\(\int \frac {e^{-3 x} (5+(-15 x-10 x^2+15 x^3) \log (x)-15 x \log (x) \log (4 \log (x)))}{x \log (x)} \, dx\) [6719]

Optimal result

Integrand size = 42, antiderivative size = 19 \[ \int \frac {e^{-3 x} \left (5+\left (-15 x-10 x^2+15 x^3\right ) \log (x)-15 x \log (x) \log (4 \log (x))\right )}{x \log (x)} \, dx=5 e^{-3 x} \left (1-x^2+\log (4 \log (x))\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {6874, 2225, 2207, 2635, 12} \[ \int \frac {e^{-3 x} \left (5+\left (-15 x-10 x^2+15 x^3\right ) \log (x)-15 x \log (x) \log (4 \log (x))\right )}{x \log (x)} \, dx=-5 e^{-3 x} x^2+5 e^{-3 x}+5 e^{-3 x} \log (4 \log (x)) \]

[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 2635

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 6874

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-3 x} \left (5+\left (-15 x-10 x^2+15 x^3\right ) \log (x)-15 x \log (x) \log (4 \log (x))\right )}{x \log (x)} \, dx=e^{-3 x} \left (5-5 x^2+5 \log (4 \log (x))\right ) \]

[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)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21

method	result	size
parallelrisch	\(-\frac {\left (-15+15 x^{2}-15 \ln \left (4 \ln \left (x \right )\right )\right ) {\mathrm e}^{-3 x}}{3}\)	\(23\)
risch	\(5 \,{\mathrm e}^{-3 x} \ln \left (4 \ln \left (x \right )\right )-5 \left (x^{2}-1\right ) {\mathrm e}^{-3 x}\)	\(24\)

[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]

-1/3*(-15+15*x^2-15*ln(4*ln(x)))/exp(3*x)

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-3 x} \left (5+\left (-15 x-10 x^2+15 x^3\right ) \log (x)-15 x \log (x) \log (4 \log (x))\right )}{x \log (x)} \, dx=-5 \, {\left (x^{2} - 1\right )} e^{\left (-3 \, x\right )} + 5 \, e^{\left (-3 \, x\right )} \log \left (4 \, \log \left (x\right )\right ) \]

[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))

Sympy [A] (verification not implemented)

Time = 3.67 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3 x} \left (5+\left (-15 x-10 x^2+15 x^3\right ) \log (x)-15 x \log (x) \log (4 \log (x))\right )}{x \log (x)} \, dx=\left (- 5 x^{2} + 5 \log {\left (4 \log {\left (x \right )} \right )} + 5\right ) e^{- 3 x} \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (18) = 36\).

Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.53 \[ \int \frac {e^{-3 x} \left (5+\left (-15 x-10 x^2+15 x^3\right ) \log (x)-15 x \log (x) \log (4 \log (x))\right )}{x \log (x)} \, dx=-\frac {5}{9} \, {\left (9 \, x^{2} + 6 \, x + 2\right )} e^{\left (-3 \, x\right )} + \frac {10}{9} \, {\left (3 \, x + 1\right )} e^{\left (-3 \, x\right )} + 5 \, {\left (2 \, \log \left (2\right ) + \log \left (\log \left (x\right )\right )\right )} e^{\left (-3 \, x\right )} + 5 \, e^{\left (-3 \, x\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-3 x} \left (5+\left (-15 x-10 x^2+15 x^3\right ) \log (x)-15 x \log (x) \log (4 \log (x))\right )}{x \log (x)} \, dx=-5 \, x^{2} e^{\left (-3 \, x\right )} + 5 \, e^{\left (-3 \, x\right )} \log \left (4 \, \log \left (x\right )\right ) + 5 \, e^{\left (-3 \, x\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 12.48 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-3 x} \left (5+\left (-15 x-10 x^2+15 x^3\right ) \log (x)-15 x \log (x) \log (4 \log (x))\right )}{x \log (x)} \, dx=5\,{\mathrm {e}}^{-3\,x}\,\left (\ln \left (4\,\ln \left (x\right )\right )-x^2+1\right ) \]

[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)