Integrand size = 94, antiderivative size = 22 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (\frac {-2 x-\log \left (1+e^{4 x}\right )}{\log \left (x^2\right )}\right ) \] Output:
ln(1/ln(x^2)*(-2*x-ln(exp(4*x)+1)))
Time = 0.45 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (2 x+\log \left (1+e^{4 x}\right )\right )-\log \left (\log \left (x^2\right )\right ) \] Input:
Integrate[(-4*x - 4*E^(4*x)*x + (-2 - 2*E^(4*x))*Log[1 + E^(4*x)] + (2*x + 6*E^(4*x)*x)*Log[x^2])/((2*x^2 + 2*E^(4*x)*x^2)*Log[x^2] + (x + E^(4*x)*x )*Log[1 + E^(4*x)]*Log[x^2]),x]
Output:
Log[2*x + Log[1 + E^(4*x)]] - Log[Log[x^2]]
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.
\(\displaystyle \int \frac {\left (6 e^{4 x} x+2 x\right ) \log \left (x^2\right )-4 e^{4 x} x-4 x+\left (-2 e^{4 x}-2\right ) \log \left (e^{4 x}+1\right )}{\left (2 e^{4 x} x^2+2 x^2\right ) \log \left (x^2\right )+\left (e^{4 x} x+x\right ) \log \left (e^{4 x}+1\right ) \log \left (x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (6 e^{4 x} x+2 x\right ) \log \left (x^2\right )-4 e^{4 x} x-4 x+\left (-2 e^{4 x}-2\right ) \log \left (e^{4 x}+1\right )}{\left (e^{4 x}+1\right ) x \left (2 x+\log \left (e^{4 x}+1\right )\right ) \log \left (x^2\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (3 x \log \left (x^2\right )-2 x-\log \left (e^{4 x}+1\right )\right )}{x \left (2 x+\log \left (e^{4 x}+1\right )\right ) \log \left (x^2\right )}-\frac {4}{\left (e^{4 x}+1\right ) \left (2 x+\log \left (e^{4 x}+1\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \text {Subst}\left (\int \frac {1}{x+\log \left (1+e^{2 x}\right )}dx,x,2 x\right )-2 \text {Subst}\left (\int \frac {1}{\left (1+e^{2 x}\right ) \left (x+\log \left (1+e^{2 x}\right )\right )}dx,x,2 x\right )-\log \left (\log \left (x^2\right )\right )\) |
Input:
Int[(-4*x - 4*E^(4*x)*x + (-2 - 2*E^(4*x))*Log[1 + E^(4*x)] + (2*x + 6*E^( 4*x)*x)*Log[x^2])/((2*x^2 + 2*E^(4*x)*x^2)*Log[x^2] + (x + E^(4*x)*x)*Log[ 1 + E^(4*x)]*Log[x^2]),x]
Output:
$Aborted
Time = 0.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(-\ln \left (\ln \left (x^{2}\right )\right )+\ln \left (x +\frac {\ln \left ({\mathrm e}^{4 x}+1\right )}{2}\right )\) | \(21\) |
risch | \(-\ln \left (\ln \left (x \right )-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \left (\operatorname {csgn}\left (i x \right )^{2}-2 \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )+\operatorname {csgn}\left (i x^{2}\right )^{2}\right )}{4}\right )+\ln \left (\ln \left ({\mathrm e}^{4 x}+1\right )+2 x \right )\) | \(62\) |
Input:
int(((-2*exp(4*x)-2)*ln(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*ln(x^2)-4*x*exp(4*x )-4*x)/((x*exp(4*x)+x)*ln(x^2)*ln(exp(4*x)+1)+(2*x^2*exp(4*x)+2*x^2)*ln(x^ 2)),x,method=_RETURNVERBOSE)
Output:
-ln(ln(x^2))+ln(x+1/2*ln(exp(4*x)+1))
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (2 \, x + \log \left (e^{\left (4 \, x\right )} + 1\right )\right ) - \log \left (\log \left (x^{2}\right )\right ) \] Input:
integrate(((-2*exp(4*x)-2)*log(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*log(x^2)-4*x *exp(4*x)-4*x)/((x*exp(4*x)+x)*log(x^2)*log(exp(4*x)+1)+(2*x^2*exp(4*x)+2* x^2)*log(x^2)),x, algorithm="fricas")
Output:
log(2*x + log(e^(4*x) + 1)) - log(log(x^2))
Time = 0.16 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log {\left (2 x + \log {\left (e^{4 x} + 1 \right )} \right )} - \log {\left (\log {\left (x^{2} \right )} \right )} \] Input:
integrate(((-2*exp(4*x)-2)*ln(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*ln(x**2)-4*x* exp(4*x)-4*x)/((x*exp(4*x)+x)*ln(x**2)*ln(exp(4*x)+1)+(2*x**2*exp(4*x)+2*x **2)*ln(x**2)),x)
Output:
log(2*x + log(exp(4*x) + 1)) - log(log(x**2))
Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (2 \, x + \log \left (e^{\left (4 \, x\right )} + 1\right )\right ) - \log \left (\log \left (x\right )\right ) \] Input:
integrate(((-2*exp(4*x)-2)*log(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*log(x^2)-4*x *exp(4*x)-4*x)/((x*exp(4*x)+x)*log(x^2)*log(exp(4*x)+1)+(2*x^2*exp(4*x)+2* x^2)*log(x^2)),x, algorithm="maxima")
Output:
log(2*x + log(e^(4*x) + 1)) - log(log(x))
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\log \left (2 \, x + \log \left (e^{\left (4 \, x\right )} + 1\right )\right ) - \log \left (\log \left (x^{2}\right )\right ) \] Input:
integrate(((-2*exp(4*x)-2)*log(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*log(x^2)-4*x *exp(4*x)-4*x)/((x*exp(4*x)+x)*log(x^2)*log(exp(4*x)+1)+(2*x^2*exp(4*x)+2* x^2)*log(x^2)),x, algorithm="giac")
Output:
log(2*x + log(e^(4*x) + 1)) - log(log(x^2))
Time = 3.98 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=\ln \left (2\,x+\ln \left ({\mathrm {e}}^{4\,x}+1\right )\right )-\ln \left (\ln \left (x^2\right )\right ) \] Input:
int(-(4*x + 4*x*exp(4*x) + log(exp(4*x) + 1)*(2*exp(4*x) + 2) - log(x^2)*( 2*x + 6*x*exp(4*x)))/(log(x^2)*(2*x^2*exp(4*x) + 2*x^2) + log(exp(4*x) + 1 )*log(x^2)*(x + x*exp(4*x))),x)
Output:
log(2*x + log(exp(4*x) + 1)) - log(log(x^2))
Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-4 x-4 e^{4 x} x+\left (-2-2 e^{4 x}\right ) \log \left (1+e^{4 x}\right )+\left (2 x+6 e^{4 x} x\right ) \log \left (x^2\right )}{\left (2 x^2+2 e^{4 x} x^2\right ) \log \left (x^2\right )+\left (x+e^{4 x} x\right ) \log \left (1+e^{4 x}\right ) \log \left (x^2\right )} \, dx=-\mathrm {log}\left (\mathrm {log}\left (x^{2}\right )\right )+\mathrm {log}\left (\mathrm {log}\left (e^{4 x}+1\right )+2 x \right ) \] Input:
int(((-2*exp(4*x)-2)*log(exp(4*x)+1)+(6*x*exp(4*x)+2*x)*log(x^2)-4*x*exp(4 *x)-4*x)/((x*exp(4*x)+x)*log(x^2)*log(exp(4*x)+1)+(2*x^2*exp(4*x)+2*x^2)*l og(x^2)),x)
Output:
- log(log(x**2)) + log(log(e**(4*x) + 1) + 2*x)