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

Optimal. Leaf size=83 \[ -\frac{1}{4} (a-b) \log \left (x^2-x+1\right )+\frac{1}{4} (a-b) \log \left (x^2+x+1\right )-\frac{(a+b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}+\frac{(a+b) \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

[Out]

-((a + b)*ArcTan[(1 - 2*x)/Sqrt[3]])/(2*Sqrt[3]) + ((a + b)*ArcTan[(1 + 2*x)/Sqrt[3]])/(2*Sqrt[3]) - ((a - b)*
Log[1 - x + x^2])/4 + ((a - b)*Log[1 + x + x^2])/4

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Rubi [A]  time = 0.0548068, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {1169, 634, 618, 204, 628} \[ -\frac{1}{4} (a-b) \log \left (x^2-x+1\right )+\frac{1}{4} (a-b) \log \left (x^2+x+1\right )-\frac{(a+b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}+\frac{(a+b) \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)/(1 + x^2 + x^4),x]

[Out]

-((a + b)*ArcTan[(1 - 2*x)/Sqrt[3]])/(2*Sqrt[3]) + ((a + b)*ArcTan[(1 + 2*x)/Sqrt[3]])/(2*Sqrt[3]) - ((a - b)*
Log[1 - x + x^2])/4 + ((a - b)*Log[1 + x + x^2])/4

Rule 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rule 634

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

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])

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]

Rubi steps

\begin{align*} \int \frac{a+b x^2}{1+x^2+x^4} \, dx &=\frac{1}{2} \int \frac{a-(a-b) x}{1-x+x^2} \, dx+\frac{1}{2} \int \frac{a+(a-b) x}{1+x+x^2} \, dx\\ &=\frac{1}{4} (a-b) \int \frac{1+2 x}{1+x+x^2} \, dx+\frac{1}{4} (-a+b) \int \frac{-1+2 x}{1-x+x^2} \, dx+\frac{1}{4} (a+b) \int \frac{1}{1-x+x^2} \, dx+\frac{1}{4} (a+b) \int \frac{1}{1+x+x^2} \, dx\\ &=-\frac{1}{4} (a-b) \log \left (1-x+x^2\right )+\frac{1}{4} (a-b) \log \left (1+x+x^2\right )+\frac{1}{2} (-a-b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac{1}{2} (-a-b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=-\frac{(a+b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}+\frac{(a+b) \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )}{2 \sqrt{3}}-\frac{1}{4} (a-b) \log \left (1-x+x^2\right )+\frac{1}{4} (a-b) \log \left (1+x+x^2\right )\\ \end{align*}

Mathematica [C]  time = 0.1246, size = 97, normalized size = 1.17 \[ \frac{\left (2 i a+\left (\sqrt{3}-i\right ) b\right ) \tan ^{-1}\left (\frac{1}{2} \left (\sqrt{3}-i\right ) x\right )}{\sqrt{6+6 i \sqrt{3}}}+\frac{\left (\left (\sqrt{3}+i\right ) b-2 i a\right ) \tan ^{-1}\left (\frac{1}{2} \left (\sqrt{3}+i\right ) x\right )}{\sqrt{6-6 i \sqrt{3}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((2*I)*a + (-I + Sqrt[3])*b)*ArcTan[((-I + Sqrt[3])*x)/2])/Sqrt[6 + (6*I)*Sqrt[3]] + (((-2*I)*a + (I + Sqrt[3
])*b)*ArcTan[((I + Sqrt[3])*x)/2])/Sqrt[6 - (6*I)*Sqrt[3]]

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Maple [A]  time = 0.047, size = 114, normalized size = 1.4 \begin{align*} -{\frac{\ln \left ({x}^{2}-x+1 \right ) a}{4}}+{\frac{\ln \left ({x}^{2}-x+1 \right ) b}{4}}+{\frac{a\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{\sqrt{3}b}{6}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{\ln \left ({x}^{2}+x+1 \right ) a}{4}}-{\frac{\ln \left ({x}^{2}+x+1 \right ) b}{4}}+{\frac{a\sqrt{3}}{6}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }+{\frac{\sqrt{3}b}{6}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*ln(x^2-x+1)*a+1/4*ln(x^2-x+1)*b+1/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))*a+1/6*3^(1/2)*arctan(1/3*(2*x-1)*
3^(1/2))*b+1/4*ln(x^2+x+1)*a-1/4*ln(x^2+x+1)*b+1/6*3^(1/2)*arctan(1/3*(1+2*x)*3^(1/2))*a+1/6*3^(1/2)*arctan(1/
3*(1+2*x)*3^(1/2))*b

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Maxima [A]  time = 1.4756, size = 93, normalized size = 1.12 \begin{align*} \frac{1}{6} \, \sqrt{3}{\left (a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{6} \, \sqrt{3}{\left (a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{1}{4} \,{\left (a - b\right )} \log \left (x^{2} + x + 1\right ) - \frac{1}{4} \,{\left (a - b\right )} \log \left (x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*(a + b)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*(a + b)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/4*(a
- b)*log(x^2 + x + 1) - 1/4*(a - b)*log(x^2 - x + 1)

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Fricas [A]  time = 1.39857, size = 223, normalized size = 2.69 \begin{align*} \frac{1}{6} \, \sqrt{3}{\left (a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{6} \, \sqrt{3}{\left (a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{1}{4} \,{\left (a - b\right )} \log \left (x^{2} + x + 1\right ) - \frac{1}{4} \,{\left (a - b\right )} \log \left (x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*(a + b)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*(a + b)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/4*(a
- b)*log(x^2 + x + 1) - 1/4*(a - b)*log(x^2 - x + 1)

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Sympy [C]  time = 0.885689, size = 740, normalized size = 8.92 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a/4 + b/4 - sqrt(3)*I*(a + b)/12)*log(x + (2*a**3*(-a/4 + b/4 - sqrt(3)*I*(a + b)/12) + 6*a**2*b*(-a/4 + b/4
- sqrt(3)*I*(a + b)/12) - 12*a*b**2*(-a/4 + b/4 - sqrt(3)*I*(a + b)/12) + 24*a*(-a/4 + b/4 - sqrt(3)*I*(a + b
)/12)**3 + 2*b**3*(-a/4 + b/4 - sqrt(3)*I*(a + b)/12) - 48*b*(-a/4 + b/4 - sqrt(3)*I*(a + b)/12)**3)/(a**4 - a
**3*b + a*b**3 - b**4)) + (-a/4 + b/4 + sqrt(3)*I*(a + b)/12)*log(x + (2*a**3*(-a/4 + b/4 + sqrt(3)*I*(a + b)/
12) + 6*a**2*b*(-a/4 + b/4 + sqrt(3)*I*(a + b)/12) - 12*a*b**2*(-a/4 + b/4 + sqrt(3)*I*(a + b)/12) + 24*a*(-a/
4 + b/4 + sqrt(3)*I*(a + b)/12)**3 + 2*b**3*(-a/4 + b/4 + sqrt(3)*I*(a + b)/12) - 48*b*(-a/4 + b/4 + sqrt(3)*I
*(a + b)/12)**3)/(a**4 - a**3*b + a*b**3 - b**4)) + (a/4 - b/4 - sqrt(3)*I*(a + b)/12)*log(x + (2*a**3*(a/4 -
b/4 - sqrt(3)*I*(a + b)/12) + 6*a**2*b*(a/4 - b/4 - sqrt(3)*I*(a + b)/12) - 12*a*b**2*(a/4 - b/4 - sqrt(3)*I*(
a + b)/12) + 24*a*(a/4 - b/4 - sqrt(3)*I*(a + b)/12)**3 + 2*b**3*(a/4 - b/4 - sqrt(3)*I*(a + b)/12) - 48*b*(a/
4 - b/4 - sqrt(3)*I*(a + b)/12)**3)/(a**4 - a**3*b + a*b**3 - b**4)) + (a/4 - b/4 + sqrt(3)*I*(a + b)/12)*log(
x + (2*a**3*(a/4 - b/4 + sqrt(3)*I*(a + b)/12) + 6*a**2*b*(a/4 - b/4 + sqrt(3)*I*(a + b)/12) - 12*a*b**2*(a/4
- b/4 + sqrt(3)*I*(a + b)/12) + 24*a*(a/4 - b/4 + sqrt(3)*I*(a + b)/12)**3 + 2*b**3*(a/4 - b/4 + sqrt(3)*I*(a
+ b)/12) - 48*b*(a/4 - b/4 + sqrt(3)*I*(a + b)/12)**3)/(a**4 - a**3*b + a*b**3 - b**4))

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Giac [A]  time = 1.12958, size = 93, normalized size = 1.12 \begin{align*} \frac{1}{6} \, \sqrt{3}{\left (a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{6} \, \sqrt{3}{\left (a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{1}{4} \,{\left (a - b\right )} \log \left (x^{2} + x + 1\right ) - \frac{1}{4} \,{\left (a - b\right )} \log \left (x^{2} - x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*(a + b)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*(a + b)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/4*(a
- b)*log(x^2 + x + 1) - 1/4*(a - b)*log(x^2 - x + 1)