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

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

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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 {gather*} -52 x^3+17 x^3 \log (x)-e^{e^x} x^2+4 x^2 \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=4 x^2-\frac {139 x^3}{3}+51 \int x^2 \log (x) \, dx+\int e^{e^x} \left (-2 x-e^x x^2\right ) \, dx\\ &=4 x^2-e^{e^x} x^2-52 x^3+17 x^3 \log (x)\\ \end {aligned} \end {gather*}

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

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 {gather*} 17 \, x^{3} \log \relax (x) - 52 \, x^{3} - x^{2} e^{\left (e^{x}\right )} + 4 \, x^{2} \end {gather*}

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 {gather*} 17 \, x^{3} \log \relax (x) - 52 \, x^{3} - x^{2} e^{\left (e^{x}\right )} + 4 \, x^{2} \end {gather*}

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	\(-{\mathrm e}^{{\mathrm e}^{x}} x^{2}+4 x^{2}-52 x^{3}+17 x^{3} \ln \relax (x )\)	\(27\)
risch	\(-{\mathrm e}^{{\mathrm e}^{x}} x^{2}+4 x^{2}-52 x^{3}+17 x^{3} \ln \relax (x )\)	\(27\)

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 {gather*} 17 \, x^{3} \log \relax (x) - 52 \, x^{3} - x^{2} e^{\left (e^{x}\right )} + 4 \, x^{2} \end {gather*}

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 {gather*} -x^2\,\left (52\,x+{\mathrm {e}}^{{\mathrm {e}}^x}-17\,x\,\ln \relax (x)-4\right ) \end {gather*}

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 {gather*} 17 x^{3} \log {\relax (x )} - 52 x^{3} - x^{2} e^{e^{x}} + 4 x^{2} \end {gather*}

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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