3.104.19 \(\int \frac {e^{-x} (-10+2 x+(6 x-x^2) \log (x^2))}{x} \, dx\)

Optimal. Leaf size=19 \[ 2-\log (259)+e^{-x} (-5+x) \log \left (x^2\right ) \]

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Rubi [A]  time = 0.23, antiderivative size = 27, normalized size of antiderivative = 1.42, number of steps used = 12, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6742, 2199, 2194, 2178, 2176, 2554, 12} \begin {gather*} e^{-x} \log \left (x^2\right )-e^{-x} (6-x) \log \left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x^2]/E^x - ((6 - x)*Log[x^2])/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 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 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} &=\int \left (\frac {2 e^{-x} (-5+x)}{x}-e^{-x} (-6+x) \log \left (x^2\right )\right ) \, dx\\ &=2 \int \frac {e^{-x} (-5+x)}{x} \, dx-\int e^{-x} (-6+x) \log \left (x^2\right ) \, dx\\ &=e^{-x} \log \left (x^2\right )-e^{-x} (6-x) \log \left (x^2\right )+2 \int \left (e^{-x}-\frac {5 e^{-x}}{x}\right ) \, dx+\int \frac {2 e^{-x} (5-x)}{x} \, dx\\ &=e^{-x} \log \left (x^2\right )-e^{-x} (6-x) \log \left (x^2\right )+2 \int e^{-x} \, dx+2 \int \frac {e^{-x} (5-x)}{x} \, dx-10 \int \frac {e^{-x}}{x} \, dx\\ &=-2 e^{-x}-10 \text {Ei}(-x)+e^{-x} \log \left (x^2\right )-e^{-x} (6-x) \log \left (x^2\right )+2 \int \left (-e^{-x}+\frac {5 e^{-x}}{x}\right ) \, dx\\ &=-2 e^{-x}-10 \text {Ei}(-x)+e^{-x} \log \left (x^2\right )-e^{-x} (6-x) \log \left (x^2\right )-2 \int e^{-x} \, dx+10 \int \frac {e^{-x}}{x} \, dx\\ &=e^{-x} \log \left (x^2\right )-e^{-x} (6-x) \log \left (x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 13, normalized size = 0.68 \begin {gather*} e^{-x} (-5+x) \log \left (x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-5 + x)*Log[x^2])/E^x

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fricas [A]  time = 0.68, size = 12, normalized size = 0.63 \begin {gather*} {\left (x - 5\right )} e^{\left (-x\right )} \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x - 5)*e^(-x)*log(x^2)

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giac [A]  time = 0.13, size = 21, normalized size = 1.11 \begin {gather*} x e^{\left (-x\right )} \log \left (x^{2}\right ) - 5 \, e^{\left (-x\right )} \log \left (x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(-x)*log(x^2) - 5*e^(-x)*log(x^2)

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maple [A]  time = 0.09, size = 19, normalized size = 1.00

method	result	size
norman	\(\left (x \ln \left (x^{2}\right )-5 \ln \left (x^{2}\right )\right ) {\mathrm e}^{-x}\)	\(19\)
default	\(\left (x \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right )-5 \ln \left (x^{2}\right )+2 x \ln \relax (x )\right ) {\mathrm e}^{-x}\)	\(29\)
risch	\(2 \left (x -5\right ) {\mathrm e}^{-x} \ln \relax (x )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (x \mathrm {csgn}\left (i x \right )^{2}-2 x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )+x \mathrm {csgn}\left (i x^{2}\right )^{2}-5 \mathrm {csgn}\left (i x \right )^{2}+10 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )-5 \mathrm {csgn}\left (i x^{2}\right )^{2}\right ) {\mathrm e}^{-x}}{2}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*ln(x^2)-5*ln(x^2))/exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 2 \, {\left (x + 1\right )} e^{\left (-x\right )} \log \relax (x) - 6 \, e^{\left (-x\right )} \log \left (x^{2}\right ) + 2 \, {\rm Ei}\left (-x\right ) - 2 \, e^{\left (-x\right )} - \int \frac {2 \, {\left (x + 1\right )} e^{\left (-x\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(x + 1)*e^(-x)*log(x) - 6*e^(-x)*log(x^2) + 2*Ei(-x) - 2*e^(-x) - integrate(2*(x + 1)*e^(-x)/x, x)

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mupad [B]  time = 7.29, size = 12, normalized size = 0.63 \begin {gather*} \ln \left (x^2\right )\,{\mathrm {e}}^{-x}\,\left (x-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^2)*exp(-x)*(x - 5)

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sympy [A]  time = 0.34, size = 15, normalized size = 0.79 \begin {gather*} \left (x \log {\left (x^{2} \right )} - 5 \log {\left (x^{2} \right )}\right ) e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+6*x)*ln(x**2)+2*x-10)/exp(x)/x,x)

[Out]

(x*log(x**2) - 5*log(x**2))*exp(-x)

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