3.4.31 \(\int \frac {e^{-x} (-4-4 e^5 x+4 x \log (x)+(-9+x^2+e^5 (-10 x+x^2)+(10 x-x^2) \log (x)) \log (\log (2)))}{x \log (\log (2))} \, dx\)

Optimal. Leaf size=29 \[ e^{-x} \left (-x+\left (-e^5+\log (x)\right ) \left (-9+x-\frac {4}{\log (\log (2))}\right )\right ) \]

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Rubi [B]  time = 0.60, antiderivative size = 88, normalized size of antiderivative = 3.03, number of steps used = 14, number of rules used = 7, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {12, 6742, 2176, 2194, 2554, 2199, 2178} \begin {gather*} \left (1+e^5\right ) \left (-e^{-x}\right ) x+e^{-x}-\left (1+e^5\right ) e^{-x}+e^{-x} \log (x)+\frac {e^{-x} \log (x) (x \log (\log (2))-2 (2+5 \log (\log (2))))}{\log (\log (2))}+\frac {2 e^{5-x} (2+5 \log (\log (2)))}{\log (\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - 4*E^5*x + 4*x*Log[x] + (-9 + x^2 + E^5*(-10*x + x^2) + (10*x - x^2)*Log[x])*Log[Log[2]])/(E^x*x*Log[
Log[2]]),x]

[Out]

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

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} &=\frac {\int \frac {e^{-x} \left (-4-4 e^5 x+4 x \log (x)+\left (-9+x^2+e^5 \left (-10 x+x^2\right )+\left (10 x-x^2\right ) \log (x)\right ) \log (\log (2))\right )}{x} \, dx}{\log (\log (2))}\\ &=\frac {\int \left (-e^{-x} \log (x) (-4-10 \log (\log (2))+x \log (\log (2)))+\frac {e^{-x} \left (-4-9 \log (\log (2))+\left (1+e^5\right ) x^2 \log (\log (2))-2 e^5 x (2+5 \log (\log (2)))\right )}{x}\right ) \, dx}{\log (\log (2))}\\ &=-\frac {\int e^{-x} \log (x) (-4-10 \log (\log (2))+x \log (\log (2))) \, dx}{\log (\log (2))}+\frac {\int \frac {e^{-x} \left (-4-9 \log (\log (2))+\left (1+e^5\right ) x^2 \log (\log (2))-2 e^5 x (2+5 \log (\log (2)))\right )}{x} \, dx}{\log (\log (2))}\\ &=e^{-x} \log (x)+\frac {e^{-x} \log (x) (x \log (\log (2))-2 (2+5 \log (\log (2))))}{\log (\log (2))}+\frac {\int \frac {e^{-x} (4+9 \log (\log (2))-x \log (\log (2)))}{x} \, dx}{\log (\log (2))}+\frac {\int \left (\frac {e^{-x} (-4-9 \log (\log (2)))}{x}+e^{-x} \left (1+e^5\right ) x \log (\log (2))-2 e^{5-x} (2+5 \log (\log (2)))\right ) \, dx}{\log (\log (2))}\\ &=e^{-x} \log (x)+\frac {e^{-x} \log (x) (x \log (\log (2))-2 (2+5 \log (\log (2))))}{\log (\log (2))}+\left (1+e^5\right ) \int e^{-x} x \, dx+\frac {\int \left (-e^{-x} \log (\log (2))+\frac {e^{-x} (4+9 \log (\log (2)))}{x}\right ) \, dx}{\log (\log (2))}+\frac {(-4-9 \log (\log (2))) \int \frac {e^{-x}}{x} \, dx}{\log (\log (2))}-\frac {(2 (2+5 \log (\log (2)))) \int e^{5-x} \, dx}{\log (\log (2))}\\ &=-e^{-x} \left (1+e^5\right ) x+e^{-x} \log (x)+\frac {2 e^{5-x} (2+5 \log (\log (2)))}{\log (\log (2))}-\frac {\text {Ei}(-x) (4+9 \log (\log (2)))}{\log (\log (2))}+\frac {e^{-x} \log (x) (x \log (\log (2))-2 (2+5 \log (\log (2))))}{\log (\log (2))}+\left (1+e^5\right ) \int e^{-x} \, dx+\frac {(4+9 \log (\log (2))) \int \frac {e^{-x}}{x} \, dx}{\log (\log (2))}-\int e^{-x} \, dx\\ &=e^{-x}-e^{-x} \left (1+e^5\right )-e^{-x} \left (1+e^5\right ) x+e^{-x} \log (x)+\frac {2 e^{5-x} (2+5 \log (\log (2)))}{\log (\log (2))}+\frac {e^{-x} \log (x) (x \log (\log (2))-2 (2+5 \log (\log (2))))}{\log (\log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 50, normalized size = 1.72 \begin {gather*} \frac {e^{-x} \left (-x \log (\log (2))+e^5 (4+9 \log (\log (2))-x \log (\log (2)))+\log (x) (-4-9 \log (\log (2))+x \log (\log (2)))\right )}{\log (\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 4*E^5*x + 4*x*Log[x] + (-9 + x^2 + E^5*(-10*x + x^2) + (10*x - x^2)*Log[x])*Log[Log[2]])/(E^x*
x*Log[Log[2]]),x]

[Out]

(-(x*Log[Log[2]]) + E^5*(4 + 9*Log[Log[2]] - x*Log[Log[2]]) + Log[x]*(-4 - 9*Log[Log[2]] + x*Log[Log[2]]))/(E^
x*Log[Log[2]])

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fricas [B]  time = 1.05, size = 54, normalized size = 1.86 \begin {gather*} -\frac {4 \, e^{\left (-x\right )} \log \relax (x) - {\left ({\left (x - 9\right )} e^{\left (-x\right )} \log \relax (x) - {\left ({\left (x - 9\right )} e^{5} + x\right )} e^{\left (-x\right )}\right )} \log \left (\log \relax (2)\right ) - 4 \, e^{\left (-x + 5\right )}}{\log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+10*x)*log(x)+(x^2-10*x)*exp(5)+x^2-9)*log(log(2))+4*x*log(x)-4*x*exp(5)-4)/x/exp(x)/log(log(
2)),x, algorithm="fricas")

[Out]

-(4*e^(-x)*log(x) - ((x - 9)*e^(-x)*log(x) - ((x - 9)*e^5 + x)*e^(-x))*log(log(2)) - 4*e^(-x + 5))/log(log(2))

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giac [B]  time = 0.33, size = 78, normalized size = 2.69 \begin {gather*} \frac {x e^{\left (-x\right )} \log \relax (x) \log \left (\log \relax (2)\right ) - x e^{\left (-x\right )} \log \left (\log \relax (2)\right ) - x e^{\left (-x + 5\right )} \log \left (\log \relax (2)\right ) - 9 \, e^{\left (-x\right )} \log \relax (x) \log \left (\log \relax (2)\right ) - 4 \, e^{\left (-x\right )} \log \relax (x) + 9 \, e^{\left (-x + 5\right )} \log \left (\log \relax (2)\right ) + 4 \, e^{\left (-x + 5\right )}}{\log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+10*x)*log(x)+(x^2-10*x)*exp(5)+x^2-9)*log(log(2))+4*x*log(x)-4*x*exp(5)-4)/x/exp(x)/log(log(
2)),x, algorithm="giac")

[Out]

(x*e^(-x)*log(x)*log(log(2)) - x*e^(-x)*log(log(2)) - x*e^(-x + 5)*log(log(2)) - 9*e^(-x)*log(x)*log(log(2)) -
4*e^(-x)*log(x) + 9*e^(-x + 5)*log(log(2)) + 4*e^(-x + 5))/log(log(2))

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maple [A]  time = 0.13, size = 50, normalized size = 1.72

method	result	size
norman	\(\left (x \ln \relax (x )+\left (-{\mathrm e}^{5}-1\right ) x +\frac {{\mathrm e}^{5} \left (9 \ln \left (\ln \relax (2)\right )+4\right )}{\ln \left (\ln \relax (2)\right )}-\frac {\left (9 \ln \left (\ln \relax (2)\right )+4\right ) \ln \relax (x )}{\ln \left (\ln \relax (2)\right )}\right ) {\mathrm e}^{-x}\)	\(50\)
risch	\(\frac {\left (x \ln \left (\ln \relax (2)\right )-9 \ln \left (\ln \relax (2)\right )-4\right ) {\mathrm e}^{-x} \ln \relax (x )}{\ln \left (\ln \relax (2)\right )}-\frac {\left ({\mathrm e}^{5} \ln \left (\ln \relax (2)\right ) x -9 \,{\mathrm e}^{5} \ln \left (\ln \relax (2)\right )+x \ln \left (\ln \relax (2)\right )-4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{\ln \left (\ln \relax (2)\right )}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^2+10*x)*ln(x)+(x^2-10*x)*exp(5)+x^2-9)*ln(ln(2))+4*x*ln(x)-4*x*exp(5)-4)/x/exp(x)/ln(ln(2)),x,method
=_RETURNVERBOSE)

[Out]

(x*ln(x)+(-exp(5)-1)*x+exp(5)*(9*ln(ln(2))+4)/ln(ln(2))-1/ln(ln(2))*(9*ln(ln(2))+4)*ln(x))/exp(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {{\left (x e^{5} + e^{5}\right )} e^{\left (-x\right )} \log \left (\log \relax (2)\right ) + {\left (x + 1\right )} e^{\left (-x\right )} \log \left (\log \relax (2)\right ) + 10 \, e^{\left (-x\right )} \log \relax (x) \log \left (\log \relax (2)\right ) + 4 \, e^{\left (-x\right )} \log \relax (x) - {\left ({\left (x + 1\right )} e^{\left (-x\right )} \log \relax (x) - {\rm Ei}\left (-x\right ) + e^{\left (-x\right )}\right )} \log \left (\log \relax (2)\right ) - {\rm Ei}\left (-x\right ) \log \left (\log \relax (2)\right ) - 10 \, e^{\left (-x + 5\right )} \log \left (\log \relax (2)\right ) - 4 \, e^{\left (-x + 5\right )}}{\log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+10*x)*log(x)+(x^2-10*x)*exp(5)+x^2-9)*log(log(2))+4*x*log(x)-4*x*exp(5)-4)/x/exp(x)/log(log(
2)),x, algorithm="maxima")

[Out]

-((x*e^5 + e^5)*e^(-x)*log(log(2)) + (x + 1)*e^(-x)*log(log(2)) + 10*e^(-x)*log(x)*log(log(2)) + 4*e^(-x)*log(
x) - ((x + 1)*e^(-x)*log(x) - integrate((x + 1)*e^(-x)/x, x))*log(log(2)) - Ei(-x)*log(log(2)) - 10*e^(-x + 5)
*log(log(2)) - 4*e^(-x + 5))/log(log(2))

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mupad [B]  time = 0.60, size = 69, normalized size = 2.38 \begin {gather*} \frac {{\mathrm {e}}^{5-x}\,\left (9\,\ln \left (\ln \relax (2)\right )+4\right )-{\mathrm {e}}^{-x}\,\ln \relax (x)\,\left (9\,\ln \left (\ln \relax (2)\right )+4\right )}{\ln \left (\ln \relax (2)\right )}-\frac {x\,\left ({\mathrm {e}}^{-x}\,\ln \left (\ln \relax (2)\right )\,\left ({\mathrm {e}}^5+1\right )-{\mathrm {e}}^{-x}\,\ln \left (\ln \relax (2)\right )\,\ln \relax (x)\right )}{\ln \left (\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(4*x*exp(5) + log(log(2))*(exp(5)*(10*x - x^2) - log(x)*(10*x - x^2) - x^2 + 9) - 4*x*log(x) + 4
))/(x*log(log(2))),x)

[Out]

(exp(5 - x)*(9*log(log(2)) + 4) - exp(-x)*log(x)*(9*log(log(2)) + 4))/log(log(2)) - (x*(exp(-x)*log(log(2))*(e
xp(5) + 1) - exp(-x)*log(log(2))*log(x)))/log(log(2))

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sympy [B]  time = 0.60, size = 65, normalized size = 2.24 \begin {gather*} \frac {\left (x \log {\relax (x )} \log {\left (\log {\relax (2 )} \right )} - x \log {\left (\log {\relax (2 )} \right )} - x e^{5} \log {\left (\log {\relax (2 )} \right )} - 4 \log {\relax (x )} - 9 \log {\relax (x )} \log {\left (\log {\relax (2 )} \right )} + 9 e^{5} \log {\left (\log {\relax (2 )} \right )} + 4 e^{5}\right ) e^{- x}}{\log {\left (\log {\relax (2 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**2+10*x)*ln(x)+(x**2-10*x)*exp(5)+x**2-9)*ln(ln(2))+4*x*ln(x)-4*x*exp(5)-4)/x/exp(x)/ln(ln(2))
,x)

[Out]

(x*log(x)*log(log(2)) - x*log(log(2)) - x*exp(5)*log(log(2)) - 4*log(x) - 9*log(x)*log(log(2)) + 9*exp(5)*log(
log(2)) + 4*exp(5))*exp(-x)/log(log(2))

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