3.41.68 $\int (8x-139x^2+e^{e^x}(-2x-e^x{x}^2)+51x^2\log (x))dx$

Optimal. Leaf size=25 $$x^2(4-e^{e^x}-x-17x(3-\log (x)))$$

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Rubi [A]  time = 0.03, antiderivative size = 28, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 2, integrand size = 34, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.059, Rules used = {2288, 2304} $$\begin{array}{c}-52x^3+17x^3\log (x)-e^{e^x}{x}^2+4x^2\end{array}$$

Antiderivative was successfully verified.

[In]

Int[8*x - 139*x^2 + E^E^x*(-2*x - E^x*x^2) + 51*x^2*Log[x],x]

[Out]

4*x^2 - E^E^x*x^2 - 52*x^3 + 17*x^3*Log[x]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =4x^2-\frac{139x^3}{3}+51\int x^2\log (x)dx+\int e^{e^x}(-2x-e^x{x}^2)dx\\ & =4x^2-e^{e^x}{x}^2-52x^3+17x^3\log (x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 21, normalized size = 0.84 $$\begin{array}{c}{x}^2(4-e^{e^x}-52x+17x\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[8*x - 139*x^2 + E^E^x*(-2*x - E^x*x^2) + 51*x^2*Log[x],x]

[Out]

x^2*(4 - E^E^x - 52*x + 17*x*Log[x])

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fricas [A]  time = 0.64, size = 26, normalized size = 1.04 $$\begin{array}{c}17x^3\log (x)-52x^3-x^2{e}^{\left(e^x\right)}+4x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(x)*x^2-2*x)*exp(exp(x))+51*x^2*log(x)-139*x^2+8*x,x, algorithm="fricas")

[Out]

17*x^3*log(x) - 52*x^3 - x^2*e^(e^x) + 4*x^2

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giac [A]  time = 0.15, size = 26, normalized size = 1.04 $$\begin{array}{c}17x^3\log (x)-52x^3-x^2{e}^{\left(e^x\right)}+4x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(x)*x^2-2*x)*exp(exp(x))+51*x^2*log(x)-139*x^2+8*x,x, algorithm="giac")

[Out]

17*x^3*log(x) - 52*x^3 - x^2*e^(e^x) + 4*x^2

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maple [A]  time = 0.03, size = 27, normalized size = 1.08

method	result	size
default	$-{\mathrm{e}}^{{\mathrm{e}}^x}{x}^2+4x^2-52x^3+17x^3\ln (x)$	$27$
risch	$-{\mathrm{e}}^{{\mathrm{e}}^x}{x}^2+4x^2-52x^3+17x^3\ln (x)$	$27$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-exp(x)*x^2-2*x)*exp(exp(x))+51*x^2*ln(x)-139*x^2+8*x,x,method=_RETURNVERBOSE)

[Out]

-exp(exp(x))*x^2+4*x^2-52*x^3+17*x^3*ln(x)

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maxima [A]  time = 0.35, size = 26, normalized size = 1.04 $$\begin{array}{c}17x^3\log (x)-52x^3-x^2{e}^{\left(e^x\right)}+4x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(x)*x^2-2*x)*exp(exp(x))+51*x^2*log(x)-139*x^2+8*x,x, algorithm="maxima")

[Out]

17*x^3*log(x) - 52*x^3 - x^2*e^(e^x) + 4*x^2

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mupad [B]  time = 3.16, size = 18, normalized size = 0.72 $$\begin{array}{c}-x^2(52x+{\mathrm{e}}^{{\mathrm{e}}^x}-17x\ln (x)-4)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*x + 51*x^2*log(x) - 139*x^2 - exp(exp(x))*(2*x + x^2*exp(x)),x)

[Out]

-x^2*(52*x + exp(exp(x)) - 17*x*log(x) - 4)

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sympy [A]  time = 0.30, size = 26, normalized size = 1.04 $$\begin{array}{c}17x^3\log (x)-52x^3-x^2{e}^{e^x}+4x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(x)*x**2-2*x)*exp(exp(x))+51*x**2*ln(x)-139*x**2+8*x,x)

[Out]

17*x**3*log(x) - 52*x**3 - x**2*exp(exp(x)) + 4*x**2

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