\(\int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+(-2 x^2+e^{\frac {-3+x^2}{x}} (3+x^2)) \log (\frac {x^4}{16}) \log (\log (\frac {x^4}{16}))}{x^2 \log (\frac {x^4}{16})} \, dx\) [171]

Optimal result

Integrand size = 78, antiderivative size = 24 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=\left (6+e^{-\frac {3}{x}+x}-2 x\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right ) \] Output:

ln(ln(1/16*x^4))*(6-2*x+exp(x-3/x))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=\left (6+e^{-\frac {3}{x}+x}-2 x\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right ) \] Input:

Integrate[(24*x + 4*E^((-3 + x^2)/x)*x - 8*x^2 + (-2*x^2 + E^((-3 + x^2)/x 
)*(3 + x^2))*Log[x^4/16]*Log[Log[x^4/16]])/(x^2*Log[x^4/16]),x]
 

Output:

(6 + E^(-3/x + x) - 2*x)*Log[Log[x^4/16]]
 

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(97\) vs. \(2(24)=48\).

Time = 1.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 4.04, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {7293, 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.

\(\displaystyle \int \frac {-8 x^2+4 e^{\frac {x^2-3}{x}} x+\left (e^{\frac {x^2-3}{x}} \left (x^2+3\right )-2 x^2\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+24 x}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^{x-\frac {3}{x}} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+4 x\right )}{x^2 \log \left (\frac {x^4}{16}\right )}-\frac {2 \left (x \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+4 x-12\right )}{x \log \left (\frac {x^4}{16}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 x \log \left (\log \left (\frac {x^4}{16}\right )\right )+6 \log \left (\log \left (\frac {x^4}{16}\right )\right )+\frac {e^{x-\frac {3}{x}} \left (3 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )+x^2 \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )\right )}{\left (\frac {3}{x^2}+1\right ) x^2 \log \left (\frac {x^4}{16}\right )}\)

Input:

Int[(24*x + 4*E^((-3 + x^2)/x)*x - 8*x^2 + (-2*x^2 + E^((-3 + x^2)/x)*(3 + 
 x^2))*Log[x^4/16]*Log[Log[x^4/16]])/(x^2*Log[x^4/16]),x]
 

Output:

6*Log[Log[x^4/16]] - 2*x*Log[Log[x^4/16]] + (E^(-3/x + x)*(3*Log[x^4/16]*L 
og[Log[x^4/16]] + x^2*Log[x^4/16]*Log[Log[x^4/16]]))/((1 + 3/x^2)*x^2*Log[ 
x^4/16])
 

Defintions of rubi rules used

rule 2009

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

rule 7293

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

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62

method	result	size
parallelrisch	\(-2 x \ln \left (\ln \left (\frac {x^{4}}{16}\right )\right )+{\mathrm e}^{\frac {x^{2}-3}{x}} \ln \left (\ln \left (\frac {x^{4}}{16}\right )\right )+6 \ln \left (\ln \left (\frac {x^{4}}{16}\right )\right )\)	\(39\)
risch	\(\left (-2 x +{\mathrm e}^{\frac {x^{2}-3}{x}}\right ) \ln \left (-4 \ln \left (2\right )+4 \ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) {\left (-\operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x \right )\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{3}\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x^{2}\right )\right ) \left (-\operatorname {csgn}\left (i x^{3}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{4}\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x^{3}\right )\right ) \left (-\operatorname {csgn}\left (i x^{4}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}\right )+6 \ln \left (\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )}{4}+\frac {i \pi \operatorname {csgn}\left (i x^{4}\right )^{2} \operatorname {csgn}\left (i x \right )}{8}-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{8}-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}}{8}+\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x^{3}\right )^{2}}{8}-\frac {i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )}{8}+\frac {i \pi \operatorname {csgn}\left (i x^{3}\right )^{2} \operatorname {csgn}\left (i x \right )}{8}-\frac {i \pi \operatorname {csgn}\left (i x^{4}\right )^{3}}{8}-\frac {i \pi \operatorname {csgn}\left (i x^{3}\right )^{3}}{8}-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{3}\right )}{8}+\frac {i \pi \,\operatorname {csgn}\left (i x^{3}\right ) \operatorname {csgn}\left (i x^{4}\right )^{2}}{8}-\ln \left (2\right )+\ln \left (x \right )\right )\)	\(348\)

Input:

int((((x^2+3)*exp((x^2-3)/x)-2*x^2)*ln(1/16*x^4)*ln(ln(1/16*x^4))+4*x*exp( 
(x^2-3)/x)-8*x^2+24*x)/x^2/ln(1/16*x^4),x,method=_RETURNVERBOSE)
 

Output:

-2*x*ln(ln(1/16*x^4))+exp((x^2-3)/x)*ln(ln(1/16*x^4))+6*ln(ln(1/16*x^4))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=-{\left (2 \, x - e^{\left (\frac {x^{2} - 3}{x}\right )} - 6\right )} \log \left (\log \left (\frac {1}{16} \, x^{4}\right )\right ) \] Input:

integrate((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4)) 
+4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/log(1/16*x^4),x, algorithm="fricas")
 

Output:

-(2*x - e^((x^2 - 3)/x) - 6)*log(log(1/16*x^4))
 

Sympy [A] (verification not implemented)

Time = 4.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=- 2 x \log {\left (\log {\left (\frac {x^{4}}{16} \right )} \right )} + e^{\frac {x^{2} - 3}{x}} \log {\left (\log {\left (\frac {x^{4}}{16} \right )} \right )} + 6 \log {\left (\log {\left (\frac {x^{4}}{16} \right )} \right )} \] Input:

integrate((((x**2+3)*exp((x**2-3)/x)-2*x**2)*ln(1/16*x**4)*ln(ln(1/16*x**4 
))+4*x*exp((x**2-3)/x)-8*x**2+24*x)/x**2/ln(1/16*x**4),x)
 

Output:

-2*x*log(log(x**4/16)) + exp((x**2 - 3)/x)*log(log(x**4/16)) + 6*log(log(x 
**4/16))
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=-4 \, x \log \left (2\right ) + 2 \, e^{\left (x - \frac {3}{x}\right )} \log \left (2\right ) - {\left (2 \, x - e^{\left (x - \frac {3}{x}\right )} - 6\right )} \log \left (-\log \left (2\right ) + \log \left (x\right )\right ) \] Input:

integrate((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4)) 
+4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/log(1/16*x^4),x, algorithm="maxima")
 

Output:

-4*x*log(2) + 2*e^(x - 3/x)*log(2) - (2*x - e^(x - 3/x) - 6)*log(-log(2) + 
 log(x))
 

Giac [F]

\[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=\int { -\frac {{\left (2 \, x^{2} - {\left (x^{2} + 3\right )} e^{\left (\frac {x^{2} - 3}{x}\right )}\right )} \log \left (\frac {1}{16} \, x^{4}\right ) \log \left (\log \left (\frac {1}{16} \, x^{4}\right )\right ) + 8 \, x^{2} - 4 \, x e^{\left (\frac {x^{2} - 3}{x}\right )} - 24 \, x}{x^{2} \log \left (\frac {1}{16} \, x^{4}\right )} \,d x } \] Input:

integrate((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4)) 
+4*x*exp((x^2-3)/x)-8*x^2+24*x)/x^2/log(1/16*x^4),x, algorithm="giac")
 

Output:

integrate(-((2*x^2 - (x^2 + 3)*e^((x^2 - 3)/x))*log(1/16*x^4)*log(log(1/16 
*x^4)) + 8*x^2 - 4*x*e^((x^2 - 3)/x) - 24*x)/(x^2*log(1/16*x^4)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=\int \frac {24\,x+4\,x\,{\mathrm {e}}^{\frac {x^2-3}{x}}-8\,x^2+\ln \left (\ln \left (\frac {x^4}{16}\right )\right )\,\ln \left (\frac {x^4}{16}\right )\,\left ({\mathrm {e}}^{\frac {x^2-3}{x}}\,\left (x^2+3\right )-2\,x^2\right )}{x^2\,\ln \left (\frac {x^4}{16}\right )} \,d x \] Input:

int((24*x + 4*x*exp((x^2 - 3)/x) - 8*x^2 + log(log(x^4/16))*log(x^4/16)*(e 
xp((x^2 - 3)/x)*(x^2 + 3) - 2*x^2))/(x^2*log(x^4/16)),x)
 

Output:

int((24*x + 4*x*exp((x^2 - 3)/x) - 8*x^2 + log(log(x^4/16))*log(x^4/16)*(e 
xp((x^2 - 3)/x)*(x^2 + 3) - 2*x^2))/(x^2*log(x^4/16)), x)
 

Reduce [B] (verification not implemented)

Time = 3.40 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.38 \[ \int \frac {24 x+4 e^{\frac {-3+x^2}{x}} x-8 x^2+\left (-2 x^2+e^{\frac {-3+x^2}{x}} \left (3+x^2\right )\right ) \log \left (\frac {x^4}{16}\right ) \log \left (\log \left (\frac {x^4}{16}\right )\right )}{x^2 \log \left (\frac {x^4}{16}\right )} \, dx=\frac {-2 e^{\frac {3}{x}} \mathrm {log}\left (\mathrm {log}\left (\frac {x^{4}}{16}\right )\right ) x +6 e^{\frac {3}{x}} \mathrm {log}\left (\mathrm {log}\left (\frac {x^{4}}{16}\right )\right )+e^{x} \mathrm {log}\left (-\mathrm {log}\left (\frac {16}{x^{4}}\right )\right )}{e^{\frac {3}{x}}} \] Input:

int((((x^2+3)*exp((x^2-3)/x)-2*x^2)*log(1/16*x^4)*log(log(1/16*x^4))+4*x*e 
xp((x^2-3)/x)-8*x^2+24*x)/x^2/log(1/16*x^4),x)
 

Output:

( - 2*e**(3/x)*log(log(x**4/16))*x + 6*e**(3/x)*log(log(x**4/16)) + e**x*l 
og( - log(16/x**4)))/e**(3/x)