\(\int (2-e^x+\log (\frac {2 x^2}{3})) \, dx\) [9810]

Optimal result

Integrand size = 15, antiderivative size = 19 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \left (-\frac {e^x}{x}+\log \left (\frac {2 x^2}{3}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2225, 2332} \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \log \left (\frac {2 x^2}{3}\right )-e^x \]

[In]

Int[2 - E^x + Log[(2*x^2)/3],x]

[Out]

-E^x + x*Log[(2*x^2)/3]

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}& = 2 x-\int e^x \, dx+\int \log \left (\frac {2 x^2}{3}\right ) \, dx \\ & = -e^x+x \log \left (\frac {2 x^2}{3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=-e^x+x \log \left (\frac {2 x^2}{3}\right ) \]

[In]

Integrate[2 - E^x + Log[(2*x^2)/3],x]

[Out]

-E^x + x*Log[(2*x^2)/3]

Maple [A] (verified)

Time = 1.79 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
norman	\(x \ln \left (\frac {2 x^{2}}{3}\right )-{\mathrm e}^{x}\)	\(14\)
default	\(2 x \ln \left (x \right )+x \left (\ln \left (\frac {2 x^{2}}{3}\right )-2 \ln \left (x \right )\right )-{\mathrm e}^{x}\)	\(24\)
parallelrisch	\(\frac {12 \,{\mathrm e}^{\ln \left (-\frac {-x \ln \left (\frac {2 x^{2}}{3}\right )+{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )} {\mathrm e}^{2 x}-24 \,{\mathrm e}^{\ln \left (-\frac {-x \ln \left (\frac {2 x^{2}}{3}\right )+{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )} {\mathrm e}^{x} \ln \left (\frac {2 x^{2}}{3}\right ) x +12 \,{\mathrm e}^{\ln \left (-\frac {-x \ln \left (\frac {2 x^{2}}{3}\right )+{\mathrm e}^{x}}{x}\right )+\ln \left (x \right )} \ln \left (\frac {2 x^{2}}{3}\right )^{2} x^{2}}{12 {\left (x \ln \left (\frac {2 x^{2}}{3}\right )-{\mathrm e}^{x}\right )}^{2}}\)	\(115\)

[In]

int((ln(2/3*x^2)-exp(x)+2)*exp(ln((x*ln(2/3*x^2)-exp(x))/x)+ln(x))/(x*ln(2/3*x^2)-exp(x)),x,method=_RETURNVERB
OSE)

[Out]

x*ln(2/3*x^2)-exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \log \left (\frac {2}{3}\right ) + 2 \, x \log \left (x\right ) - e^{x} \]

[In]

integrate((log(2/3*x^2)-exp(x)+2)*exp(log((x*log(2/3*x^2)-exp(x))/x)+log(x))/(x*log(2/3*x^2)-exp(x)),x, algori
thm="fricas")

[Out]

x*log(2/3) + 2*x*log(x) - e^x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.63 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \log {\left (\frac {2 x^{2}}{3} \right )} - e^{x} \]

[In]

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

[Out]

x*log(2*x**2/3) - exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x \log \left (\frac {2}{3} \, x^{2}\right ) - e^{x} \]

[In]

integrate((log(2/3*x^2)-exp(x)+2)*exp(log((x*log(2/3*x^2)-exp(x))/x)+log(x))/(x*log(2/3*x^2)-exp(x)),x, algori
thm="maxima")

[Out]

x*log(2/3*x^2) - e^x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=-x \log \left (3\right ) + x \log \left (2\right ) + 2 \, x \log \left (x \mathrm {sgn}\left (x\right )\right ) - e^{x} \]

[In]

integrate((log(2/3*x^2)-exp(x)+2)*exp(log((x*log(2/3*x^2)-exp(x))/x)+log(x))/(x*log(2/3*x^2)-exp(x)),x, algori
thm="giac")

[Out]

-x*log(3) + x*log(2) + 2*x*log(x*sgn(x)) - e^x

Mupad [B] (verification not implemented)

Time = 15.38 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \left (2-e^x+\log \left (\frac {2 x^2}{3}\right )\right ) \, dx=x\,\ln \left (x^2\right )-{\mathrm {e}}^x+x\,\ln \left (2\right )-x\,\ln \left (3\right ) \]

[In]

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

[Out]

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