3.50.41 \(\int \frac {e^{-16+x} (-13+39 x+11 x^2-x^3)-13 e^{-16+x} x \log (x)}{13 x} \, dx\)

Optimal. Leaf size=20 \[ e^{-16+x} \left (2+x-\frac {x^2}{13}-\log (x)\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 36, normalized size of antiderivative = 1.80, number of steps used = 12, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 14, 2194, 2178, 2176, 2554} \begin {gather*} -\frac {1}{13} e^{x-16} x^2+e^{x-16} x+2 e^{x-16}-e^{x-16} \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-16 + x)*(-13 + 39*x + 11*x^2 - x^3) - 13*E^(-16 + x)*x*Log[x])/(13*x),x]

[Out]

2*E^(-16 + x) + E^(-16 + x)*x - (E^(-16 + x)*x^2)/13 - E^(-16 + x)*Log[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2554

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} &=\frac {1}{13} \int \frac {e^{-16+x} \left (-13+39 x+11 x^2-x^3\right )-13 e^{-16+x} x \log (x)}{x} \, dx\\ &=\frac {1}{13} \int \left (39 e^{-16+x}-\frac {13 e^{-16+x}}{x}+11 e^{-16+x} x-e^{-16+x} x^2-13 e^{-16+x} \log (x)\right ) \, dx\\ &=-\left (\frac {1}{13} \int e^{-16+x} x^2 \, dx\right )+\frac {11}{13} \int e^{-16+x} x \, dx+3 \int e^{-16+x} \, dx-\int \frac {e^{-16+x}}{x} \, dx-\int e^{-16+x} \log (x) \, dx\\ &=3 e^{-16+x}+\frac {11}{13} e^{-16+x} x-\frac {1}{13} e^{-16+x} x^2-\frac {\text {Ei}(x)}{e^{16}}-e^{-16+x} \log (x)+\frac {2}{13} \int e^{-16+x} x \, dx-\frac {11}{13} \int e^{-16+x} \, dx+\int \frac {e^{-16+x}}{x} \, dx\\ &=\frac {28 e^{-16+x}}{13}+e^{-16+x} x-\frac {1}{13} e^{-16+x} x^2-e^{-16+x} \log (x)-\frac {2}{13} \int e^{-16+x} \, dx\\ &=2 e^{-16+x}+e^{-16+x} x-\frac {1}{13} e^{-16+x} x^2-e^{-16+x} \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 21, normalized size = 1.05 \begin {gather*} -\frac {1}{13} e^{-16+x} \left (-26-13 x+x^2+13 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-16 + x)*(-13 + 39*x + 11*x^2 - x^3) - 13*E^(-16 + x)*x*Log[x])/(13*x),x]

[Out]

-1/13*(E^(-16 + x)*(-26 - 13*x + x^2 + 13*Log[x]))

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fricas [A]  time = 1.05, size = 23, normalized size = 1.15 \begin {gather*} -\frac {1}{13} \, {\left (x^{2} - 13 \, x - 26\right )} e^{\left (x - 16\right )} - e^{\left (x - 16\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(-13*x*exp(x-16)*log(x)+(-x^3+11*x^2+39*x-13)*exp(x-16))/x,x, algorithm="fricas")

[Out]

-1/13*(x^2 - 13*x - 26)*e^(x - 16) - e^(x - 16)*log(x)

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giac [A]  time = 0.17, size = 26, normalized size = 1.30 \begin {gather*} -\frac {1}{13} \, {\left (x^{2} e^{x} - 13 \, x e^{x} + 13 \, e^{x} \log \relax (x) - 26 \, e^{x}\right )} e^{\left (-16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(-13*x*exp(x-16)*log(x)+(-x^3+11*x^2+39*x-13)*exp(x-16))/x,x, algorithm="giac")

[Out]

-1/13*(x^2*e^x - 13*x*e^x + 13*e^x*log(x) - 26*e^x)*e^(-16)

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maple [A]  time = 0.03, size = 24, normalized size = 1.20

method	result	size
risch	\(-{\mathrm e}^{x -16} \ln \relax (x )-\frac {\left (x^{2}-13 x -26\right ) {\mathrm e}^{x -16}}{13}\)	\(24\)
norman	\(x \,{\mathrm e}^{x -16}-\frac {x^{2} {\mathrm e}^{x -16}}{13}-{\mathrm e}^{x -16} \ln \relax (x )+2 \,{\mathrm e}^{x -16}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/13*(-13*x*exp(x-16)*ln(x)+(-x^3+11*x^2+39*x-13)*exp(x-16))/x,x,method=_RETURNVERBOSE)

[Out]

-exp(x-16)*ln(x)-1/13*(x^2-13*x-26)*exp(x-16)

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maxima [B]  time = 0.39, size = 38, normalized size = 1.90 \begin {gather*} -\frac {1}{13} \, {\left (x^{2} - 2 \, x + 2\right )} e^{\left (x - 16\right )} + \frac {11}{13} \, {\left (x - 1\right )} e^{\left (x - 16\right )} - e^{\left (x - 16\right )} \log \relax (x) + 3 \, e^{\left (x - 16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(-13*x*exp(x-16)*log(x)+(-x^3+11*x^2+39*x-13)*exp(x-16))/x,x, algorithm="maxima")

[Out]

-1/13*(x^2 - 2*x + 2)*e^(x - 16) + 11/13*(x - 1)*e^(x - 16) - e^(x - 16)*log(x) + 3*e^(x - 16)

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mupad [B]  time = 3.71, size = 17, normalized size = 0.85 \begin {gather*} {\mathrm {e}}^{x-16}\,\left (x-\ln \relax (x)-\frac {x^2}{13}+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x - 16)*(39*x + 11*x^2 - x^3 - 13))/13 - x*exp(x - 16)*log(x))/x,x)

[Out]

exp(x - 16)*(x - log(x) - x^2/13 + 2)

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sympy [A]  time = 0.35, size = 19, normalized size = 0.95 \begin {gather*} \frac {\left (- x^{2} + 13 x - 13 \log {\relax (x )} + 26\right ) e^{x - 16}}{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(-13*x*exp(x-16)*ln(x)+(-x**3+11*x**2+39*x-13)*exp(x-16))/x,x)

[Out]

(-x**2 + 13*x - 13*log(x) + 26)*exp(x - 16)/13

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