3.2298 \(\int \frac{a+b x}{(1+x) (1-x+x^2)} \, dx\)

Optimal. Leaf size=54 \[ -\frac{1}{6} (a-b) \log \left (x^2-x+1\right )+\frac{1}{3} (a-b) \log (x+1)-\frac{(a+b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

-(((a + b)*ArcTan[(1 - 2*x)/Sqrt[3]])/Sqrt[3]) + ((a - b)*Log[1 + x])/3 - ((a - b)*Log[1 - x + x^2])/6

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Rubi [A]  time = 0.0662246, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {800, 634, 618, 204, 628} \[ -\frac{1}{6} (a-b) \log \left (x^2-x+1\right )+\frac{1}{3} (a-b) \log (x+1)-\frac{(a+b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b)*ArcTan[(1 - 2*x)/Sqrt[3]])/Sqrt[3]) + ((a - b)*Log[1 + x])/3 - ((a - b)*Log[1 - x + x^2])/6

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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}{(1+x) \left (1-x+x^2\right )} \, dx &=\int \left (\frac{a-b}{3 (1+x)}+\frac{2 a+b-(a-b) x}{3 \left (1-x+x^2\right )}\right ) \, dx\\ &=\frac{1}{3} (a-b) \log (1+x)+\frac{1}{3} \int \frac{2 a+b-(a-b) x}{1-x+x^2} \, dx\\ &=\frac{1}{3} (a-b) \log (1+x)+\frac{1}{6} (-a+b) \int \frac{-1+2 x}{1-x+x^2} \, dx+\frac{1}{2} (a+b) \int \frac{1}{1-x+x^2} \, dx\\ &=\frac{1}{3} (a-b) \log (1+x)-\frac{1}{6} (a-b) \log \left (1-x+x^2\right )+(-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 )}{\sqrt{3}}+\frac{1}{3} (a-b) \log (1+x)-\frac{1}{6} (a-b) \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0251866, size = 49, normalized size = 0.91 \[ \frac{1}{6} (a-b) \left (2 \log (x+1)-\log \left (x^2-x+1\right )\right )+\frac{(a+b) \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)*ArcTan[(-1 + 2*x)/Sqrt[3]])/Sqrt[3] + ((a - b)*(2*Log[1 + x] - Log[1 - x + x^2]))/6

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Maple [A]  time = 0.006, size = 74, normalized size = 1.4 \begin{align*}{\frac{\ln \left ( 1+x \right ) a}{3}}-{\frac{\ln \left ( 1+x \right ) b}{3}}-{\frac{\ln \left ({x}^{2}-x+1 \right ) a}{6}}+{\frac{\ln \left ({x}^{2}-x+1 \right ) b}{6}}+{\frac{\sqrt{3}a}{3}\arctan \left ({\frac{ \left ( -1+2\,x \right ) \sqrt{3}}{3}} \right ) }+{\frac{b\sqrt{3}}{3}\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+a)/(1+x)/(x^2-x+1),x)

[Out]

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

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Maxima [A]  time = 1.52289, size = 63, normalized size = 1.17 \begin{align*} \frac{1}{3} \, \sqrt{3}{\left (a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{6} \,{\left (a - b\right )} \log \left (x^{2} - x + 1\right ) + \frac{1}{3} \,{\left (a - b\right )} \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.30048, size = 144, normalized size = 2.67 \begin{align*} \frac{1}{3} \, \sqrt{3}{\left (a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{6} \,{\left (a - b\right )} \log \left (x^{2} - x + 1\right ) + \frac{1}{3} \,{\left (a - b\right )} \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 0.378216, size = 201, normalized size = 3.72 \begin{align*} \frac{\left (a - b\right ) \log{\left (x + \frac{a^{2} \left (a - b\right ) + 2 a b^{2} + b \left (a - b\right )^{2}}{a^{3} + b^{3}} \right )}}{3} + \left (- \frac{a}{6} + \frac{b}{6} - \frac{\sqrt{3} i \left (a + b\right )}{6}\right ) \log{\left (x + \frac{3 a^{2} \left (- \frac{a}{6} + \frac{b}{6} - \frac{\sqrt{3} i \left (a + b\right )}{6}\right ) + 2 a b^{2} + 9 b \left (- \frac{a}{6} + \frac{b}{6} - \frac{\sqrt{3} i \left (a + b\right )}{6}\right )^{2}}{a^{3} + b^{3}} \right )} + \left (- \frac{a}{6} + \frac{b}{6} + \frac{\sqrt{3} i \left (a + b\right )}{6}\right ) \log{\left (x + \frac{3 a^{2} \left (- \frac{a}{6} + \frac{b}{6} + \frac{\sqrt{3} i \left (a + b\right )}{6}\right ) + 2 a b^{2} + 9 b \left (- \frac{a}{6} + \frac{b}{6} + \frac{\sqrt{3} i \left (a + b\right )}{6}\right )^{2}}{a^{3} + b^{3}} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a - b)*log(x + (a**2*(a - b) + 2*a*b**2 + b*(a - b)**2)/(a**3 + b**3))/3 + (-a/6 + b/6 - sqrt(3)*I*(a + b)/6)
*log(x + (3*a**2*(-a/6 + b/6 - sqrt(3)*I*(a + b)/6) + 2*a*b**2 + 9*b*(-a/6 + b/6 - sqrt(3)*I*(a + b)/6)**2)/(a
**3 + b**3)) + (-a/6 + b/6 + sqrt(3)*I*(a + b)/6)*log(x + (3*a**2*(-a/6 + b/6 + sqrt(3)*I*(a + b)/6) + 2*a*b**
2 + 9*b*(-a/6 + b/6 + sqrt(3)*I*(a + b)/6)**2)/(a**3 + b**3))

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Giac [A]  time = 1.12466, size = 72, normalized size = 1.33 \begin{align*} \frac{1}{3} \,{\left (\sqrt{3} a + \sqrt{3} b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{6} \,{\left (a - b\right )} \log \left (x^{2} - x + 1\right ) + \frac{1}{3} \,{\left (a - b\right )} \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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