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

Optimal. Leaf size=19 \[ x \left (-\frac {e^x}{x}+\log \left (\frac {2 x^2}{3}\right )\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2194, 2295} \begin {gather*} x \log \left (\frac {2 x^2}{3}\right )-e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2 - E^x + Log[(2*x^2)/3],x]

[Out]

-E^x + x*Log[(2*x^2)/3]

Rule 2194

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 2295

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.84 \begin {gather*} -e^x+x \log \left (\frac {2 x^2}{3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2 - E^x + Log[(2*x^2)/3],x]

[Out]

-E^x + x*Log[(2*x^2)/3]

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fricas [A]  time = 0.80, size = 14, normalized size = 0.74 \begin {gather*} x \log \left (\frac {2}{3}\right ) + 2 \, x \log \relax (x) - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.21, size = 22, normalized size = 1.16 \begin {gather*} -x \log \relax (3) + x \log \relax (2) + 2 \, x \log \left (x \mathrm {sgn}\relax (x)\right ) - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.11, size = 14, normalized size = 0.74

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.37, size = 13, normalized size = 0.68 \begin {gather*} x \log \left (\frac {2}{3} \, x^{2}\right ) - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 5.71, size = 20, normalized size = 1.05 \begin {gather*} x\,\ln \left (x^2\right )-{\mathrm {e}}^x+x\,\ln \relax (2)-x\,\ln \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.24, size = 12, normalized size = 0.63 \begin {gather*} x \log {\left (\frac {2 x^{2}}{3} \right )} - e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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