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

Optimal. Leaf size=66 \[ \frac {\log \left (\sqrt {4-b} x+2 x^2+1\right )}{2 \sqrt {4-b}}-\frac {\log \left (-\sqrt {4-b} x+2 x^2+1\right )}{2 \sqrt {4-b}} \]

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Rubi [A]  time = 0.03, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1164, 628} \begin {gather*} \frac {\log \left (\sqrt {4-b} x+2 x^2+1\right )}{2 \sqrt {4-b}}-\frac {\log \left (-\sqrt {4-b} x+2 x^2+1\right )}{2 \sqrt {4-b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[1 - Sqrt[4 - b]*x + 2*x^2]/(2*Sqrt[4 - b]) + Log[1 + Sqrt[4 - b]*x + 2*x^2]/(2*Sqrt[4 - b])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1164

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*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/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[b^2
- 4*a*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 127, normalized size = 1.92 \begin {gather*} \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}} \end {gather*}

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]*Sqr
t[-16 + b^2])

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-2 x^2}{1+b x^2+4 x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.81, size = 109, normalized size = 1.65 \begin {gather*} \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 ] \end {gather*}

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 = 73, normalized size = 1.11 \begin {gather*} -\frac {\sqrt {b - 4} b \arctan \left (\frac {4 \, \sqrt {\frac {1}{2}} x}{\sqrt {b + \sqrt {b^{2} - 16}}}\right )}{b^{2} - 4 \, b} - \frac {\sqrt {b - 4} b \arctan \left (\frac {4 \, \sqrt {\frac {1}{2}} x}{\sqrt {b - \sqrt {b^{2} - 16}}}\right )}{b^{2} - 4 \, b} \end {gather*}

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*arctan(4*sqrt(1/2)*x/sqrt(b + sqrt(b^2 - 16)))/(b^2 - 4*b) - sqrt(b - 4)*b*arctan(4*sqrt(1/2)*x
/sqrt(b - sqrt(b^2 - 16)))/(b^2 - 4*b)

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maple [B]  time = 0.02, size = 279, normalized size = 4.23 \begin {gather*} \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 )}}} \end {gather*}

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

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

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 = 0.07, size = 63, normalized size = 0.95 \begin {gather*} -\frac {\mathrm {atan}\left (\frac {2\,x}{\sqrt {b-4}}\right )-\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-4\right )}^{3/2}\,\left (b+4\right )}\right )}{\sqrt {b-4}} \end {gather*}

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

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sympy [A]  time = 0.38, size = 94, normalized size = 1.42 \begin {gather*} \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} \end {gather*}

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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