\(\int \frac {a+b x}{(1+x)^2 (1-x+x^2)^2} \, dx\) [2299]

Optimal result

Integrand size = 21, antiderivative size = 79 \[ \int \frac {a+b x}{(1+x)^2 \left (1-x+x^2\right )^2} \, dx=\frac {x (a+b x)}{3 \left (1+x^3\right )}-\frac {(2 a+b) \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{9} (2 a-b) \log (1+x)-\frac {1}{18} (2 a-b) \log \left (1-x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {823, 1869, 1874, 31, 648, 632, 210, 642} \[ \int \frac {a+b x}{(1+x)^2 \left (1-x+x^2\right )^2} \, dx=-\frac {(2 a+b) \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {x (a+b x)}{3 \left (x^3+1\right )}-\frac {1}{18} (2 a-b) \log \left (x^2-x+1\right )+\frac {1}{9} (2 a-b) \log (x+1) \]

[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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 632

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 642

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 648

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 823

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 1869

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-x)*Pq*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] +
Dist[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 1874

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*} \text {integral}& = \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) \text {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] (verified)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91 \[ \int \frac {a+b x}{(1+x)^2 \left (1-x+x^2\right )^2} \, dx=\frac {1}{18} \left (\frac {6 x (a+b x)}{1+x^3}+2 \sqrt {3} (2 a+b) \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+2 (2 a-b) \log (1+x)+(-2 a+b) \log \left (1-x+x^2\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.20

method	result	size
default	\(-\frac {\left (-a -2 b \right ) x -a +b}{9 \left (x^{2}-x +1\right )}-\frac {\left (2 a -b \right ) \ln \left (x^{2}-x +1\right )}{18}-\frac {2 \left (-3 a -\frac {3 b}{2}\right ) \sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{27}-\frac {\frac {a}{9}-\frac {b}{9}}{1+x}+\left (-\frac {b}{9}+\frac {2 a}{9}\right ) \ln \left (1+x \right )\)	\(95\)
risch	\(\frac {\frac {1}{3} b \,x^{2}+\frac {1}{3} a x}{\left (1+x \right ) \left (x^{2}-x +1\right )}-\frac {a \ln \left (16 a^{2} x^{2}-8 a b \,x^{2}+4 b^{2} x^{2}-16 a^{2} x +8 a b x -4 b^{2} x +16 a^{2}-8 b a +4 b^{2}\right )}{9}+\frac {b \ln \left (16 a^{2} x^{2}-8 a b \,x^{2}+4 b^{2} x^{2}-16 a^{2} x +8 a b x -4 b^{2} x +16 a^{2}-8 b a +4 b^{2}\right )}{18}+\frac {2 \ln \left (1+x \right ) a}{9}-\frac {\ln \left (1+x \right ) b}{9}+\frac {2 \sqrt {3}\, a \arctan \left (\frac {8 a^{2} \sqrt {3}\, x}{3 \left (4 a^{2}-2 b a +b^{2}\right )}-\frac {4 a^{2} \sqrt {3}}{3 \left (4 a^{2}-2 b a +b^{2}\right )}+\frac {2 a b \sqrt {3}}{3 \left (4 a^{2}-2 b a +b^{2}\right )}-\frac {4 \sqrt {3}\, a b x}{3 \left (4 a^{2}-2 b a +b^{2}\right )}+\frac {2 \sqrt {3}\, b^{2} x}{3 \left (4 a^{2}-2 b a +b^{2}\right )}-\frac {\sqrt {3}\, b^{2}}{3 \left (4 a^{2}-2 b a +b^{2}\right )}\right )}{9}+\frac {2 \sqrt {3}\, a \arctan \left (-\frac {\sqrt {3}}{3}+\frac {\sqrt {3}\, b}{3 a}\right )}{9}+\frac {\sqrt {3}\, b \arctan \left (\frac {8 a^{2} \sqrt {3}\, x}{3 \left (4 a^{2}-2 b a +b^{2}\right )}-\frac {4 a^{2} \sqrt {3}}{3 \left (4 a^{2}-2 b a +b^{2}\right )}+\frac {2 a b \sqrt {3}}{3 \left (4 a^{2}-2 b a +b^{2}\right )}-\frac {4 \sqrt {3}\, a b x}{3 \left (4 a^{2}-2 b a +b^{2}\right )}+\frac {2 \sqrt {3}\, b^{2} x}{3 \left (4 a^{2}-2 b a +b^{2}\right )}-\frac {\sqrt {3}\, b^{2}}{3 \left (4 a^{2}-2 b a +b^{2}\right )}\right )}{9}+\frac {\sqrt {3}\, b \arctan \left (-\frac {\sqrt {3}}{3}+\frac {\sqrt {3}\, b}{3 a}\right )}{9}\)	\(499\)

[In]

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

[Out]

-1/9*((-a-2*b)*x-a+b)/(x^2-x+1)-1/18*(2*a-b)*ln(x^2-x+1)-2/27*(-3*a-3/2*b)*3^(1/2)*arctan(1/3*(-1+2*x)*3^(1/2)
)-(1/9*a-1/9*b)/(1+x)+(-1/9*b+2/9*a)*ln(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.30 \[ \int \frac {a+b x}{(1+x)^2 \left (1-x+x^2\right )^2} \, dx=\frac {6 \, b x^{2} + 2 \, \sqrt {3} {\left ({\left (2 \, a + b\right )} x^{3} + 2 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 6 \, a x - {\left ({\left (2 \, a - b\right )} x^{3} + 2 \, a - b\right )} \log \left (x^{2} - x + 1\right ) + 2 \, {\left ({\left (2 \, a - b\right )} x^{3} + 2 \, a - b\right )} \log \left (x + 1\right )}{18 \, {\left (x^{3} + 1\right )}} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 238, normalized size of antiderivative = 3.01 \[ \int \frac {a+b x}{(1+x)^2 \left (1-x+x^2\right )^2} \, dx=\frac {\left (2 a - b\right ) \log {\left (x + \frac {4 a^{2} \cdot \left (2 a - b\right ) + 4 a b^{2} + b \left (2 a - b\right )^{2}}{8 a^{3} + b^{3}} \right )}}{9} + \left (- \frac {a}{9} + \frac {b}{18} - \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right ) \log {\left (x + \frac {36 a^{2} \left (- \frac {a}{9} + \frac {b}{18} - \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right ) + 4 a b^{2} + 81 b \left (- \frac {a}{9} + \frac {b}{18} - \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right )^{2}}{8 a^{3} + b^{3}} \right )} + \left (- \frac {a}{9} + \frac {b}{18} + \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right ) \log {\left (x + \frac {36 a^{2} \left (- \frac {a}{9} + \frac {b}{18} + \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right ) + 4 a b^{2} + 81 b \left (- \frac {a}{9} + \frac {b}{18} + \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right )^{2}}{8 a^{3} + b^{3}} \right )} + \frac {a x + b x^{2}}{3 x^{3} + 3} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \frac {a+b x}{(1+x)^2 \left (1-x+x^2\right )^2} \, dx=\frac {1}{9} \, \sqrt {3} {\left (2 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{18} \, {\left (2 \, a - b\right )} \log \left (x^{2} - x + 1\right ) + \frac {1}{9} \, {\left (2 \, a - b\right )} \log \left (x + 1\right ) + \frac {b x^{2} + a x}{3 \, {\left (x^{3} + 1\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.28 \[ \int \frac {a+b x}{(1+x)^2 \left (1-x+x^2\right )^2} \, dx=\frac {1}{9} \, \sqrt {3} {\left (2 \, a + b\right )} \arctan \left (-\sqrt {3} {\left (\frac {2}{x + 1} - 1\right )}\right ) - \frac {1}{18} \, {\left (2 \, a - b\right )} \log \left (-\frac {3}{x + 1} + \frac {3}{{\left (x + 1\right )}^{2}} + 1\right ) - \frac {a}{9 \, {\left (x + 1\right )}} + \frac {b}{9 \, {\left (x + 1\right )}} - \frac {b + \frac {a - b}{x + 1}}{9 \, {\left (\frac {3}{x + 1} - \frac {3}{{\left (x + 1\right )}^{2}} - 1\right )}} \]

[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)

Mupad [B] (verification not implemented)

Time = 11.07 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.23 \[ \int \frac {a+b x}{(1+x)^2 \left (1-x+x^2\right )^2} \, dx=\frac {\frac {b\,x^2}{3}+\frac {a\,x}{3}}{x^3+1}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {a}{9}-\frac {b}{18}+\frac {\sqrt {3}\,a\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{18}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {b}{18}-\frac {a}{9}+\frac {\sqrt {3}\,a\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{18}\right )+\ln \left (x+1\right )\,\left (\frac {2\,a}{9}-\frac {b}{9}\right ) \]

[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)