\(\int \frac {1+2 x^2}{1+3 x^2+4 x^4} \, dx\) [44]

Optimal result

Integrand size = 22, antiderivative size = 38 \[ \int \frac {1+2 x^2}{1+3 x^2+4 x^4} \, dx=-\frac {\arctan \left (\frac {1-4 x}{\sqrt {7}}\right )}{\sqrt {7}}+\frac {\arctan \left (\frac {1+4 x}{\sqrt {7}}\right )}{\sqrt {7}} \]

[Out]

-1/7*arctan(1/7*(1-4*x)*7^(1/2))*7^(1/2)+1/7*arctan(1/7*(1+4*x)*7^(1/2))*7^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1175, 632, 210} \[ \int \frac {1+2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {\arctan \left (\frac {4 x+1}{\sqrt {7}}\right )}{\sqrt {7}}-\frac {\arctan \left (\frac {1-4 x}{\sqrt {7}}\right )}{\sqrt {7}} \]

[In]

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

[Out]

-(ArcTan[(1 - 4*x)/Sqrt[7]]/Sqrt[7]) + ArcTan[(1 + 4*x)/Sqrt[7]]/Sqrt[7]

Rule 210

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1175

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e) - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], 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[2*(d/e) - b/c, 0] || ( !Lt
Q[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \frac {1}{\frac {1}{2}-\frac {x}{2}+x^2} \, dx+\frac {1}{4} \int \frac {1}{\frac {1}{2}+\frac {x}{2}+x^2} \, dx \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{-\frac {7}{4}-x^2} \, dx,x,-\frac {1}{2}+2 x\right )\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{-\frac {7}{4}-x^2} \, dx,x,\frac {1}{2}+2 x\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {1-4 x}{\sqrt {7}}\right )}{\sqrt {7}}+\frac {\tan ^{-1}\left (\frac {1+4 x}{\sqrt {7}}\right )}{\sqrt {7}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.55 \[ \int \frac {1+2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {\left (-i+\sqrt {7}\right ) \arctan \left (\frac {2 x}{\sqrt {\frac {1}{2} \left (3-i \sqrt {7}\right )}}\right )}{\sqrt {42-14 i \sqrt {7}}}+\frac {\left (i+\sqrt {7}\right ) \arctan \left (\frac {2 x}{\sqrt {\frac {1}{2} \left (3+i \sqrt {7}\right )}}\right )}{\sqrt {42+14 i \sqrt {7}}} \]

[In]

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

[Out]

((-I + Sqrt[7])*ArcTan[(2*x)/Sqrt[(3 - I*Sqrt[7])/2]])/Sqrt[42 - (14*I)*Sqrt[7]] + ((I + Sqrt[7])*ArcTan[(2*x)
/Sqrt[(3 + I*Sqrt[7])/2]])/Sqrt[42 + (14*I)*Sqrt[7]]

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\sqrt {7}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {7}}{7}\right )}{7}+\frac {\arctan \left (\frac {\left (1+4 x \right ) \sqrt {7}}{7}\right ) \sqrt {7}}{7}\)	\(34\)
risch	\(\frac {\sqrt {7}\, \arctan \left (\frac {2 x \sqrt {7}}{7}\right )}{7}+\frac {\sqrt {7}\, \arctan \left (\frac {4 x^{3} \sqrt {7}}{7}+\frac {5 x \sqrt {7}}{7}\right )}{7}\)	\(35\)

[In]

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

[Out]

1/7*7^(1/2)*arctan(1/7*(4*x-1)*7^(1/2))+1/7*arctan(1/7*(1+4*x)*7^(1/2))*7^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {1+2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {1}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (4 \, x^{3} + 5 \, x\right )}\right ) + \frac {1}{7} \, \sqrt {7} \arctan \left (\frac {2}{7} \, \sqrt {7} x\right ) \]

[In]

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

[Out]

1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x^3 + 5*x)) + 1/7*sqrt(7)*arctan(2/7*sqrt(7)*x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {1+2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {\sqrt {7} \cdot \left (2 \operatorname {atan}{\left (\frac {2 \sqrt {7} x}{7} \right )} + 2 \operatorname {atan}{\left (\frac {4 \sqrt {7} x^{3}}{7} + \frac {5 \sqrt {7} x}{7} \right )}\right )}{14} \]

[In]

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

[Out]

sqrt(7)*(2*atan(2*sqrt(7)*x/7) + 2*atan(4*sqrt(7)*x**3/7 + 5*sqrt(7)*x/7))/14

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {1+2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {1}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (4 \, x + 1\right )}\right ) + \frac {1}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (4 \, x - 1\right )}\right ) \]

[In]

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

[Out]

1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x + 1)) + 1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x - 1))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87 \[ \int \frac {1+2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {1}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (4 \, x + 1\right )}\right ) + \frac {1}{7} \, \sqrt {7} \arctan \left (\frac {1}{7} \, \sqrt {7} {\left (4 \, x - 1\right )}\right ) \]

[In]

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

[Out]

1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x + 1)) + 1/7*sqrt(7)*arctan(1/7*sqrt(7)*(4*x - 1))

Mupad [B] (verification not implemented)

Time = 13.75 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {1+2 x^2}{1+3 x^2+4 x^4} \, dx=\frac {\sqrt {7}\,\left (\mathrm {atan}\left (\frac {4\,\sqrt {7}\,x^3}{7}+\frac {5\,\sqrt {7}\,x}{7}\right )+\mathrm {atan}\left (\frac {2\,\sqrt {7}\,x}{7}\right )\right )}{7} \]

[In]

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

[Out]

(7^(1/2)*(atan((5*7^(1/2)*x)/7 + (4*7^(1/2)*x^3)/7) + atan((2*7^(1/2)*x)/7)))/7