\(\int \frac {a+b x}{(1+x^3)^2} \, dx\) [6]

Optimal result

Integrand size = 13, antiderivative size = 79 \[ \int \frac {a+b x}{\left (1+x^3\right )^2} \, dx=\frac {x (a+b x)}{3 \left (1+x^3\right )}-\frac {(2 a+b) \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{9} (2 a-b) \log (1+x)-\frac {1}{18} (2 a-b) \log \left (1-x+x^2\right ) \] Output:

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

Mathematica [A] (verified)

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

Integrate[(a + b*x)/(1 + x^3)^2,x]
 

Output:

((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
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {2394, 25, 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.

\(\displaystyle \int \frac {a+b x}{\left (x^3+1\right )^2} \, dx\)

\(\Big \downarrow \) 2394

\(\displaystyle \frac {x (a+b x)}{3 \left (x^3+1\right )}-\frac {1}{3} \int -\frac {2 a+b x}{x^3+1}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \int \frac {2 a+b x}{x^3+1}dx+\frac {x (a+b x)}{3 \left (x^3+1\right )}\)

\(\Big \downarrow \) 2399

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \int \frac {4 a+b-(2 a-b) x}{x^2-x+1}dx+\frac {1}{3} (2 a-b) \int \frac {1}{x+1}dx\right )+\frac {x (a+b x)}{3 \left (x^3+1\right )}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \int \frac {4 a+b-(2 a-b) x}{x^2-x+1}dx+\frac {1}{3} (2 a-b) \log (x+1)\right )+\frac {x (a+b x)}{3 \left (x^3+1\right )}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {3}{2} (2 a+b) \int \frac {1}{x^2-x+1}dx-\frac {1}{2} (2 a-b) \int -\frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{3} (2 a-b) \log (x+1)\right )+\frac {x (a+b x)}{3 \left (x^3+1\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {3}{2} (2 a+b) \int \frac {1}{x^2-x+1}dx+\frac {1}{2} (2 a-b) \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{3} (2 a-b) \log (x+1)\right )+\frac {x (a+b x)}{3 \left (x^3+1\right )}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{2} (2 a-b) \int \frac {1-2 x}{x^2-x+1}dx-3 (2 a+b) \int \frac {1}{-(2 x-1)^2-3}d(2 x-1)\right )+\frac {1}{3} (2 a-b) \log (x+1)\right )+\frac {x (a+b x)}{3 \left (x^3+1\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{2} (2 a-b) \int \frac {1-2 x}{x^2-x+1}dx+\sqrt {3} (2 a+b) \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )\right )+\frac {1}{3} (2 a-b) \log (x+1)\right )+\frac {x (a+b x)}{3 \left (x^3+1\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{3} \left (\frac {1}{3} \left (\sqrt {3} (2 a+b) \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )-\frac {1}{2} (2 a-b) \log \left (x^2-x+1\right )\right )+\frac {1}{3} (2 a-b) \log (x+1)\right )+\frac {x (a+b x)}{3 \left (x^3+1\right )}\)

Input:

Int[(a + b*x)/(1 + x^3)^2,x]
 

Output:

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

Defintions of rubi rules used

rule 16

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

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 217

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 1083

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]
 

rule 1103

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]
 

rule 1142

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]
 

rule 2394

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]
 

rule 2399

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]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.10

method	result	size
risch	\(\frac {\frac {1}{3} b \,x^{2}+\frac {1}{3} a x}{x^{3}+1}-\frac {\ln \left (x +1\right ) b}{9}+\frac {2 \ln \left (x +1\right ) a}{9}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{2}+\left (2 a -b \right ) \textit {\_Z} +4 a^{2}+2 a b +b^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (2 \textit {\_R} a +b^{2}\right ) x +\textit {\_R}^{2}+2 a b \right )\right )}{9}\)	\(87\)
default	\(-\frac {\left (-2 b -a \right ) x +b -a}{9 \left (x^{2}-x +1\right )}-\frac {\left (2 a -b \right ) \ln \left (x^{2}-x +1\right )}{18}-\frac {2 \left (-3 a -\frac {3 b}{2}\right ) \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{27}+\left (-\frac {b}{9}+\frac {2 a}{9}\right ) \ln \left (x +1\right )-\frac {-\frac {b}{9}+\frac {a}{9}}{x +1}\)	\(95\)
meijerg	\(\frac {b \left (\frac {3 x^{2}}{3 x^{3}+3}-\frac {x^{2} \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}\right )}{3}+\frac {a \left (\frac {3 x}{3 x^{3}+3}+\frac {2 x \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}-\frac {x \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {2 x \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}\right )}{3}\)	\(186\)

Input:

int((b*x+a)/(x^3+1)^2,x,method=_RETURNVERBOSE)
 

Output:

(1/3*b*x^2+1/3*a*x)/(x^3+1)-1/9*ln(x+1)*b+2/9*ln(x+1)*a+1/9*sum(_R*ln((2*_ 
R*a+b^2)*x+_R^2+2*a*b),_R=RootOf(_Z^2+(2*a-b)*_Z+4*a^2+2*a*b+b^2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.30 \[ \int \frac {a+b x}{\left (1+x^3\right )^2} \, dx=\frac {6 \, b x^{2} + 2 \, \sqrt {3} {\left ({\left (2 \, a + b\right )} x^{3} + 2 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 6 \, a x - {\left ({\left (2 \, a - b\right )} x^{3} + 2 \, a - b\right )} \log \left (x^{2} - x + 1\right ) + 2 \, {\left ({\left (2 \, a - b\right )} x^{3} + 2 \, a - b\right )} \log \left (x + 1\right )}{18 \, {\left (x^{3} + 1\right )}} \] Input:

integrate((b*x+a)/(x^3+1)^2,x, algorithm="fricas")
 

Output:

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 = 0.37 (sec) , antiderivative size = 238, normalized size of antiderivative = 3.01 \[ \int \frac {a+b x}{\left (1+x^3\right )^2} \, dx=\frac {\left (2 a - b\right ) \log {\left (x + \frac {4 a^{2} \cdot \left (2 a - b\right ) + 4 a b^{2} + b \left (2 a - b\right )^{2}}{8 a^{3} + b^{3}} \right )}}{9} + \left (- \frac {a}{9} + \frac {b}{18} - \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right ) \log {\left (x + \frac {36 a^{2} \left (- \frac {a}{9} + \frac {b}{18} - \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right ) + 4 a b^{2} + 81 b \left (- \frac {a}{9} + \frac {b}{18} - \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right )^{2}}{8 a^{3} + b^{3}} \right )} + \left (- \frac {a}{9} + \frac {b}{18} + \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right ) \log {\left (x + \frac {36 a^{2} \left (- \frac {a}{9} + \frac {b}{18} + \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right ) + 4 a b^{2} + 81 b \left (- \frac {a}{9} + \frac {b}{18} + \frac {\sqrt {3} i \left (2 a + b\right )}{18}\right )^{2}}{8 a^{3} + b^{3}} \right )} + \frac {a x + b x^{2}}{3 x^{3} + 3} \] Input:

integrate((b*x+a)/(x**3+1)**2,x)
 

Output:

(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 + 8 
1*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)

Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \frac {a+b x}{\left (1+x^3\right )^2} \, dx=\frac {1}{9} \, \sqrt {3} {\left (2 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{18} \, {\left (2 \, a - b\right )} \log \left (x^{2} - x + 1\right ) + \frac {1}{9} \, {\left (2 \, a - b\right )} \log \left (x + 1\right ) + \frac {b x^{2} + a x}{3 \, {\left (x^{3} + 1\right )}} \] Input:

integrate((b*x+a)/(x^3+1)^2,x, algorithm="maxima")
 

Output:

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)

Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91 \[ \int \frac {a+b x}{\left (1+x^3\right )^2} \, dx=\frac {1}{9} \, \sqrt {3} {\left (2 \, a + b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{18} \, {\left (2 \, a - b\right )} \log \left (x^{2} - x + 1\right ) + \frac {1}{9} \, {\left (2 \, a - b\right )} \log \left ({\left | x + 1 \right |}\right ) + \frac {b x^{2} + a x}{3 \, {\left (x^{3} + 1\right )}} \] Input:

integrate((b*x+a)/(x^3+1)^2,x, algorithm="giac")
 

Output:

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(abs(x + 1)) + 1/3*(b*x^2 + a*x)/(x^3 + 1)
 

Mupad [B] (verification not implemented)

Time = 5.67 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.23 \[ \int \frac {a+b x}{\left (1+x^3\right )^2} \, dx=\frac {\frac {b\,x^2}{3}+\frac {a\,x}{3}}{x^3+1}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {a}{9}-\frac {b}{18}+\frac {\sqrt {3}\,a\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{18}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {b}{18}-\frac {a}{9}+\frac {\sqrt {3}\,a\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{18}\right )+\ln \left (x+1\right )\,\left (\frac {2\,a}{9}-\frac {b}{9}\right ) \] Input:

int((a + b*x)/(x^3 + 1)^2,x)
 

Output:

((a*x)/3 + (b*x^2)/3)/(x^3 + 1) - log(x - (3^(1/2)*1i)/2 - 1/2)*(a/9 - b/1 
8 + (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)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.24 \[ \int \frac {a+b x}{\left (1+x^3\right )^2} \, dx=\frac {4 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) a \,x^{3}+4 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) a +2 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) b \,x^{3}+2 \sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) b -2 \,\mathrm {log}\left (x^{2}-x +1\right ) a \,x^{3}-2 \,\mathrm {log}\left (x^{2}-x +1\right ) a +\mathrm {log}\left (x^{2}-x +1\right ) b \,x^{3}+\mathrm {log}\left (x^{2}-x +1\right ) b +4 \,\mathrm {log}\left (x +1\right ) a \,x^{3}+4 \,\mathrm {log}\left (x +1\right ) a -2 \,\mathrm {log}\left (x +1\right ) b \,x^{3}-2 \,\mathrm {log}\left (x +1\right ) b +6 a x +6 b \,x^{2}}{18 x^{3}+18} \] Input:

int((b*x+a)/(x^3+1)^2,x)
 

Output:

(4*sqrt(3)*atan((2*x - 1)/sqrt(3))*a*x**3 + 4*sqrt(3)*atan((2*x - 1)/sqrt( 
3))*a + 2*sqrt(3)*atan((2*x - 1)/sqrt(3))*b*x**3 + 2*sqrt(3)*atan((2*x - 1 
)/sqrt(3))*b - 2*log(x**2 - x + 1)*a*x**3 - 2*log(x**2 - x + 1)*a + log(x* 
*2 - x + 1)*b*x**3 + log(x**2 - x + 1)*b + 4*log(x + 1)*a*x**3 + 4*log(x + 
 1)*a - 2*log(x + 1)*b*x**3 - 2*log(x + 1)*b + 6*a*x + 6*b*x**2)/(18*(x**3 
 + 1))