3.3.28 \(\int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx\) [228]

3.3.28.1 Optimal result

Integrand size = 20, antiderivative size = 137 \[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\frac {x \sqrt {1+x^2+x^4}}{1+x^2}+\frac {1}{2} \arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {1+x^2+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{4 \sqrt {1+x^2+x^4}} \]

output

1/2*arctan(x/(x^4+x^2+1)^(1/2))+x*(x^4+x^2+1)^(1/2)/(x^2+1)-(x^2+1)*(cos(2 
*arctan(x))^2)^(1/2)/cos(2*arctan(x))*EllipticE(sin(2*arctan(x)),1/2)*((x^ 
4+x^2+1)/(x^2+1)^2)^(1/2)/(x^4+x^2+1)^(1/2)+3/4*(x^2+1)*(cos(2*arctan(x))^ 
2)^(1/2)/cos(2*arctan(x))*EllipticF(sin(2*arctan(x)),1/2)*((x^4+x^2+1)/(x^ 
2+1)^2)^(1/2)/(x^4+x^2+1)^(1/2)
 

3.3.28.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 8.90 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\frac {\sqrt [3]{-1} \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} \left (E\left (i \text {arcsinh}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )+\sqrt [3]{-1} \operatorname {EllipticPi}\left (\sqrt [3]{-1},i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )\right )}{\sqrt {1+x^2+x^4}} \]

input

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

output

((-1)^(1/3)*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^(2/3)*x^2]*(EllipticE[I 
*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)] - EllipticF[I*ArcSinh[(-1)^(5/6)*x], ( 
-1)^(2/3)] + (-1)^(1/3)*EllipticPi[(-1)^(1/3), I*ArcSinh[(-1)^(5/6)*x], (- 
1)^(2/3)]))/Sqrt[1 + x^2 + x^4]
 

3.3.28.3 Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1520, 1509, 2214, 1416, 2212, 216}

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.

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

\(\Big \downarrow \) 1520

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

\(\Big \downarrow \) 1509

\(\displaystyle \int \frac {x^2+2}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {x^4+x^2+1}}+\frac {\sqrt {x^4+x^2+1} x}{x^2+1}\)

\(\Big \downarrow \) 2214

\(\displaystyle \frac {3}{2} \int \frac {1}{\sqrt {x^4+x^2+1}}dx+\frac {1}{2} \int \frac {1-x^2}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {x^4+x^2+1}}+\frac {\sqrt {x^4+x^2+1} x}{x^2+1}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {1}{2} \int \frac {1-x^2}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {x^4+x^2+1}}+\frac {\sqrt {x^4+x^2+1} x}{x^2+1}\)

\(\Big \downarrow \) 2212

\(\displaystyle \frac {1}{2} \int \frac {1}{\frac {x^2}{x^4+x^2+1}+1}d\frac {x}{\sqrt {x^4+x^2+1}}+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {x^4+x^2+1}}+\frac {\sqrt {x^4+x^2+1} x}{x^2+1}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2} \arctan \left (\frac {x}{\sqrt {x^4+x^2+1}}\right )+\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{4 \sqrt {x^4+x^2+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{\sqrt {x^4+x^2+1}}+\frac {\sqrt {x^4+x^2+1} x}{x^2+1}\)

input

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

output

(x*Sqrt[1 + x^2 + x^4])/(1 + x^2) + ArcTan[x/Sqrt[1 + x^2 + x^4]]/2 - ((1 
+ x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/Sqrt 
[1 + x^2 + x^4] + (3*(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticF 
[2*ArcTan[x], 1/4])/(4*Sqrt[1 + x^2 + x^4])
 

3.3.28.3.1 Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1509

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

rule 1520

Int[Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]/((d_) + (e_.)*(x_)^2), x_Symbo 
l] :> Simp[-c/e^2   Int[(d - e*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] + Int[( 
2*a + b*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x] /; FreeQ[{a, b, c, d 
, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d 
^2 - a*e^2, 0]
 

rule 2212

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + 
 (c_.)*(x_)^4]), x_Symbol] :> Simp[A   Subst[Int[1/(d - (b*d - 2*a*e)*x^2), 
 x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B}, x] & 
& EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]
 

rule 2214

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + 
 (c_.)*(x_)^4]), x_Symbol] :> Simp[(B*d + A*e)/(2*d*e)   Int[1/Sqrt[a + b*x 
^2 + c*x^4], x], x] - Simp[(B*d - A*e)/(2*d*e)   Int[(d - e*x^2)/((d + e*x^ 
2)*Sqrt[a + b*x^2 + c*x^4]), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && E 
qQ[c*d^2 - a*e^2, 0] && NeQ[B*d + A*e, 0]
 

3.3.28.4 Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.49 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.14

method	result	size
default	\(-\frac {4 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {4 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {\sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \Pi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(293\)
elliptic	\(-\frac {4 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {4 \sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}+\frac {\sqrt {1+\frac {x^{2}}{2}-\frac {i x^{2} \sqrt {3}}{2}}\, \sqrt {1+\frac {x^{2}}{2}+\frac {i x^{2} \sqrt {3}}{2}}\, \Pi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , -\frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(293\)

input

int((x^4+x^2+1)^(1/2)/(x^2+1),x,method=_RETURNVERBOSE)
 

output

-4/(-2+2*I*3^(1/2))^(1/2)*(1+1/2*x^2-1/2*I*x^2*3^(1/2))^(1/2)*(1+1/2*x^2+1 
/2*I*x^2*3^(1/2))^(1/2)/(x^4+x^2+1)^(1/2)/(1+I*3^(1/2))*EllipticF(1/2*x*(- 
2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2))+4/(-2+2*I*3^(1/2))^(1/2)* 
(1+1/2*x^2-1/2*I*x^2*3^(1/2))^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(1/2)/(x 
^4+x^2+1)^(1/2)/(1+I*3^(1/2))*EllipticE(1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*( 
-2+2*I*3^(1/2))^(1/2))+1/(-1/2+1/2*I*3^(1/2))^(1/2)*(1+1/2*x^2-1/2*I*x^2*3 
^(1/2))^(1/2)*(1+1/2*x^2+1/2*I*x^2*3^(1/2))^(1/2)/(x^4+x^2+1)^(1/2)*Ellipt 
icPi((-1/2+1/2*I*3^(1/2))^(1/2)*x,-1/(-1/2+1/2*I*3^(1/2)),(-1/2-1/2*I*3^(1 
/2))^(1/2)/(-1/2+1/2*I*3^(1/2))^(1/2))
 

3.3.28.5 Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\frac {2 \, \sqrt {2} {\left (\sqrt {-3} x - x\right )} \sqrt {\sqrt {-3} - 1} E(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) - \sqrt {2} {\left (\sqrt {-3} x - 3 \, x\right )} \sqrt {\sqrt {-3} - 1} F(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) + 4 \, x \arctan \left (\frac {x}{\sqrt {x^{4} + x^{2} + 1}}\right ) + 8 \, \sqrt {x^{4} + x^{2} + 1}}{8 \, x} \]

input

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

output

1/8*(2*sqrt(2)*(sqrt(-3)*x - x)*sqrt(sqrt(-3) - 1)*elliptic_e(arcsin(1/2*s 
qrt(2)*sqrt(sqrt(-3) - 1)/x), 1/2*sqrt(-3) - 1/2) - sqrt(2)*(sqrt(-3)*x - 
3*x)*sqrt(sqrt(-3) - 1)*elliptic_f(arcsin(1/2*sqrt(2)*sqrt(sqrt(-3) - 1)/x 
), 1/2*sqrt(-3) - 1/2) + 4*x*arctan(x/sqrt(x^4 + x^2 + 1)) + 8*sqrt(x^4 + 
x^2 + 1))/x
 

3.3.28.6 Sympy [F]

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

input

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

output

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

3.3.28.7 Maxima [F]

\[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\int { \frac {\sqrt {x^{4} + x^{2} + 1}}{x^{2} + 1} \,d x } \]

input

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

output

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

3.3.28.8 Giac [F]

\[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\int { \frac {\sqrt {x^{4} + x^{2} + 1}}{x^{2} + 1} \,d x } \]

input

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

output

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

3.3.28.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {1+x^2+x^4}}{1+x^2} \, dx=\int \frac {\sqrt {x^4+x^2+1}}{x^2+1} \,d x \]

input

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

output

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