3.92.92 \(\int (16+3 x+6 x^2+x^3+(-2 x-3 x^2) \log (e^{-x} x)) \, dx\) [9192]

Optimal. Leaf size=28 \[ 4 x \left (4+\left (\frac {1}{4}+\frac {x}{4}\right ) x \left (2-\log \left (e^{-x} x\right )\right )\right ) \]

[Out]

4*(4+(1/4+1/4*x)*x*(2-ln(x/exp(x))))*x

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Rubi [A]
time = 0.04, antiderivative size = 40, normalized size of antiderivative = 1.43, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {1607, 45, 2634, 14} \begin {gather*} 2 x^3+x^3 \left (-\log \left (e^{-x} x\right )\right )+2 x^2-x^2 \log \left (e^{-x} x\right )+16 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[16 + 3*x + 6*x^2 + x^3 + (-2*x - 3*x^2)*Log[x/E^x],x]

[Out]

16*x + 2*x^2 + 2*x^3 - x^2*Log[x/E^x] - x^3*Log[x/E^x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2634

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*(D[u, x]
/u), x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=16 x+\frac {3 x^2}{2}+2 x^3+\frac {x^4}{4}+\int \left (-2 x-3 x^2\right ) \log \left (e^{-x} x\right ) \, dx\\ &=16 x+\frac {3 x^2}{2}+2 x^3+\frac {x^4}{4}+\int (-2-3 x) x \log \left (e^{-x} x\right ) \, dx\\ &=16 x+\frac {3 x^2}{2}+2 x^3+\frac {x^4}{4}-x^2 \log \left (e^{-x} x\right )-x^3 \log \left (e^{-x} x\right )-\int x \left (-1+x^2\right ) \, dx\\ &=16 x+\frac {3 x^2}{2}+2 x^3+\frac {x^4}{4}-x^2 \log \left (e^{-x} x\right )-x^3 \log \left (e^{-x} x\right )-\int \left (-x+x^3\right ) \, dx\\ &=16 x+2 x^2+2 x^3-x^2 \log \left (e^{-x} x\right )-x^3 \log \left (e^{-x} x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.02, size = 40, normalized size = 1.43 \begin {gather*} 16 x+2 x^2+2 x^3-x^2 \log \left (e^{-x} x\right )-x^3 \log \left (e^{-x} x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[16 + 3*x + 6*x^2 + x^3 + (-2*x - 3*x^2)*Log[x/E^x],x]

[Out]

16*x + 2*x^2 + 2*x^3 - x^2*Log[x/E^x] - x^3*Log[x/E^x]

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Maple [A]
time = 10.49, size = 39, normalized size = 1.39

method	result	size
default	\(16 x -\ln \left (x \,{\mathrm e}^{-x}\right ) x^{3}-x^{2} \ln \left (x \,{\mathrm e}^{-x}\right )+2 x^{2}+2 x^{3}\)	\(39\)
norman	\(16 x -\ln \left (x \,{\mathrm e}^{-x}\right ) x^{3}-x^{2} \ln \left (x \,{\mathrm e}^{-x}\right )+2 x^{2}+2 x^{3}\)	\(39\)
risch	\(\left (x^{3}+x^{2}\right ) \ln \left ({\mathrm e}^{x}\right )-x^{3} \ln \left (x \right )-x^{2} \ln \left (x \right )+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )}{2}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}}{2}-\frac {i \pi \,x^{3} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}}{2}+\frac {i \pi \,x^{3} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}}{2}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}}{2}+2 x^{2}+2 x^{3}+16 x\)	\(232\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x^2-2*x)*ln(x/exp(x))+x^3+6*x^2+3*x+16,x,method=_RETURNVERBOSE)

[Out]

16*x-ln(x/exp(x))*x^3-x^2*ln(x/exp(x))+2*x^2+2*x^3

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Maxima [A]
time = 0.27, size = 30, normalized size = 1.07 \begin {gather*} 2 \, x^{3} + 2 \, x^{2} - {\left (x^{3} + x^{2}\right )} \log \left (x e^{\left (-x\right )}\right ) + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2-2*x)*log(x/exp(x))+x^3+6*x^2+3*x+16,x, algorithm="maxima")

[Out]

2*x^3 + 2*x^2 - (x^3 + x^2)*log(x*e^(-x)) + 16*x

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Fricas [A]
time = 0.40, size = 30, normalized size = 1.07 \begin {gather*} 2 \, x^{3} + 2 \, x^{2} - {\left (x^{3} + x^{2}\right )} \log \left (x e^{\left (-x\right )}\right ) + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2-2*x)*log(x/exp(x))+x^3+6*x^2+3*x+16,x, algorithm="fricas")

[Out]

2*x^3 + 2*x^2 - (x^3 + x^2)*log(x*e^(-x)) + 16*x

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Sympy [A]
time = 0.07, size = 27, normalized size = 0.96 \begin {gather*} 2 x^{3} + 2 x^{2} + 16 x + \left (- x^{3} - x^{2}\right ) \log {\left (x e^{- x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x**2-2*x)*ln(x/exp(x))+x**3+6*x**2+3*x+16,x)

[Out]

2*x**3 + 2*x**2 + 16*x + (-x**3 - x**2)*log(x*exp(-x))

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Giac [A]
time = 0.41, size = 31, normalized size = 1.11 \begin {gather*} x^{4} - x^{3} \log \left (x\right ) + 3 \, x^{3} - x^{2} \log \left (x\right ) + 2 \, x^{2} + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x^2-2*x)*log(x/exp(x))+x^3+6*x^2+3*x+16,x, algorithm="giac")

[Out]

x^4 - x^3*log(x) + 3*x^3 - x^2*log(x) + 2*x^2 + 16*x

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Mupad [B]
time = 6.85, size = 31, normalized size = 1.11 \begin {gather*} 16\,x-x^2\,\ln \left (x\right )-x^3\,\ln \left (x\right )+2\,x^2+3\,x^3+x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(3*x - log(x*exp(-x))*(2*x + 3*x^2) + 6*x^2 + x^3 + 16,x)

[Out]

16*x - x^2*log(x) - x^3*log(x) + 2*x^2 + 3*x^3 + x^4

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