3.17.48 \(\int (-4+2 x+\log (\frac {4}{9} e^{2 e^3} x)) \, dx\)

Optimal. Leaf size=18 \[ x \left (-5+x+\log \left (\frac {4}{9} e^{2 e^3} x\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.22, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {2295} \begin {gather*} x^2-5 x+x \log \left (\frac {4}{9} e^{2 e^3} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-4 + 2*x + Log[(4*E^(2*E^3)*x)/9],x]

[Out]

-5*x + x^2 + x*Log[(4*E^(2*E^3)*x)/9]

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} &=-4 x+x^2+\int \log \left (\frac {4}{9} e^{2 e^3} x\right ) \, dx\\ &=-5 x+x^2+x \log \left (\frac {4}{9} e^{2 e^3} x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.17 \begin {gather*} -5 x+2 e^3 x+x^2+x \log \left (\frac {4 x}{9}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-4 + 2*x + Log[(4*E^(2*E^3)*x)/9],x]

[Out]

-5*x + 2*E^3*x + x^2 + x*Log[(4*x)/9]

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fricas [A]  time = 0.78, size = 18, normalized size = 1.00 \begin {gather*} x^{2} + x \log \left (\frac {4}{9} \, x e^{\left (2 \, e^{3}\right )}\right ) - 5 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(4/9*x*exp(exp(3))^2)+2*x-4,x, algorithm="fricas")

[Out]

x^2 + x*log(4/9*x*e^(2*e^3)) - 5*x

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giac [B]  time = 0.14, size = 38, normalized size = 2.11 \begin {gather*} x^{2} + {\left (x e^{\left (2 \, e^{3}\right )} \log \left (\frac {4}{9} \, x e^{\left (2 \, e^{3}\right )}\right ) - x e^{\left (2 \, e^{3}\right )}\right )} e^{\left (-2 \, e^{3}\right )} - 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(4/9*x*exp(exp(3))^2)+2*x-4,x, algorithm="giac")

[Out]

x^2 + (x*e^(2*e^3)*log(4/9*x*e^(2*e^3)) - x*e^(2*e^3))*e^(-2*e^3) - 4*x

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maple [A]  time = 0.03, size = 19, normalized size = 1.06

method	result	size
default	\(x^{2}-5 x +\ln \left (\frac {4 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}}}{9}\right ) x\)	\(19\)
norman	\(x^{2}-5 x +\ln \left (\frac {4 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}}}{9}\right ) x\)	\(19\)
risch	\(x^{2}-5 x +\ln \left (\frac {4 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}}}{9}\right ) x\)	\(19\)
derivativedivides	\(\frac {9 \,{\mathrm e}^{-2 \,{\mathrm e}^{3}} \left (-\frac {40 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}}}{9}+\frac {8 x^{2} {\mathrm e}^{2 \,{\mathrm e}^{3}}}{9}+\frac {8 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}} \ln \left (\frac {4 x \,{\mathrm e}^{2 \,{\mathrm e}^{3}}}{9}\right )}{9}\right )}{8}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(4/9*x*exp(exp(3))^2)+2*x-4,x,method=_RETURNVERBOSE)

[Out]

x^2-5*x+ln(4/9*x*exp(exp(3))^2)*x

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maxima [B]  time = 0.37, size = 38, normalized size = 2.11 \begin {gather*} x^{2} + {\left (x e^{\left (2 \, e^{3}\right )} \log \left (\frac {4}{9} \, x e^{\left (2 \, e^{3}\right )}\right ) - x e^{\left (2 \, e^{3}\right )}\right )} e^{\left (-2 \, e^{3}\right )} - 4 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(4/9*x*exp(exp(3))^2)+2*x-4,x, algorithm="maxima")

[Out]

x^2 + (x*e^(2*e^3)*log(4/9*x*e^(2*e^3)) - x*e^(2*e^3))*e^(-2*e^3) - 4*x

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mupad [B]  time = 1.03, size = 13, normalized size = 0.72 \begin {gather*} x\,\left (x+\ln \left (\frac {4\,x}{9}\right )+2\,{\mathrm {e}}^3-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x + log((4*x*exp(2*exp(3)))/9) - 4,x)

[Out]

x*(x + log((4*x)/9) + 2*exp(3) - 5)

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sympy [A]  time = 0.09, size = 20, normalized size = 1.11 \begin {gather*} x^{2} + x \log {\left (\frac {4 x e^{2 e^{3}}}{9} \right )} - 5 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(4/9*x*exp(exp(3))**2)+2*x-4,x)

[Out]

x**2 + x*log(4*x*exp(2*exp(3))/9) - 5*x

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