Integrand size = 104, antiderivative size = 30 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=e^3-\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (4 (4+x))}{x} \] Output:
exp(3)-ln(4*x+16)/x*ln(5-5/2/exp(x)/x)
Time = 0.51 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=\log (4+x)-\frac {\left (x+\log \left (5-\frac {5 e^{-x}}{2 x}\right )\right ) \log (4 (4+x))}{x} \] Input:
Integrate[((-4 - 5*x - x^2)*Log[16 + 4*x] + (x - 2*E^x*x^2 + (-4 - x + E^x *(8*x + 2*x^2))*Log[16 + 4*x])*Log[(-5 + 10*E^x*x)/(2*E^x*x)])/(-4*x^2 - x ^3 + E^x*(8*x^3 + 2*x^4)),x]
Output:
Log[4 + x] - ((x + Log[5 - 5/(2*E^x*x)])*Log[4*(4 + x)])/x
Time = 3.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {7292, 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.
\(\displaystyle \int \frac {\left (-x^2-5 x-4\right ) \log (4 x+16)+\left (-2 e^x x^2+\left (e^x \left (2 x^2+8 x\right )-x-4\right ) \log (4 x+16)+x\right ) \log \left (\frac {e^{-x} \left (10 e^x x-5\right )}{2 x}\right )}{-x^3-4 x^2+e^x \left (2 x^4+8 x^3\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (\left (-x^2-5 x-4\right ) \log (4 x+16)\right )-\left (-2 e^x x^2+\left (e^x \left (2 x^2+8 x\right )-x-4\right ) \log (4 x+16)+x\right ) \log \left (\frac {e^{-x} \left (10 e^x x-5\right )}{2 x}\right )}{x^2 (x+4) \left (1-2 e^x x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) (-x+x \log (4 (x+4))+4 \log (4 (x+4)))}{x^2 (x+4)}+\frac {(x+1) \log (4 x+16)}{x^2 \left (1-2 e^x x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log \left (5-\frac {5 e^{-x}}{2 x}\right ) \log (4 x+16)}{x}\) |
Input:
Int[((-4 - 5*x - x^2)*Log[16 + 4*x] + (x - 2*E^x*x^2 + (-4 - x + E^x*(8*x + 2*x^2))*Log[16 + 4*x])*Log[(-5 + 10*E^x*x)/(2*E^x*x)])/(-4*x^2 - x^3 + E ^x*(8*x^3 + 2*x^4)),x]
Output:
-((Log[5 - 5/(2*E^x*x)]*Log[16 + 4*x])/x)
Time = 7.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(-\frac {\ln \left (4 x +16\right ) \ln \left (\frac {5 \left (2 \,{\mathrm e}^{x} x -1\right ) {\mathrm e}^{-x}}{2 x}\right )}{x}\) | \(29\) |
risch | \(\frac {\ln \left (4 x +16\right ) \ln \left ({\mathrm e}^{x}\right )}{x}+\frac {\ln \left (4 x +16\right ) \left (i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )}^{2}+i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )^{2}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )^{2}+i \pi \operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )^{3}-i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )}^{2}+i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x -\frac {1}{2}\right ) {\mathrm e}^{-x}}{x}\right )}^{3}-2 \ln \left (5\right )+2 \ln \left (x \right )-2 \ln \left ({\mathrm e}^{x} x -\frac {1}{2}\right )\right )}{2 x}\) | \(290\) |
Input:
int(((((2*x^2+8*x)*exp(x)-x-4)*ln(4*x+16)-2*exp(x)*x^2+x)*ln(1/2*(10*exp(x )*x-5)/exp(x)/x)+(-x^2-5*x-4)*ln(4*x+16))/((2*x^4+8*x^3)*exp(x)-x^3-4*x^2) ,x,method=_RETURNVERBOSE)
Output:
-1/x*ln(4*x+16)*ln(5/2*(2*exp(x)*x-1)/exp(x)/x)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=-\frac {\log \left (4 \, x + 16\right ) \log \left (\frac {5 \, {\left (2 \, x e^{x} - 1\right )} e^{\left (-x\right )}}{2 \, x}\right )}{x} \] Input:
integrate(((((2*x^2+8*x)*exp(x)-x-4)*log(4*x+16)-2*exp(x)*x^2+x)*log(1/2*( 10*exp(x)*x-5)/exp(x)/x)+(-x^2-5*x-4)*log(4*x+16))/((2*x^4+8*x^3)*exp(x)-x ^3-4*x^2),x, algorithm="fricas")
Output:
-log(4*x + 16)*log(5/2*(2*x*e^x - 1)*e^(-x)/x)/x
Time = 0.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=- \frac {\log {\left (\frac {\left (5 x e^{x} - \frac {5}{2}\right ) e^{- x}}{x} \right )} \log {\left (4 x + 16 \right )}}{x} \] Input:
integrate(((((2*x**2+8*x)*exp(x)-x-4)*ln(4*x+16)-2*exp(x)*x**2+x)*ln(1/2*( 10*exp(x)*x-5)/exp(x)/x)+(-x**2-5*x-4)*ln(4*x+16))/((2*x**4+8*x**3)*exp(x) -x**3-4*x**2),x)
Output:
-log((5*x*exp(x) - 5/2)*exp(-x)/x)*log(4*x + 16)/x
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (26) = 52\).
Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.93 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=-\frac {2 \, \log \left (5\right ) \log \left (2\right ) - 2 \, \log \left (2\right )^{2} + {\left (2 \, \log \left (2\right ) + \log \left (x + 4\right )\right )} \log \left (2 \, x e^{x} - 1\right ) - {\left (x - \log \left (5\right ) + \log \left (2\right ) + \log \left (x\right )\right )} \log \left (x + 4\right ) - 2 \, \log \left (2\right ) \log \left (x\right )}{x} \] Input:
integrate(((((2*x^2+8*x)*exp(x)-x-4)*log(4*x+16)-2*exp(x)*x^2+x)*log(1/2*( 10*exp(x)*x-5)/exp(x)/x)+(-x^2-5*x-4)*log(4*x+16))/((2*x^4+8*x^3)*exp(x)-x ^3-4*x^2),x, algorithm="maxima")
Output:
-(2*log(5)*log(2) - 2*log(2)^2 + (2*log(2) + log(x + 4))*log(2*x*e^x - 1) - (x - log(5) + log(2) + log(x))*log(x + 4) - 2*log(2)*log(x))/x
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (26) = 52\).
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.30 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=\frac {2 \, \log \left (2\right )^{2} - 2 \, \log \left (2\right ) \log \left (5 \, {\left (2 \, x e^{x} - 1\right )} e^{\left (-x\right )}\right ) + \log \left (2\right ) \log \left (x + 4\right ) - \log \left (5 \, {\left (2 \, x e^{x} - 1\right )} e^{\left (-x\right )}\right ) \log \left (x + 4\right ) + 2 \, \log \left (2\right ) \log \left (x\right ) + \log \left (x + 4\right ) \log \left (x\right )}{x} \] Input:
integrate(((((2*x^2+8*x)*exp(x)-x-4)*log(4*x+16)-2*exp(x)*x^2+x)*log(1/2*( 10*exp(x)*x-5)/exp(x)/x)+(-x^2-5*x-4)*log(4*x+16))/((2*x^4+8*x^3)*exp(x)-x ^3-4*x^2),x, algorithm="giac")
Output:
(2*log(2)^2 - 2*log(2)*log(5*(2*x*e^x - 1)*e^(-x)) + log(2)*log(x + 4) - l og(5*(2*x*e^x - 1)*e^(-x))*log(x + 4) + 2*log(2)*log(x) + log(x + 4)*log(x ))/x
Time = 2.97 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=\ln \left (x+4\right )-\frac {\ln \left (4\,x+16\right )\,\ln \left (\frac {5\,\left (2\,x\,{\mathrm {e}}^x-1\right )}{2\,x}\right )}{x} \] Input:
int((log(4*x + 16)*(5*x + x^2 + 4) + log((exp(-x)*(5*x*exp(x) - 5/2))/x)*( 2*x^2*exp(x) - x + log(4*x + 16)*(x - exp(x)*(8*x + 2*x^2) + 4)))/(4*x^2 - exp(x)*(8*x^3 + 2*x^4) + x^3),x)
Output:
log(x + 4) - (log(4*x + 16)*log((5*(2*x*exp(x) - 1))/(2*x)))/x
Time = 0.17 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-4-5 x-x^2\right ) \log (16+4 x)+\left (x-2 e^x x^2+\left (-4-x+e^x \left (8 x+2 x^2\right )\right ) \log (16+4 x)\right ) \log \left (\frac {e^{-x} \left (-5+10 e^x x\right )}{2 x}\right )}{-4 x^2-x^3+e^x \left (8 x^3+2 x^4\right )} \, dx=-\frac {\mathrm {log}\left (4 x +16\right ) \mathrm {log}\left (\frac {10 e^{x} x -5}{2 e^{x} x}\right )}{x} \] Input:
int(((((2*x^2+8*x)*exp(x)-x-4)*log(4*x+16)-2*exp(x)*x^2+x)*log(1/2*(10*exp (x)*x-5)/exp(x)/x)+(-x^2-5*x-4)*log(4*x+16))/((2*x^4+8*x^3)*exp(x)-x^3-4*x ^2),x)
Output:
( - log(4*x + 16)*log((10*e**x*x - 5)/(2*e**x*x)))/x