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

Optimal. Leaf size=62 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {4-b}+4 x}{\sqrt {b+4}}\right )}{\sqrt {b+4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {4-b}-4 x}{\sqrt {b+4}}\right )}{\sqrt {b+4}} \]

[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)

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Rubi [A]  time = 0.06, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1161, 618, 204} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {4-b}+4 x}{\sqrt {b+4}}\right )}{\sqrt {b+4}}-\frac {\tan ^{-1}\left (\frac {\sqrt {4-b}-4 x}{\sqrt {b+4}}\right )}{\sqrt {b+4}} \]

Antiderivative was successfully verified.

[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 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])

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 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]))

Rubi steps

\begin {align*} \int \frac {1+2 x^2}{1+b x^2+4 x^4} \, dx &=\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} \operatorname {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} \operatorname {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*}

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Mathematica [B]  time = 0.06, size = 126, normalized size = 2.03 \[ \frac {\frac {\left (\sqrt {b^2-16}-b+4\right ) \tan ^{-1}\left (\frac {2 \sqrt {2} x}{\sqrt {b-\sqrt {b^2-16}}}\right )}{\sqrt {b-\sqrt {b^2-16}}}+\frac {\left (\sqrt {b^2-16}+b-4\right ) \tan ^{-1}\left (\frac {2 \sqrt {2} x}{\sqrt {\sqrt {b^2-16}+b}}\right )}{\sqrt {\sqrt {b^2-16}+b}}}{\sqrt {2} \sqrt {b^2-16}} \]

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.42, size = 110, normalized size = 1.77 \[ \left [-\frac {\sqrt {-b - 4} \log \left (\frac {4 \, x^{4} - {\left (b + 8\right )} x^{2} - 2 \, {\left (2 \, x^{3} - x\right )} \sqrt {-b - 4} + 1}{4 \, x^{4} + b x^{2} + 1}\right )}{2 \, {\left (b + 4\right )}}, \frac {\sqrt {b + 4} \arctan \left (\frac {4 \, x^{3} + {\left (b + 2\right )} x}{\sqrt {b + 4}}\right ) + \sqrt {b + 4} \arctan \left (\frac {2 \, x}{\sqrt {b + 4}}\right )}{b + 4}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.31, size = 77, normalized size = 1.24 \[ \frac {\sqrt {b + 4} {\left (b - 8\right )} \arctan \left (\frac {4 \, \sqrt {\frac {1}{2}} x}{\sqrt {b + \sqrt {b^{2} - 16}}}\right )}{b^{2} - 4 \, b - 32} + \frac {\sqrt {b + 4} {\left (b - 8\right )} \arctan \left (\frac {4 \, \sqrt {\frac {1}{2}} x}{\sqrt {b - \sqrt {b^{2} - 16}}}\right )}{b^{2} - 4 \, b - 32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(b + 4)*(b - 8)*arctan(4*sqrt(1/2)*x/sqrt(b + sqrt(b^2 - 16)))/(b^2 - 4*b - 32) + sqrt(b + 4)*(b - 8)*arct
an(4*sqrt(1/2)*x/sqrt(b - sqrt(b^2 - 16)))/(b^2 - 4*b - 32)

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maple [B]  time = 0.04, size = 277, normalized size = 4.47 \[ -\frac {b \arctan \left (\frac {4 x}{\sqrt {2 b -2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}\right )}{\sqrt {\left (b -4\right ) \left (b +4\right )}\, \sqrt {2 b -2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}+\frac {b \arctan \left (\frac {4 x}{\sqrt {2 b +2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}\right )}{\sqrt {\left (b -4\right ) \left (b +4\right )}\, \sqrt {2 b +2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}+\frac {4 \arctan \left (\frac {4 x}{\sqrt {2 b -2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}\right )}{\sqrt {\left (b -4\right ) \left (b +4\right )}\, \sqrt {2 b -2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}+\frac {\arctan \left (\frac {4 x}{\sqrt {2 b -2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}\right )}{\sqrt {2 b -2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}-\frac {4 \arctan \left (\frac {4 x}{\sqrt {2 b +2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}\right )}{\sqrt {\left (b -4\right ) \left (b +4\right )}\, \sqrt {2 b +2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}+\frac {\arctan \left (\frac {4 x}{\sqrt {2 b +2 \sqrt {\left (b -4\right ) \left (b +4\right )}}}\right )}{\sqrt {2 b +2 \sqrt {\left (b -4\right ) \left (b +4\right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {2 \, x^{2} + 1}{4 \, x^{4} + b x^{2} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.39, size = 66, normalized size = 1.06 \[ -\frac {\mathrm {atan}\left (\frac {-b^3\,x-4\,b^2\,x^3-2\,b^2\,x+16\,b\,x+64\,x^3+32\,x}{\left (b^2-16\right )\,\sqrt {b+4}}\right )-\mathrm {atan}\left (\frac {2\,x}{\sqrt {b+4}}\right )}{\sqrt {b+4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.38, size = 95, normalized size = 1.53 \[ - \frac {\sqrt {- \frac {1}{b + 4}} \log {\left (x^{2} + x \left (- \frac {b \sqrt {- \frac {1}{b + 4}}}{2} - 2 \sqrt {- \frac {1}{b + 4}}\right ) - \frac {1}{2} \right )}}{2} + \frac {\sqrt {- \frac {1}{b + 4}} \log {\left (x^{2} + x \left (\frac {b \sqrt {- \frac {1}{b + 4}}}{2} + 2 \sqrt {- \frac {1}{b + 4}}\right ) - \frac {1}{2} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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