3.101.42 \(\int \frac {\frac {6}{e^3}+(6+\frac {12 x}{e^3}) \log (x)+\frac {6 \log (x) \log (\log (x))}{e^3}}{2 \log (x)} \, dx\)

Optimal. Leaf size=17 \[ \frac {3}{2} x \left (2+\frac {2 (x+\log (\log (x)))}{e^3}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 21, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 6688, 6742, 2298, 2520} \begin {gather*} \frac {3 x^2}{e^3}+3 x+\frac {3 x \log (\log (x))}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6/E^3 + (6 + (12*x)/E^3)*Log[x] + (6*Log[x]*Log[Log[x]])/E^3)/(2*Log[x]),x]

[Out]

3*x + (3*x^2)/E^3 + (3*x*Log[Log[x]])/E^3

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]]

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 {1}{2} \int \frac {\frac {6}{e^3}+\left (6+\frac {12 x}{e^3}\right ) \log (x)+\frac {6 \log (x) \log (\log (x))}{e^3}}{\log (x)} \, dx\\ &=\frac {1}{2} \int \frac {6+6 \log (x) \left (e^3+2 x+\log (\log (x))\right )}{e^3 \log (x)} \, dx\\ &=\frac {\int \frac {6+6 \log (x) \left (e^3+2 x+\log (\log (x))\right )}{\log (x)} \, dx}{2 e^3}\\ &=\frac {\int \left (\frac {6 \left (1+e^3 \log (x)+2 x \log (x)\right )}{\log (x)}+6 \log (\log (x))\right ) \, dx}{2 e^3}\\ &=\frac {3 \int \frac {1+e^3 \log (x)+2 x \log (x)}{\log (x)} \, dx}{e^3}+\frac {3 \int \log (\log (x)) \, dx}{e^3}\\ &=\frac {3 x \log (\log (x))}{e^3}+\frac {3 \int \left (e^3+2 x+\frac {1}{\log (x)}\right ) \, dx}{e^3}-\frac {3 \int \frac {1}{\log (x)} \, dx}{e^3}\\ &=3 x+\frac {3 x^2}{e^3}+\frac {3 x \log (\log (x))}{e^3}-\frac {3 \text {li}(x)}{e^3}+\frac {3 \int \frac {1}{\log (x)} \, dx}{e^3}\\ &=3 x+\frac {3 x^2}{e^3}+\frac {3 x \log (\log (x))}{e^3}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 22, normalized size = 1.29 \begin {gather*} \frac {3 e^3 x+3 x^2+3 x \log (\log (x))}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6/E^3 + (6 + (12*x)/E^3)*Log[x] + (6*Log[x]*Log[Log[x]])/E^3)/(2*Log[x]),x]

[Out]

(3*E^3*x + 3*x^2 + 3*x*Log[Log[x]])/E^3

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 25, normalized size = 1.47 \begin {gather*} \frac {3}{2} \, x^{2} e^{\left (\log \relax (2) - 3\right )} + \frac {3}{2} \, x e^{\left (\log \relax (2) - 3\right )} \log \left (\log \relax (x)\right ) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(3*exp(log(2)-3)*log(x)*log(log(x))+(6*x*exp(log(2)-3)+6)*log(x)+3*exp(log(2)-3))/log(x),x, algo
rithm="fricas")

[Out]

3/2*x^2*e^(log(2) - 3) + 3/2*x*e^(log(2) - 3)*log(log(x)) + 3*x

________________________________________________________________________________________

giac [A]  time = 0.14, size = 19, normalized size = 1.12 \begin {gather*} 3 \, x^{2} e^{\left (-3\right )} + 3 \, x e^{\left (-3\right )} \log \left (\log \relax (x)\right ) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(3*exp(log(2)-3)*log(x)*log(log(x))+(6*x*exp(log(2)-3)+6)*log(x)+3*exp(log(2)-3))/log(x),x, algo
rithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 20, normalized size = 1.18

method	result	size
default	\(3 x +3 x^{2} {\mathrm e}^{-3}+3 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{-3} x\)	\(20\)
risch	\(3 x +3 x^{2} {\mathrm e}^{-3}+3 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{-3} x\)	\(20\)
norman	\(3 x +3 x^{2} {\mathrm e}^{-3}+3 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{-3} x\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(3*exp(ln(2)-3)*ln(x)*ln(ln(x))+(6*x*exp(ln(2)-3)+6)*ln(x)+3*exp(ln(2)-3))/ln(x),x,method=_RETURNVERBO
SE)

[Out]

3*x+3*x^2*exp(-3)+3*ln(ln(x))*exp(-3)*x

________________________________________________________________________________________

maxima [C]  time = 0.39, size = 33, normalized size = 1.94 \begin {gather*} 3 \, x^{2} e^{\left (-3\right )} + 3 \, {\left (x \log \left (\log \relax (x)\right ) - {\rm Ei}\left (\log \relax (x)\right )\right )} e^{\left (-3\right )} + 3 \, {\rm Ei}\left (\log \relax (x)\right ) e^{\left (-3\right )} + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(3*exp(log(2)-3)*log(x)*log(log(x))+(6*x*exp(log(2)-3)+6)*log(x)+3*exp(log(2)-3))/log(x),x, algo
rithm="maxima")

[Out]

3*x^2*e^(-3) + 3*(x*log(log(x)) - Ei(log(x)))*e^(-3) + 3*Ei(log(x))*e^(-3) + 3*x

________________________________________________________________________________________

mupad [B]  time = 6.29, size = 12, normalized size = 0.71 \begin {gather*} 3\,x\,{\mathrm {e}}^{-3}\,\left (x+\ln \left (\ln \relax (x)\right )+{\mathrm {e}}^3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*exp(log(2) - 3))/2 + (log(x)*(6*x*exp(log(2) - 3) + 6))/2 + (3*log(log(x))*exp(log(2) - 3)*log(x))/2)/
log(x),x)

[Out]

3*x*exp(-3)*(x + log(log(x)) + exp(3))

________________________________________________________________________________________

sympy [A]  time = 0.46, size = 22, normalized size = 1.29 \begin {gather*} \frac {3 x^{2}}{e^{3}} + \frac {3 x \log {\left (\log {\relax (x )} \right )}}{e^{3}} + 3 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(3*exp(ln(2)-3)*ln(x)*ln(ln(x))+(6*x*exp(ln(2)-3)+6)*ln(x)+3*exp(ln(2)-3))/ln(x),x)

[Out]

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

________________________________________________________________________________________