\(\int \frac {1+x^2}{1+5 x^2+x^4} \, dx\) [69]

Optimal result

Integrand size = 18, antiderivative size = 49 \[ \int \frac {1+x^2}{1+5 x^2+x^4} \, dx=\frac {\arctan \left (\sqrt {\frac {2}{5+\sqrt {21}}} x\right )}{\sqrt {7}}+\frac {\arctan \left (\sqrt {\frac {1}{2} \left (5+\sqrt {21}\right )} x\right )}{\sqrt {7}} \]

[Out]

1/7*arctan(x*2^(1/2)/(5+21^(1/2))^(1/2))*7^(1/2)+1/7*arctan(x*(1/2*7^(1/2)+1/2*3^(1/2)))*7^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1177, 209} \[ \int \frac {1+x^2}{1+5 x^2+x^4} \, dx=\frac {\arctan \left (\sqrt {\frac {2}{5+\sqrt {21}}} x\right )}{\sqrt {7}}+\frac {\arctan \left (\sqrt {\frac {1}{2} \left (5+\sqrt {21}\right )} x\right )}{\sqrt {7}} \]

[In]

Int[(1 + x^2)/(1 + 5*x^2 + x^4),x]

[Out]

ArcTan[Sqrt[2/(5 + Sqrt[21])]*x]/Sqrt[7] + ArcTan[Sqrt[(5 + Sqrt[21])/2]*x]/Sqrt[7]

Rule 209

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

Rule 1177

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{14} \left (7-\sqrt {21}\right ) \int \frac {1}{\frac {5}{2}-\frac {\sqrt {21}}{2}+x^2} \, dx+\frac {1}{14} \left (7+\sqrt {21}\right ) \int \frac {1}{\frac {5}{2}+\frac {\sqrt {21}}{2}+x^2} \, dx \\ & = \frac {\tan ^{-1}\left (\sqrt {\frac {2}{5+\sqrt {21}}} x\right )}{\sqrt {7}}+\frac {\tan ^{-1}\left (\sqrt {\frac {1}{2} \left (5+\sqrt {21}\right )} x\right )}{\sqrt {7}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.69 \[ \int \frac {1+x^2}{1+5 x^2+x^4} \, dx=\frac {\left (-3+\sqrt {21}\right ) \arctan \left (\sqrt {\frac {2}{5-\sqrt {21}}} x\right )}{\sqrt {42 \left (5-\sqrt {21}\right )}}+\frac {\left (3+\sqrt {21}\right ) \arctan \left (\sqrt {\frac {2}{5+\sqrt {21}}} x\right )}{\sqrt {42 \left (5+\sqrt {21}\right )}} \]

[In]

Integrate[(1 + x^2)/(1 + 5*x^2 + x^4),x]

[Out]

((-3 + Sqrt[21])*ArcTan[Sqrt[2/(5 - Sqrt[21])]*x])/Sqrt[42*(5 - Sqrt[21])] + ((3 + Sqrt[21])*ArcTan[Sqrt[2/(5
+ Sqrt[21])]*x])/Sqrt[42*(5 + Sqrt[21])]

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.71

method	result	size
risch	\(\frac {\sqrt {7}\, \arctan \left (\frac {x \sqrt {7}}{7}\right )}{7}+\frac {\sqrt {7}\, \arctan \left (\frac {x^{3} \sqrt {7}}{7}+\frac {6 x \sqrt {7}}{7}\right )}{7}\)	\(35\)
default	\(\frac {2 \left (3+\sqrt {21}\right ) \sqrt {21}\, \arctan \left (\frac {4 x}{2 \sqrt {7}+2 \sqrt {3}}\right )}{21 \left (2 \sqrt {7}+2 \sqrt {3}\right )}+\frac {2 \left (-3+\sqrt {21}\right ) \sqrt {21}\, \arctan \left (\frac {4 x}{2 \sqrt {7}-2 \sqrt {3}}\right )}{21 \left (2 \sqrt {7}-2 \sqrt {3}\right )}\)	\(82\)

[In]

int((x^2+1)/(x^4+5*x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/7*7^(1/2)*arctan(1/7*x*7^(1/2))+1/7*7^(1/2)*arctan(1/7*x^3*7^(1/2)+6/7*x*7^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.63 \[ \int \frac {1+x^2}{1+5 x^2+x^4} \, dx=\frac {1}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (x^{3} + 6 \, x\right )}\right ) + \frac {1}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} x\right ) \]

[In]

integrate((x^2+1)/(x^4+5*x^2+1),x, algorithm="fricas")

[Out]

1/7*sqrt(7)*arctan(1/7*sqrt(7)*(x^3 + 6*x)) + 1/7*sqrt(7)*arctan(1/7*sqrt(7)*x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {1+x^2}{1+5 x^2+x^4} \, dx=\frac {\sqrt {7} \cdot \left (2 \operatorname {atan}{\left (\frac {\sqrt {7} x}{7} \right )} + 2 \operatorname {atan}{\left (\frac {\sqrt {7} x^{3}}{7} + \frac {6 \sqrt {7} x}{7} \right )}\right )}{14} \]

[In]

integrate((x**2+1)/(x**4+5*x**2+1),x)

[Out]

sqrt(7)*(2*atan(sqrt(7)*x/7) + 2*atan(sqrt(7)*x**3/7 + 6*sqrt(7)*x/7))/14

Maxima [F]

\[ \int \frac {1+x^2}{1+5 x^2+x^4} \, dx=\int { \frac {x^{2} + 1}{x^{4} + 5 \, x^{2} + 1} \,d x } \]

[In]

integrate((x^2+1)/(x^4+5*x^2+1),x, algorithm="maxima")

[Out]

integrate((x^2 + 1)/(x^4 + 5*x^2 + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.53 \[ \int \frac {1+x^2}{1+5 x^2+x^4} \, dx=\frac {1}{14} \, \sqrt {7} {\left (\pi \mathrm {sgn}\left (x\right ) + 2 \, \arctan \left (\frac {\sqrt {7} {\left (x^{2} - 1\right )}}{7 \, x}\right )\right )} \]

[In]

integrate((x^2+1)/(x^4+5*x^2+1),x, algorithm="giac")

[Out]

1/14*sqrt(7)*(pi*sgn(x) + 2*arctan(1/7*sqrt(7)*(x^2 - 1)/x))

Mupad [B] (verification not implemented)

Time = 13.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.59 \[ \int \frac {1+x^2}{1+5 x^2+x^4} \, dx=\frac {\sqrt {7}\,\left (\mathrm {atan}\left (\frac {\sqrt {7}\,x^3}{7}+\frac {6\,\sqrt {7}\,x}{7}\right )+\mathrm {atan}\left (\frac {\sqrt {7}\,x}{7}\right )\right )}{7} \]

[In]

int((x^2 + 1)/(5*x^2 + x^4 + 1),x)

[Out]

(7^(1/2)*(atan((6*7^(1/2)*x)/7 + (7^(1/2)*x^3)/7) + atan((7^(1/2)*x)/7)))/7