Integrand size = 21, antiderivative size = 101 \[ \int \frac {a+b x}{(1+x)^3 \left (1-x+x^2\right )^3} \, dx=\frac {x (a+b x)}{6 \left (1+x^3\right )^2}+\frac {x (5 a+4 b x)}{18 \left (1+x^3\right )}-\frac {(5 a+2 b) \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{9 \sqrt {3}}+\frac {1}{27} (5 a-2 b) \log (1+x)-\frac {1}{54} (5 a-2 b) \log \left (1-x+x^2\right ) \]
1/6*x*(b*x+a)/(x^3+1)^2+1/18*x*(4*b*x+5*a)/(x^3+1)+1/27*(5*a-2*b)*ln(1+x)- 1/54*(5*a-2*b)*ln(x^2-x+1)-1/27*(5*a+2*b)*arctan(1/3*(1-2*x)*3^(1/2))*3^(1 /2)
Time = 0.05 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.93 \[ \int \frac {a+b x}{(1+x)^3 \left (1-x+x^2\right )^3} \, dx=\frac {1}{54} \left (\frac {9 x (a+b x)}{\left (1+x^3\right )^2}+\frac {3 x (5 a+4 b x)}{1+x^3}+2 \sqrt {3} (5 a+2 b) \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+2 (5 a-2 b) \log (1+x)+(-5 a+2 b) \log \left (1-x+x^2\right )\right ) \]
((9*x*(a + b*x))/(1 + x^3)^2 + (3*x*(5*a + 4*b*x))/(1 + x^3) + 2*Sqrt[3]*( 5*a + 2*b)*ArcTan[(-1 + 2*x)/Sqrt[3]] + 2*(5*a - 2*b)*Log[1 + x] + (-5*a + 2*b)*Log[1 - x + x^2])/54
Time = 0.36 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.12, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {1209, 2394, 25, 2394, 27, 2399, 16, 1142, 25, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {a+b x}{(x+1)^3 \left (x^2-x+1\right )^3} \, dx\) |
| \(\Big \downarrow \) 1209 |
| \(\displaystyle \int \frac {a+b x}{\left (x^3+1\right )^3}dx\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {x (a+b x)}{6 \left (x^3+1\right )^2}-\frac {1}{6} \int -\frac {5 a+4 b x}{\left (x^3+1\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{6} \int \frac {5 a+4 b x}{\left (x^3+1\right )^2}dx+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {1}{6} \left (\frac {x (5 a+4 b x)}{3 \left (x^3+1\right )}-\frac {1}{3} \int -\frac {2 (5 a+2 b x)}{x^3+1}dx\right )+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {2}{3} \int \frac {5 a+2 b x}{x^3+1}dx+\frac {x (5 a+4 b x)}{3 \left (x^3+1\right )}\right )+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
| \(\Big \downarrow \) 2399 |
| \(\displaystyle \frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \int \frac {2 (5 a+b)-(5 a-2 b) x}{x^2-x+1}dx+\frac {1}{3} (5 a-2 b) \int \frac {1}{x+1}dx\right )+\frac {x (5 a+4 b x)}{3 \left (x^3+1\right )}\right )+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \int \frac {2 (5 a+b)-(5 a-2 b) x}{x^2-x+1}dx+\frac {1}{3} (5 a-2 b) \log (x+1)\right )+\frac {x (5 a+4 b x)}{3 \left (x^3+1\right )}\right )+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {3}{2} (5 a+2 b) \int \frac {1}{x^2-x+1}dx-\frac {1}{2} (5 a-2 b) \int -\frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{3} (5 a-2 b) \log (x+1)\right )+\frac {x (5 a+4 b x)}{3 \left (x^3+1\right )}\right )+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {3}{2} (5 a+2 b) \int \frac {1}{x^2-x+1}dx+\frac {1}{2} (5 a-2 b) \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{3} (5 a-2 b) \log (x+1)\right )+\frac {x (5 a+4 b x)}{3 \left (x^3+1\right )}\right )+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{2} (5 a-2 b) \int \frac {1-2 x}{x^2-x+1}dx-3 (5 a+2 b) \int \frac {1}{-(2 x-1)^2-3}d(2 x-1)\right )+\frac {1}{3} (5 a-2 b) \log (x+1)\right )+\frac {x (5 a+4 b x)}{3 \left (x^3+1\right )}\right )+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{2} (5 a-2 b) \int \frac {1-2 x}{x^2-x+1}dx+\sqrt {3} (5 a+2 b) \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )\right )+\frac {1}{3} (5 a-2 b) \log (x+1)\right )+\frac {x (5 a+4 b x)}{3 \left (x^3+1\right )}\right )+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{6} \left (\frac {2}{3} \left (\frac {1}{3} \left (\sqrt {3} (5 a+2 b) \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )-\frac {1}{2} (5 a-2 b) \log \left (x^2-x+1\right )\right )+\frac {1}{3} (5 a-2 b) \log (x+1)\right )+\frac {x (5 a+4 b x)}{3 \left (x^3+1\right )}\right )+\frac {x (a+b x)}{6 \left (x^3+1\right )^2}\) |
(x*(a + b*x))/(6*(1 + x^3)^2) + ((x*(5*a + 4*b*x))/(3*(1 + x^3)) + (2*(((5 *a - 2*b)*Log[1 + x])/3 + (Sqrt[3]*(5*a + 2*b)*ArcTan[(-1 + 2*x)/Sqrt[3]] - ((5*a - 2*b)*Log[1 - x + x^2])/2)/3))/3)/6
3.23.100.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[(f + g*x)^n*(a*d + c*e*x^3)^p, x] / ; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b *e, 0] && EqQ[m, p] && ILtQ[p, -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-x)*Pq*((a + b *x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[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]
Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numer ator[Rt[a/b, 3]], s = Denominator[Rt[a/b, 3]]}, Simp[(-r)*((B*r - A*s)/(3*a *s)) Int[1/(r + s*x), x], x] + Simp[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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.40 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.10
| method | result | size |
| risch | \(\frac {\frac {2}{9} b \,x^{5}+\frac {5}{18} a \,x^{4}+\frac {7}{18} b \,x^{2}+\frac {4}{9} a x}{\left (1+x \right )^{2} \left (x^{2}-x +1\right )^{2}}-\frac {2 \ln \left (1+x \right ) b}{27}+\frac {5 \ln \left (1+x \right ) a}{27}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{2}+\left (5 a -2 b \right ) \textit {\_Z} +25 a^{2}+10 b a +4 b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (5 \textit {\_R} a +4 b^{2}\right ) x +\textit {\_R}^{2}+10 b a \right )\right )}{27}\) | \(111\) |
| default | \(-\frac {\left (-3 a -4 b \right ) x^{3}+\left (a +\frac {13 b}{2}\right ) x^{2}+\left (-a -8 b \right ) x -\frac {7 a}{2}+\frac {5 b}{2}}{27 \left (x^{2}-x +1\right )^{2}}-\frac {\left (5 a -2 b \right ) \ln \left (x^{2}-x +1\right )}{54}-\frac {2 \left (-\frac {15 a}{2}-3 b \right ) \sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{81}-\frac {\frac {a}{27}-\frac {b}{27}}{2 \left (1+x \right )^{2}}+\left (-\frac {2 b}{27}+\frac {5 a}{27}\right ) \ln \left (1+x \right )-\frac {-\frac {2 b}{27}+\frac {a}{9}}{1+x}\) | \(131\) |
(2/9*b*x^5+5/18*a*x^4+7/18*b*x^2+4/9*a*x)/(1+x)^2/(x^2-x+1)^2-2/27*ln(1+x) *b+5/27*ln(1+x)*a+1/27*sum(_R*ln((5*_R*a+4*b^2)*x+_R^2+10*b*a),_R=RootOf(_ Z^2+(5*a-2*b)*_Z+25*a^2+10*b*a+4*b^2))
Time = 0.29 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.58 \[ \int \frac {a+b x}{(1+x)^3 \left (1-x+x^2\right )^3} \, dx=\frac {12 \, b x^{5} + 15 \, a x^{4} + 21 \, b x^{2} + 2 \, \sqrt {3} {\left ({\left (5 \, a + 2 \, b\right )} x^{6} + 2 \, {\left (5 \, a + 2 \, b\right )} x^{3} + 5 \, a + 2 \, b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 24 \, a x - {\left ({\left (5 \, a - 2 \, b\right )} x^{6} + 2 \, {\left (5 \, a - 2 \, b\right )} x^{3} + 5 \, a - 2 \, b\right )} \log \left (x^{2} - x + 1\right ) + 2 \, {\left ({\left (5 \, a - 2 \, b\right )} x^{6} + 2 \, {\left (5 \, a - 2 \, b\right )} x^{3} + 5 \, a - 2 \, b\right )} \log \left (x + 1\right )}{54 \, {\left (x^{6} + 2 \, x^{3} + 1\right )}} \]
1/54*(12*b*x^5 + 15*a*x^4 + 21*b*x^2 + 2*sqrt(3)*((5*a + 2*b)*x^6 + 2*(5*a + 2*b)*x^3 + 5*a + 2*b)*arctan(1/3*sqrt(3)*(2*x - 1)) + 24*a*x - ((5*a - 2*b)*x^6 + 2*(5*a - 2*b)*x^3 + 5*a - 2*b)*log(x^2 - x + 1) + 2*((5*a - 2*b )*x^6 + 2*(5*a - 2*b)*x^3 + 5*a - 2*b)*log(x + 1))/(x^6 + 2*x^3 + 1)
Result contains complex when optimal does not.
Time = 0.39 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.89 \[ \int \frac {a+b x}{(1+x)^3 \left (1-x+x^2\right )^3} \, dx=\frac {\left (5 a - 2 b\right ) \log {\left (x + \frac {25 a^{2} \cdot \left (5 a - 2 b\right ) + 40 a b^{2} + 2 b \left (5 a - 2 b\right )^{2}}{125 a^{3} + 8 b^{3}} \right )}}{27} + \left (- \frac {5 a}{54} + \frac {b}{27} - \frac {\sqrt {3} i \left (5 a + 2 b\right )}{54}\right ) \log {\left (x + \frac {675 a^{2} \left (- \frac {5 a}{54} + \frac {b}{27} - \frac {\sqrt {3} i \left (5 a + 2 b\right )}{54}\right ) + 40 a b^{2} + 1458 b \left (- \frac {5 a}{54} + \frac {b}{27} - \frac {\sqrt {3} i \left (5 a + 2 b\right )}{54}\right )^{2}}{125 a^{3} + 8 b^{3}} \right )} + \left (- \frac {5 a}{54} + \frac {b}{27} + \frac {\sqrt {3} i \left (5 a + 2 b\right )}{54}\right ) \log {\left (x + \frac {675 a^{2} \left (- \frac {5 a}{54} + \frac {b}{27} + \frac {\sqrt {3} i \left (5 a + 2 b\right )}{54}\right ) + 40 a b^{2} + 1458 b \left (- \frac {5 a}{54} + \frac {b}{27} + \frac {\sqrt {3} i \left (5 a + 2 b\right )}{54}\right )^{2}}{125 a^{3} + 8 b^{3}} \right )} + \frac {5 a x^{4} + 8 a x + 4 b x^{5} + 7 b x^{2}}{18 x^{6} + 36 x^{3} + 18} \]
(5*a - 2*b)*log(x + (25*a**2*(5*a - 2*b) + 40*a*b**2 + 2*b*(5*a - 2*b)**2) /(125*a**3 + 8*b**3))/27 + (-5*a/54 + b/27 - sqrt(3)*I*(5*a + 2*b)/54)*log (x + (675*a**2*(-5*a/54 + b/27 - sqrt(3)*I*(5*a + 2*b)/54) + 40*a*b**2 + 1 458*b*(-5*a/54 + b/27 - sqrt(3)*I*(5*a + 2*b)/54)**2)/(125*a**3 + 8*b**3)) + (-5*a/54 + b/27 + sqrt(3)*I*(5*a + 2*b)/54)*log(x + (675*a**2*(-5*a/54 + b/27 + sqrt(3)*I*(5*a + 2*b)/54) + 40*a*b**2 + 1458*b*(-5*a/54 + b/27 + sqrt(3)*I*(5*a + 2*b)/54)**2)/(125*a**3 + 8*b**3)) + (5*a*x**4 + 8*a*x + 4 *b*x**5 + 7*b*x**2)/(18*x**6 + 36*x**3 + 18)
Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.91 \[ \int \frac {a+b x}{(1+x)^3 \left (1-x+x^2\right )^3} \, dx=\frac {1}{27} \, \sqrt {3} {\left (5 \, a + 2 \, b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{54} \, {\left (5 \, a - 2 \, b\right )} \log \left (x^{2} - x + 1\right ) + \frac {1}{27} \, {\left (5 \, a - 2 \, b\right )} \log \left (x + 1\right ) + \frac {4 \, b x^{5} + 5 \, a x^{4} + 7 \, b x^{2} + 8 \, a x}{18 \, {\left (x^{6} + 2 \, x^{3} + 1\right )}} \]
1/27*sqrt(3)*(5*a + 2*b)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/54*(5*a - 2*b)* log(x^2 - x + 1) + 1/27*(5*a - 2*b)*log(x + 1) + 1/18*(4*b*x^5 + 5*a*x^4 + 7*b*x^2 + 8*a*x)/(x^6 + 2*x^3 + 1)
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.87 \[ \int \frac {a+b x}{(1+x)^3 \left (1-x+x^2\right )^3} \, dx=\frac {1}{27} \, \sqrt {3} {\left (5 \, a + 2 \, b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{54} \, {\left (5 \, a - 2 \, b\right )} \log \left (x^{2} - x + 1\right ) + \frac {1}{27} \, {\left (5 \, a - 2 \, b\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac {4 \, b x^{5} + 5 \, a x^{4} + 7 \, b x^{2} + 8 \, a x}{18 \, {\left (x^{3} + 1\right )}^{2}} \]
1/27*sqrt(3)*(5*a + 2*b)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/54*(5*a - 2*b)* log(x^2 - x + 1) + 1/27*(5*a - 2*b)*log(abs(x + 1)) + 1/18*(4*b*x^5 + 5*a* x^4 + 7*b*x^2 + 8*a*x)/(x^3 + 1)^2
Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13 \[ \int \frac {a+b x}{(1+x)^3 \left (1-x+x^2\right )^3} \, dx=\ln \left (x+1\right )\,\left (\frac {5\,a}{27}-\frac {2\,b}{27}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {b}{27}-\frac {5\,a}{54}+\frac {\sqrt {3}\,a\,5{}\mathrm {i}}{54}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{27}\right )-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {5\,a}{54}-\frac {b}{27}+\frac {\sqrt {3}\,a\,5{}\mathrm {i}}{54}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{27}\right )+\frac {\frac {2\,b\,x^5}{9}+\frac {5\,a\,x^4}{18}+\frac {7\,b\,x^2}{18}+\frac {4\,a\,x}{9}}{x^6+2\,x^3+1} \]