Integrand size = 114, antiderivative size = 24 \[ \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx=\log ^2\left (4+\left (-e^{2^{8 x}}+\frac {x}{5}\right )^2+\log (2)\right ) \] Output:
ln((1/5*x-exp(exp(8*x*ln(2))))^2+4+ln(2))^2
\[ \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx=\int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx \] Input:
Integrate[((4*x + 25*2^(5 + 8*x)*E^2^(1 + 8*x)*Log[2] + E^2^(8*x)*(-20 - 5 *2^(5 + 8*x)*x*Log[2]))*Log[(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])/25])/(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log [2]),x]
Output:
Integrate[((4*x + 25*2^(5 + 8*x)*E^2^(1 + 8*x)*Log[2] + E^2^(8*x)*(-20 - 5 *2^(5 + 8*x)*x*Log[2]))*Log[(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2])/25])/(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log [2]), x]
Time = 0.74 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {7237}
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.
\(\displaystyle \int \frac {\left (4 x+e^{2^{8 x}} \left (-5\ 2^{8 x+5} x \log (2)-20\right )+25\ 2^{8 x+5} e^{2^{8 x+1}} \log (2)\right ) \log \left (\frac {1}{25} \left (x^2-10 e^{2^{8 x}} x+25 e^{2^{8 x+1}}+100+25 \log (2)\right )\right )}{x^2-10 e^{2^{8 x}} x+25 e^{2^{8 x+1}}+100+25 \log (2)} \, dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \log ^2\left (\frac {1}{25} \left (x^2-10 e^{2^{8 x}} x+25 e^{2^{8 x+1}}+25 (4+\log (2))\right )\right )\) |
Input:
Int[((4*x + 25*2^(5 + 8*x)*E^2^(1 + 8*x)*Log[2] + E^2^(8*x)*(-20 - 5*2^(5 + 8*x)*x*Log[2]))*Log[(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25* Log[2])/25])/(100 + 25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*Log[2]),x ]
Output:
Log[(25*E^2^(1 + 8*x) - 10*E^2^(8*x)*x + x^2 + 25*(4 + Log[2]))/25]^2
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Time = 5.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08
method | result | size |
risch | \(\ln \left ({\mathrm e}^{2 \,256^{x}}-\frac {2 x \,{\mathrm e}^{256^{x}}}{5}+\ln \left (2\right )+\frac {x^{2}}{25}+4\right )^{2}\) | \(26\) |
parallelrisch | \(\ln \left ({\mathrm e}^{2 \,{\mathrm e}^{8 x \ln \left (2\right )}}-\frac {2 x \,{\mathrm e}^{{\mathrm e}^{8 x \ln \left (2\right )}}}{5}+\ln \left (2\right )+\frac {x^{2}}{25}+4\right )^{2}\) | \(32\) |
Input:
int((800*ln(2)*exp(8*x*ln(2))*exp(exp(8*x*ln(2)))^2+(-160*x*ln(2)*exp(8*x* ln(2))-20)*exp(exp(8*x*ln(2)))+4*x)*ln(exp(exp(8*x*ln(2)))^2-2/5*x*exp(exp (8*x*ln(2)))+ln(2)+1/25*x^2+4)/(25*exp(exp(8*x*ln(2)))^2-10*x*exp(exp(8*x* ln(2)))+25*ln(2)+x^2+100),x,method=_RETURNVERBOSE)
Output:
ln(exp(2*256^x)-2/5*x*exp(256^x)+ln(2)+1/25*x^2+4)^2
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx=\log \left (\frac {1}{25} \, x^{2} - \frac {2}{5} \, x e^{\left (2^{8 \, x}\right )} + e^{\left (2 \cdot 2^{8 \, x}\right )} + \log \left (2\right ) + 4\right )^{2} \] Input:
integrate((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2 )*exp(8*x*log(2))-20)*exp(exp(8*x*log(2)))+4*x)*log(exp(exp(8*x*log(2)))^2 -2/5*x*exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^2- 10*x*exp(exp(8*x*log(2)))+25*log(2)+x^2+100),x, algorithm="fricas")
Output:
log(1/25*x^2 - 2/5*x*e^(2^(8*x)) + e^(2*2^(8*x)) + log(2) + 4)^2
Time = 0.98 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx=\log {\left (\frac {x^{2}}{25} - \frac {2 x e^{e^{8 x \log {\left (2 \right )}}}}{5} + e^{2 e^{8 x \log {\left (2 \right )}}} + \log {\left (2 \right )} + 4 \right )}^{2} \] Input:
integrate((800*ln(2)*exp(8*x*ln(2))*exp(exp(8*x*ln(2)))**2+(-160*x*ln(2)*e xp(8*x*ln(2))-20)*exp(exp(8*x*ln(2)))+4*x)*ln(exp(exp(8*x*ln(2)))**2-2/5*x *exp(exp(8*x*ln(2)))+ln(2)+1/25*x**2+4)/(25*exp(exp(8*x*ln(2)))**2-10*x*ex p(exp(8*x*ln(2)))+25*ln(2)+x**2+100),x)
Output:
log(x**2/25 - 2*x*exp(exp(8*x*log(2)))/5 + exp(2*exp(8*x*log(2))) + log(2) + 4)**2
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (21) = 42\).
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.83 \[ \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx=-\log \left (x^{2} - 10 \, x e^{\left (2^{8 \, x}\right )} + 25 \, e^{\left (2 \cdot 2^{8 \, x}\right )} + 25 \, \log \left (2\right ) + 100\right )^{2} + 2 \, \log \left (x^{2} - 10 \, x e^{\left (2^{8 \, x}\right )} + 25 \, e^{\left (2 \cdot 2^{8 \, x}\right )} + 25 \, \log \left (2\right ) + 100\right ) \log \left (\frac {1}{25} \, x^{2} - \frac {2}{5} \, x e^{\left (2^{8 \, x}\right )} + e^{\left (2^{8 \, x + 1}\right )} + \log \left (2\right ) + 4\right ) \] Input:
integrate((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2 )*exp(8*x*log(2))-20)*exp(exp(8*x*log(2)))+4*x)*log(exp(exp(8*x*log(2)))^2 -2/5*x*exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^2- 10*x*exp(exp(8*x*log(2)))+25*log(2)+x^2+100),x, algorithm="maxima")
Output:
-log(x^2 - 10*x*e^(2^(8*x)) + 25*e^(2*2^(8*x)) + 25*log(2) + 100)^2 + 2*lo g(x^2 - 10*x*e^(2^(8*x)) + 25*e^(2*2^(8*x)) + 25*log(2) + 100)*log(1/25*x^ 2 - 2/5*x*e^(2^(8*x)) + e^(2^(8*x + 1)) + log(2) + 4)
\[ \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx=\int { \frac {4 \, {\left (200 \cdot 2^{8 \, x} e^{\left (2 \cdot 2^{8 \, x}\right )} \log \left (2\right ) - 5 \, {\left (8 \cdot 2^{8 \, x} x \log \left (2\right ) + 1\right )} e^{\left (2^{8 \, x}\right )} + x\right )} \log \left (\frac {1}{25} \, x^{2} - \frac {2}{5} \, x e^{\left (2^{8 \, x}\right )} + e^{\left (2 \cdot 2^{8 \, x}\right )} + \log \left (2\right ) + 4\right )}{x^{2} - 10 \, x e^{\left (2^{8 \, x}\right )} + 25 \, e^{\left (2 \cdot 2^{8 \, x}\right )} + 25 \, \log \left (2\right ) + 100} \,d x } \] Input:
integrate((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2 )*exp(8*x*log(2))-20)*exp(exp(8*x*log(2)))+4*x)*log(exp(exp(8*x*log(2)))^2 -2/5*x*exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^2- 10*x*exp(exp(8*x*log(2)))+25*log(2)+x^2+100),x, algorithm="giac")
Output:
integrate(4*(200*2^(8*x)*e^(2*2^(8*x))*log(2) - 5*(8*2^(8*x)*x*log(2) + 1) *e^(2^(8*x)) + x)*log(1/25*x^2 - 2/5*x*e^(2^(8*x)) + e^(2*2^(8*x)) + log(2 ) + 4)/(x^2 - 10*x*e^(2^(8*x)) + 25*e^(2*2^(8*x)) + 25*log(2) + 100), x)
Time = 4.94 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx={\ln \left (\ln \left (2\right )+{\mathrm {e}}^{2\,2^{8\,x}}-\frac {2\,x\,{\mathrm {e}}^{2^{8\,x}}}{5}+\frac {x^2}{25}+4\right )}^2 \] Input:
int((log(log(2) + exp(2*exp(8*x*log(2))) + x^2/25 - (2*x*exp(exp(8*x*log(2 ))))/5 + 4)*(4*x - exp(exp(8*x*log(2)))*(160*x*exp(8*x*log(2))*log(2) + 20 ) + 800*exp(2*exp(8*x*log(2)))*exp(8*x*log(2))*log(2)))/(25*log(2) + 25*ex p(2*exp(8*x*log(2))) + x^2 - 10*x*exp(exp(8*x*log(2))) + 100),x)
Output:
log(log(2) + exp(2*2^(8*x)) - (2*x*exp(2^(8*x)))/5 + x^2/25 + 4)^2
Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (4 x+25\ 2^{5+8 x} e^{2^{1+8 x}} \log (2)+e^{2^{8 x}} \left (-20-5\ 2^{5+8 x} x \log (2)\right )\right ) \log \left (\frac {1}{25} \left (100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)\right )\right )}{100+25 e^{2^{1+8 x}}-10 e^{2^{8 x}} x+x^2+25 \log (2)} \, dx=\mathrm {log}\left (e^{2 \,2^{8 x}}-\frac {2 e^{2^{8 x}} x}{5}+\mathrm {log}\left (2\right )+\frac {x^{2}}{25}+4\right )^{2} \] Input:
int((800*log(2)*exp(8*x*log(2))*exp(exp(8*x*log(2)))^2+(-160*x*log(2)*exp( 8*x*log(2))-20)*exp(exp(8*x*log(2)))+4*x)*log(exp(exp(8*x*log(2)))^2-2/5*x *exp(exp(8*x*log(2)))+log(2)+1/25*x^2+4)/(25*exp(exp(8*x*log(2)))^2-10*x*e xp(exp(8*x*log(2)))+25*log(2)+x^2+100),x)
Output:
log((25*e**(2*2**(8*x)) - 10*e**(2**(8*x))*x + 25*log(2) + x**2 + 100)/25) **2