3.102.39 \(\int \frac {e^{-x} (e^x (-5-8 x)-6 x+2 x^2+(-6 x-8 e^x x+10 x^2-2 x^3) \log (x))}{2 x} \, dx\)

Optimal. Leaf size=23 \[ \left (-\frac {5}{2}-4 x+e^{-x} \left (-3 x+x^2\right )\right ) \log (x) \]

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Rubi [A]  time = 0.59, antiderivative size = 33, normalized size of antiderivative = 1.43, number of steps used = 27, number of rules used = 10, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {12, 6742, 14, 43, 2295, 2194, 2176, 2554, 2178, 2199} \begin {gather*} e^{-x} x^2 \log (x)-3 e^{-x} x \log (x)-4 x \log (x)-\frac {5 \log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(-5 - 8*x) - 6*x + 2*x^2 + (-6*x - 8*E^x*x + 10*x^2 - 2*x^3)*Log[x])/(2*E^x*x),x]

[Out]

(-5*Log[x])/2 - 4*x*Log[x] - (3*x*Log[x])/E^x + (x^2*Log[x])/E^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 43

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 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 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {e^{-x} \left (e^x (-5-8 x)-6 x+2 x^2+\left (-6 x-8 e^x x+10 x^2-2 x^3\right ) \log (x)\right )}{x} \, dx\\ &=\frac {1}{2} \int \left (-\frac {5+8 x+8 x \log (x)}{x}-2 e^{-x} \left (3-x+3 \log (x)-5 x \log (x)+x^2 \log (x)\right )\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {5+8 x+8 x \log (x)}{x} \, dx\right )-\int e^{-x} \left (3-x+3 \log (x)-5 x \log (x)+x^2 \log (x)\right ) \, dx\\ &=-\left (\frac {1}{2} \int \left (\frac {5+8 x}{x}+8 \log (x)\right ) \, dx\right )-\int \left (3 e^{-x}-e^{-x} x+3 e^{-x} \log (x)-5 e^{-x} x \log (x)+e^{-x} x^2 \log (x)\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {5+8 x}{x} \, dx\right )-3 \int e^{-x} \, dx-3 \int e^{-x} \log (x) \, dx-4 \int \log (x) \, dx+5 \int e^{-x} x \log (x) \, dx+\int e^{-x} x \, dx-\int e^{-x} x^2 \log (x) \, dx\\ &=3 e^{-x}+4 x-e^{-x} x-4 x \log (x)-3 e^{-x} x \log (x)+e^{-x} x^2 \log (x)-\frac {1}{2} \int \left (8+\frac {5}{x}\right ) \, dx-3 \int \frac {e^{-x}}{x} \, dx-5 \int \frac {e^{-x} (-1-x)}{x} \, dx+\int e^{-x} \, dx+\int \frac {e^{-x} \left (-2-2 x-x^2\right )}{x} \, dx\\ &=2 e^{-x}-e^{-x} x-3 \text {Ei}(-x)-\frac {5 \log (x)}{2}-4 x \log (x)-3 e^{-x} x \log (x)+e^{-x} x^2 \log (x)-5 \int \left (-e^{-x}-\frac {e^{-x}}{x}\right ) \, dx+\int \left (-2 e^{-x}-\frac {2 e^{-x}}{x}-e^{-x} x\right ) \, dx\\ &=2 e^{-x}-e^{-x} x-3 \text {Ei}(-x)-\frac {5 \log (x)}{2}-4 x \log (x)-3 e^{-x} x \log (x)+e^{-x} x^2 \log (x)-2 \int e^{-x} \, dx-2 \int \frac {e^{-x}}{x} \, dx+5 \int e^{-x} \, dx+5 \int \frac {e^{-x}}{x} \, dx-\int e^{-x} x \, dx\\ &=-e^{-x}-\frac {5 \log (x)}{2}-4 x \log (x)-3 e^{-x} x \log (x)+e^{-x} x^2 \log (x)-\int e^{-x} \, dx\\ &=-\frac {5 \log (x)}{2}-4 x \log (x)-3 e^{-x} x \log (x)+e^{-x} x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 27, normalized size = 1.17 \begin {gather*} -\frac {5 \log (x)}{2}+x \left (-4-3 e^{-x}+e^{-x} x\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-5 - 8*x) - 6*x + 2*x^2 + (-6*x - 8*E^x*x + 10*x^2 - 2*x^3)*Log[x])/(2*E^x*x),x]

[Out]

(-5*Log[x])/2 + x*(-4 - 3/E^x + x/E^x)*Log[x]

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fricas [A]  time = 0.53, size = 26, normalized size = 1.13 \begin {gather*} \frac {1}{2} \, {\left (2 \, x^{2} - {\left (8 \, x + 5\right )} e^{x} - 6 \, x\right )} e^{\left (-x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-8*exp(x)*x-2*x^3+10*x^2-6*x)*log(x)+(-8*x-5)*exp(x)+2*x^2-6*x)/exp(x)/x,x, algorithm="fricas"
)

[Out]

1/2*(2*x^2 - (8*x + 5)*e^x - 6*x)*e^(-x)*log(x)

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giac [A]  time = 0.19, size = 29, normalized size = 1.26 \begin {gather*} x^{2} e^{\left (-x\right )} \log \relax (x) - 3 \, x e^{\left (-x\right )} \log \relax (x) - 4 \, x \log \relax (x) - \frac {5}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-8*exp(x)*x-2*x^3+10*x^2-6*x)*log(x)+(-8*x-5)*exp(x)+2*x^2-6*x)/exp(x)/x,x, algorithm="giac")

[Out]

x^2*e^(-x)*log(x) - 3*x*e^(-x)*log(x) - 4*x*log(x) - 5/2*log(x)

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maple [A]  time = 0.04, size = 21, normalized size = 0.91

method	result	size
risch	\(x \left (-4 \,{\mathrm e}^{x}+x -3\right ) {\mathrm e}^{-x} \ln \relax (x )-\frac {5 \ln \relax (x )}{2}\)	\(21\)
default	\(\frac {\left (-6 x \ln \relax (x )+2 x^{2} \ln \relax (x )\right ) {\mathrm e}^{-x}}{2}-\frac {5 \ln \relax (x )}{2}-4 x \ln \relax (x )\)	\(30\)
norman	\(\left (x^{2} \ln \relax (x )-\frac {5 \,{\mathrm e}^{x} \ln \relax (x )}{2}-3 x \ln \relax (x )-4 x \,{\mathrm e}^{x} \ln \relax (x )\right ) {\mathrm e}^{-x}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-8*exp(x)*x-2*x^3+10*x^2-6*x)*ln(x)+(-8*x-5)*exp(x)+2*x^2-6*x)/exp(x)/x,x,method=_RETURNVERBOSE)

[Out]

x*(-4*exp(x)+x-3)*exp(-x)*ln(x)-5/2*ln(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} {\left (x^{2} - 3 \, x - 3\right )} e^{\left (-x\right )} \log \relax (x) - {\left (x + 1\right )} e^{\left (-x\right )} - 4 \, x \log \relax (x) + 3 \, e^{\left (-x\right )} \log \relax (x) - 3 \, {\rm Ei}\left (-x\right ) + 3 \, e^{\left (-x\right )} - \frac {1}{2} \, \int \frac {2 \, {\left (x^{2} - 3 \, x - 3\right )} e^{\left (-x\right )}}{x}\,{d x} - \frac {5}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-8*exp(x)*x-2*x^3+10*x^2-6*x)*log(x)+(-8*x-5)*exp(x)+2*x^2-6*x)/exp(x)/x,x, algorithm="maxima"
)

[Out]

(x^2 - 3*x - 3)*e^(-x)*log(x) - (x + 1)*e^(-x) - 4*x*log(x) + 3*e^(-x)*log(x) - 3*Ei(-x) + 3*e^(-x) - 1/2*inte
grate(2*(x^2 - 3*x - 3)*e^(-x)/x, x) - 5/2*log(x)

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mupad [B]  time = 8.52, size = 25, normalized size = 1.09 \begin {gather*} -\frac {\ln \relax (x)\,\left (8\,x+6\,x\,{\mathrm {e}}^{-x}-2\,x^2\,{\mathrm {e}}^{-x}+5\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(3*x + (exp(x)*(8*x + 5))/2 + (log(x)*(6*x + 8*x*exp(x) - 10*x^2 + 2*x^3))/2 - x^2))/x,x)

[Out]

-(log(x)*(8*x + 6*x*exp(-x) - 2*x^2*exp(-x) + 5))/2

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sympy [A]  time = 0.35, size = 29, normalized size = 1.26 \begin {gather*} - 4 x \log {\relax (x )} + \left (x^{2} \log {\relax (x )} - 3 x \log {\relax (x )}\right ) e^{- x} - \frac {5 \log {\relax (x )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-8*exp(x)*x-2*x**3+10*x**2-6*x)*ln(x)+(-8*x-5)*exp(x)+2*x**2-6*x)/exp(x)/x,x)

[Out]

-4*x*log(x) + (x**2*log(x) - 3*x*log(x))*exp(-x) - 5*log(x)/2

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