\(\int \frac {12+23 e^4-3 e^4 \log (10)+(12+23 e^4-3 e^4 \log (10)) \log (x \log (\log (2)))}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx\) [810]

Optimal result

Integrand size = 53, antiderivative size = 23 \[ \int \frac {12+23 e^4-3 e^4 \log (10)+\left (12+23 e^4-3 e^4 \log (10)\right ) \log (x \log (\log (2)))}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx=\frac {-\frac {23}{3}-\frac {4}{e^4}+\log (10)}{x \log (x \log (\log (2)))} \] Output:

(ln(10)-4/exp(4)-23/3)/x/ln(x*ln(ln(2)))
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {12+23 e^4-3 e^4 \log (10)+\left (12+23 e^4-3 e^4 \log (10)\right ) \log (x \log (\log (2)))}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx=\frac {-12+e^4 (-23+\log (1000))}{3 e^4 x \log (x \log (\log (2)))} \] Input:

Integrate[(12 + 23*E^4 - 3*E^4*Log[10] + (12 + 23*E^4 - 3*E^4*Log[10])*Log 
[x*Log[Log[2]]])/(3*E^4*x^2*Log[x*Log[Log[2]]]^2),x]
 

Output:

(-12 + E^4*(-23 + Log[1000]))/(3*E^4*x*Log[x*Log[Log[2]]])
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.47 (sec) , antiderivative size = 146, normalized size of antiderivative = 6.35, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {27, 2813, 25, 2010, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (12+23 e^4-3 e^4 \log (10)\right ) \log (x \log (\log (2)))+23 e^4+12-3 e^4 \log (10)}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (12+e^4 (23-3 \log (10))\right ) \log (x \log (\log (2)))+e^4 (23-3 \log (10))+12}{x^2 \log ^2(x \log (\log (2)))}dx}{3 e^4}\)

\(\Big \downarrow \) 2813

\(\displaystyle \frac {-\left (12+e^4 (23-3 \log (10))\right ) \int -\frac {x \log (\log (2)) \operatorname {ExpIntegralEi}(-\log (x \log (\log (2))))+\frac {1}{\log (x \log (\log (2)))}}{x^2}dx-\log (\log (2)) \left (\left (12+e^4 (23-3 \log (10))\right ) \log (x \log (\log (2)))+12+e^4 (23-3 \log (10))\right ) \operatorname {ExpIntegralEi}(-\log (x \log (\log (2))))-\frac {\left (12+e^4 (23-3 \log (10))\right ) \log (x \log (\log (2)))+12+e^4 (23-3 \log (10))}{x \log (x \log (\log (2)))}}{3 e^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\left (12+e^4 (23-3 \log (10))\right ) \int \frac {x \log (\log (2)) \operatorname {ExpIntegralEi}(-\log (x \log (\log (2))))+\frac {1}{\log (x \log (\log (2)))}}{x^2}dx-\log (\log (2)) \left (\left (12+e^4 (23-3 \log (10))\right ) \log (x \log (\log (2)))+12+e^4 (23-3 \log (10))\right ) \operatorname {ExpIntegralEi}(-\log (x \log (\log (2))))-\frac {\left (12+e^4 (23-3 \log (10))\right ) \log (x \log (\log (2)))+12+e^4 (23-3 \log (10))}{x \log (x \log (\log (2)))}}{3 e^4}\)

\(\Big \downarrow \) 2010

\(\displaystyle \frac {\left (12+e^4 (23-3 \log (10))\right ) \int \left (\frac {\log (\log (2)) \operatorname {ExpIntegralEi}(-\log (x \log (\log (2))))}{x}+\frac {1}{x^2 \log (x \log (\log (2)))}\right )dx-\log (\log (2)) \left (\left (12+e^4 (23-3 \log (10))\right ) \log (x \log (\log (2)))+12+e^4 (23-3 \log (10))\right ) \operatorname {ExpIntegralEi}(-\log (x \log (\log (2))))-\frac {\left (12+e^4 (23-3 \log (10))\right ) \log (x \log (\log (2)))+12+e^4 (23-3 \log (10))}{x \log (x \log (\log (2)))}}{3 e^4}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\log (\log (2)) \left (\left (12+e^4 (23-3 \log (10))\right ) \log (x \log (\log (2)))+12+e^4 (23-3 \log (10))\right ) \operatorname {ExpIntegralEi}(-\log (x \log (\log (2))))+\left (12+e^4 (23-3 \log (10))\right ) \left (\log (\log (2)) \log (x \log (\log (2))) \operatorname {ExpIntegralEi}(-\log (x \log (\log (2))))+\log (\log (2)) \operatorname {ExpIntegralEi}(-\log (x \log (\log (2))))+\frac {1}{x}\right )-\frac {\left (12+e^4 (23-3 \log (10))\right ) \log (x \log (\log (2)))+12+e^4 (23-3 \log (10))}{x \log (x \log (\log (2)))}}{3 e^4}\)

Input:

Int[(12 + 23*E^4 - 3*E^4*Log[10] + (12 + 23*E^4 - 3*E^4*Log[10])*Log[x*Log 
[Log[2]]])/(3*E^4*x^2*Log[x*Log[Log[2]]]^2),x]
 

Output:

(-(ExpIntegralEi[-Log[x*Log[Log[2]]]]*Log[Log[2]]*(12 + E^4*(23 - 3*Log[10 
]) + (12 + E^4*(23 - 3*Log[10]))*Log[x*Log[Log[2]]])) - (12 + E^4*(23 - 3* 
Log[10]) + (12 + E^4*(23 - 3*Log[10]))*Log[x*Log[Log[2]]])/(x*Log[x*Log[Lo 
g[2]]]) + (12 + E^4*(23 - 3*Log[10]))*(x^(-1) + ExpIntegralEi[-Log[x*Log[L 
og[2]]]]*Log[Log[2]] + ExpIntegralEi[-Log[x*Log[Log[2]]]]*Log[Log[2]]*Log[ 
x*Log[Log[2]]]))/(3*E^4)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2010

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 2813

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_ 
.)]*(e_.))*((g_.)*(x_))^(m_.), x_Symbol] :> With[{u = IntHide[(g*x)^m*(a + 
b*Log[c*x^n])^p, x]}, Simp[(d + e*Log[f*x^r])   u, x] - Simp[e*r   Int[Simp 
lifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, 
x] &&  !(EqQ[p, 1] && EqQ[a, 0] && NeQ[d, 0])
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30

method	result	size
norman	\(\frac {{\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \left (10\right )-23 \,{\mathrm e}^{4}-12\right )}{3 x \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}\)	\(30\)
parallelrisch	\(\frac {{\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \left (10\right )-23 \,{\mathrm e}^{4}-12\right )}{3 x \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}\)	\(30\)
risch	\(\frac {{\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \left (2\right )+3 \,{\mathrm e}^{4} \ln \left (5\right )-23 \,{\mathrm e}^{4}-12\right )}{3 x \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}\)	\(34\)
parts	\(-\frac {{\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \left (10\right )-23 \,{\mathrm e}^{4}-12\right ) \ln \left (\ln \left (2\right )\right ) \left (-\frac {1}{x \ln \left (\ln \left (2\right )\right ) \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}+\operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )\right )}{3}+\frac {{\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \left (10\right )-23 \,{\mathrm e}^{4}-12\right ) \ln \left (\ln \left (2\right )\right ) \operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )}{3}\)	\(79\)
derivativedivides	\(\frac {\ln \left (\ln \left (2\right )\right ) {\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \left (10\right ) \operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )-3 \,{\mathrm e}^{4} \ln \left (10\right ) \left (-\frac {1}{x \ln \left (\ln \left (2\right )\right ) \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}+\operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )\right )-23 \,{\mathrm e}^{4} \operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )+23 \,{\mathrm e}^{4} \left (-\frac {1}{x \ln \left (\ln \left (2\right )\right ) \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}+\operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )\right )-\frac {12}{x \ln \left (\ln \left (2\right )\right ) \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}\right )}{3}\)	\(119\)
default	\(\frac {\ln \left (\ln \left (2\right )\right ) {\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \left (10\right ) \operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )-3 \,{\mathrm e}^{4} \ln \left (10\right ) \left (-\frac {1}{x \ln \left (\ln \left (2\right )\right ) \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}+\operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )\right )-23 \,{\mathrm e}^{4} \operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )+23 \,{\mathrm e}^{4} \left (-\frac {1}{x \ln \left (\ln \left (2\right )\right ) \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}+\operatorname {expIntegral}_{1}\left (\ln \left (x \ln \left (\ln \left (2\right )\right )\right )\right )\right )-\frac {12}{x \ln \left (\ln \left (2\right )\right ) \ln \left (x \ln \left (\ln \left (2\right )\right )\right )}\right )}{3}\)	\(119\)

Input:

int(1/3*((-3*exp(4)*ln(10)+23*exp(4)+12)*ln(x*ln(ln(2)))-3*exp(4)*ln(10)+2 
3*exp(4)+12)/x^2/exp(4)/ln(x*ln(ln(2)))^2,x,method=_RETURNVERBOSE)
 

Output:

1/3/exp(4)*(3*exp(4)*ln(10)-23*exp(4)-12)/x/ln(x*ln(ln(2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {12+23 e^4-3 e^4 \log (10)+\left (12+23 e^4-3 e^4 \log (10)\right ) \log (x \log (\log (2)))}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx=\frac {{\left (3 \, e^{4} \log \left (10\right ) - 23 \, e^{4} - 12\right )} e^{\left (-4\right )}}{3 \, x \log \left (x \log \left (\log \left (2\right )\right )\right )} \] Input:

integrate(1/3*((-3*exp(4)*log(10)+23*exp(4)+12)*log(x*log(log(2)))-3*exp(4 
)*log(10)+23*exp(4)+12)/x^2/exp(4)/log(x*log(log(2)))^2,x, algorithm="fric 
as")
 

Output:

1/3*(3*e^4*log(10) - 23*e^4 - 12)*e^(-4)/(x*log(x*log(log(2))))
 

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {12+23 e^4-3 e^4 \log (10)+\left (12+23 e^4-3 e^4 \log (10)\right ) \log (x \log (\log (2)))}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx=\frac {- 23 e^{4} - 12 + 3 e^{4} \log {\left (10 \right )}}{3 x e^{4} \log {\left (x \log {\left (\log {\left (2 \right )} \right )} \right )}} \] Input:

integrate(1/3*((-3*exp(4)*ln(10)+23*exp(4)+12)*ln(x*ln(ln(2)))-3*exp(4)*ln 
(10)+23*exp(4)+12)/x**2/exp(4)/ln(x*ln(ln(2)))**2,x)
 

Output:

(-23*exp(4) - 12 + 3*exp(4)*log(10))*exp(-4)/(3*x*log(x*log(log(2))))
 

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 3.61 \[ \int \frac {12+23 e^4-3 e^4 \log (10)+\left (12+23 e^4-3 e^4 \log (10)\right ) \log (x \log (\log (2)))}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx=-\frac {1}{3} \, {\left (3 \, {\rm Ei}\left (-\log \left (x \log \left (\log \left (2\right )\right )\right )\right ) e^{4} \log \left (10\right ) - 3 \, e^{4} \Gamma \left (-1, \log \left (x \log \left (\log \left (2\right )\right )\right )\right ) \log \left (10\right ) - 23 \, {\rm Ei}\left (-\log \left (x \log \left (\log \left (2\right )\right )\right )\right ) e^{4} + 23 \, e^{4} \Gamma \left (-1, \log \left (x \log \left (\log \left (2\right )\right )\right )\right ) - 12 \, {\rm Ei}\left (-\log \left (x \log \left (\log \left (2\right )\right )\right )\right ) + 12 \, \Gamma \left (-1, \log \left (x \log \left (\log \left (2\right )\right )\right )\right )\right )} e^{\left (-4\right )} \log \left (\log \left (2\right )\right ) \] Input:

integrate(1/3*((-3*exp(4)*log(10)+23*exp(4)+12)*log(x*log(log(2)))-3*exp(4 
)*log(10)+23*exp(4)+12)/x^2/exp(4)/log(x*log(log(2)))^2,x, algorithm="maxi 
ma")
 

Output:

-1/3*(3*Ei(-log(x*log(log(2))))*e^4*log(10) - 3*e^4*gamma(-1, log(x*log(lo 
g(2))))*log(10) - 23*Ei(-log(x*log(log(2))))*e^4 + 23*e^4*gamma(-1, log(x* 
log(log(2)))) - 12*Ei(-log(x*log(log(2)))) + 12*gamma(-1, log(x*log(log(2) 
))))*e^(-4)*log(log(2))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {12+23 e^4-3 e^4 \log (10)+\left (12+23 e^4-3 e^4 \log (10)\right ) \log (x \log (\log (2)))}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx=\frac {{\left (3 \, e^{4} \log \left (5\right ) + 3 \, e^{4} \log \left (2\right ) - 23 \, e^{4} - 12\right )} e^{\left (-4\right )}}{3 \, x \log \left (x \log \left (\log \left (2\right )\right )\right )} \] Input:

integrate(1/3*((-3*exp(4)*log(10)+23*exp(4)+12)*log(x*log(log(2)))-3*exp(4 
)*log(10)+23*exp(4)+12)/x^2/exp(4)/log(x*log(log(2)))^2,x, algorithm="giac 
")
 

Output:

1/3*(3*e^4*log(5) + 3*e^4*log(2) - 23*e^4 - 12)*e^(-4)/(x*log(x*log(log(2) 
)))
 

Mupad [B] (verification not implemented)

Time = 2.59 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {12+23 e^4-3 e^4 \log (10)+\left (12+23 e^4-3 e^4 \log (10)\right ) \log (x \log (\log (2)))}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx=-\frac {4\,{\mathrm {e}}^{-4}-\ln \left (10\right )+\frac {23}{3}}{x\,\ln \left (x\,\ln \left (\ln \left (2\right )\right )\right )} \] Input:

int((exp(-4)*((23*exp(4))/3 - exp(4)*log(10) + (log(x*log(log(2)))*(23*exp 
(4) - 3*exp(4)*log(10) + 12))/3 + 4))/(x^2*log(x*log(log(2)))^2),x)
 

Output:

-(4*exp(-4) - log(10) + 23/3)/(x*log(x*log(log(2))))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {12+23 e^4-3 e^4 \log (10)+\left (12+23 e^4-3 e^4 \log (10)\right ) \log (x \log (\log (2)))}{3 e^4 x^2 \log ^2(x \log (\log (2)))} \, dx=\frac {3 \,\mathrm {log}\left (10\right ) e^{4}-23 e^{4}-12}{3 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x \right ) e^{4} x} \] Input:

int(1/3*((-3*exp(4)*log(10)+23*exp(4)+12)*log(x*log(log(2)))-3*exp(4)*log( 
10)+23*exp(4)+12)/x^2/exp(4)/log(x*log(log(2)))^2,x)
 

Output:

(3*log(10)*e**4 - 23*e**4 - 12)/(3*log(log(log(2))*x)*e**4*x)