\(\int \frac {(1+x^2)^3}{\sqrt {1+x^2+x^4}} \, dx\) [232]

Optimal result

Integrand size = 20, antiderivative size = 159 \[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\frac {11}{15} x \sqrt {1+x^2+x^4}+\frac {1}{5} x^3 \sqrt {1+x^2+x^4}+\frac {14 x \sqrt {1+x^2+x^4}}{15 \left (1+x^2\right )}-\frac {14 \left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{15 \sqrt {1+x^2+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{5 \sqrt {1+x^2+x^4}} \]

[Out]

11/15*x*(x^4+x^2+1)^(1/2)+1/5*x^3*(x^4+x^2+1)^(1/2)+14/15*x*(x^4+x^2+1)^(1/2)/(x^2+1)-14/15*(x^2+1)*(cos(2*arc
tan(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)+3/5*(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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1220, 1693, 1211, 1117, 1209} \[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\frac {3 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{4}\right )}{5 \sqrt {x^4+x^2+1}}-\frac {14 \left (x^2+1\right ) \sqrt {\frac {x^4+x^2+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{4}\right .\right )}{15 \sqrt {x^4+x^2+1}}+\frac {14 \sqrt {x^4+x^2+1} x}{15 \left (x^2+1\right )}+\frac {11}{15} \sqrt {x^4+x^2+1} x+\frac {1}{5} \sqrt {x^4+x^2+1} x^3 \]

[In]

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

[Out]

(11*x*Sqrt[1 + x^2 + x^4])/15 + (x^3*Sqrt[1 + x^2 + x^4])/5 + (14*x*Sqrt[1 + x^2 + x^4])/(15*(1 + x^2)) - (14*
(1 + x^2)*Sqrt[(1 + x^2 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/4])/(15*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])/(5*Sqrt[1 + x^2 + x^4])

Rule 1117

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 1209

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> 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 1211

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

Rule 1220

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[e^q*x^(2*q - 3)*((
a + b*x^2 + c*x^4)^(p + 1)/(c*(4*p + 2*q + 1))), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + b*x^2 + c*x^4)^p*Ex
pandToSum[c*(4*p + 2*q + 1)*(d + e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - b*(2*p + 2*q - 1)*e^q*x^(2*q - 2) -
c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && IGtQ[q, 1]

Rule 1693

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{q = Expon[Pq, x^2], e = Coeff[Pq, x^2,
Expon[Pq, x^2]]}, Simp[e*x^(2*q - 3)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(2*q + 4*p + 1))), x] + Dist[1/(c*(2*q +
4*p + 1)), Int[(a + b*x^2 + c*x^4)^p*ExpandToSum[c*(2*q + 4*p + 1)*Pq - a*e*(2*q - 3)*x^(2*q - 4) - b*e*(2*q
+ 2*p - 1)*x^(2*q - 2) - c*e*(2*q + 4*p + 1)*x^(2*q), x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2]
&& Expon[Pq, x^2] > 1 && NeQ[b^2 - 4*a*c, 0] &&  !LtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^3 \sqrt {1+x^2+x^4}+\frac {1}{5} \int \frac {5+12 x^2+11 x^4}{\sqrt {1+x^2+x^4}} \, dx \\ & = \frac {11}{15} x \sqrt {1+x^2+x^4}+\frac {1}{5} x^3 \sqrt {1+x^2+x^4}+\frac {1}{15} \int \frac {4+14 x^2}{\sqrt {1+x^2+x^4}} \, dx \\ & = \frac {11}{15} x \sqrt {1+x^2+x^4}+\frac {1}{5} x^3 \sqrt {1+x^2+x^4}-\frac {14}{15} \int \frac {1-x^2}{\sqrt {1+x^2+x^4}} \, dx+\frac {6}{5} \int \frac {1}{\sqrt {1+x^2+x^4}} \, dx \\ & = \frac {11}{15} x \sqrt {1+x^2+x^4}+\frac {1}{5} x^3 \sqrt {1+x^2+x^4}+\frac {14 x \sqrt {1+x^2+x^4}}{15 \left (1+x^2\right )}-\frac {14 \left (1+x^2\right ) \sqrt {\frac {1+x^2+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{15 \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}} F\left (2 \tan ^{-1}(x)|\frac {1}{4}\right )}{5 \sqrt {1+x^2+x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.15 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.99 \[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\frac {x \left (11+14 x^2+14 x^4+3 x^6\right )+14 \sqrt [3]{-1} \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} E\left (i \text {arcsinh}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )+2 \sqrt [3]{-1} \left (-7+2 \sqrt [3]{-1}\right ) \sqrt {1+\sqrt [3]{-1} x^2} \sqrt {1-(-1)^{2/3} x^2} \operatorname {EllipticF}\left (i \text {arcsinh}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )}{15 \sqrt {1+x^2+x^4}} \]

[In]

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

[Out]

(x*(11 + 14*x^2 + 14*x^4 + 3*x^6) + 14*(-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)] + 2*(-1)^(1/3)*(-7 + 2*(-1)^(1/3))*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)
^(2/3)*x^2]*EllipticF[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)])/(15*Sqrt[1 + x^2 + x^4])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.12 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.42

method	result	size
risch	\(\frac {x \left (3 x^{2}+11\right ) \sqrt {x^{4}+x^{2}+1}}{15}+\frac {8 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}-\frac {56 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\)	\(225\)
default	\(\frac {8 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {x^{3} \sqrt {x^{4}+x^{2}+1}}{5}+\frac {11 x \sqrt {x^{4}+x^{2}+1}}{15}-\frac {56 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\)	\(233\)
elliptic	\(\frac {8 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {x^{3} \sqrt {x^{4}+x^{2}+1}}{5}+\frac {11 x \sqrt {x^{4}+x^{2}+1}}{15}-\frac {56 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \left (F\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )-E\left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )\right )}{15 \sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}\, \left (1+i \sqrt {3}\right )}\)	\(233\)

[In]

int((x^2+1)^3/(x^4+x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/15*x*(3*x^2+11)*(x^4+x^2+1)^(1/2)+8/15/(-2+2*I*3^(1/2))^(1/2)*(1-(-1/2+1/2*I*3^(1/2))*x^2)^(1/2)*(1-(-1/2-1/
2*I*3^(1/2))*x^2)^(1/2)/(x^4+x^2+1)^(1/2)*EllipticF(1/2*x*(-2+2*I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2))-5
6/15/(-2+2*I*3^(1/2))^(1/2)*(1-(-1/2+1/2*I*3^(1/2))*x^2)^(1/2)*(1-(-1/2-1/2*I*3^(1/2))*x^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))-EllipticE(1/2*x*(-2+2*
I*3^(1/2))^(1/2),1/2*(-2+2*I*3^(1/2))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.76 \[ \int \frac {\left (1+x^2\right )^3}{\sqrt {1+x^2+x^4}} \, dx=\frac {7 \, \sqrt {2} {\left (\sqrt {-3} x - x\right )} \sqrt {\sqrt {-3} - 1} E(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) - \sqrt {2} {\left (5 \, \sqrt {-3} x - 9 \, x\right )} \sqrt {\sqrt {-3} - 1} F(\arcsin \left (\frac {\sqrt {2} \sqrt {\sqrt {-3} - 1}}{2 \, x}\right )\,|\,\frac {1}{2} \, \sqrt {-3} - \frac {1}{2}) + 2 \, {\left (3 \, x^{4} + 11 \, x^{2} + 14\right )} \sqrt {x^{4} + x^{2} + 1}}{30 \, x} \]

[In]

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

[Out]

1/30*(7*sqrt(2)*(sqrt(-3)*x - x)*sqrt(sqrt(-3) - 1)*elliptic_e(arcsin(1/2*sqrt(2)*sqrt(sqrt(-3) - 1)/x), 1/2*s
qrt(-3) - 1/2) - sqrt(2)*(5*sqrt(-3)*x - 9*x)*sqrt(sqrt(-3) - 1)*elliptic_f(arcsin(1/2*sqrt(2)*sqrt(sqrt(-3) -
1)/x), 1/2*sqrt(-3) - 1/2) + 2*(3*x^4 + 11*x^2 + 14)*sqrt(x^4 + x^2 + 1))/x

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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