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

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 31

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

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 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 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 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]

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 = 0.04, size = 72, normalized size = 0.91 \begin {gather*} \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 ) \end {gather*}

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

________________________________________________________________________________________

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

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 103, normalized size = 1.30 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

giac [A]  time = 0.19, size = 101, normalized size = 1.28 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 116, normalized size = 1.47 \begin {gather*} \frac {2 \sqrt {3}\, a \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}+\frac {2 a \ln \left (x +1\right )}{9}-\frac {a \ln \left (x^{2}-x +1\right )}{9}+\frac {\sqrt {3}\, b \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}-\frac {b \ln \left (x +1\right )}{9}+\frac {b \ln \left (x^{2}-x +1\right )}{18}-\frac {a}{9 \left (x +1\right )}+\frac {b}{9 x +9}-\frac {-a +b +\left (-a -2 b \right ) x}{9 \left (x^{2}-x +1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.36, size = 71, normalized size = 0.90 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

mupad [B]  time = 2.29, size = 97, normalized size = 1.23 \begin {gather*} \frac {\frac {b\,x^2}{3}+\frac {a\,x}{3}}{x^3+1}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {a}{9}-\frac {b}{18}+\frac {\sqrt {3}\,a\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{18}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {b}{18}-\frac {a}{9}+\frac {\sqrt {3}\,a\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{18}\right )+\ln \left (x+1\right )\,\left (\frac {2\,a}{9}-\frac {b}{9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 0.59, size = 238, normalized size = 3.01 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________