3.2299 \(\int \frac{a+b x}{(1+x)^2 (1-x+x^2)^2} \, dx\)

Optimal. Leaf size=79 \[ \frac{x (a+b x)}{3 \left (x^3+1\right )}-\frac{1}{18} (2 a-b) \log \left (x^2-x+1\right )+\frac{1}{9} (2 a-b) \log (x+1)-\frac{(2 a+b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

(x*(a + b*x))/(3*(1 + x^3)) - ((2*a + b)*ArcTan[(1 - 2*x)/Sqrt[3]])/(3*Sqrt[3]) + ((2*a - b)*Log[1 + x])/9 - (
(2*a - b)*Log[1 - x + x^2])/18

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Rubi [A]  time = 0.067177, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {809, 1855, 1860, 31, 634, 618, 204, 628} \[ \frac{x (a+b x)}{3 \left (x^3+1\right )}-\frac{1}{18} (2 a-b) \log \left (x^2-x+1\right )+\frac{1}{9} (2 a-b) \log (x+1)-\frac{(2 a+b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x))/(3*(1 + x^3)) - ((2*a + b)*ArcTan[(1 - 2*x)/Sqrt[3]])/(3*Sqrt[3]) + ((2*a - b)*Log[1 + x])/9 - (
(2*a - b)*Log[1 - x + x^2])/18

Rule 809

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[
((d + e*x)^FracPart[p]*(a + b*x + c*x^2)^FracPart[p])/(a*d + c*e*x^3)^FracPart[p], Int[(f + g*x)*(a*d + c*e*x^
3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[m, p] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0]

Rule 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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)^2 \left (1-x+x^2\right )^2} \, dx &=\int \frac{a+b x}{\left (1+x^3\right )^2} \, dx\\ &=\frac{x (a+b x)}{3 \left (1+x^3\right )}-\frac{1}{3} \int \frac{-2 a-b x}{1+x^3} \, dx\\ &=\frac{x (a+b x)}{3 \left (1+x^3\right )}-\frac{1}{9} \int \frac{-4 a-b+(2 a-b) x}{1-x+x^2} \, dx-\frac{1}{9} (-2 a+b) \int \frac{1}{1+x} \, dx\\ &=\frac{x (a+b x)}{3 \left (1+x^3\right )}+\frac{1}{9} (2 a-b) \log (1+x)-\frac{1}{6} (-2 a-b) \int \frac{1}{1-x+x^2} \, dx-\frac{1}{18} (2 a-b) \int \frac{-1+2 x}{1-x+x^2} \, dx\\ &=\frac{x (a+b x)}{3 \left (1+x^3\right )}+\frac{1}{9} (2 a-b) \log (1+x)-\frac{1}{18} (2 a-b) \log \left (1-x+x^2\right )-\frac{1}{3} (2 a+b) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=\frac{x (a+b x)}{3 \left (1+x^3\right )}-\frac{(2 a+b) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}+\frac{1}{9} (2 a-b) \log (1+x)-\frac{1}{18} (2 a-b) \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.044839, size = 72, normalized size = 0.91 \[ \frac{1}{18} \left (\frac{6 x (a+b x)}{x^3+1}+(b-2 a) \log \left (x^2-x+1\right )+2 (2 a-b) \log (x+1)+2 \sqrt{3} (2 a+b) \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*x*(a + b*x))/(1 + x^3) + 2*Sqrt[3]*(2*a + b)*ArcTan[(-1 + 2*x)/Sqrt[3]] + 2*(2*a - b)*Log[1 + x] + (-2*a +
b)*Log[1 - x + x^2])/18

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Maple [A]  time = 0.011, size = 116, normalized size = 1.5 \begin{align*} -{\frac{a}{9+9\,x}}+{\frac{b}{9+9\,x}}+{\frac{2\,\ln \left ( 1+x \right ) a}{9}}-{\frac{\ln \left ( 1+x \right ) b}{9}}-{\frac{ \left ( -a-2\,b \right ) x-a+b}{9\,{x}^{2}-9\,x+9}}-{\frac{\ln \left ({x}^{2}-x+1 \right ) a}{9}}+{\frac{\ln \left ({x}^{2}-x+1 \right ) b}{18}}+{\frac{2\,\sqrt{3}a}{9}\arctan \left ({\frac{ \left ( -1+2\,x \right ) \sqrt{3}}{3}} \right ) }+{\frac{b\sqrt{3}}{9}\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)^2/(x^2-x+1)^2,x)

[Out]

-1/9/(1+x)*a+1/9/(1+x)*b+2/9*ln(1+x)*a-1/9*ln(1+x)*b-1/9*((-a-2*b)*x-a+b)/(x^2-x+1)-1/9*ln(x^2-x+1)*a+1/18*ln(
x^2-x+1)*b+2/9*3^(1/2)*arctan(1/3*(-1+2*x)*3^(1/2))*a+1/9*3^(1/2)*arctan(1/3*(-1+2*x)*3^(1/2))*b

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Maxima [A]  time = 1.52832, size = 96, normalized size = 1.22 \begin{align*} \frac{1}{9} \, \sqrt{3}{\left (2 \, a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{1}{18} \,{\left (2 \, a - b\right )} \log \left (x^{2} - x + 1\right ) + \frac{1}{9} \,{\left (2 \, a - b\right )} \log \left (x + 1\right ) + \frac{b x^{2} + a x}{3 \,{\left (x^{3} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*sqrt(3)*(2*a + b)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/18*(2*a - b)*log(x^2 - x + 1) + 1/9*(2*a - b)*log(x +
1) + 1/3*(b*x^2 + a*x)/(x^3 + 1)

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Fricas [A]  time = 1.29217, size = 254, normalized size = 3.22 \begin{align*} \frac{6 \, b x^{2} + 2 \, \sqrt{3}{\left ({\left (2 \, a + b\right )} x^{3} + 2 \, a + b\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 6 \, a x -{\left ({\left (2 \, a - b\right )} x^{3} + 2 \, a - b\right )} \log \left (x^{2} - x + 1\right ) + 2 \,{\left ({\left (2 \, a - b\right )} x^{3} + 2 \, a - b\right )} \log \left (x + 1\right )}{18 \,{\left (x^{3} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(6*b*x^2 + 2*sqrt(3)*((2*a + b)*x^3 + 2*a + b)*arctan(1/3*sqrt(3)*(2*x - 1)) + 6*a*x - ((2*a - b)*x^3 + 2
*a - b)*log(x^2 - x + 1) + 2*((2*a - b)*x^3 + 2*a - b)*log(x + 1))/(x^3 + 1)

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Sympy [C]  time = 0.643432, size = 238, normalized size = 3.01 \begin{align*} \frac{\left (2 a - b\right ) \log{\left (x + \frac{4 a^{2} \left (2 a - b\right ) + 4 a b^{2} + b \left (2 a - b\right )^{2}}{8 a^{3} + b^{3}} \right )}}{9} + \left (- \frac{a}{9} + \frac{b}{18} - \frac{\sqrt{3} i \left (2 a + b\right )}{18}\right ) \log{\left (x + \frac{36 a^{2} \left (- \frac{a}{9} + \frac{b}{18} - \frac{\sqrt{3} i \left (2 a + b\right )}{18}\right ) + 4 a b^{2} + 81 b \left (- \frac{a}{9} + \frac{b}{18} - \frac{\sqrt{3} i \left (2 a + b\right )}{18}\right )^{2}}{8 a^{3} + b^{3}} \right )} + \left (- \frac{a}{9} + \frac{b}{18} + \frac{\sqrt{3} i \left (2 a + b\right )}{18}\right ) \log{\left (x + \frac{36 a^{2} \left (- \frac{a}{9} + \frac{b}{18} + \frac{\sqrt{3} i \left (2 a + b\right )}{18}\right ) + 4 a b^{2} + 81 b \left (- \frac{a}{9} + \frac{b}{18} + \frac{\sqrt{3} i \left (2 a + b\right )}{18}\right )^{2}}{8 a^{3} + b^{3}} \right )} + \frac{a x + b x^{2}}{3 x^{3} + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a - b)*log(x + (4*a**2*(2*a - b) + 4*a*b**2 + b*(2*a - b)**2)/(8*a**3 + b**3))/9 + (-a/9 + b/18 - sqrt(3)*I
*(2*a + b)/18)*log(x + (36*a**2*(-a/9 + b/18 - sqrt(3)*I*(2*a + b)/18) + 4*a*b**2 + 81*b*(-a/9 + b/18 - sqrt(3
)*I*(2*a + b)/18)**2)/(8*a**3 + b**3)) + (-a/9 + b/18 + sqrt(3)*I*(2*a + b)/18)*log(x + (36*a**2*(-a/9 + b/18
+ sqrt(3)*I*(2*a + b)/18) + 4*a*b**2 + 81*b*(-a/9 + b/18 + sqrt(3)*I*(2*a + b)/18)**2)/(8*a**3 + b**3)) + (a*x
+ b*x**2)/(3*x**3 + 3)

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Giac [A]  time = 1.09771, size = 136, normalized size = 1.72 \begin{align*} \frac{1}{9} \, \sqrt{3}{\left (2 \, a + b\right )} \arctan \left (-\sqrt{3}{\left (\frac{2}{x + 1} - 1\right )}\right ) - \frac{1}{18} \,{\left (2 \, a - b\right )} \log \left (-\frac{3}{x + 1} + \frac{3}{{\left (x + 1\right )}^{2}} + 1\right ) - \frac{a}{9 \,{\left (x + 1\right )}} + \frac{b}{9 \,{\left (x + 1\right )}} - \frac{b + \frac{a - b}{x + 1}}{9 \,{\left (\frac{3}{x + 1} - \frac{3}{{\left (x + 1\right )}^{2}} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*sqrt(3)*(2*a + b)*arctan(-sqrt(3)*(2/(x + 1) - 1)) - 1/18*(2*a - b)*log(-3/(x + 1) + 3/(x + 1)^2 + 1) - 1/
9*a/(x + 1) + 1/9*b/(x + 1) - 1/9*(b + (a - b)/(x + 1))/(3/(x + 1) - 3/(x + 1)^2 - 1)