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

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

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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} $$\begin{array}{c}{x}^2-5x+x\log \left(\frac{4}{9}{e}^{2e^3}x\right)\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =-4x+x^2+\int \log \left(\frac{4}{9}{e}^{2e^3}x\right)dx\\ & =-5x+x^2+x\log \left(\frac{4}{9}{e}^{2e^3}x\right)\end{array}\end{array}$$

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

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 $$\begin{array}{c}{x}^2+x\log \left(\frac{4}{9}x{e}^{\left(2e^3\right)}\right)-5x\end{array}$$

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 $$\begin{array}{c}{x}^2+(x{e}^{\left(2e^3\right)}\log \left(\frac{4}{9}x{e}^{\left(2e^3\right)}\right)-x{e}^{\left(2e^3\right)})e^{(-2e^3)}-4x\end{array}$$

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

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 $$\begin{array}{c}{x}^2+(x{e}^{\left(2e^3\right)}\log \left(\frac{4}{9}x{e}^{\left(2e^3\right)}\right)-x{e}^{\left(2e^3\right)})e^{(-2e^3)}-4x\end{array}$$

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 $$\begin{array}{c}x(x+\ln \left(\frac{4x}{9}\right)+2{\mathrm{e}}^3-5)\end{array}$$

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 $$\begin{array}{c}{x}^2+x\log \left(\frac{4x{e}^{2e^3}}{9}\right)-5x\end{array}$$

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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