3.29.5 \(\int \frac {e^{-x} (2+(-1-x) \log (x^2))}{5 x^2} \, dx\) [2805]

Optimal. Leaf size=18 \[ 1+\frac {e^{-x} \log \left (x^2\right )}{5 x} \]

[Out]

1/5*ln(x^2)/exp(x)/x+1

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Rubi [A]
time = 0.16, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 6874, 2208, 2209, 2228, 2634} \begin {gather*} \frac {e^{-x} \log \left (x^2\right )}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + (-1 - x)*Log[x^2])/(5*E^x*x^2),x]

[Out]

Log[x^2]/(5*E^x*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 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), 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] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

Rule 2209

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

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

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]

Rule 6874

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

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 16, normalized size = 0.89 \begin {gather*} \frac {e^{-x} \log \left (x^2\right )}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + (-1 - x)*Log[x^2])/(5*E^x*x^2),x]

[Out]

Log[x^2]/(5*E^x*x)

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Maple [A]
time = 0.21, size = 14, normalized size = 0.78

method	result	size
default	\(\frac {\ln \left (x^{2}\right ) {\mathrm e}^{-x}}{5 x}\)	\(14\)
norman	\(\frac {\ln \left (x^{2}\right ) {\mathrm e}^{-x}}{5 x}\)	\(14\)
risch	\(\frac {2 \ln \left (x \right ) {\mathrm e}^{-x}}{5 x}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right ) {\mathrm e}^{-x}}{10 x}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*ln(x^2)/exp(x)/x

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*e^(-x)*log(x)/x - 2/5*gamma(-1, x) - 2/5*integrate(e^(-x)/x^2, x)

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Fricas [A]
time = 0.37, size = 13, normalized size = 0.72 \begin {gather*} \frac {e^{\left (-x\right )} \log \left (x^{2}\right )}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.06, size = 10, normalized size = 0.56 \begin {gather*} \frac {e^{- x} \log {\left (x^{2} \right )}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-1-x)*ln(x**2)+2)/exp(x)/x**2,x)

[Out]

exp(-x)*log(x**2)/(5*x)

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Giac [A]
time = 0.39, size = 13, normalized size = 0.72 \begin {gather*} \frac {e^{\left (-x\right )} \log \left (x^{2}\right )}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.78, size = 13, normalized size = 0.72 \begin {gather*} \frac {\ln \left (x^2\right )\,{\mathrm {e}}^{-x}}{5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*((log(x^2)*(x + 1))/5 - 2/5))/x^2,x)

[Out]

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

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