b_.)*( 
x_)^2), x_Symbol] :> Simp[b^2/(b*c - a*d)^2   Int[(c + d*x^2)^(q + 2)*((e + 
 f*x^2)^r/(a + b*x^2)), x], x] - Simp[d/(b*c - a*d)^2   Int[(c + d*x^2)^q*( 
e + f*x^2)^r*(2*b*c - a*d + b*d*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, r} 
, x] && LtQ[q, -1]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 6.85 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.58

method	result	size
elliptic	\(\frac {\sqrt {\left (x^{2}+1\right ) \left (x^{2}+2\right )}\, \left (\frac {x \sqrt {x^{4}+3 x^{2}+2}}{3 \left (a -b \right ) \left (x^{2}+1\right )^{2}}+\frac {\left (x^{2}+2\right ) x \left (a -2 b \right )}{\left (a -b \right )^{2} \sqrt {\left (x^{2}+1\right ) \left (x^{2}+2\right )}}-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{6 \sqrt {x^{4}+3 x^{2}+2}\, \left (a -b \right )}+\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, a \operatorname {EllipticE}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{2 \left (a -b \right )^{2} \sqrt {x^{4}+3 x^{2}+2}}-\frac {i \sqrt {2}\, \sqrt {2 x^{2}+4}\, \sqrt {x^{2}+1}\, b \operatorname {EllipticE}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right )}{\left (a -b \right )^{2} \sqrt {x^{4}+3 x^{2}+2}}+\frac {i b \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \operatorname {EllipticPi}\left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{\left (a -b \right )^{2} \sqrt {x^{4}+3 x^{2}+2}}-\frac {2 i b^{2} \sqrt {2}\, \sqrt {1+\frac {x^{2}}{2}}\, \sqrt {x^{2}+1}\, \operatorname {EllipticPi}\left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right )}{\left (a -b \right )^{2} a \sqrt {x^{4}+3 x^{2}+2}}\right )}{\sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}\)	\(372\)
default	\(\frac {3 i \operatorname {EllipticE}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a^{2} x^{2} \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}-i \operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a^{2} x^{2} \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}+3 i \operatorname {EllipticPi}\left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) a b \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}-i \operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a^{2} \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}+i \operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a b \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}-6 i \operatorname {EllipticE}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a b \,x^{2} \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}+3 a^{2} x^{5}-6 a b \,x^{5}+i \operatorname {EllipticF}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a b \,x^{2} \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}+3 i \operatorname {EllipticPi}\left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) a b \,x^{2} \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}+3 i \operatorname {EllipticE}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a^{2} \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}-6 i \operatorname {EllipticPi}\left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) b^{2} x^{2} \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}-6 i \operatorname {EllipticPi}\left (\frac {i x \sqrt {2}}{2}, \frac {2 b}{a}, \sqrt {2}\right ) b^{2} \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}-6 i \operatorname {EllipticE}\left (\frac {i x \sqrt {2}}{2}, \sqrt {2}\right ) a b \sqrt {x^{2}+1}\, \sqrt {x^{2}+2}+10 a^{2} x^{3}-19 a b \,x^{3}+8 a^{2} x -14 a b x}{3 \sqrt {x^{2}+2}\, \left (a -b \right )^{2} a \left (x^{2}+1\right )^{\frac {3}{2}}}\)	\(477\)

Input:

int((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

((x^2+1)*(x^2+2))^(1/2)/(x^2+1)^(1/2)/(x^2+2)^(1/2)*(1/3*x/(a-b)*(x^4+3*x^ 
2+2)^(1/2)/(x^2+1)^2+(x^2+2)*x*(a-2*b)/(a-b)^2/((x^2+1)*(x^2+2))^(1/2)-1/6 
*I*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticF(1/2 
*I*x*2^(1/2),2^(1/2))/(a-b)+1/2*I/(a-b)^2*2^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^ 
(1/2)/(x^4+3*x^2+2)^(1/2)*a*EllipticE(1/2*I*x*2^(1/2),2^(1/2))-I/(a-b)^2*2 
^(1/2)*(2*x^2+4)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*b*EllipticE(1/2*I 
*x*2^(1/2),2^(1/2))+I/(a-b)^2*b*2^(1/2)*(1+1/2*x^2)^(1/2)*(x^2+1)^(1/2)/(x 
^4+3*x^2+2)^(1/2)*EllipticPi(1/2*I*x*2^(1/2),2*b/a,2^(1/2))-2*I/(a-b)^2*b^ 
2/a*2^(1/2)*(1+1/2*x^2)^(1/2)*(x^2+1)^(1/2)/(x^4+3*x^2+2)^(1/2)*EllipticPi 
(1/2*I*x*2^(1/2),2*b/a,2^(1/2)))
 

Fricas [F]

\[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx=\int { \frac {\sqrt {x^{2} + 2}}{{\left (b x^{2} + a\right )} {\left (x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x, algorithm="fricas")
 

Output:

integral(sqrt(x^2 + 2)*sqrt(x^2 + 1)/(b*x^8 + (a + 3*b)*x^6 + 3*(a + b)*x^ 
4 + (3*a + b)*x^2 + a), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx=\text {Timed out} \] Input:

integrate((x**2+2)**(1/2)/(x**2+1)**(5/2)/(b*x**2+a),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx=\int { \frac {\sqrt {x^{2} + 2}}{{\left (b x^{2} + a\right )} {\left (x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x, algorithm="maxima")
 

Output:

integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(5/2)), x)
 

Giac [F]

\[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx=\int { \frac {\sqrt {x^{2} + 2}}{{\left (b x^{2} + a\right )} {\left (x^{2} + 1\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x, algorithm="giac")
 

Output:

integrate(sqrt(x^2 + 2)/((b*x^2 + a)*(x^2 + 1)^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx=\int \frac {\sqrt {x^2+2}}{{\left (x^2+1\right )}^{5/2}\,\left (b\,x^2+a\right )} \,d x \] Input:

int((x^2 + 2)^(1/2)/((x^2 + 1)^(5/2)*(a + b*x^2)),x)
 

Output:

int((x^2 + 2)^(1/2)/((x^2 + 1)^(5/2)*(a + b*x^2)), x)
 

Reduce [F]

\[ \int \frac {\sqrt {2+x^2}}{\left (1+x^2\right )^{5/2} \left (a+b x^2\right )} \, dx=\int \frac {\sqrt {x^{2}+1}\, \sqrt {x^{2}+2}}{b \,x^{8}+a \,x^{6}+3 b \,x^{6}+3 a \,x^{4}+3 b \,x^{4}+3 a \,x^{2}+b \,x^{2}+a}d x \] Input:

int((x^2+2)^(1/2)/(x^2+1)^(5/2)/(b*x^2+a),x)
 

Output:

int((sqrt(x**2 + 1)*sqrt(x**2 + 2))/(a*x**6 + 3*a*x**4 + 3*a*x**2 + a + b* 
x**8 + 3*b*x**6 + 3*b*x**4 + b*x**2),x)