\(\int \frac {3-x+e^x (-3-5 x-x^2+x^3) \log (16)+(-x+e^x (1+x^2) \log (16)) \log (x+e^x (-1-x^2) \log (16)) \log (\log (x+e^x (-1-x^2) \log (16)))}{(-x+e^x (1+x^2) \log (16)) \log (x+e^x (-1-x^2) \log (16))} \, dx\) [117]

Optimal result

Integrand size = 109, antiderivative size = 23 \[ \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=(-3+x) \log \left (\log \left (x-\left (e^x+e^x x^2\right ) \log (16)\right )\right ) \] Output:

(-3+x)*ln(ln(x-4*ln(2)*(exp(x)*x^2+exp(x))))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=-3 \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right )+x \log \left (\log \left (x-e^x \left (1+x^2\right ) \log (16)\right )\right ) \] Input:

Integrate[(3 - x + E^x*(-3 - 5*x - x^2 + x^3)*Log[16] + (-x + E^x*(1 + x^2 
)*Log[16])*Log[x + E^x*(-1 - x^2)*Log[16]]*Log[Log[x + E^x*(-1 - x^2)*Log[ 
16]]])/((-x + E^x*(1 + x^2)*Log[16])*Log[x + E^x*(-1 - x^2)*Log[16]]),x]
 

Output:

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

Rubi [F]

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.

\(\displaystyle \int \frac {\left (e^x \left (x^2+1\right ) \log (16)-x\right ) \log \left (e^x \left (-x^2-1\right ) \log (16)+x\right ) \log \left (\log \left (e^x \left (-x^2-1\right ) \log (16)+x\right )\right )+e^x \left (x^3-x^2-5 x-3\right ) \log (16)-x+3}{\left (e^x \left (x^2+1\right ) \log (16)-x\right ) \log \left (e^x \left (-x^2-1\right ) \log (16)+x\right )} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {x^3-x^2+x^2 \log \left (x-e^x \left (x^2+1\right ) \log (16)\right ) \log \left (\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )\right )+\log \left (x-e^x \left (x^2+1\right ) \log (16)\right ) \log \left (\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )\right )-5 x-3}{\left (x^2+1\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}+\frac {x^4-2 x^3-2 x^2-4 x+3}{\left (x^2+1\right ) \left (e^x x^2 \log (16)-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\int \frac {1}{\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+(3-i) \int \frac {1}{(i-x) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+\int \frac {x}{\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx-(3+i) \int \frac {1}{(x+i) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx-3 \int \frac {1}{\left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+(1+3 i) \int \frac {1}{(i-x) \left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx-2 \int \frac {x}{\left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+\int \frac {x^2}{\left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx-(1-3 i) \int \frac {1}{(x+i) \left (e^x \log (16) x^2-x+e^x \log (16)\right ) \log \left (x-e^x \left (x^2+1\right ) \log (16)\right )}dx+\int \log \left (\log \left (x-e^x \left (x^2+1\right ) \log (16)\right )\right )dx\)

Input:

Int[(3 - x + E^x*(-3 - 5*x - x^2 + x^3)*Log[16] + (-x + E^x*(1 + x^2)*Log[ 
16])*Log[x + E^x*(-1 - x^2)*Log[16]]*Log[Log[x + E^x*(-1 - x^2)*Log[16]]]) 
/((-x + E^x*(1 + x^2)*Log[16])*Log[x + E^x*(-1 - x^2)*Log[16]]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 21.87 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74

method	result	size
risch	\(\ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \left (2\right ) {\mathrm e}^{x}+x \right )\right ) x -3 \ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \left (2\right ) {\mathrm e}^{x}+x \right )\right )\)	\(40\)
parallelrisch	\(\ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \left (2\right ) {\mathrm e}^{x}+x \right )\right ) x -3 \ln \left (\ln \left (4 \left (-x^{2}-1\right ) \ln \left (2\right ) {\mathrm e}^{x}+x \right )\right )\)	\(40\)

Input:

int(((4*(x^2+1)*ln(2)*exp(x)-x)*ln(4*(-x^2-1)*ln(2)*exp(x)+x)*ln(ln(4*(-x^ 
2-1)*ln(2)*exp(x)+x))+4*(x^3-x^2-5*x-3)*ln(2)*exp(x)+3-x)/(4*(x^2+1)*ln(2) 
*exp(x)-x)/ln(4*(-x^2-1)*ln(2)*exp(x)+x),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx={\left (x - 3\right )} \log \left (\log \left (-4 \, {\left (x^{2} + 1\right )} e^{x} \log \left (2\right ) + x\right )\right ) \] Input:

integrate(((4*(x^2+1)*log(2)*exp(x)-x)*log(4*(-x^2-1)*log(2)*exp(x)+x)*log 
(log(4*(-x^2-1)*log(2)*exp(x)+x))+4*(x^3-x^2-5*x-3)*log(2)*exp(x)+3-x)/(4* 
(x^2+1)*log(2)*exp(x)-x)/log(4*(-x^2-1)*log(2)*exp(x)+x),x, algorithm="fri 
cas")
 

Output:

(x - 3)*log(log(-4*(x^2 + 1)*e^x*log(2) + x))
 

Sympy [A] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=\left (x - 1\right ) \log {\left (\log {\left (x + \left (- 4 x^{2} - 4\right ) e^{x} \log {\left (2 \right )} \right )} \right )} - 2 \log {\left (\log {\left (x + \left (- 4 x^{2} - 4\right ) e^{x} \log {\left (2 \right )} \right )} \right )} \] Input:

integrate(((4*(x**2+1)*ln(2)*exp(x)-x)*ln(4*(-x**2-1)*ln(2)*exp(x)+x)*ln(l 
n(4*(-x**2-1)*ln(2)*exp(x)+x))+4*(x**3-x**2-5*x-3)*ln(2)*exp(x)+3-x)/(4*(x 
**2+1)*ln(2)*exp(x)-x)/ln(4*(-x**2-1)*ln(2)*exp(x)+x),x)
 

Output:

(x - 1)*log(log(x + (-4*x**2 - 4)*exp(x)*log(2))) - 2*log(log(x + (-4*x**2 
 - 4)*exp(x)*log(2)))
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx={\left (x - 3\right )} \log \left (\log \left (-4 \, {\left (x^{2} \log \left (2\right ) + \log \left (2\right )\right )} e^{x} + x\right )\right ) \] Input:

integrate(((4*(x^2+1)*log(2)*exp(x)-x)*log(4*(-x^2-1)*log(2)*exp(x)+x)*log 
(log(4*(-x^2-1)*log(2)*exp(x)+x))+4*(x^3-x^2-5*x-3)*log(2)*exp(x)+3-x)/(4* 
(x^2+1)*log(2)*exp(x)-x)/log(4*(-x^2-1)*log(2)*exp(x)+x),x, algorithm="max 
ima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (21) = 42\).

Time = 0.35 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.87 \[ \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=x \log \left (\log \left (-4 \, x^{2} e^{x} \log \left (2\right ) - 4 \, e^{x} \log \left (2\right ) + x\right )\right ) - 3 \, \log \left (\log \left (-4 \, x^{2} e^{x} \log \left (2\right ) - 4 \, e^{x} \log \left (2\right ) + x\right )\right ) \] Input:

integrate(((4*(x^2+1)*log(2)*exp(x)-x)*log(4*(-x^2-1)*log(2)*exp(x)+x)*log 
(log(4*(-x^2-1)*log(2)*exp(x)+x))+4*(x^3-x^2-5*x-3)*log(2)*exp(x)+3-x)/(4* 
(x^2+1)*log(2)*exp(x)-x)/log(4*(-x^2-1)*log(2)*exp(x)+x),x, algorithm="gia 
c")
 

Output:

x*log(log(-4*x^2*e^x*log(2) - 4*e^x*log(2) + x)) - 3*log(log(-4*x^2*e^x*lo 
g(2) - 4*e^x*log(2) + x))
 

Mupad [B] (verification not implemented)

Time = 4.61 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx=\ln \left (\ln \left (x-4\,{\mathrm {e}}^x\,\ln \left (2\right )\,\left (x^2+1\right )\right )\right )\,\left (x-3\right ) \] Input:

int((x + log(x - 4*exp(x)*log(2)*(x^2 + 1))*log(log(x - 4*exp(x)*log(2)*(x 
^2 + 1)))*(x - 4*exp(x)*log(2)*(x^2 + 1)) + 4*exp(x)*log(2)*(5*x + x^2 - x 
^3 + 3) - 3)/(log(x - 4*exp(x)*log(2)*(x^2 + 1))*(x - 4*exp(x)*log(2)*(x^2 
 + 1))),x)
 

Output:

log(log(x - 4*exp(x)*log(2)*(x^2 + 1)))*(x - 3)
 

Reduce [F]

\[ \int \frac {3-x+e^x \left (-3-5 x-x^2+x^3\right ) \log (16)+\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right ) \log \left (\log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )\right )}{\left (-x+e^x \left (1+x^2\right ) \log (16)\right ) \log \left (x+e^x \left (-1-x^2\right ) \log (16)\right )} \, dx =\text {Too large to display} \] Input:

int(((4*(x^2+1)*log(2)*exp(x)-x)*log(4*(-x^2-1)*log(2)*exp(x)+x)*log(log(4 
*(-x^2-1)*log(2)*exp(x)+x))+4*(x^3-x^2-5*x-3)*log(2)*exp(x)+3-x)/(4*(x^2+1 
)*log(2)*exp(x)-x)/log(4*(-x^2-1)*log(2)*exp(x)+x),x)
 

Output:

 - 8*int(e**x/(4*e**x*log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*log(2 
)*x**2 + 4*e**x*log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*log(2) - lo 
g( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*x),x)*log(2) + 4*int((e**x*x* 
*3)/(4*e**x*log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*log(2)*x**2 + 4 
*e**x*log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*log(2) - log( - 4*e** 
x*log(2)*x**2 - 4*e**x*log(2) + x)*x),x)*log(2) + 4*int((e**x*log(log( - 4 
*e**x*log(2)*x**2 - 4*e**x*log(2) + x))*x**2)/(4*e**x*log(2)*x**2 + 4*e**x 
*log(2) - x),x)*log(2) + 4*int((e**x*log(log( - 4*e**x*log(2)*x**2 - 4*e** 
x*log(2) + x)))/(4*e**x*log(2)*x**2 + 4*e**x*log(2) - x),x)*log(2) - 12*in 
t((e**x*x)/(4*e**x*log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*log(2)*x 
**2 + 4*e**x*log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*log(2) - log( 
- 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*x),x)*log(2) - int((log(log( - 4 
*e**x*log(2)*x**2 - 4*e**x*log(2) + x))*x)/(4*e**x*log(2)*x**2 + 4*e**x*lo 
g(2) - x),x) - int(x/(4*e**x*log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x 
)*log(2)*x**2 + 4*e**x*log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*log( 
2) - log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*x),x) + 2*int(1/(4*e** 
x*log( - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*log(2)*x**2 + 4*e**x*log( 
 - 4*e**x*log(2)*x**2 - 4*e**x*log(2) + x)*log(2) - log( - 4*e**x*log(2)*x 
**2 - 4*e**x*log(2) + x)*x),x) - log(log( - 4*e**x*log(2)*x**2 - 4*e**x*lo 
g(2) + x))