3.31.21 \(\int \frac {e^{-x} (2+2 x+x^2-x^2 \log (x^2)+(2+x-x \log (x^2)) \log (\log (3)))}{x} \, dx\)

Optimal. Leaf size=18 \[ e^{-x} \left (-1+\log \left (x^2\right )\right ) (1+x+\log (\log (3))) \]

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Rubi [B]  time = 0.42, antiderivative size = 51, normalized size of antiderivative = 2.83, number of steps used = 14, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6742, 2176, 2194, 2554, 12, 2199, 2178} \begin {gather*} e^{-x} \log \left (x^2\right )+e^{-x} \log \left (x^2\right ) (x+\log (\log (3)))-e^{-x} x+e^{-x}-e^{-x} (2+\log (\log (3))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-x) - x/E^x + Log[x^2]/E^x - (2 + Log[Log[3]])/E^x + (Log[x^2]*(x + Log[Log[3]]))/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 (-e^{-x} \log \left (x^2\right ) (x+\log (\log (3)))+\frac {e^{-x} \left (x^2+2 (1+\log (\log (3)))+x (2+\log (\log (3)))\right )}{x}\right ) \, dx\\ &=-\int e^{-x} \log \left (x^2\right ) (x+\log (\log (3))) \, dx+\int \frac {e^{-x} \left (x^2+2 (1+\log (\log (3)))+x (2+\log (\log (3)))\right )}{x} \, dx\\ &=e^{-x} \log \left (x^2\right )+e^{-x} \log \left (x^2\right ) (x+\log (\log (3)))+\int \frac {2 e^{-x} (-1-x-\log (\log (3)))}{x} \, dx+\int \left (e^{-x} x+\frac {2 e^{-x} (1+\log (\log (3)))}{x}+e^{-x} (2+\log (\log (3)))\right ) \, dx\\ &=e^{-x} \log \left (x^2\right )+e^{-x} \log \left (x^2\right ) (x+\log (\log (3)))+2 \int \frac {e^{-x} (-1-x-\log (\log (3)))}{x} \, dx+(2 (1+\log (\log (3)))) \int \frac {e^{-x}}{x} \, dx+(2+\log (\log (3))) \int e^{-x} \, dx+\int e^{-x} x \, dx\\ &=-e^{-x} x+e^{-x} \log \left (x^2\right )+2 \text {Ei}(-x) (1+\log (\log (3)))-e^{-x} (2+\log (\log (3)))+e^{-x} \log \left (x^2\right ) (x+\log (\log (3)))+2 \int \left (-e^{-x}+\frac {e^{-x} (-1-\log (\log (3)))}{x}\right ) \, dx+\int e^{-x} \, dx\\ &=-e^{-x}-e^{-x} x+e^{-x} \log \left (x^2\right )+2 \text {Ei}(-x) (1+\log (\log (3)))-e^{-x} (2+\log (\log (3)))+e^{-x} \log \left (x^2\right ) (x+\log (\log (3)))-2 \int e^{-x} \, dx-(2 (1+\log (\log (3)))) \int \frac {e^{-x}}{x} \, dx\\ &=e^{-x}-e^{-x} x+e^{-x} \log \left (x^2\right )-e^{-x} (2+\log (\log (3)))+e^{-x} \log \left (x^2\right ) (x+\log (\log (3)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 18, normalized size = 1.00 \begin {gather*} e^{-x} \left (-1+\log \left (x^2\right )\right ) (1+x+\log (\log (3))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 + Log[x^2])*(1 + x + Log[Log[3]]))/E^x

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fricas [B]  time = 0.49, size = 42, normalized size = 2.33 \begin {gather*} {\left (x + 1\right )} e^{\left (-x\right )} \log \left (x^{2}\right ) - {\left (x + 1\right )} e^{\left (-x\right )} + {\left (e^{\left (-x\right )} \log \left (x^{2}\right ) - e^{\left (-x\right )}\right )} \log \left (\log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x^2)+2+x)*log(log(3))-x^2*log(x^2)+x^2+2*x+2)/exp(x)/x,x, algorithm="fricas")

[Out]

(x + 1)*e^(-x)*log(x^2) - (x + 1)*e^(-x) + (e^(-x)*log(x^2) - e^(-x))*log(log(3))

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giac [B]  time = 0.24, size = 54, normalized size = 3.00 \begin {gather*} x e^{\left (-x\right )} \log \left (x^{2}\right ) + e^{\left (-x\right )} \log \left (x^{2}\right ) \log \left (\log \relax (3)\right ) - x e^{\left (-x\right )} + e^{\left (-x\right )} \log \left (x^{2}\right ) - e^{\left (-x\right )} \log \left (\log \relax (3)\right ) - e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x^2)+2+x)*log(log(3))-x^2*log(x^2)+x^2+2*x+2)/exp(x)/x,x, algorithm="giac")

[Out]

x*e^(-x)*log(x^2) + e^(-x)*log(x^2)*log(log(3)) - x*e^(-x) + e^(-x)*log(x^2) - e^(-x)*log(log(3)) - e^(-x)

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maple [A]  time = 0.12, size = 32, normalized size = 1.78

method	result	size
norman	\(\left (x \ln \left (x^{2}\right )+\left (1+\ln \left (\ln \relax (3)\right )\right ) \ln \left (x^{2}\right )-x -1-\ln \left (\ln \relax (3)\right )\right ) {\mathrm e}^{-x}\)	\(32\)
default	\(\left (\left (\ln \left (x^{2}\right )-2 \ln \relax (x )-1\right ) x +\left (2 \ln \left (\ln \relax (3)\right )+2\right ) \ln \relax (x )+2 x \ln \relax (x )-1+\ln \left (x^{2}\right )-2 \ln \relax (x )+\ln \left (\ln \relax (3)\right ) \left (\ln \left (x^{2}\right )-2 \ln \relax (x )\right )-\ln \left (\ln \relax (3)\right )\right ) {\mathrm e}^{-x}\)	\(61\)
risch	\(2 \left (x +1+\ln \left (\ln \relax (3)\right )\right ) {\mathrm e}^{-x} \ln \relax (x )-\frac {i \left (\ln \left (\ln \relax (3)\right ) \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \ln \left (\ln \relax (3)\right ) \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\ln \left (\ln \relax (3)\right ) \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+x \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+x \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 i \ln \left (\ln \relax (3)\right )-2 i x -2 i\right ) {\mathrm e}^{-x}}{2}\)	\(180\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*ln(x^2)+2+x)*ln(ln(3))-x^2*ln(x^2)+x^2+2*x+2)/exp(x)/x,x,method=_RETURNVERBOSE)

[Out]

(x*ln(x^2)+(1+ln(ln(3)))*ln(x^2)-x-1-ln(ln(3)))/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) + e^{\left (-x\right )} \log \left (x^{2}\right ) \log \left (\log \relax (3)\right ) - {\left (x + 1\right )} e^{\left (-x\right )} - e^{\left (-x\right )} \log \left (\log \relax (3)\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*log(x^2)+2+x)*log(log(3))-x^2*log(x^2)+x^2+2*x+2)/exp(x)/x,x, algorithm="maxima")

[Out]

2*(x + 1)*e^(-x)*log(x) + e^(-x)*log(x^2)*log(log(3)) - (x + 1)*e^(-x) - e^(-x)*log(log(3)) + 2*Ei(-x) - 2*e^(
-x) - integrate(2*(x + 1)*e^(-x)/x, x)

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mupad [B]  time = 1.79, size = 17, normalized size = 0.94 \begin {gather*} {\mathrm {e}}^{-x}\,\left (\ln \left (x^2\right )-1\right )\,\left (x+\ln \left (\ln \relax (3)\right )+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(2*x + log(log(3))*(x - x*log(x^2) + 2) - x^2*log(x^2) + x^2 + 2))/x,x)

[Out]

exp(-x)*(log(x^2) - 1)*(x + log(log(3)) + 1)

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sympy [A]  time = 0.41, size = 32, normalized size = 1.78 \begin {gather*} \left (x \log {\left (x^{2} \right )} - x + \log {\left (x^{2} \right )} \log {\left (\log {\relax (3 )} \right )} + \log {\left (x^{2} \right )} - 1 - \log {\left (\log {\relax (3 )} \right )}\right ) e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*ln(x**2)+2+x)*ln(ln(3))-x**2*ln(x**2)+x**2+2*x+2)/exp(x)/x,x)

[Out]

(x*log(x**2) - x + log(x**2)*log(log(3)) + log(x**2) - 1 - log(log(3)))*exp(-x)

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