\(\int \frac {2+3 \sqrt {2}+2 (3+\sqrt {2}) x^2}{(3+2 x^2) \sqrt {1+2 x^2+2 x^4}} \, dx\) [188]

Optimal result

Integrand size = 47, antiderivative size = 148 \[ \int \frac {2+3 \sqrt {2}+2 \left (3+\sqrt {2}\right ) x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=-\frac {7 \arctan \left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {1+2 x^2+2 x^4}}\right )}{2 \sqrt {15}}+\frac {\left (3+\sqrt {2}\right )^2 \left (1+\sqrt {2} x^2\right ) \sqrt {\frac {1+2 x^2+2 x^4}{\left (1+\sqrt {2} x^2\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{24} \left (12-11 \sqrt {2}\right ),2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{12 \sqrt [4]{2} \sqrt {1+2 x^2+2 x^4}} \] Output:

-7/30*arctan(1/3*15^(1/2)*x/(2*x^4+2*x^2+1)^(1/2))*15^(1/2)+1/24*(3+2^(1/2 
))^2*(1+x^2*2^(1/2))*((2*x^4+2*x^2+1)/(1+x^2*2^(1/2))^2)^(1/2)*EllipticPi( 
sin(2*arctan(2^(1/4)*x)),1/2-11/24*2^(1/2),1/2*(2-2^(1/2))^(1/2))*2^(3/4)/ 
(2*x^4+2*x^2+1)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.51 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.73 \[ \int \frac {2+3 \sqrt {2}+2 \left (3+\sqrt {2}\right ) x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\frac {(1-i)^{3/2} \sqrt {1+(1-i) x^2} \sqrt {1+(1+i) x^2} \left (3 \left (3+\sqrt {2}\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-i} x\right ),i\right )-7 \operatorname {EllipticPi}\left (\frac {1}{3}+\frac {i}{3},i \text {arcsinh}\left (\sqrt {1-i} x\right ),i\right )\right )}{6 \sqrt {1+2 x^2+2 x^4}} \] Input:

Integrate[(2 + 3*Sqrt[2] + 2*(3 + Sqrt[2])*x^2)/((3 + 2*x^2)*Sqrt[1 + 2*x^ 
2 + 2*x^4]),x]
 

Output:

((1 - I)^(3/2)*Sqrt[1 + (1 - I)*x^2]*Sqrt[1 + (1 + I)*x^2]*(3*(3 + Sqrt[2] 
)*EllipticF[I*ArcSinh[Sqrt[1 - I]*x], I] - 7*EllipticPi[1/3 + I/3, I*ArcSi 
nh[Sqrt[1 - I]*x], I]))/(6*Sqrt[1 + 2*x^2 + 2*x^4])
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {2220}

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.

\(\displaystyle \int \frac {2 \left (3+\sqrt {2}\right ) x^2+3 \sqrt {2}+2}{\left (2 x^2+3\right ) \sqrt {2 x^4+2 x^2+1}} \, dx\)

\(\Big \downarrow \) 2220

\(\displaystyle \frac {\left (3+\sqrt {2}\right )^2 \left (\sqrt {2} x^2+1\right ) \sqrt {\frac {2 x^4+2 x^2+1}{\left (\sqrt {2} x^2+1\right )^2}} \operatorname {EllipticPi}\left (\frac {1}{24} \left (12-11 \sqrt {2}\right ),2 \arctan \left (\sqrt [4]{2} x\right ),\frac {1}{4} \left (2-\sqrt {2}\right )\right )}{12 \sqrt [4]{2} \sqrt {2 x^4+2 x^2+1}}-\frac {7 \arctan \left (\frac {\sqrt {\frac {5}{3}} x}{\sqrt {2 x^4+2 x^2+1}}\right )}{2 \sqrt {15}}\)

Input:

Int[(2 + 3*Sqrt[2] + 2*(3 + Sqrt[2])*x^2)/((3 + 2*x^2)*Sqrt[1 + 2*x^2 + 2* 
x^4]),x]
 

Output:

(-7*ArcTan[(Sqrt[5/3]*x)/Sqrt[1 + 2*x^2 + 2*x^4]])/(2*Sqrt[15]) + ((3 + Sq 
rt[2])^2*(1 + Sqrt[2]*x^2)*Sqrt[(1 + 2*x^2 + 2*x^4)/(1 + Sqrt[2]*x^2)^2]*E 
llipticPi[(12 - 11*Sqrt[2])/24, 2*ArcTan[2^(1/4)*x], (2 - Sqrt[2])/4])/(12 
*2^(1/4)*Sqrt[1 + 2*x^2 + 2*x^4])
 

Defintions of rubi rules used

rule 2220

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + 
 (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(A 
rcTan[Rt[-b + c*(d/e) + a*(e/d), 2]*(x/Sqrt[a + b*x^2 + c*x^4])]/(2*d*e*Rt[ 
-b + c*(d/e) + a*(e/d), 2])), x] + Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a 
+ b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(4*d*e*q*Sqrt[a + b*x^2 + c*x^4]))*El 
lipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*ArcTan[q*x], 1/2 - b/(4*a*q^2)], x]] 
 /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] & 
& EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && PosQ[-b + c*(d/e) + a*(e/d)]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.58 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.34

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {1+\left (1-i\right ) x^{2}}\, \sqrt {1+\left (1+i\right ) x^{2}}\, \operatorname {EllipticF}\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {3 \sqrt {1+\left (1-i\right ) x^{2}}\, \sqrt {1+\left (1+i\right ) x^{2}}\, \operatorname {EllipticF}\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {7 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \operatorname {EllipticPi}\left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{3 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\)	\(199\)
elliptic	\(\frac {3 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}+\frac {\sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-1+i}, \frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {2}}{\sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}-\frac {7 \sqrt {-i x^{2}+x^{2}+1}\, \sqrt {i x^{2}+x^{2}+1}\, \operatorname {EllipticPi}\left (x \sqrt {-1+i}, \frac {1}{3}+\frac {i}{3}, \frac {\sqrt {-1-i}}{\sqrt {-1+i}}\right )}{3 \sqrt {-1+i}\, \sqrt {2 x^{4}+2 x^{2}+1}}\)	\(207\)

Input:

int((2+3*2^(1/2)+2*(3+2^(1/2))*x^2)/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2),x,meth 
od=_RETURNVERBOSE)
 

Output:

2^(1/2)/(-1+I)^(1/2)*(1+(1-I)*x^2)^(1/2)*(1+(1+I)*x^2)^(1/2)/(2*x^4+2*x^2+ 
1)^(1/2)*EllipticF(x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))+3/(-1+I)^(1/2 
)*(1+(1-I)*x^2)^(1/2)*(1+(1+I)*x^2)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticF( 
x*(-1+I)^(1/2),1/2*2^(1/2)+1/2*I*2^(1/2))-7/3/(-1+I)^(1/2)*(1+x^2-I*x^2)^( 
1/2)*(1+x^2+I*x^2)^(1/2)/(2*x^4+2*x^2+1)^(1/2)*EllipticPi(x*(-1+I)^(1/2),1 
/3+1/3*I,(-1-I)^(1/2)/(-1+I)^(1/2))
 

Fricas [F]

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

integrate((2+3*2^(1/2)+2*(3+2^(1/2))*x^2)/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2), 
x, algorithm="fricas")
 

Output:

integral(sqrt(2*x^4 + 2*x^2 + 1)*(6*x^2 + sqrt(2)*(2*x^2 + 3) + 2)/(4*x^6 
+ 10*x^4 + 8*x^2 + 3), x)
 

Sympy [F]

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

integrate((2+3*2**(1/2)+2*(3+2**(1/2))*x**2)/(2*x**2+3)/(2*x**4+2*x**2+1)* 
*(1/2),x)
 

Output:

Integral((2*sqrt(2)*x**2 + 6*x**2 + 2 + 3*sqrt(2))/((2*x**2 + 3)*sqrt(2*x* 
*4 + 2*x**2 + 1)), x)
                                                                                    
                                                                                    
 

Maxima [F]

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

integrate((2+3*2^(1/2)+2*(3+2^(1/2))*x^2)/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2), 
x, algorithm="maxima")
 

Output:

integrate((2*x^2*(sqrt(2) + 3) + 3*sqrt(2) + 2)/(sqrt(2*x^4 + 2*x^2 + 1)*( 
2*x^2 + 3)), x)
 

Giac [F]

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

integrate((2+3*2^(1/2)+2*(3+2^(1/2))*x^2)/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2), 
x, algorithm="giac")
 

Output:

integrate((2*x^2*(sqrt(2) + 3) + 3*sqrt(2) + 2)/(sqrt(2*x^4 + 2*x^2 + 1)*( 
2*x^2 + 3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {2+3 \sqrt {2}+2 \left (3+\sqrt {2}\right ) x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=\int \frac {3\,\sqrt {2}+2\,x^2\,\left (\sqrt {2}+3\right )+2}{\left (2\,x^2+3\right )\,\sqrt {2\,x^4+2\,x^2+1}} \,d x \] Input:

int((3*2^(1/2) + 2*x^2*(2^(1/2) + 3) + 2)/((2*x^2 + 3)*(2*x^2 + 2*x^4 + 1) 
^(1/2)),x)
 

Output:

int((3*2^(1/2) + 2*x^2*(2^(1/2) + 3) + 2)/((2*x^2 + 3)*(2*x^2 + 2*x^4 + 1) 
^(1/2)), x)
 

Reduce [F]

\[ \int \frac {2+3 \sqrt {2}+2 \left (3+\sqrt {2}\right ) x^2}{\left (3+2 x^2\right ) \sqrt {1+2 x^2+2 x^4}} \, dx=3 \sqrt {2}\, \left (\int \frac {\sqrt {2 x^{4}+2 x^{2}+1}}{4 x^{6}+10 x^{4}+8 x^{2}+3}d x \right )+2 \sqrt {2}\, \left (\int \frac {\sqrt {2 x^{4}+2 x^{2}+1}\, x^{2}}{4 x^{6}+10 x^{4}+8 x^{2}+3}d x \right )+2 \left (\int \frac {\sqrt {2 x^{4}+2 x^{2}+1}}{4 x^{6}+10 x^{4}+8 x^{2}+3}d x \right )+6 \left (\int \frac {\sqrt {2 x^{4}+2 x^{2}+1}\, x^{2}}{4 x^{6}+10 x^{4}+8 x^{2}+3}d x \right ) \] Input:

int((2+3*2^(1/2)+2*(3+2^(1/2))*x^2)/(2*x^2+3)/(2*x^4+2*x^2+1)^(1/2),x)
 

Output:

3*sqrt(2)*int(sqrt(2*x**4 + 2*x**2 + 1)/(4*x**6 + 10*x**4 + 8*x**2 + 3),x) 
 + 2*sqrt(2)*int((sqrt(2*x**4 + 2*x**2 + 1)*x**2)/(4*x**6 + 10*x**4 + 8*x* 
*2 + 3),x) + 2*int(sqrt(2*x**4 + 2*x**2 + 1)/(4*x**6 + 10*x**4 + 8*x**2 + 
3),x) + 6*int((sqrt(2*x**4 + 2*x**2 + 1)*x**2)/(4*x**6 + 10*x**4 + 8*x**2 
+ 3),x)