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

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

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Rubi [A]  time = 0.23, 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, 6742, 2177, 2178, 2197, 2554} \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 2177

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] &&  !$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 2197

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 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}{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 = 0.07, 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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fricas [A]  time = 1.25, 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*((-x-1)*log(x^2)+2)/exp(x)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, 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*((-x-1)*log(x^2)+2)/exp(x)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.10, 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	\(\frac {2 \ln \relax (x ) {\mathrm e}^{-x}}{5 x}-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (\mathrm {csgn}\left (i x \right )^{2}-2 \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )+\mathrm {csgn}\left (i x^{2}\right )^{2}\right ) {\mathrm e}^{-x}}{10 x}\)	\(62\)

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*} \frac {2 \, e^{\left (-x\right )} \log \relax (x)}{5 \, x} - \frac {2}{5} \, \Gamma \left (-1, x\right ) - \frac {2}{5} \, \int \frac {e^{\left (-x\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-x-1)*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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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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sympy [A]  time = 0.24, 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*((-x-1)*ln(x**2)+2)/exp(x)/x**2,x)

[Out]

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

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