\(\int \frac {3^{-1/x} \sqrt [x]{\log (x^4)} (-48+12 \log (x^4) \log (\frac {\log (x^4)}{3}))}{(625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)) \log (x^4)+3^{-1/x} (-50 x^2-2 x^2 \log (4)) \log ^{1+\frac {1}{x}}(x^4)+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}(x^4)} \, dx\) [121]

Optimal result
Mathematica [F]
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [F]
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 116, antiderivative size = 26 \[ \int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (-48+12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{\left (625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)\right ) \log \left (x^4\right )+3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{1+\frac {1}{x}}\left (x^4\right )+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}\left (x^4\right )} \, dx=\frac {12}{-25-\log (4)+3^{-1/x} \sqrt [x]{\log \left (x^4\right )}} \] Output:

12/(-25-2*ln(2)+exp(ln(1/3*ln(x^4))/x))
 

Mathematica [F]

\[ \int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (-48+12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{\left (625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)\right ) \log \left (x^4\right )+3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{1+\frac {1}{x}}\left (x^4\right )+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}\left (x^4\right )} \, dx=\int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (-48+12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{\left (625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)\right ) \log \left (x^4\right )+3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{1+\frac {1}{x}}\left (x^4\right )+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}\left (x^4\right )} \, dx \] Input:

Integrate[(Log[x^4]^x^(-1)*(-48 + 12*Log[x^4]*Log[Log[x^4]/3]))/(3^x^(-1)* 
((625*x^2 + 50*x^2*Log[4] + x^2*Log[4]^2)*Log[x^4] + ((-50*x^2 - 2*x^2*Log 
[4])*Log[x^4]^(1 + x^(-1)))/3^x^(-1) + (x^2*Log[x^4]^(1 + 2/x))/3^(2/x))), 
x]
 

Output:

Integrate[(Log[x^4]^x^(-1)*(-48 + 12*Log[x^4]*Log[Log[x^4]/3]))/(3^x^(-1)* 
((625*x^2 + 50*x^2*Log[4] + x^2*Log[4]^2)*Log[x^4] + ((-50*x^2 - 2*x^2*Log 
[4])*Log[x^4]^(1 + x^(-1)))/3^x^(-1) + (x^2*Log[x^4]^(1 + 2/x))/3^(2/x))), 
 x]
 

Rubi [A] (verified)

Time = 1.69 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {7239, 27, 25, 7262, 17}

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.

\(\displaystyle \int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )-48\right )}{3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{\frac {1}{x}+1}\left (x^4\right )+3^{-2/x} x^2 \log ^{\frac {2}{x}+1}\left (x^4\right )+\left (625 x^2+x^2 \log ^2(4)+50 x^2 \log (4)\right ) \log \left (x^4\right )} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {4\ 3^{\frac {1}{x}+1} \log ^{\frac {1}{x}-1}\left (x^4\right ) \left (\log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )-4\right )}{x^2 \left (3^{\frac {1}{x}} (25+\log (4))-\sqrt [x]{\log \left (x^4\right )}\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 4 \int -\frac {3^{1+\frac {1}{x}} \log ^{\frac {1}{x}-1}\left (x^4\right ) \left (4-\log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{x^2 \left (3^{\frac {1}{x}} (25+\log (4))-\sqrt [x]{\log \left (x^4\right )}\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -4 \int \frac {3^{1+\frac {1}{x}} \log ^{\frac {1}{x}-1}\left (x^4\right ) \left (4-\log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{x^2 \left (3^{\frac {1}{x}} (25+\log (4))-\sqrt [x]{\log \left (x^4\right )}\right )^2}dx\)

\(\Big \downarrow \) 7262

\(\displaystyle 12 \int \frac {1}{\left (3^{\frac {1}{x}} (25+\log (4)) \log ^{-\frac {1}{x}}\left (x^4\right )-1\right )^2}d\left (3^{\frac {1}{x}} \log ^{-\frac {1}{x}}\left (x^4\right )\right )\)

\(\Big \downarrow \) 17

\(\displaystyle \frac {12}{(25+\log (4)) \left (1-3^{\frac {1}{x}} (25+\log (4)) \log ^{-\frac {1}{x}}\left (x^4\right )\right )}\)

Input:

Int[(Log[x^4]^x^(-1)*(-48 + 12*Log[x^4]*Log[Log[x^4]/3]))/(3^x^(-1)*((625* 
x^2 + 50*x^2*Log[4] + x^2*Log[4]^2)*Log[x^4] + ((-50*x^2 - 2*x^2*Log[4])*L 
og[x^4]^(1 + x^(-1)))/3^x^(-1) + (x^2*Log[x^4]^(1 + 2/x))/3^(2/x))),x]
 

Output:

12/((25 + Log[4])*(1 - (3^x^(-1)*(25 + Log[4]))/Log[x^4]^x^(-1)))
 

Defintions of rubi rules used

rule 17
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

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 7239
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7262
Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c 
= Simplify[u/(p*w*D[v, x] - q*v*D[w, x])]}, Simp[c*p   Subst[Int[(b + a*x^p 
)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}, x] 
 && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]
 
Maple [A] (verified)

Time = 11.48 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method result size
parallelrisch \(-\frac {12}{25+2 \ln \left (2\right )-{\mathrm e}^{\frac {\ln \left (\frac {\ln \left (x^{4}\right )}{3}\right )}{x}}}\) \(25\)

Input:

int((12*ln(x^4)*ln(1/3*ln(x^4))-48)*exp(ln(1/3*ln(x^4))/x)/(x^2*ln(x^4)*ex 
p(ln(1/3*ln(x^4))/x)^2+(-4*x^2*ln(2)-50*x^2)*ln(x^4)*exp(ln(1/3*ln(x^4))/x 
)+(4*x^2*ln(2)^2+100*x^2*ln(2)+625*x^2)*ln(x^4)),x,method=_RETURNVERBOSE)
 

Output:

-12/(25+2*ln(2)-exp(ln(1/3*ln(x^4))/x))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (-48+12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{\left (625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)\right ) \log \left (x^4\right )+3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{1+\frac {1}{x}}\left (x^4\right )+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}\left (x^4\right )} \, dx=\frac {12}{\left (\frac {1}{3} \, \log \left (x^{4}\right )\right )^{\left (\frac {1}{x}\right )} - 2 \, \log \left (2\right ) - 25} \] Input:

integrate((12*log(x^4)*log(1/3*log(x^4))-48)*exp(log(1/3*log(x^4))/x)/(x^2 
*log(x^4)*exp(log(1/3*log(x^4))/x)^2+(-4*x^2*log(2)-50*x^2)*log(x^4)*exp(l 
og(1/3*log(x^4))/x)+(4*x^2*log(2)^2+100*x^2*log(2)+625*x^2)*log(x^4)),x, a 
lgorithm="fricas")
 

Output:

12/((1/3*log(x^4))^(1/x) - 2*log(2) - 25)
 

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (-48+12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{\left (625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)\right ) \log \left (x^4\right )+3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{1+\frac {1}{x}}\left (x^4\right )+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}\left (x^4\right )} \, dx=\frac {12}{e^{\frac {\log {\left (\frac {\log {\left (x^{4} \right )}}{3} \right )}}{x}} - 25 - 2 \log {\left (2 \right )}} \] Input:

integrate((12*ln(x**4)*ln(1/3*ln(x**4))-48)*exp(ln(1/3*ln(x**4))/x)/(x**2* 
ln(x**4)*exp(ln(1/3*ln(x**4))/x)**2+(-4*x**2*ln(2)-50*x**2)*ln(x**4)*exp(l 
n(1/3*ln(x**4))/x)+(4*x**2*ln(2)**2+100*x**2*ln(2)+625*x**2)*ln(x**4)),x)
 

Output:

12/(exp(log(log(x**4)/3)/x) - 25 - 2*log(2))
 

Maxima [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (-48+12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{\left (625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)\right ) \log \left (x^4\right )+3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{1+\frac {1}{x}}\left (x^4\right )+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}\left (x^4\right )} \, dx=-\frac {12 \cdot 3^{\left (\frac {1}{x}\right )}}{3^{\left (\frac {1}{x}\right )} {\left (2 \, \log \left (2\right ) + 25\right )} - e^{\left (\frac {2 \, \log \left (2\right )}{x} + \frac {\log \left (\log \left (x\right )\right )}{x}\right )}} \] Input:

integrate((12*log(x^4)*log(1/3*log(x^4))-48)*exp(log(1/3*log(x^4))/x)/(x^2 
*log(x^4)*exp(log(1/3*log(x^4))/x)^2+(-4*x^2*log(2)-50*x^2)*log(x^4)*exp(l 
og(1/3*log(x^4))/x)+(4*x^2*log(2)^2+100*x^2*log(2)+625*x^2)*log(x^4)),x, a 
lgorithm="maxima")
 

Output:

-12*3^(1/x)/(3^(1/x)*(2*log(2) + 25) - e^(2*log(2)/x + log(log(x))/x))
 

Giac [F]

\[ \int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (-48+12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{\left (625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)\right ) \log \left (x^4\right )+3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{1+\frac {1}{x}}\left (x^4\right )+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}\left (x^4\right )} \, dx=\int { \frac {12 \, {\left (\log \left (x^{4}\right ) \log \left (\frac {1}{3} \, \log \left (x^{4}\right )\right ) - 4\right )} \left (\frac {1}{3} \, \log \left (x^{4}\right )\right )^{\left (\frac {1}{x}\right )}}{x^{2} \left (\frac {1}{3} \, \log \left (x^{4}\right )\right )^{\frac {2}{x}} \log \left (x^{4}\right ) - 2 \, {\left (2 \, x^{2} \log \left (2\right ) + 25 \, x^{2}\right )} \left (\frac {1}{3} \, \log \left (x^{4}\right )\right )^{\left (\frac {1}{x}\right )} \log \left (x^{4}\right ) + {\left (4 \, x^{2} \log \left (2\right )^{2} + 100 \, x^{2} \log \left (2\right ) + 625 \, x^{2}\right )} \log \left (x^{4}\right )} \,d x } \] Input:

integrate((12*log(x^4)*log(1/3*log(x^4))-48)*exp(log(1/3*log(x^4))/x)/(x^2 
*log(x^4)*exp(log(1/3*log(x^4))/x)^2+(-4*x^2*log(2)-50*x^2)*log(x^4)*exp(l 
og(1/3*log(x^4))/x)+(4*x^2*log(2)^2+100*x^2*log(2)+625*x^2)*log(x^4)),x, a 
lgorithm="giac")
 

Output:

integrate(12*(log(x^4)*log(1/3*log(x^4)) - 4)*(1/3*log(x^4))^(1/x)/(x^2*(1 
/3*log(x^4))^(2/x)*log(x^4) - 2*(2*x^2*log(2) + 25*x^2)*(1/3*log(x^4))^(1/ 
x)*log(x^4) + (4*x^2*log(2)^2 + 100*x^2*log(2) + 625*x^2)*log(x^4)), x)
 

Mupad [B] (verification not implemented)

Time = 3.76 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42 \[ \int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (-48+12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{\left (625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)\right ) \log \left (x^4\right )+3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{1+\frac {1}{x}}\left (x^4\right )+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}\left (x^4\right )} \, dx=-\frac {24\,\mathrm {atanh}\left (\frac {\ln \left (16\right )-\frac {2\,{\ln \left (x^4\right )}^{1/x}}{3^{1/x}}+50}{\sqrt {\ln \left (16\right )-2\,\ln \left (4\right )}\,\sqrt {2\,\ln \left (4\right )+\ln \left (16\right )+100}}\right )}{\sqrt {\ln \left (16\right )-2\,\ln \left (4\right )}\,\sqrt {2\,\ln \left (4\right )+\ln \left (16\right )+100}} \] Input:

int((exp(log(log(x^4)/3)/x)*(12*log(log(x^4)/3)*log(x^4) - 48))/(log(x^4)* 
(4*x^2*log(2)^2 + 100*x^2*log(2) + 625*x^2) + x^2*log(x^4)*exp((2*log(log( 
x^4)/3))/x) - log(x^4)*exp(log(log(x^4)/3)/x)*(4*x^2*log(2) + 50*x^2)),x)
 

Output:

-(24*atanh((log(16) - (2*log(x^4)^(1/x))/3^(1/x) + 50)/((log(16) - 2*log(4 
))^(1/2)*(2*log(4) + log(16) + 100)^(1/2))))/((log(16) - 2*log(4))^(1/2)*( 
2*log(4) + log(16) + 100)^(1/2))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.35 \[ \int \frac {3^{-1/x} \sqrt [x]{\log \left (x^4\right )} \left (-48+12 \log \left (x^4\right ) \log \left (\frac {\log \left (x^4\right )}{3}\right )\right )}{\left (625 x^2+50 x^2 \log (4)+x^2 \log ^2(4)\right ) \log \left (x^4\right )+3^{-1/x} \left (-50 x^2-2 x^2 \log (4)\right ) \log ^{1+\frac {1}{x}}\left (x^4\right )+3^{-2/x} x^2 \log ^{1+\frac {2}{x}}\left (x^4\right )} \, dx=\frac {12 e^{\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (x^{4}\right )}{3}\right )}{x}}}{2 e^{\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (x^{4}\right )}{3}\right )}{x}} \mathrm {log}\left (2\right )+25 e^{\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (x^{4}\right )}{3}\right )}{x}}-4 \mathrm {log}\left (2\right )^{2}-100 \,\mathrm {log}\left (2\right )-625} \] Input:

int((12*log(x^4)*log(1/3*log(x^4))-48)*exp(log(1/3*log(x^4))/x)/(x^2*log(x 
^4)*exp(log(1/3*log(x^4))/x)^2+(-4*x^2*log(2)-50*x^2)*log(x^4)*exp(log(1/3 
*log(x^4))/x)+(4*x^2*log(2)^2+100*x^2*log(2)+625*x^2)*log(x^4)),x)
 

Output:

(12*e**(log(log(x**4)/3)/x))/(2*e**(log(log(x**4)/3)/x)*log(2) + 25*e**(lo 
g(log(x**4)/3)/x) - 4*log(2)**2 - 100*log(2) - 625)