3.68.19 \(\int \frac {e^{-3 x} (5+(-15 x-10 x^2+15 x^3) \log (x)-15 x \log (x) \log (4 \log (x)))}{x \log (x)} \, dx\)

Optimal. Leaf size=19 \[ 5 e^{-3 x} \left (1-x^2+\log (4 \log (x))\right ) \]

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Rubi [A]  time = 0.71, antiderivative size = 30, normalized size of antiderivative = 1.58, number of steps used = 12, number of rules used = 5, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {6742, 2194, 2176, 2555, 12} \begin {gather*} -5 e^{-3 x} x^2+5 e^{-3 x}+5 e^{-3 x} \log (4 \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + (-15*x - 10*x^2 + 15*x^3)*Log[x] - 15*x*Log[x]*Log[4*Log[x]])/(E^(3*x)*x*Log[x]),x]

[Out]

5/E^(3*x) - (5*x^2)/E^(3*x) + (5*Log[4*Log[x]])/E^(3*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 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 2555

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[w*Simplify
[D[u, x]/u], x], x] /; InverseFunctionFreeQ[w, x]] /; ProductQ[u]

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 {5 e^{-3 x} \left (1-3 x \log (x)-2 x^2 \log (x)+3 x^3 \log (x)\right )}{x \log (x)}-15 e^{-3 x} \log (4 \log (x))\right ) \, dx\\ &=5 \int \frac {e^{-3 x} \left (1-3 x \log (x)-2 x^2 \log (x)+3 x^3 \log (x)\right )}{x \log (x)} \, dx-15 \int e^{-3 x} \log (4 \log (x)) \, dx\\ &=5 e^{-3 x} \log (4 \log (x))+5 \int \left (-3 e^{-3 x}-2 e^{-3 x} x+3 e^{-3 x} x^2+\frac {e^{-3 x}}{x \log (x)}\right ) \, dx+15 \int -\frac {e^{-3 x}}{3 x \log (x)} \, dx\\ &=5 e^{-3 x} \log (4 \log (x))-10 \int e^{-3 x} x \, dx-15 \int e^{-3 x} \, dx+15 \int e^{-3 x} x^2 \, dx\\ &=5 e^{-3 x}+\frac {10}{3} e^{-3 x} x-5 e^{-3 x} x^2+5 e^{-3 x} \log (4 \log (x))-\frac {10}{3} \int e^{-3 x} \, dx+10 \int e^{-3 x} x \, dx\\ &=\frac {55 e^{-3 x}}{9}-5 e^{-3 x} x^2+5 e^{-3 x} \log (4 \log (x))+\frac {10}{3} \int e^{-3 x} \, dx\\ &=5 e^{-3 x}-5 e^{-3 x} x^2+5 e^{-3 x} \log (4 \log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 20, normalized size = 1.05 \begin {gather*} e^{-3 x} \left (5-5 x^2+5 \log (4 \log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + (-15*x - 10*x^2 + 15*x^3)*Log[x] - 15*x*Log[x]*Log[4*Log[x]])/(E^(3*x)*x*Log[x]),x]

[Out]

(5 - 5*x^2 + 5*Log[4*Log[x]])/E^(3*x)

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fricas [A]  time = 0.65, size = 23, normalized size = 1.21 \begin {gather*} -5 \, {\left (x^{2} - 1\right )} e^{\left (-3 \, x\right )} + 5 \, e^{\left (-3 \, x\right )} \log \left (4 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x*log(x)*log(4*log(x))+(15*x^3-10*x^2-15*x)*log(x)+5)/x/exp(3*x)/log(x),x, algorithm="fricas")

[Out]

-5*(x^2 - 1)*e^(-3*x) + 5*e^(-3*x)*log(4*log(x))

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giac [A]  time = 0.16, size = 27, normalized size = 1.42 \begin {gather*} -5 \, x^{2} e^{\left (-3 \, x\right )} + 5 \, e^{\left (-3 \, x\right )} \log \left (4 \, \log \relax (x)\right ) + 5 \, e^{\left (-3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x*log(x)*log(4*log(x))+(15*x^3-10*x^2-15*x)*log(x)+5)/x/exp(3*x)/log(x),x, algorithm="giac")

[Out]

-5*x^2*e^(-3*x) + 5*e^(-3*x)*log(4*log(x)) + 5*e^(-3*x)

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maple [A]  time = 0.02, size = 24, normalized size = 1.26

method	result	size
risch	\(5 \,{\mathrm e}^{-3 x} \ln \left (4 \ln \relax (x )\right )-5 \left (x^{2}-1\right ) {\mathrm e}^{-3 x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-15*x*ln(x)*ln(4*ln(x))+(15*x^3-10*x^2-15*x)*ln(x)+5)/x/exp(3*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

5*exp(-3*x)*ln(4*ln(x))-5*(x^2-1)*exp(-3*x)

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maxima [B]  time = 0.49, size = 48, normalized size = 2.53 \begin {gather*} -\frac {5}{9} \, {\left (9 \, x^{2} + 6 \, x + 2\right )} e^{\left (-3 \, x\right )} + \frac {10}{9} \, {\left (3 \, x + 1\right )} e^{\left (-3 \, x\right )} + 5 \, {\left (2 \, \log \relax (2) + \log \left (\log \relax (x)\right )\right )} e^{\left (-3 \, x\right )} + 5 \, e^{\left (-3 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x*log(x)*log(4*log(x))+(15*x^3-10*x^2-15*x)*log(x)+5)/x/exp(3*x)/log(x),x, algorithm="maxima")

[Out]

-5/9*(9*x^2 + 6*x + 2)*e^(-3*x) + 10/9*(3*x + 1)*e^(-3*x) + 5*(2*log(2) + log(log(x)))*e^(-3*x) + 5*e^(-3*x)

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mupad [B]  time = 4.18, size = 18, normalized size = 0.95 \begin {gather*} 5\,{\mathrm {e}}^{-3\,x}\,\left (\ln \left (4\,\ln \relax (x)\right )-x^2+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-3*x)*(log(x)*(15*x + 10*x^2 - 15*x^3) + 15*x*log(4*log(x))*log(x) - 5))/(x*log(x)),x)

[Out]

5*exp(-3*x)*(log(4*log(x)) - x^2 + 1)

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sympy [A]  time = 9.23, size = 19, normalized size = 1.00 \begin {gather*} \left (- 5 x^{2} + 5 \log {\left (4 \log {\relax (x )} \right )} + 5\right ) e^{- 3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-15*x*ln(x)*ln(4*ln(x))+(15*x**3-10*x**2-15*x)*ln(x)+5)/x/exp(3*x)/ln(x),x)

[Out]

(-5*x**2 + 5*log(4*log(x)) + 5)*exp(-3*x)

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