Integrand size = 82, antiderivative size = 19 \[ \int \frac {e^{-2 x} \left (576+36 x+e^{2 x} \left (288 x+19 x^2\right )+\left (576-1112 x-72 x^2\right ) \log (x)+\left (64+4 x+e^{2 x} \left (32 x+2 x^2\right )+\left (64-124 x-8 x^2\right ) \log (x)\right ) \log (16+x)\right )}{16+x} \, dx=x \left (x+4 e^{-2 x} \log (x)\right ) (9+\log (16+x)) \] Output:
(ln(x+16)+9)*x*(x+4*ln(x)/exp(2*x))
Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32 \[ \int \frac {e^{-2 x} \left (576+36 x+e^{2 x} \left (288 x+19 x^2\right )+\left (576-1112 x-72 x^2\right ) \log (x)+\left (64+4 x+e^{2 x} \left (32 x+2 x^2\right )+\left (64-124 x-8 x^2\right ) \log (x)\right ) \log (16+x)\right )}{16+x} \, dx=e^{-2 x} x \left (e^{2 x} x+4 \log (x)\right ) (9+\log (16+x)) \] Input:
Integrate[(576 + 36*x + E^(2*x)*(288*x + 19*x^2) + (576 - 1112*x - 72*x^2) *Log[x] + (64 + 4*x + E^(2*x)*(32*x + 2*x^2) + (64 - 124*x - 8*x^2)*Log[x] )*Log[16 + x])/(E^(2*x)*(16 + x)),x]
Output:
(x*(E^(2*x)*x + 4*Log[x])*(9 + Log[16 + x]))/E^(2*x)
Time = 3.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, 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.
\(\displaystyle \int \frac {e^{-2 x} \left (e^{2 x} \left (19 x^2+288 x\right )+\left (-72 x^2-1112 x+576\right ) \log (x)+\left (e^{2 x} \left (2 x^2+32 x\right )+\left (-8 x^2-124 x+64\right ) \log (x)+4 x+64\right ) \log (x+16)+36 x+576\right )}{x+16} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x (19 x+2 x \log (x+16)+32 \log (x+16)+288)}{x+16}-\frac {4 e^{-2 x} \left (18 x^2 \log (x)+2 x^2 \log (x) \log (x+16)-9 x+278 x \log (x)+31 x \log (x) \log (x+16)-x \log (x+16)-144 \log (x)-16 \log (x) \log (x+16)-16 \log (x+16)-144\right )}{x+16}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 9 x^2+x^2 \log (x+16)+36 e^{-2 x} x \log (x)+4 e^{-2 x} x \log (x) \log (x+16)\) |
Input:
Int[(576 + 36*x + E^(2*x)*(288*x + 19*x^2) + (576 - 1112*x - 72*x^2)*Log[x ] + (64 + 4*x + E^(2*x)*(32*x + 2*x^2) + (64 - 124*x - 8*x^2)*Log[x])*Log[ 16 + x])/(E^(2*x)*(16 + x)),x]
Output:
9*x^2 + (36*x*Log[x])/E^(2*x) + x^2*Log[16 + x] + (4*x*Log[x]*Log[16 + x]) /E^(2*x)
Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.95
\[\ln \left (x +16\right ) x^{2}+4 \ln \left (x \right ) \ln \left (x +16\right ) {\mathrm e}^{-2 x} x +9 x^{2}+36 \ln \left (x \right ) {\mathrm e}^{-2 x} x\]
Input:
int((((-8*x^2-124*x+64)*ln(x)+(2*x^2+32*x)*exp(2*x)+4*x+64)*ln(x+16)+(-72* x^2-1112*x+576)*ln(x)+(19*x^2+288*x)*exp(2*x)+36*x+576)/(x+16)/exp(2*x),x)
Output:
ln(x+16)*x^2+4*ln(x)*ln(x+16)*exp(-2*x)*x+9*x^2+36*ln(x)*exp(-2*x)*x
Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (18) = 36\).
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.05 \[ \int \frac {e^{-2 x} \left (576+36 x+e^{2 x} \left (288 x+19 x^2\right )+\left (576-1112 x-72 x^2\right ) \log (x)+\left (64+4 x+e^{2 x} \left (32 x+2 x^2\right )+\left (64-124 x-8 x^2\right ) \log (x)\right ) \log (16+x)\right )}{16+x} \, dx={\left (9 \, x^{2} e^{\left (2 \, x\right )} + {\left (x^{2} e^{\left (2 \, x\right )} + 4 \, x \log \left (x\right )\right )} \log \left (x + 16\right ) + 36 \, x \log \left (x\right )\right )} e^{\left (-2 \, x\right )} \] Input:
integrate((((-8*x^2-124*x+64)*log(x)+(2*x^2+32*x)*exp(2*x)+4*x+64)*log(x+1 6)+(-72*x^2-1112*x+576)*log(x)+(19*x^2+288*x)*exp(2*x)+36*x+576)/(x+16)/ex p(2*x),x, algorithm="fricas")
Output:
(9*x^2*e^(2*x) + (x^2*e^(2*x) + 4*x*log(x))*log(x + 16) + 36*x*log(x))*e^( -2*x)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (19) = 38\).
Time = 7.98 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.53 \[ \int \frac {e^{-2 x} \left (576+36 x+e^{2 x} \left (288 x+19 x^2\right )+\left (576-1112 x-72 x^2\right ) \log (x)+\left (64+4 x+e^{2 x} \left (32 x+2 x^2\right )+\left (64-124 x-8 x^2\right ) \log (x)\right ) \log (16+x)\right )}{16+x} \, dx=9 x^{2} + \left (x^{2} - \frac {256}{3}\right ) \log {\left (x + 16 \right )} + \left (4 x \log {\left (x \right )} \log {\left (x + 16 \right )} + 36 x \log {\left (x \right )}\right ) e^{- 2 x} + \frac {256 \log {\left (x + 16 \right )}}{3} \] Input:
integrate((((-8*x**2-124*x+64)*ln(x)+(2*x**2+32*x)*exp(2*x)+4*x+64)*ln(x+1 6)+(-72*x**2-1112*x+576)*ln(x)+(19*x**2+288*x)*exp(2*x)+36*x+576)/(x+16)/e xp(2*x),x)
Output:
9*x**2 + (x**2 - 256/3)*log(x + 16) + (4*x*log(x)*log(x + 16) + 36*x*log(x ))*exp(-2*x) + 256*log(x + 16)/3
\[ \int \frac {e^{-2 x} \left (576+36 x+e^{2 x} \left (288 x+19 x^2\right )+\left (576-1112 x-72 x^2\right ) \log (x)+\left (64+4 x+e^{2 x} \left (32 x+2 x^2\right )+\left (64-124 x-8 x^2\right ) \log (x)\right ) \log (16+x)\right )}{16+x} \, dx=\int { \frac {{\left ({\left (19 \, x^{2} + 288 \, x\right )} e^{\left (2 \, x\right )} + 2 \, {\left ({\left (x^{2} + 16 \, x\right )} e^{\left (2 \, x\right )} - 2 \, {\left (2 \, x^{2} + 31 \, x - 16\right )} \log \left (x\right ) + 2 \, x + 32\right )} \log \left (x + 16\right ) - 8 \, {\left (9 \, x^{2} + 139 \, x - 72\right )} \log \left (x\right ) + 36 \, x + 576\right )} e^{\left (-2 \, x\right )}}{x + 16} \,d x } \] Input:
integrate((((-8*x^2-124*x+64)*log(x)+(2*x^2+32*x)*exp(2*x)+4*x+64)*log(x+1 6)+(-72*x^2-1112*x+576)*log(x)+(19*x^2+288*x)*exp(2*x)+36*x+576)/(x+16)/ex p(2*x),x, algorithm="maxima")
Output:
36*x*e^(-2*x)*log(x) + 9*x^2 - 576*e^32*exp_integral_e(1, 2*x + 32) + (4*x *e^(-2*x)*log(x) + x^2)*log(x + 16) - 576*integrate(e^(-2*x)/(x + 16), x)
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89 \[ \int \frac {e^{-2 x} \left (576+36 x+e^{2 x} \left (288 x+19 x^2\right )+\left (576-1112 x-72 x^2\right ) \log (x)+\left (64+4 x+e^{2 x} \left (32 x+2 x^2\right )+\left (64-124 x-8 x^2\right ) \log (x)\right ) \log (16+x)\right )}{16+x} \, dx=4 \, x e^{\left (-2 \, x\right )} \log \left (x + 16\right ) \log \left (x\right ) + x^{2} \log \left (x + 16\right ) + 40 \, x e^{\left (-2 \, x\right )} \log \left (x\right ) + 9 \, x^{2} \] Input:
integrate((((-8*x^2-124*x+64)*log(x)+(2*x^2+32*x)*exp(2*x)+4*x+64)*log(x+1 6)+(-72*x^2-1112*x+576)*log(x)+(19*x^2+288*x)*exp(2*x)+36*x+576)/(x+16)/ex p(2*x),x, algorithm="giac")
Output:
4*x*e^(-2*x)*log(x + 16)*log(x) + x^2*log(x + 16) + 40*x*e^(-2*x)*log(x) + 9*x^2
Timed out. \[ \int \frac {e^{-2 x} \left (576+36 x+e^{2 x} \left (288 x+19 x^2\right )+\left (576-1112 x-72 x^2\right ) \log (x)+\left (64+4 x+e^{2 x} \left (32 x+2 x^2\right )+\left (64-124 x-8 x^2\right ) \log (x)\right ) \log (16+x)\right )}{16+x} \, dx=\int \frac {{\mathrm {e}}^{-2\,x}\,\left (36\,x+{\mathrm {e}}^{2\,x}\,\left (19\,x^2+288\,x\right )+\ln \left (x+16\right )\,\left (4\,x+{\mathrm {e}}^{2\,x}\,\left (2\,x^2+32\,x\right )-\ln \left (x\right )\,\left (8\,x^2+124\,x-64\right )+64\right )-\ln \left (x\right )\,\left (72\,x^2+1112\,x-576\right )+576\right )}{x+16} \,d x \] Input:
int((exp(-2*x)*(36*x + exp(2*x)*(288*x + 19*x^2) + log(x + 16)*(4*x + exp( 2*x)*(32*x + 2*x^2) - log(x)*(124*x + 8*x^2 - 64) + 64) - log(x)*(1112*x + 72*x^2 - 576) + 576))/(x + 16),x)
Output:
int((exp(-2*x)*(36*x + exp(2*x)*(288*x + 19*x^2) + log(x + 16)*(4*x + exp( 2*x)*(32*x + 2*x^2) - log(x)*(124*x + 8*x^2 - 64) + 64) - log(x)*(1112*x + 72*x^2 - 576) + 576))/(x + 16), x)
Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16 \[ \int \frac {e^{-2 x} \left (576+36 x+e^{2 x} \left (288 x+19 x^2\right )+\left (576-1112 x-72 x^2\right ) \log (x)+\left (64+4 x+e^{2 x} \left (32 x+2 x^2\right )+\left (64-124 x-8 x^2\right ) \log (x)\right ) \log (16+x)\right )}{16+x} \, dx=\frac {x \left (e^{2 x} \mathrm {log}\left (x +16\right ) x +9 e^{2 x} x +4 \,\mathrm {log}\left (x +16\right ) \mathrm {log}\left (x \right )+36 \,\mathrm {log}\left (x \right )\right )}{e^{2 x}} \] Input:
int((((-8*x^2-124*x+64)*log(x)+(2*x^2+32*x)*exp(2*x)+4*x+64)*log(x+16)+(-7 2*x^2-1112*x+576)*log(x)+(19*x^2+288*x)*exp(2*x)+36*x+576)/(x+16)/exp(2*x) ,x)
Output:
(x*(e**(2*x)*log(x + 16)*x + 9*e**(2*x)*x + 4*log(x + 16)*log(x) + 36*log( x)))/e**(2*x)