\(\int \frac {1+2 x^2}{1+b x^2+4 x^4} \, dx\) [39]

Optimal result

Integrand size = 22, antiderivative size = 62 \[ \int \frac {1+2 x^2}{1+b x^2+4 x^4} \, dx=-\frac {\arctan \left (\frac {\sqrt {4-b}-4 x}{\sqrt {4+b}}\right )}{\sqrt {4+b}}+\frac {\arctan \left (\frac {\sqrt {4-b}+4 x}{\sqrt {4+b}}\right )}{\sqrt {4+b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 62, 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} \[ \int \frac {1+2 x^2}{1+b x^2+4 x^4} \, dx=\frac {\arctan \left (\frac {\sqrt {4-b}+4 x}{\sqrt {b+4}}\right )}{\sqrt {b+4}}-\frac {\arctan \left (\frac {\sqrt {4-b}-4 x}{\sqrt {b+4}}\right )}{\sqrt {b+4}} \]

[In]

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

[Out]

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(126\) vs. \(2(62)=124\).

Time = 0.05 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.03 \[ \int \frac {1+2 x^2}{1+b x^2+4 x^4} \, dx=\frac {\frac {\left (4-b+\sqrt {-16+b^2}\right ) \arctan \left (\frac {2 \sqrt {2} x}{\sqrt {b-\sqrt {-16+b^2}}}\right )}{\sqrt {b-\sqrt {-16+b^2}}}+\frac {\left (-4+b+\sqrt {-16+b^2}\right ) \arctan \left (\frac {2 \sqrt {2} x}{\sqrt {b+\sqrt {-16+b^2}}}\right )}{\sqrt {b+\sqrt {-16+b^2}}}}{\sqrt {2} \sqrt {-16+b^2}} \]

[In]

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

[Out]

(((4 - b + Sqrt[-16 + b^2])*ArcTan[(2*Sqrt[2]*x)/Sqrt[b - Sqrt[-16 + b^2]]])/Sqrt[b - Sqrt[-16 + b^2]] + ((-4
+ b + Sqrt[-16 + b^2])*ArcTan[(2*Sqrt[2]*x)/Sqrt[b + Sqrt[-16 + b^2]]])/Sqrt[b + Sqrt[-16 + b^2]])/(Sqrt[2]*Sq
rt[-16 + b^2])

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.19

method	result	size
risch	\(-\frac {\ln \left (-2 x^{2} \sqrt {-4-b}+x \left (4+b \right )+\sqrt {-4-b}\right )}{2 \sqrt {-4-b}}+\frac {\ln \left (-2 x^{2} \sqrt {-4-b}+\left (-4-b \right ) x +\sqrt {-4-b}\right )}{2 \sqrt {-4-b}}\)	\(74\)
default	\(\frac {\left (-4+\sqrt {\left (b -4\right ) \left (4+b \right )}+b \right ) \arctan \left (\frac {4 x}{\sqrt {2 \sqrt {\left (b -4\right ) \left (4+b \right )}+2 b}}\right )}{\sqrt {\left (b -4\right ) \left (4+b \right )}\, \sqrt {2 \sqrt {\left (b -4\right ) \left (4+b \right )}+2 b}}+\frac {\left (4+\sqrt {\left (b -4\right ) \left (4+b \right )}-b \right ) \arctan \left (\frac {4 x}{\sqrt {-2 \sqrt {\left (b -4\right ) \left (4+b \right )}+2 b}}\right )}{\sqrt {\left (b -4\right ) \left (4+b \right )}\, \sqrt {-2 \sqrt {\left (b -4\right ) \left (4+b \right )}+2 b}}\)	\(124\)

[In]

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

[Out]

-1/2/(-4-b)^(1/2)*ln(-2*x^2*(-4-b)^(1/2)+x*(4+b)+(-4-b)^(1/2))+1/2/(-4-b)^(1/2)*ln(-2*x^2*(-4-b)^(1/2)+(-4-b)*
x+(-4-b)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.77 \[ \int \frac {1+2 x^2}{1+b x^2+4 x^4} \, dx=\left [-\frac {\sqrt {-b - 4} \log \left (\frac {4 \, x^{4} - {\left (b + 8\right )} x^{2} - 2 \, {\left (2 \, x^{3} - x\right )} \sqrt {-b - 4} + 1}{4 \, x^{4} + b x^{2} + 1}\right )}{2 \, {\left (b + 4\right )}}, \frac {\sqrt {b + 4} \arctan \left (\frac {4 \, x^{3} + {\left (b + 2\right )} x}{\sqrt {b + 4}}\right ) + \sqrt {b + 4} \arctan \left (\frac {2 \, x}{\sqrt {b + 4}}\right )}{b + 4}\right ] \]

[In]

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

[Out]

[-1/2*sqrt(-b - 4)*log((4*x^4 - (b + 8)*x^2 - 2*(2*x^3 - x)*sqrt(-b - 4) + 1)/(4*x^4 + b*x^2 + 1))/(b + 4), (s
qrt(b + 4)*arctan((4*x^3 + (b + 2)*x)/sqrt(b + 4)) + sqrt(b + 4)*arctan(2*x/sqrt(b + 4)))/(b + 4)]

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.53 \[ \int \frac {1+2 x^2}{1+b x^2+4 x^4} \, dx=- \frac {\sqrt {- \frac {1}{b + 4}} \log {\left (x^{2} + x \left (- \frac {b \sqrt {- \frac {1}{b + 4}}}{2} - 2 \sqrt {- \frac {1}{b + 4}}\right ) - \frac {1}{2} \right )}}{2} + \frac {\sqrt {- \frac {1}{b + 4}} \log {\left (x^{2} + x \left (\frac {b \sqrt {- \frac {1}{b + 4}}}{2} + 2 \sqrt {- \frac {1}{b + 4}}\right ) - \frac {1}{2} \right )}}{2} \]

[In]

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

[Out]

-sqrt(-1/(b + 4))*log(x**2 + x*(-b*sqrt(-1/(b + 4))/2 - 2*sqrt(-1/(b + 4))) - 1/2)/2 + sqrt(-1/(b + 4))*log(x*
*2 + x*(b*sqrt(-1/(b + 4))/2 + 2*sqrt(-1/(b + 4))) - 1/2)/2

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 13.58 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06 \[ \int \frac {1+2 x^2}{1+b x^2+4 x^4} \, dx=-\frac {\mathrm {atan}\left (\frac {-b^3\,x-4\,b^2\,x^3-2\,b^2\,x+16\,b\,x+64\,x^3+32\,x}{\left (b^2-16\right )\,\sqrt {b+4}}\right )-\mathrm {atan}\left (\frac {2\,x}{\sqrt {b+4}}\right )}{\sqrt {b+4}} \]

[In]

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

[Out]

-(atan((32*x + 16*b*x - 2*b^2*x - b^3*x + 64*x^3 - 4*b^2*x^3)/((b^2 - 16)*(b + 4)^(1/2))) - atan((2*x)/(b + 4)
^(1/2)))/(b + 4)^(1/2)