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

Optimal. Leaf size=24 \[ 2+\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x} \]

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Rubi [A]  time = 0.44, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 16, number of rules used = 8, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used = {6742, 6688, 14, 43, 2309, 2178, 30, 2555} \begin {gather*} \frac {5 \log \left (-\frac {e^{x-2} \log (5) \log (4 x)}{x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, 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 (\frac {5 (1-\log (4 x)+x \log (4 x))}{x^2 \log (4 x)}-\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x^2}\right ) \, dx\\ &=5 \int \frac {1-\log (4 x)+x \log (4 x)}{x^2 \log (4 x)} \, dx-5 \int \frac {\log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x^2} \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}+5 \int \frac {-1+x+\frac {1}{\log (4 x)}}{x^2} \, dx+5 \int \frac {-1-(-1+x) \log (4 x)}{x^2 \log (4 x)} \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}+5 \int \left (\frac {1-x}{x^2}-\frac {1}{x^2 \log (4 x)}\right ) \, dx+5 \int \left (\frac {-1+x}{x^2}+\frac {1}{x^2 \log (4 x)}\right ) \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}+5 \int \frac {1-x}{x^2} \, dx+5 \int \frac {-1+x}{x^2} \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}+5 \int \left (\frac {1}{x^2}-\frac {1}{x}\right ) \, dx+5 \int \left (-\frac {1}{x^2}+\frac {1}{x}\right ) \, dx\\ &=\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 24, normalized size = 1.00 \begin {gather*} -5+\frac {5 \log \left (-\frac {e^{-2+x} \log (5) \log (4 x)}{x}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 21, normalized size = 0.88 \begin {gather*} \frac {5 \, \log \left (-\frac {e^{\left (x - 2\right )} \log \relax (5) \log \left (4 \, x\right )}{x}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*log(-e^(x - 2)*log(5)*log(4*x)/x)/x

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giac [A]  time = 0.17, size = 33, normalized size = 1.38 \begin {gather*} \frac {5 \, {\left (\log \left (\log \relax (5)\right ) - 2\right )}}{x} - \frac {5 \, \log \relax (x)}{x} + \frac {5 \, \log \left (-2 \, \log \relax (2) - \log \relax (x)\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {-5 \ln \left (4 x \right ) \ln \left (-\frac {\ln \relax (5) \ln \left (4 x \right ) {\mathrm e}^{x -2}}{x}\right )+\left (5 x -5\right ) \ln \left (4 x \right )+5}{x^{2} \ln \left (4 x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(4*x)*ln(-ln(5)*ln(4*x)/x/exp(2-x))+(5*x-5)*ln(4*x)+5)/x^2/ln(4*x),x)

[Out]

int((-5*ln(4*x)*ln(-ln(5)*ln(4*x)/x/exp(2-x))+(5*x-5)*ln(4*x)+5)/x^2/ln(4*x),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {5 \, {\left ({\left (x + 1\right )} \log \relax (x) - \log \left (-2 \, \log \relax (2) - \log \relax (x)\right ) - \log \left (\log \relax (5)\right ) + 3\right )}}{x} + \frac {5}{x} + 20 \, {\rm Ei}\left (-\log \left (4 \, x\right )\right ) - 5 \, \int \frac {1}{2 \, x^{2} \log \relax (2) + x^{2} \log \relax (x)}\,{d x} + 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*((x + 1)*log(x) - log(-2*log(2) - log(x)) - log(log(5)) + 3)/x + 5/x + 20*Ei(-log(4*x)) - 5*integrate(1/(2*
x^2*log(2) + x^2*log(x)), x) + 5*log(x)

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mupad [B]  time = 5.69, size = 20, normalized size = 0.83 \begin {gather*} \frac {5\,\ln \left (-\frac {\ln \left (4\,x\right )\,\ln \relax (5)}{x}\right )-10}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*log(-(log(4*x)*log(5))/x) - 10)/x

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sympy [A]  time = 0.42, size = 20, normalized size = 0.83 \begin {gather*} \frac {5 \log {\left (- \frac {e^{x - 2} \log {\relax (5 )} \log {\left (4 x \right )}}{x} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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