3.12.36 \(\int \frac {-1-2 x \log (2 x)+\log (2 x) \log (\frac {5}{\log (4) \log (2 x)})}{192 e^6 \log (2 x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {x \left (-x+\log \left (\frac {5}{\log (4) \log (2 x)}\right )\right )}{192 e^6} \]

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Rubi [A]  time = 0.09, antiderivative size = 32, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {12, 6688, 2298, 2520} \begin {gather*} \frac {x \log \left (\frac {5}{\log (4) \log (2 x)}\right )}{192 e^6}-\frac {x^2}{192 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 2*x*Log[2*x] + Log[2*x]*Log[5/(Log[4]*Log[2*x])])/(192*E^6*Log[2*x]),x]

[Out]

-1/192*x^2/E^6 + (x*Log[5/(Log[4]*Log[2*x])])/(192*E^6)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2520

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-1-2 x \log (2 x)+\log (2 x) \log \left (\frac {5}{\log (4) \log (2 x)}\right )}{\log (2 x)} \, dx}{192 e^6}\\ &=\frac {\int \left (-2 x-\frac {1}{\log (2 x)}+\log \left (\frac {5}{\log (4) \log (2 x)}\right )\right ) \, dx}{192 e^6}\\ &=-\frac {x^2}{192 e^6}-\frac {\int \frac {1}{\log (2 x)} \, dx}{192 e^6}+\frac {\int \log \left (\frac {5}{\log (4) \log (2 x)}\right ) \, dx}{192 e^6}\\ &=-\frac {x^2}{192 e^6}+\frac {x \log \left (\frac {5}{\log (4) \log (2 x)}\right )}{192 e^6}-\frac {\text {li}(2 x)}{384 e^6}+\frac {\int \frac {1}{\log (2 x)} \, dx}{192 e^6}\\ &=-\frac {x^2}{192 e^6}+\frac {x \log \left (\frac {5}{\log (4) \log (2 x)}\right )}{192 e^6}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 28, normalized size = 1.12 \begin {gather*} \frac {-x^2+x \log \left (\frac {5}{\log (4) \log (2 x)}\right )}{192 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 2*x*Log[2*x] + Log[2*x]*Log[5/(Log[4]*Log[2*x])])/(192*E^6*Log[2*x]),x]

[Out]

(-x^2 + x*Log[5/(Log[4]*Log[2*x])])/(192*E^6)

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fricas [A]  time = 0.61, size = 24, normalized size = 0.96 \begin {gather*} -\frac {1}{192} \, {\left (x^{2} - x \log \left (\frac {5}{2 \, \log \relax (2) \log \left (2 \, x\right )}\right )\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/192*(log(2*x)*log(5/2/log(2)/log(2*x))-2*x*log(2*x)-1)/exp(6)/log(2*x),x, algorithm="fricas")

[Out]

-1/192*(x^2 - x*log(5/2/(log(2)*log(2*x))))*e^(-6)

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giac [A]  time = 0.48, size = 29, normalized size = 1.16 \begin {gather*} -\frac {1}{192} \, {\left (x^{2} - x \log \relax (5) + x \log \relax (2) + x \log \left (\log \relax (2)\right ) + x \log \left (\log \left (2 \, x\right )\right )\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/192*(log(2*x)*log(5/2/log(2)/log(2*x))-2*x*log(2*x)-1)/exp(6)/log(2*x),x, algorithm="giac")

[Out]

-1/192*(x^2 - x*log(5) + x*log(2) + x*log(log(2)) + x*log(log(2*x)))*e^(-6)

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maple [A]  time = 0.24, size = 31, normalized size = 1.24

method	result	size
norman	\(-\frac {{\mathrm e}^{-6} x^{2}}{192}+\frac {x \,{\mathrm e}^{-6} \ln \left (\frac {5}{2 \ln \relax (2) \ln \left (2 x \right )}\right )}{192}\)	\(31\)
risch	\(-\frac {{\mathrm e}^{-6} x \ln \left (\ln \left (2 x \right )\right )}{192}+\frac {{\mathrm e}^{-6} x \ln \relax (5)}{192}-\frac {{\mathrm e}^{-6} x \ln \relax (2)}{192}-\frac {{\mathrm e}^{-6} x \ln \left (\ln \relax (2)\right )}{192}-\frac {{\mathrm e}^{-6} x^{2}}{192}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/192*(ln(2*x)*ln(5/2/ln(2)/ln(2*x))-2*x*ln(2*x)-1)/exp(6)/ln(2*x),x,method=_RETURNVERBOSE)

[Out]

-1/192/exp(6)*x^2+1/192*x/exp(6)*ln(5/2/ln(2)/ln(2*x))

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maxima [A]  time = 0.55, size = 24, normalized size = 0.96 \begin {gather*} -\frac {1}{192} \, {\left (x^{2} - x \log \left (\frac {5}{2 \, \log \relax (2) \log \left (2 \, x\right )}\right )\right )} e^{\left (-6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/192*(log(2*x)*log(5/2/log(2)/log(2*x))-2*x*log(2*x)-1)/exp(6)/log(2*x),x, algorithm="maxima")

[Out]

-1/192*(x^2 - x*log(5/2/(log(2)*log(2*x))))*e^(-6)

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mupad [B]  time = 1.03, size = 22, normalized size = 0.88 \begin {gather*} -\frac {x\,{\mathrm {e}}^{-6}\,\left (x-\ln \left (\frac {5}{2\,\ln \left (2\,x\right )\,\ln \relax (2)}\right )\right )}{192} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-6)*((x*log(2*x))/96 - (log(5/(2*log(2*x)*log(2)))*log(2*x))/192 + 1/192))/log(2*x),x)

[Out]

-(x*exp(-6)*(x - log(5/(2*log(2*x)*log(2)))))/192

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sympy [A]  time = 0.33, size = 27, normalized size = 1.08 \begin {gather*} - \frac {x^{2}}{192 e^{6}} + \frac {x \log {\left (\frac {5}{2 \log {\relax (2 )} \log {\left (2 x \right )}} \right )}}{192 e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/192*(ln(2*x)*ln(5/2/ln(2)/ln(2*x))-2*x*ln(2*x)-1)/exp(6)/ln(2*x),x)

[Out]

-x**2*exp(-6)/192 + x*exp(-6)*log(5/(2*log(2)*log(2*x)))/192

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