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

Optimal. Leaf size=62 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{2-b}+2 x}{\sqrt{b+2}}\right )}{\sqrt{b+2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2-b}-2 x}{\sqrt{b+2}}\right )}{\sqrt{b+2}} \]

[Out]

-(ArcTan[(Sqrt[2 - b] - 2*x)/Sqrt[2 + b]]/Sqrt[2 + b]) + ArcTan[(Sqrt[2 - b] + 2*x)/Sqrt[2 + b]]/Sqrt[2 + b]

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Rubi [A]  time = 0.0556889, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1161, 618, 204} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{2-b}+2 x}{\sqrt{b+2}}\right )}{\sqrt{b+2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{2-b}-2 x}{\sqrt{b+2}}\right )}{\sqrt{b+2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)/(1 + b*x^2 + x^4),x]

[Out]

-(ArcTan[(Sqrt[2 - b] - 2*x)/Sqrt[2 + b]]/Sqrt[2 + b]) + ArcTan[(Sqrt[2 - b] + 2*x)/Sqrt[2 + b]]/Sqrt[2 + b]

Rule 1161

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]))

Rule 618

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 204

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

Rubi steps

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

Mathematica [A]  time = 0.0564094, size = 124, normalized size = 2. \[ \frac{\frac{\left (\sqrt{b^2-4}-b+2\right ) \tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{b-\sqrt{b^2-4}}}\right )}{\sqrt{b-\sqrt{b^2-4}}}+\frac{\left (\sqrt{b^2-4}+b-2\right ) \tan ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{\sqrt{b^2-4}+b}}\right )}{\sqrt{\sqrt{b^2-4}+b}}}{\sqrt{2} \sqrt{b^2-4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.138, size = 277, normalized size = 4.5 \begin{align*} -2\,{\frac{1}{\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}\arctan \left ( 2\,{\frac{x}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ) }+{\arctan \left ( 2\,{\frac{x}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ){\frac{1}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}}}+{b\arctan \left ( 2\,{\frac{x}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ){\frac{1}{\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }}}{\frac{1}{\sqrt{2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}}}+2\,{\frac{1}{\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}\arctan \left ( 2\,{\frac{x}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ) }+{\arctan \left ( 2\,{\frac{x}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ){\frac{1}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}}}-{b\arctan \left ( 2\,{\frac{x}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}} \right ){\frac{1}{\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }}}{\frac{1}{\sqrt{-2\,\sqrt{ \left ( -2+b \right ) \left ( 2+b \right ) }+2\,b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} + 1}{x^{4} + b x^{2} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.32819, size = 273, normalized size = 4.4 \begin{align*} \left [-\frac{\sqrt{-b - 2} \log \left (\frac{x^{4} -{\left (b + 4\right )} x^{2} - 2 \,{\left (x^{3} - x\right )} \sqrt{-b - 2} + 1}{x^{4} + b x^{2} + 1}\right )}{2 \,{\left (b + 2\right )}}, \frac{\sqrt{b + 2} \arctan \left (\frac{x^{3} +{\left (b + 1\right )} x}{\sqrt{b + 2}}\right ) + \sqrt{b + 2} \arctan \left (\frac{x}{\sqrt{b + 2}}\right )}{b + 2}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.258486, size = 88, normalized size = 1.42 \begin{align*} - \frac{\sqrt{- \frac{1}{b + 2}} \log{\left (x^{2} + x \left (- b \sqrt{- \frac{1}{b + 2}} - 2 \sqrt{- \frac{1}{b + 2}}\right ) - 1 \right )}}{2} + \frac{\sqrt{- \frac{1}{b + 2}} \log{\left (x^{2} + x \left (b \sqrt{- \frac{1}{b + 2}} + 2 \sqrt{- \frac{1}{b + 2}}\right ) - 1 \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError