\(\int \frac {x}{3+3 e^x+e^{2 x}} \, dx\) [436]

Optimal result

Integrand size = 16, antiderivative size = 204 \[ \int \frac {x}{3+3 e^x+e^{2 x}} \, dx=-\frac {x^2}{\sqrt {3} \left (3 i-\sqrt {3}\right )}+\frac {x^2}{\sqrt {3} \left (3 i+\sqrt {3}\right )}-\frac {2 x \log \left (1+\frac {2 e^x}{3-i \sqrt {3}}\right )}{\sqrt {3} \left (3 i+\sqrt {3}\right )}+\frac {2 x \log \left (1+\frac {2 e^x}{3+i \sqrt {3}}\right )}{\sqrt {3} \left (3 i-\sqrt {3}\right )}-\frac {2 \operatorname {PolyLog}\left (2,-\frac {2 e^x}{3-i \sqrt {3}}\right )}{\sqrt {3} \left (3 i+\sqrt {3}\right )}+\frac {2 \operatorname {PolyLog}\left (2,-\frac {2 e^x}{3+i \sqrt {3}}\right )}{\sqrt {3} \left (3 i-\sqrt {3}\right )} \] Output:

-1/3*x^2*3^(1/2)/(3*I-3^(1/2))+1/3*x^2*3^(1/2)/(3*I+3^(1/2))-2/3*x*ln(1+2* 
exp(x)/(3-I*3^(1/2)))*3^(1/2)/(3*I+3^(1/2))+2/3*x*ln(1+2*exp(x)/(3+I*3^(1/ 
2)))*3^(1/2)/(3*I-3^(1/2))-2/3*polylog(2,-2*exp(x)/(3-I*3^(1/2)))*3^(1/2)/ 
(3*I+3^(1/2))+2/3*polylog(2,-2*exp(x)/(3+I*3^(1/2)))*3^(1/2)/(3*I-3^(1/2))
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.71 \[ \int \frac {x}{3+3 e^x+e^{2 x}} \, dx=\frac {-x \left (\left (-3 i+\sqrt {3}\right ) \log \left (1+\frac {1}{2} \left (3-i \sqrt {3}\right ) e^{-x}\right )+\left (3 i+\sqrt {3}\right ) \log \left (1+\frac {1}{2} \left (3+i \sqrt {3}\right ) e^{-x}\right )\right )+\left (3 i+\sqrt {3}\right ) \operatorname {PolyLog}\left (2,-\frac {1}{2} i \left (-3 i+\sqrt {3}\right ) e^{-x}\right )+\left (-3 i+\sqrt {3}\right ) \operatorname {PolyLog}\left (2,\frac {1}{2} i \left (3 i+\sqrt {3}\right ) e^{-x}\right )}{6 \sqrt {3}} \] Input:

Integrate[x/(3 + 3*E^x + E^(2*x)),x]
 

Output:

(-(x*((-3*I + Sqrt[3])*Log[1 + (3 - I*Sqrt[3])/(2*E^x)] + (3*I + Sqrt[3])* 
Log[1 + (3 + I*Sqrt[3])/(2*E^x)])) + (3*I + Sqrt[3])*PolyLog[2, ((-1/2*I)* 
(-3*I + Sqrt[3]))/E^x] + (-3*I + Sqrt[3])*PolyLog[2, ((I/2)*(3*I + Sqrt[3] 
))/E^x])/(6*Sqrt[3])
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2693, 2615, 2620, 2715, 2838}

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.

\(\displaystyle \int \frac {x}{3 e^x+e^{2 x}+3} \, dx\)

\(\Big \downarrow \) 2693

\(\displaystyle \frac {2 i \int \frac {x}{3+i \sqrt {3}+2 e^x}dx}{\sqrt {3}}-\frac {2 i \int \frac {x}{3-i \sqrt {3}+2 e^x}dx}{\sqrt {3}}\)

\(\Big \downarrow \) 2615

\(\displaystyle \frac {2 i \left (\frac {x^2}{2 \left (3+i \sqrt {3}\right )}-\frac {2 \int \frac {e^x x}{3+i \sqrt {3}+2 e^x}dx}{3+i \sqrt {3}}\right )}{\sqrt {3}}-\frac {2 i \left (\frac {x^2}{2 \left (3-i \sqrt {3}\right )}-\frac {2 \int \frac {e^x x}{3-i \sqrt {3}+2 e^x}dx}{3-i \sqrt {3}}\right )}{\sqrt {3}}\)

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

\(\displaystyle \frac {2 i \left (\frac {x^2}{2 \left (3+i \sqrt {3}\right )}-\frac {2 \left (\frac {1}{2} x \log \left (1+\frac {2 e^x}{3+i \sqrt {3}}\right )-\frac {1}{2} \int e^{-x} \log \left (1+\frac {2 e^x}{3+i \sqrt {3}}\right )de^x\right )}{3+i \sqrt {3}}\right )}{\sqrt {3}}-\frac {2 i \left (\frac {x^2}{2 \left (3-i \sqrt {3}\right )}-\frac {2 \left (\frac {1}{2} x \log \left (1+\frac {2 e^x}{3-i \sqrt {3}}\right )-\frac {1}{2} \int e^{-x} \log \left (1+\frac {2 e^x}{3-i \sqrt {3}}\right )de^x\right )}{3-i \sqrt {3}}\right )}{\sqrt {3}}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {2 i \left (\frac {x^2}{2 \left (3+i \sqrt {3}\right )}-\frac {2 \left (\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {2 e^x}{3+i \sqrt {3}}\right )+\frac {1}{2} x \log \left (1+\frac {2 e^x}{3+i \sqrt {3}}\right )\right )}{3+i \sqrt {3}}\right )}{\sqrt {3}}-\frac {2 i \left (\frac {x^2}{2 \left (3-i \sqrt {3}\right )}-\frac {2 \left (\frac {1}{2} \operatorname {PolyLog}\left (2,-\frac {2 e^x}{3-i \sqrt {3}}\right )+\frac {1}{2} x \log \left (1+\frac {2 e^x}{3-i \sqrt {3}}\right )\right )}{3-i \sqrt {3}}\right )}{\sqrt {3}}\)

Input:

Int[x/(3 + 3*E^x + E^(2*x)),x]
 

Output:

((-2*I)*(x^2/(2*(3 - I*Sqrt[3])) - (2*((x*Log[1 + (2*E^x)/(3 - I*Sqrt[3])] 
)/2 + PolyLog[2, (-2*E^x)/(3 - I*Sqrt[3])]/2))/(3 - I*Sqrt[3])))/Sqrt[3] + 
 ((2*I)*(x^2/(2*(3 + I*Sqrt[3])) - (2*((x*Log[1 + (2*E^x)/(3 + I*Sqrt[3])] 
)/2 + PolyLog[2, (-2*E^x)/(3 + I*Sqrt[3])]/2))/(3 + I*Sqrt[3])))/Sqrt[3]
 

Defintions of rubi rules used

rule 2615

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x 
_))))^(n_.)), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(a*d*(m + 1)), x] - Simp[ 
b/a   Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n)), x] 
, x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2693

Int[((f_.) + (g_.)*(x_))^(m_.)/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), 
 x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q)   Int[(f + g*x)^m 
/(b - q + 2*c*F^u), x], x] - Simp[2*(c/q)   Int[(f + g*x)^m/(b + q + 2*c*F^ 
u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] 
 && NeQ[b^2 - 4*a*c, 0] && IGtQ[m, 0]
 

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {x^{2}}{6}+\frac {i \sqrt {3}\, x \ln \left (\frac {i \sqrt {3}-2 \,{\mathrm e}^{x}-3}{i \sqrt {3}-3}\right )}{6}-\frac {x \ln \left (\frac {i \sqrt {3}-2 \,{\mathrm e}^{x}-3}{i \sqrt {3}-3}\right )}{6}-\frac {i \sqrt {3}\, x \ln \left (\frac {i \sqrt {3}+2 \,{\mathrm e}^{x}+3}{3+i \sqrt {3}}\right )}{6}-\frac {x \ln \left (\frac {i \sqrt {3}+2 \,{\mathrm e}^{x}+3}{3+i \sqrt {3}}\right )}{6}+\frac {i \sqrt {3}\, \operatorname {dilog}\left (\frac {i \sqrt {3}-2 \,{\mathrm e}^{x}-3}{i \sqrt {3}-3}\right )}{6}-\frac {\operatorname {dilog}\left (\frac {i \sqrt {3}-2 \,{\mathrm e}^{x}-3}{i \sqrt {3}-3}\right )}{6}-\frac {i \sqrt {3}\, \operatorname {dilog}\left (\frac {i \sqrt {3}+2 \,{\mathrm e}^{x}+3}{3+i \sqrt {3}}\right )}{6}-\frac {\operatorname {dilog}\left (\frac {i \sqrt {3}+2 \,{\mathrm e}^{x}+3}{3+i \sqrt {3}}\right )}{6}\)	\(235\)
risch	\(\frac {x^{2}}{6}+\frac {i \sqrt {3}\, x \ln \left (\frac {i \sqrt {3}-2 \,{\mathrm e}^{x}-3}{i \sqrt {3}-3}\right )}{6}-\frac {x \ln \left (\frac {i \sqrt {3}-2 \,{\mathrm e}^{x}-3}{i \sqrt {3}-3}\right )}{6}-\frac {i \sqrt {3}\, x \ln \left (\frac {i \sqrt {3}+2 \,{\mathrm e}^{x}+3}{3+i \sqrt {3}}\right )}{6}-\frac {x \ln \left (\frac {i \sqrt {3}+2 \,{\mathrm e}^{x}+3}{3+i \sqrt {3}}\right )}{6}+\frac {i \sqrt {3}\, \operatorname {dilog}\left (\frac {i \sqrt {3}-2 \,{\mathrm e}^{x}-3}{i \sqrt {3}-3}\right )}{6}-\frac {\operatorname {dilog}\left (\frac {i \sqrt {3}-2 \,{\mathrm e}^{x}-3}{i \sqrt {3}-3}\right )}{6}-\frac {i \sqrt {3}\, \operatorname {dilog}\left (\frac {i \sqrt {3}+2 \,{\mathrm e}^{x}+3}{3+i \sqrt {3}}\right )}{6}-\frac {\operatorname {dilog}\left (\frac {i \sqrt {3}+2 \,{\mathrm e}^{x}+3}{3+i \sqrt {3}}\right )}{6}\)	\(235\)

Input:

int(x/(3+3*exp(x)+exp(2*x)),x,method=_RETURNVERBOSE)
 

Output:

1/6*x^2+1/6*I*3^(1/2)*x*ln((I*3^(1/2)-2*exp(x)-3)/(I*3^(1/2)-3))-1/6*x*ln( 
(I*3^(1/2)-2*exp(x)-3)/(I*3^(1/2)-3))-1/6*I*3^(1/2)*x*ln((I*3^(1/2)+2*exp( 
x)+3)/(3+I*3^(1/2)))-1/6*x*ln((I*3^(1/2)+2*exp(x)+3)/(3+I*3^(1/2)))+1/6*I* 
3^(1/2)*dilog((I*3^(1/2)-2*exp(x)-3)/(I*3^(1/2)-3))-1/6*dilog((I*3^(1/2)-2 
*exp(x)-3)/(I*3^(1/2)-3))-1/6*I*3^(1/2)*dilog((I*3^(1/2)+2*exp(x)+3)/(3+I* 
3^(1/2)))-1/6*dilog((I*3^(1/2)+2*exp(x)+3)/(3+I*3^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.44 \[ \int \frac {x}{3+3 e^x+e^{2 x}} \, dx=\frac {1}{6} \, x^{2} + \frac {1}{6} \, {\left (3 \, \sqrt {-\frac {1}{3}} - 1\right )} {\rm Li}_2\left (-\frac {1}{2} \, {\left (\sqrt {-\frac {1}{3}} + 1\right )} e^{x}\right ) - \frac {1}{6} \, {\left (3 \, \sqrt {-\frac {1}{3}} + 1\right )} {\rm Li}_2\left (\frac {1}{2} \, {\left (\sqrt {-\frac {1}{3}} - 1\right )} e^{x}\right ) + \frac {1}{6} \, {\left (3 \, \sqrt {-\frac {1}{3}} x - x\right )} \log \left (\frac {1}{2} \, {\left (\sqrt {-\frac {1}{3}} + 1\right )} e^{x} + 1\right ) - \frac {1}{6} \, {\left (3 \, \sqrt {-\frac {1}{3}} x + x\right )} \log \left (-\frac {1}{2} \, {\left (\sqrt {-\frac {1}{3}} - 1\right )} e^{x} + 1\right ) \] Input:

integrate(x/(3+3*exp(x)+exp(2*x)),x, algorithm="fricas")
 

Output:

1/6*x^2 + 1/6*(3*sqrt(-1/3) - 1)*dilog(-1/2*(sqrt(-1/3) + 1)*e^x) - 1/6*(3 
*sqrt(-1/3) + 1)*dilog(1/2*(sqrt(-1/3) - 1)*e^x) + 1/6*(3*sqrt(-1/3)*x - x 
)*log(1/2*(sqrt(-1/3) + 1)*e^x + 1) - 1/6*(3*sqrt(-1/3)*x + x)*log(-1/2*(s 
qrt(-1/3) - 1)*e^x + 1)
 

Sympy [F]

\[ \int \frac {x}{3+3 e^x+e^{2 x}} \, dx=\int \frac {x}{e^{2 x} + 3 e^{x} + 3}\, dx \] Input:

integrate(x/(3+3*exp(x)+exp(2*x)),x)
 

Output:

Integral(x/(exp(2*x) + 3*exp(x) + 3), x)
 

Maxima [F]

\[ \int \frac {x}{3+3 e^x+e^{2 x}} \, dx=\int { \frac {x}{e^{\left (2 \, x\right )} + 3 \, e^{x} + 3} \,d x } \] Input:

integrate(x/(3+3*exp(x)+exp(2*x)),x, algorithm="maxima")
 

Output:

integrate(x/(e^(2*x) + 3*e^x + 3), x)
 

Giac [F]

\[ \int \frac {x}{3+3 e^x+e^{2 x}} \, dx=\int { \frac {x}{e^{\left (2 \, x\right )} + 3 \, e^{x} + 3} \,d x } \] Input:

integrate(x/(3+3*exp(x)+exp(2*x)),x, algorithm="giac")
 

Output:

integrate(x/(e^(2*x) + 3*e^x + 3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{3+3 e^x+e^{2 x}} \, dx=\int \frac {x}{{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^x+3} \,d x \] Input:

int(x/(exp(2*x) + 3*exp(x) + 3),x)
 

Output:

int(x/(exp(2*x) + 3*exp(x) + 3), x)
 

Reduce [F]

\[ \int \frac {x}{3+3 e^x+e^{2 x}} \, dx=\int \frac {x}{e^{2 x}+3 e^{x}+3}d x \] Input:

int(x/(3+3*exp(x)+exp(2*x)),x)
 

Output:

int(x/(e**(2*x) + 3*e**x + 3),x)