Integrand size = 96, antiderivative size = 26 \[ \int \frac {100 e^{x/2}+e^{x/4} (100+25 x)+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+\left (4 e^{x/2}+e^{x/4} (4+x)\right ) \log ^2\left (x^4\right )}{1+2 e^{x/4}+e^{x/2}} \, dx=4 \left (x-\frac {x}{1+e^{x/4}}\right ) \left (25+\log ^2\left (x^4\right )\right ) \] Output:
4*(x-x/(exp(1/4*x)+1))*(ln(x^4)^2+25)
Time = 2.86 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {100 e^{x/2}+e^{x/4} (100+25 x)+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+\left (4 e^{x/2}+e^{x/4} (4+x)\right ) \log ^2\left (x^4\right )}{1+2 e^{x/4}+e^{x/2}} \, dx=\frac {4 e^{x/4} x \left (25+\log ^2\left (x^4\right )\right )}{1+e^{x/4}} \] Input:
Integrate[(100*E^(x/2) + E^(x/4)*(100 + 25*x) + (32*E^(x/4) + 32*E^(x/2))* Log[x^4] + (4*E^(x/2) + E^(x/4)*(4 + x))*Log[x^4]^2)/(1 + 2*E^(x/4) + E^(x /2)),x]
Output:
(4*E^(x/4)*x*(25 + Log[x^4]^2))/(1 + E^(x/4))
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.
\(\displaystyle \int \frac {\left (e^{x/4} (x+4)+4 e^{x/2}\right ) \log ^2\left (x^4\right )+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+100 e^{x/2}+e^{x/4} (25 x+100)}{2 e^{x/4}+e^{x/2}+1} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{x/4} \left (4 e^{x/4} \log ^2\left (x^4\right )+x \log ^2\left (x^4\right )+4 \log ^2\left (x^4\right )+32 e^{x/4} \log \left (x^4\right )+32 \log \left (x^4\right )+100 e^{x/4}+25 x+100\right )}{\left (e^{x/4}+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x/4} x \left (\log ^2\left (x^4\right )+25\right )}{\left (e^{x/4}+1\right )^2}+\frac {4 e^{x/4} \left (\log ^2\left (x^4\right )+8 \log \left (x^4\right )+25\right )}{e^{x/4}+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 16 \text {Subst}\left (\int \frac {e^x \log ^2\left (256 x^4\right )}{1+e^x}dx,x,\frac {x}{4}\right )+128 \text {Subst}\left (\int \frac {e^x \log \left (256 x^4\right )}{1+e^x}dx,x,\frac {x}{4}\right )+\int \frac {e^{x/4} x \log ^2\left (x^4\right )}{\left (1+e^{x/4}\right )^2}dx-\frac {100 x}{e^{x/4}+1}+100 x\) |
Input:
Int[(100*E^(x/2) + E^(x/4)*(100 + 25*x) + (32*E^(x/4) + 32*E^(x/2))*Log[x^ 4] + (4*E^(x/2) + E^(x/4)*(4 + x))*Log[x^4]^2)/(1 + 2*E^(x/4) + E^(x/2)),x ]
Output:
$Aborted
Time = 0.83 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19
method | result | size |
norman | \(\frac {100 x \,{\mathrm e}^{\frac {x}{4}}+4 \ln \left (x^{4}\right )^{2} x \,{\mathrm e}^{\frac {x}{4}}}{{\mathrm e}^{\frac {x}{4}}+1}\) | \(31\) |
parallelrisch | \(\frac {16 \ln \left (x^{4}\right )^{2} x \,{\mathrm e}^{\frac {x}{4}}+400 x \,{\mathrm e}^{\frac {x}{4}}}{4 \,{\mathrm e}^{\frac {x}{4}}+4}\) | \(32\) |
default | \(\frac {\left (4 {\left (\ln \left (x^{4}\right )-4 \ln \left (x \right )\right )}^{2}+100\right ) x \,{\mathrm e}^{\frac {x}{4}}+64 \,{\mathrm e}^{\frac {x}{4}} \ln \left (x \right )^{2} x +32 \,{\mathrm e}^{\frac {x}{4}} \ln \left (x \right ) x \left (\ln \left (x^{4}\right )-4 \ln \left (x \right )\right )}{{\mathrm e}^{\frac {x}{4}}+1}\) | \(61\) |
risch | \(\text {Expression too large to display}\) | \(1515\) |
Input:
int(((4*exp(1/4*x)^2+(4+x)*exp(1/4*x))*ln(x^4)^2+(32*exp(1/4*x)^2+32*exp(1 /4*x))*ln(x^4)+100*exp(1/4*x)^2+(25*x+100)*exp(1/4*x))/(exp(1/4*x)^2+2*exp (1/4*x)+1),x,method=_RETURNVERBOSE)
Output:
(100*x*exp(1/4*x)+4*ln(x^4)^2*x*exp(1/4*x))/(exp(1/4*x)+1)
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {100 e^{x/2}+e^{x/4} (100+25 x)+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+\left (4 e^{x/2}+e^{x/4} (4+x)\right ) \log ^2\left (x^4\right )}{1+2 e^{x/4}+e^{x/2}} \, dx=\frac {4 \, {\left (x e^{\left (\frac {1}{4} \, x\right )} \log \left (x^{4}\right )^{2} + 25 \, x e^{\left (\frac {1}{4} \, x\right )}\right )}}{e^{\left (\frac {1}{4} \, x\right )} + 1} \] Input:
integrate(((4*exp(1/4*x)^2+(4+x)*exp(1/4*x))*log(x^4)^2+(32*exp(1/4*x)^2+3 2*exp(1/4*x))*log(x^4)+100*exp(1/4*x)^2+(25*x+100)*exp(1/4*x))/(exp(1/4*x) ^2+2*exp(1/4*x)+1),x, algorithm="fricas")
Output:
4*(x*e^(1/4*x)*log(x^4)^2 + 25*x*e^(1/4*x))/(e^(1/4*x) + 1)
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {100 e^{x/2}+e^{x/4} (100+25 x)+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+\left (4 e^{x/2}+e^{x/4} (4+x)\right ) \log ^2\left (x^4\right )}{1+2 e^{x/4}+e^{x/2}} \, dx=4 x \log {\left (x^{4} \right )}^{2} + 100 x + \frac {- 4 x \log {\left (x^{4} \right )}^{2} - 100 x}{e^{\frac {x}{4}} + 1} \] Input:
integrate(((4*exp(1/4*x)**2+(4+x)*exp(1/4*x))*ln(x**4)**2+(32*exp(1/4*x)** 2+32*exp(1/4*x))*ln(x**4)+100*exp(1/4*x)**2+(25*x+100)*exp(1/4*x))/(exp(1/ 4*x)**2+2*exp(1/4*x)+1),x)
Output:
4*x*log(x**4)**2 + 100*x + (-4*x*log(x**4)**2 - 100*x)/(exp(x/4) + 1)
Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {100 e^{x/2}+e^{x/4} (100+25 x)+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+\left (4 e^{x/2}+e^{x/4} (4+x)\right ) \log ^2\left (x^4\right )}{1+2 e^{x/4}+e^{x/2}} \, dx=\frac {64 \, x e^{\left (\frac {1}{4} \, x\right )} \log \left (x\right )^{2}}{e^{\left (\frac {1}{4} \, x\right )} + 1} + \frac {100 \, x e^{\left (\frac {1}{4} \, x\right )}}{e^{\left (\frac {1}{4} \, x\right )} + 1} \] Input:
integrate(((4*exp(1/4*x)^2+(4+x)*exp(1/4*x))*log(x^4)^2+(32*exp(1/4*x)^2+3 2*exp(1/4*x))*log(x^4)+100*exp(1/4*x)^2+(25*x+100)*exp(1/4*x))/(exp(1/4*x) ^2+2*exp(1/4*x)+1),x, algorithm="maxima")
Output:
64*x*e^(1/4*x)*log(x)^2/(e^(1/4*x) + 1) + 100*x*e^(1/4*x)/(e^(1/4*x) + 1)
Time = 0.15 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {100 e^{x/2}+e^{x/4} (100+25 x)+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+\left (4 e^{x/2}+e^{x/4} (4+x)\right ) \log ^2\left (x^4\right )}{1+2 e^{x/4}+e^{x/2}} \, dx=\frac {4 \, {\left (x e^{\left (\frac {1}{4} \, x\right )} \log \left (x^{4}\right )^{2} + 25 \, x e^{\left (\frac {1}{4} \, x\right )}\right )}}{e^{\left (\frac {1}{4} \, x\right )} + 1} \] Input:
integrate(((4*exp(1/4*x)^2+(4+x)*exp(1/4*x))*log(x^4)^2+(32*exp(1/4*x)^2+3 2*exp(1/4*x))*log(x^4)+100*exp(1/4*x)^2+(25*x+100)*exp(1/4*x))/(exp(1/4*x) ^2+2*exp(1/4*x)+1),x, algorithm="giac")
Output:
4*(x*e^(1/4*x)*log(x^4)^2 + 25*x*e^(1/4*x))/(e^(1/4*x) + 1)
Time = 2.66 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {100 e^{x/2}+e^{x/4} (100+25 x)+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+\left (4 e^{x/2}+e^{x/4} (4+x)\right ) \log ^2\left (x^4\right )}{1+2 e^{x/4}+e^{x/2}} \, dx=\frac {4\,x\,{\mathrm {e}}^{x/4}\,\left ({\ln \left (x^4\right )}^2+25\right )}{{\mathrm {e}}^{x/4}+1} \] Input:
int((100*exp(x/2) + log(x^4)*(32*exp(x/2) + 32*exp(x/4)) + log(x^4)^2*(4*e xp(x/2) + exp(x/4)*(x + 4)) + exp(x/4)*(25*x + 100))/(exp(x/2) + 2*exp(x/4 ) + 1),x)
Output:
(4*x*exp(x/4)*(log(x^4)^2 + 25))/(exp(x/4) + 1)
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {100 e^{x/2}+e^{x/4} (100+25 x)+\left (32 e^{x/4}+32 e^{x/2}\right ) \log \left (x^4\right )+\left (4 e^{x/2}+e^{x/4} (4+x)\right ) \log ^2\left (x^4\right )}{1+2 e^{x/4}+e^{x/2}} \, dx=\frac {4 e^{\frac {x}{4}} x \left (\mathrm {log}\left (x^{4}\right )^{2}+25\right )}{e^{\frac {x}{4}}+1} \] Input:
int(((4*exp(1/4*x)^2+(4+x)*exp(1/4*x))*log(x^4)^2+(32*exp(1/4*x)^2+32*exp( 1/4*x))*log(x^4)+100*exp(1/4*x)^2+(25*x+100)*exp(1/4*x))/(exp(1/4*x)^2+2*e xp(1/4*x)+1),x)
Output:
(4*e**(x/4)*x*(log(x**4)**2 + 25))/(e**(x/4) + 1)