\(\int \frac {x^{2016} (2017+2018 x)}{1+x^{4034} (1+x)^2} \, dx\) [2]

Optimal result

Integrand size = 22, antiderivative size = 8 \[ \int \frac {x^{2016} (2017+2018 x)}{1+x^{4034} (1+x)^2} \, dx=\arctan \left (x^{2017} (1+x)\right ) \] Output:

arctan(x^2017*(1+x))
 

Mathematica [F(-1)]

Timed out. \[ \int \frac {x^{2016} (2017+2018 x)}{1+x^{4034} (1+x)^2} \, dx=\text {\$Aborted} \] Input:

Integrate[(x^2016*(2017 + 2018*x))/(1 + x^4034*(1 + x)^2),x]
 

Output:

$Aborted
 

Rubi [A] (verified)

Time = 32.60 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {7260, 216}

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.

\(\displaystyle \int \frac {x^{2016} (2018 x+2017)}{(x+1)^2 x^{4034}+1} \, dx\)

\(\Big \downarrow \) 7260

\(\displaystyle \int \frac {1}{(x+1)^2 x^{4034}+1}d\left (x^{2017} (x+1)\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \arctan \left (x^{2017} (x+1)\right )\)

Input:

Int[(x^2016*(2017 + 2018*x))/(1 + x^4034*(1 + x)^2),x]
 

Output:

ArcTan[x^2017*(1 + x)]
 

Defintions of rubi rules used

rule 216

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

rule 7260

Int[(u_)*(v_)^(r_.)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(q_.))^(m_.), x_Symbol] : 
> With[{c = Simplify[u/(p*w*D[v, x] + q*v*D[w, x])]}, Simp[c*(p/(r + 1)) 
Subst[Int[(a + b*x^(p/(r + 1)))^m, x], x, v^(r + 1)*w], x] /; FreeQ[c, x]] 
/; FreeQ[{a, b, m, p, q, r}, x] && EqQ[p, q*(r + 1)] && NeQ[r, -1] && Integ 
erQ[p/(r + 1)]
 

Maple [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.12

\[\arctan \left (x^{2018}+x^{2017}\right )\]

Input:

int(x^2016*(2017+2018*x)/(1+x^4034*(x+1)^2),x)
 

Output:

arctan(x^2018+x^2017)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {x^{2016} (2017+2018 x)}{1+x^{4034} (1+x)^2} \, dx=\text {Timed out} \] Input:

integrate(x^2016*(2017+2018*x)/(1+x^4034*(1+x)^2),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{2016} (2017+2018 x)}{1+x^{4034} (1+x)^2} \, dx=\text {Timed out} \] Input:

integrate(x**2016*(2017+2018*x)/(1+x**4034*(1+x)**2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{2016} (2017+2018 x)}{1+x^{4034} (1+x)^2} \, dx=\int { \frac {{\left (2018 \, x + 2017\right )} x^{2016}}{{\left (x + 1\right )}^{2} x^{4034} + 1} \,d x } \] Input:

integrate(x^2016*(2017+2018*x)/(1+x^4034*(1+x)^2),x, algorithm="maxima")
 

Output:

integrate((2018*x + 2017)*x^2016/((x + 1)^2*x^4034 + 1), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {x^{2016} (2017+2018 x)}{1+x^{4034} (1+x)^2} \, dx=\text {Timed out} \] Input:

integrate(x^2016*(2017+2018*x)/(1+x^4034*(1+x)^2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{2016} (2017+2018 x)}{1+x^{4034} (1+x)^2} \, dx=\int \frac {x^{2016}\,\left (2018\,x+2017\right )}{x^{4034}\,{\left (x+1\right )}^2+1} \,d x \] Input:

int((x^2016*(2018*x + 2017))/(x^4034*(x + 1)^2 + 1),x)
 

Output:

int((x^2016*(2018*x + 2017))/(x^4034*(x + 1)^2 + 1), x)
 

Reduce [F]

\[ \int \frac {x^{2016} (2017+2018 x)}{1+x^{4034} (1+x)^2} \, dx=2018 \left (\int \frac {x^{2017}}{x^{4036}+2 x^{4035}+x^{4034}+1}d x \right )+2017 \left (\int \frac {x^{2016}}{x^{4036}+2 x^{4035}+x^{4034}+1}d x \right ) \] Input:

int(x^2016*(2017+2018*x)/(1+x^4034*(1+x)^2),x)
 

Output:

2018*int(x**2017/(x**4036 + 2*x**4035 + x**4034 + 1),x) + 2017*int(x**2016 
/(x**4036 + 2*x**4035 + x**4034 + 1),x)