\(\int \frac {1}{3} (3+e^3-e^{1+x}-2 e x+e^3 \log (\frac {x}{4})) \, dx\) [7686]

Optimal result

Integrand size = 30, antiderivative size = 33 \[ \int \frac {1}{3} \left (3+e^3-e^{1+x}-2 e x+e^3 \log \left (\frac {x}{4}\right )\right ) \, dx=x+\frac {1}{3} e \left (2-e^x-x^2+e^2 \left (5+x \log \left (\frac {x}{4}\right )\right )\right ) \]

[Out]

1/3*(2-exp(x)+exp(2)*(x*ln(1/4*x)+5)-x^2)*exp(1)+x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.52, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 2225, 2332} \[ \int \frac {1}{3} \left (3+e^3-e^{1+x}-2 e x+e^3 \log \left (\frac {x}{4}\right )\right ) \, dx=-\frac {e x^2}{3}+\frac {1}{3} \left (3+e^3\right ) x-\frac {e^3 x}{3}-\frac {e^{x+1}}{3}+\frac {1}{3} e^3 x \log \left (\frac {x}{4}\right ) \]

[In]

Int[(3 + E^3 - E^(1 + x) - 2*E*x + E^3*Log[x/4])/3,x]

[Out]

-1/3*E^(1 + x) - (E^3*x)/3 + ((3 + E^3)*x)/3 - (E*x^2)/3 + (E^3*x*Log[x/4])/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \left (3+e^3-e^{1+x}-2 e x+e^3 \log \left (\frac {x}{4}\right )\right ) \, dx \\ & = \frac {1}{3} \left (3+e^3\right ) x-\frac {e x^2}{3}-\frac {1}{3} \int e^{1+x} \, dx+\frac {1}{3} e^3 \int \log \left (\frac {x}{4}\right ) \, dx \\ & = -\frac {e^{1+x}}{3}-\frac {e^3 x}{3}+\frac {1}{3} \left (3+e^3\right ) x-\frac {e x^2}{3}+\frac {1}{3} e^3 x \log \left (\frac {x}{4}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {1}{3} \left (3+e^3-e^{1+x}-2 e x+e^3 \log \left (\frac {x}{4}\right )\right ) \, dx=\frac {1}{3} \left (-e^{1+x}+3 x-e x^2+e^3 x \log \left (\frac {x}{4}\right )\right ) \]

[In]

Integrate[(3 + E^3 - E^(1 + x) - 2*E*x + E^3*Log[x/4])/3,x]

[Out]

(-E^(1 + x) + 3*x - E*x^2 + E^3*x*Log[x/4])/3

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76

method	result	size
risch	\(\frac {{\mathrm e}^{3} x \ln \left (\frac {x}{4}\right )}{3}-\frac {{\mathrm e}^{1+x}}{3}-\frac {x^{2} {\mathrm e}}{3}+x\)	\(25\)
norman	\(x -\frac {x^{2} {\mathrm e}}{3}-\frac {{\mathrm e} \,{\mathrm e}^{x}}{3}+\frac {x \,{\mathrm e} \,{\mathrm e}^{2} \ln \left (\frac {x}{4}\right )}{3}\)	\(27\)
default	\(x +\frac {x \,{\mathrm e} \,{\mathrm e}^{2}}{3}+\frac {{\mathrm e} \,{\mathrm e}^{2} \left (x \ln \left (\frac {x}{4}\right )-x \right )}{3}-\frac {x^{2} {\mathrm e}}{3}-\frac {{\mathrm e} \,{\mathrm e}^{x}}{3}\)	\(39\)
parts	\(x +\frac {x \,{\mathrm e} \,{\mathrm e}^{2}}{3}+\frac {{\mathrm e} \,{\mathrm e}^{2} \left (x \ln \left (\frac {x}{4}\right )-x \right )}{3}-\frac {x^{2} {\mathrm e}}{3}-\frac {{\mathrm e} \,{\mathrm e}^{x}}{3}\)	\(39\)
parallelrisch	\(\frac {x \,{\mathrm e} \,{\mathrm e}^{2} \ln \left (\frac {x}{4}\right )}{3}-\frac {x \,{\mathrm e} \,{\mathrm e}^{2}}{3}-\frac {x^{2} {\mathrm e}}{3}-\frac {{\mathrm e} \,{\mathrm e}^{x}}{3}+\left (1+\frac {{\mathrm e} \,{\mathrm e}^{2}}{3}\right ) x\)	\(43\)

[In]

int(1/3*exp(1)*exp(2)*ln(1/4*x)-1/3*exp(1)*exp(x)+1/3*exp(1)*exp(2)-2/3*x*exp(1)+1,x,method=_RETURNVERBOSE)

[Out]

1/3*exp(3)*x*ln(1/4*x)-1/3*exp(1+x)-1/3*x^2*exp(1)+x

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.73 \[ \int \frac {1}{3} \left (3+e^3-e^{1+x}-2 e x+e^3 \log \left (\frac {x}{4}\right )\right ) \, dx=-\frac {1}{3} \, x^{2} e + \frac {1}{3} \, x e^{3} \log \left (\frac {1}{4} \, x\right ) + x - \frac {1}{3} \, e^{\left (x + 1\right )} \]

[In]

integrate(1/3*exp(1)*exp(2)*log(1/4*x)-1/3*exp(1)*exp(x)+1/3*exp(1)*exp(2)-2/3*x*exp(1)+1,x, algorithm="fricas
")

[Out]

-1/3*x^2*e + 1/3*x*e^3*log(1/4*x) + x - 1/3*e^(x + 1)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {1}{3} \left (3+e^3-e^{1+x}-2 e x+e^3 \log \left (\frac {x}{4}\right )\right ) \, dx=- \frac {e x^{2}}{3} + \frac {x e^{3} \log {\left (\frac {x}{4} \right )}}{3} + x - \frac {e e^{x}}{3} \]

[In]

integrate(1/3*exp(1)*exp(2)*ln(1/4*x)-1/3*exp(1)*exp(x)+1/3*exp(1)*exp(2)-2/3*x*exp(1)+1,x)

[Out]

-E*x**2/3 + x*exp(3)*log(x/4)/3 + x - E*exp(x)/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {1}{3} \left (3+e^3-e^{1+x}-2 e x+e^3 \log \left (\frac {x}{4}\right )\right ) \, dx=-\frac {1}{3} \, x^{2} e + \frac {1}{3} \, {\left (x \log \left (\frac {1}{4} \, x\right ) - x\right )} e^{3} + \frac {1}{3} \, x e^{3} + x - \frac {1}{3} \, e^{\left (x + 1\right )} \]

[In]

integrate(1/3*exp(1)*exp(2)*log(1/4*x)-1/3*exp(1)*exp(x)+1/3*exp(1)*exp(2)-2/3*x*exp(1)+1,x, algorithm="maxima
")

[Out]

-1/3*x^2*e + 1/3*(x*log(1/4*x) - x)*e^3 + 1/3*x*e^3 + x - 1/3*e^(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {1}{3} \left (3+e^3-e^{1+x}-2 e x+e^3 \log \left (\frac {x}{4}\right )\right ) \, dx=-\frac {1}{3} \, x^{2} e + \frac {1}{3} \, {\left (x \log \left (\frac {1}{4} \, x\right ) - x\right )} e^{3} + \frac {1}{3} \, x e^{3} + x - \frac {1}{3} \, e^{\left (x + 1\right )} \]

[In]

integrate(1/3*exp(1)*exp(2)*log(1/4*x)-1/3*exp(1)*exp(x)+1/3*exp(1)*exp(2)-2/3*x*exp(1)+1,x, algorithm="giac")

[Out]

-1/3*x^2*e + 1/3*(x*log(1/4*x) - x)*e^3 + 1/3*x*e^3 + x - 1/3*e^(x + 1)

Mupad [B] (verification not implemented)

Time = 13.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {1}{3} \left (3+e^3-e^{1+x}-2 e x+e^3 \log \left (\frac {x}{4}\right )\right ) \, dx=x-\frac {{\mathrm {e}}^{x+1}}{3}-\frac {x^2\,\mathrm {e}}{3}+\frac {x\,{\mathrm {e}}^3\,\ln \left (x\right )}{3}-\frac {2\,x\,{\mathrm {e}}^3\,\ln \left (2\right )}{3} \]

[In]

int(exp(3)/3 + (log(x/4)*exp(3))/3 - (2*x*exp(1))/3 - (exp(1)*exp(x))/3 + 1,x)

[Out]

x - exp(x + 1)/3 - (x^2*exp(1))/3 + (x*exp(3)*log(x))/3 - (2*x*exp(3)*log(2))/3