Integrand size = 198, antiderivative size = 26 \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=2+64^x \log ^x\left (\left (5-e^{e^x}\right )^2 (1+x)^2\right ) \]
\[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=\int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx \]
Integrate[(64^x*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^( 2*E^x)*(1 + 2*x + x^2)]^(-1 + x)*(-10*x + E^E^x*(2*x + E^x*(2*x + 2*x^2)) + (-5 - 5*x + E^E^x*(1 + x))*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^(2*E^x)*(1 + 2*x + x^2)]*Log[64*Log[25 + 50*x + 25*x^2 + E^E^x *(-10 - 20*x - 10*x^2) + E^(2*E^x)*(1 + 2*x + x^2)]]))/(-5 - 5*x + E^E^x*( 1 + x)),x]
Integrate[(64^x*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^( 2*E^x)*(1 + 2*x + x^2)]^(-1 + x)*(-10*x + E^E^x*(2*x + E^x*(2*x + 2*x^2)) + (-5 - 5*x + E^E^x*(1 + x))*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^(2*E^x)*(1 + 2*x + x^2)]*Log[64*Log[25 + 50*x + 25*x^2 + E^E^x *(-10 - 20*x - 10*x^2) + E^(2*E^x)*(1 + 2*x + x^2)]]))/(-5 - 5*x + E^E^x*( 1 + x)), x]
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.
| \(\displaystyle \int \frac {64^x \log ^{x-1}\left (25 x^2+e^{e^x} \left (-10 x^2-20 x-10\right )+e^{2 e^x} \left (x^2+2 x+1\right )+50 x+25\right ) \left (e^{e^x} \left (e^x \left (2 x^2+2 x\right )+2 x\right )+\left (-5 x+e^{e^x} (x+1)-5\right ) \log \left (25 x^2+e^{e^x} \left (-10 x^2-20 x-10\right )+e^{2 e^x} \left (x^2+2 x+1\right )+50 x+25\right ) \log \left (64 \log \left (25 x^2+e^{e^x} \left (-10 x^2-20 x-10\right )+e^{2 e^x} \left (x^2+2 x+1\right )+50 x+25\right )\right )-10 x\right )}{-5 x+e^{e^x} (x+1)-5} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {64^x \log ^{x-1}\left (\left (e^{e^x}-5\right )^2 (x+1)^2\right ) \left (-2 e^{e^x} \left (e^x (x+1)+1\right ) x+10 x-\left (e^{e^x}-5\right ) (x+1) \log \left (\left (e^{e^x}-5\right )^2 (x+1)^2\right ) \log \left (64 \log \left (\left (e^{e^x}-5\right )^2 (x+1)^2\right )\right )\right )}{\left (5-e^{e^x}\right ) (x+1)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2^{6 x+1} e^{x+e^x} x \log ^{x-1}\left (\left (e^{e^x}-5\right )^2 (x+1)^2\right )}{e^{e^x}-5}+\frac {64^x \left (2 x+x \log \left (\left (e^{e^x}-5\right )^2 (x+1)^2\right ) \log \left (64 \log \left (\left (e^{e^x}-5\right )^2 (x+1)^2\right )\right )+\log \left (\left (e^{e^x}-5\right )^2 (x+1)^2\right ) \log \left (64 \log \left (\left (e^{e^x}-5\right )^2 (x+1)^2\right )\right )\right ) \log ^{x-1}\left (\left (e^{e^x}-5\right )^2 (x+1)^2\right )}{x+1}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \int 2^{6 x+1} \log ^{x-1}\left (\left (-5+e^{e^x}\right )^2 (x+1)^2\right )dx+\int \frac {2^{6 x+1} \log ^{x-1}\left (\left (-5+e^{e^x}\right )^2 (x+1)^2\right )}{-x-1}dx+\int \frac {2^{6 x+1} e^{x+e^x} x \log ^{x-1}\left (\left (-5+e^{e^x}\right )^2 (x+1)^2\right )}{-5+e^{e^x}}dx+\int 64^x \log ^x\left (\left (-5+e^{e^x}\right )^2 (x+1)^2\right ) \log \left (64 \log \left (\left (-5+e^{e^x}\right )^2 (x+1)^2\right )\right )dx\) |
Int[(64^x*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^(2*E^x) *(1 + 2*x + x^2)]^(-1 + x)*(-10*x + E^E^x*(2*x + E^x*(2*x + 2*x^2)) + (-5 - 5*x + E^E^x*(1 + x))*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2 ) + E^(2*E^x)*(1 + 2*x + x^2)]*Log[64*Log[25 + 50*x + 25*x^2 + E^E^x*(-10 - 20*x - 10*x^2) + E^(2*E^x)*(1 + 2*x + x^2)]]))/(-5 - 5*x + E^E^x*(1 + x) ),x]
3.14.91.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.19 (sec) , antiderivative size = 170, normalized size of antiderivative = 6.54
\[{\left (128 \ln \left (1+x \right )+128 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )-32 i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )\right )}^{2}-32 i \pi \,\operatorname {csgn}\left (i \left (1+x \right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \left (1+x \right )^{2}\right )+\operatorname {csgn}\left (i \left (1+x \right )\right )\right )}^{2}-32 i \pi \,\operatorname {csgn}\left (i \left (1+x \right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right ) \left (-\operatorname {csgn}\left (i \left (1+x \right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )+\operatorname {csgn}\left (i \left (1+x \right )^{2}\right )\right ) \left (-\operatorname {csgn}\left (i \left (1+x \right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )+\operatorname {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )^{2}\right )\right )\right )}^{x}\]
int((((1+x)*exp(exp(x))-5*x-5)*ln((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x- 10)*exp(exp(x))+25*x^2+50*x+25)*ln(64*ln((x^2+2*x+1)*exp(exp(x))^2+(-10*x^ 2-20*x-10)*exp(exp(x))+25*x^2+50*x+25))+((2*x^2+2*x)*exp(x)+2*x)*exp(exp(x ))-10*x)*exp(x*ln(64*ln((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(ex p(x))+25*x^2+50*x+25)))/((1+x)*exp(exp(x))-5*x-5)/ln((x^2+2*x+1)*exp(exp(x ))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25),x)
(128*ln(1+x)+128*ln(exp(exp(x))-5)-32*I*Pi*csgn(I*(exp(exp(x))-5)^2)*(-csg n(I*(exp(exp(x))-5)^2)+csgn(I*(exp(exp(x))-5)))^2-32*I*Pi*csgn(I*(1+x)^2)* (-csgn(I*(1+x)^2)+csgn(I*(1+x)))^2-32*I*Pi*csgn(I*(1+x)^2*(exp(exp(x))-5)^ 2)*(-csgn(I*(1+x)^2*(exp(exp(x))-5)^2)+csgn(I*(1+x)^2))*(-csgn(I*(1+x)^2*( exp(exp(x))-5)^2)+csgn(I*(exp(exp(x))-5)^2)))^x
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).
Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=\left (64 \, \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )\right )^{x} \]
integrate((((1+x)*exp(exp(x))-5*x-5)*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^ 2-20*x-10)*exp(exp(x))+25*x^2+50*x+25)*log(64*log((x^2+2*x+1)*exp(exp(x))^ 2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25))+((2*x^2+2*x)*exp(x)+2*x)* exp(exp(x))-10*x)*exp(x*log(64*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x -10)*exp(exp(x))+25*x^2+50*x+25)))/((1+x)*exp(exp(x))-5*x-5)/log((x^2+2*x+ 1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25),x, algorith m=\
Timed out. \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=\text {Timed out} \]
integrate((((1+x)*exp(exp(x))-5*x-5)*ln((x**2+2*x+1)*exp(exp(x))**2+(-10*x **2-20*x-10)*exp(exp(x))+25*x**2+50*x+25)*ln(64*ln((x**2+2*x+1)*exp(exp(x) )**2+(-10*x**2-20*x-10)*exp(exp(x))+25*x**2+50*x+25))+((2*x**2+2*x)*exp(x) +2*x)*exp(exp(x))-10*x)*exp(x*ln(64*ln((x**2+2*x+1)*exp(exp(x))**2+(-10*x* *2-20*x-10)*exp(exp(x))+25*x**2+50*x+25)))/((1+x)*exp(exp(x))-5*x-5)/ln((x **2+2*x+1)*exp(exp(x))**2+(-10*x**2-20*x-10)*exp(exp(x))+25*x**2+50*x+25), x)
Time = 0.56 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=e^{\left (7 \, x \log \left (2\right ) + x \log \left (\log \left (x + 1\right ) + \log \left (e^{\left (e^{x}\right )} - 5\right )\right )\right )} \]
integrate((((1+x)*exp(exp(x))-5*x-5)*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^ 2-20*x-10)*exp(exp(x))+25*x^2+50*x+25)*log(64*log((x^2+2*x+1)*exp(exp(x))^ 2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25))+((2*x^2+2*x)*exp(x)+2*x)* exp(exp(x))-10*x)*exp(x*log(64*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x -10)*exp(exp(x))+25*x^2+50*x+25)))/((1+x)*exp(exp(x))-5*x-5)/log((x^2+2*x+ 1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25),x, algorith m=\
\[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=\int { \frac {{\left ({\left ({\left (x + 1\right )} e^{\left (e^{x}\right )} - 5 \, x - 5\right )} \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right ) \log \left (64 \, \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )\right ) + 2 \, {\left ({\left (x^{2} + x\right )} e^{x} + x\right )} e^{\left (e^{x}\right )} - 10 \, x\right )} \left (64 \, \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )\right )^{x}}{{\left ({\left (x + 1\right )} e^{\left (e^{x}\right )} - 5 \, x - 5\right )} \log \left (25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, e^{x}\right )} - 10 \, {\left (x^{2} + 2 \, x + 1\right )} e^{\left (e^{x}\right )} + 50 \, x + 25\right )} \,d x } \]
integrate((((1+x)*exp(exp(x))-5*x-5)*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^ 2-20*x-10)*exp(exp(x))+25*x^2+50*x+25)*log(64*log((x^2+2*x+1)*exp(exp(x))^ 2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25))+((2*x^2+2*x)*exp(x)+2*x)* exp(exp(x))-10*x)*exp(x*log(64*log((x^2+2*x+1)*exp(exp(x))^2+(-10*x^2-20*x -10)*exp(exp(x))+25*x^2+50*x+25)))/((1+x)*exp(exp(x))-5*x-5)/log((x^2+2*x+ 1)*exp(exp(x))^2+(-10*x^2-20*x-10)*exp(exp(x))+25*x^2+50*x+25),x, algorith m=\
integrate((((x + 1)*e^(e^x) - 5*x - 5)*log(25*x^2 + (x^2 + 2*x + 1)*e^(2*e ^x) - 10*(x^2 + 2*x + 1)*e^(e^x) + 50*x + 25)*log(64*log(25*x^2 + (x^2 + 2 *x + 1)*e^(2*e^x) - 10*(x^2 + 2*x + 1)*e^(e^x) + 50*x + 25)) + 2*((x^2 + x )*e^x + x)*e^(e^x) - 10*x)*(64*log(25*x^2 + (x^2 + 2*x + 1)*e^(2*e^x) - 10 *(x^2 + 2*x + 1)*e^(e^x) + 50*x + 25))^x/(((x + 1)*e^(e^x) - 5*x - 5)*log( 25*x^2 + (x^2 + 2*x + 1)*e^(2*e^x) - 10*(x^2 + 2*x + 1)*e^(e^x) + 50*x + 2 5)), x)
Time = 11.85 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.31 \[ \int \frac {64^x \log ^{-1+x}\left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \left (-10 x+e^{e^x} \left (2 x+e^x \left (2 x+2 x^2\right )\right )+\left (-5-5 x+e^{e^x} (1+x)\right ) \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right ) \log \left (64 \log \left (25+50 x+25 x^2+e^{e^x} \left (-10-20 x-10 x^2\right )+e^{2 e^x} \left (1+2 x+x^2\right )\right )\right )\right )}{-5-5 x+e^{e^x} (1+x)} \, dx=2^{6\,x}\,{\ln \left (50\,x-10\,{\mathrm {e}}^{{\mathrm {e}}^x}+{\mathrm {e}}^{2\,{\mathrm {e}}^x}-20\,x\,{\mathrm {e}}^{{\mathrm {e}}^x}+2\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}-10\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}+25\,x^2+x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+25\right )}^x \]
int((exp(x*log(64*log(50*x + exp(2*exp(x))*(2*x + x^2 + 1) - exp(exp(x))*( 20*x + 10*x^2 + 10) + 25*x^2 + 25)))*(10*x - exp(exp(x))*(2*x + exp(x)*(2* x + 2*x^2)) + log(64*log(50*x + exp(2*exp(x))*(2*x + x^2 + 1) - exp(exp(x) )*(20*x + 10*x^2 + 10) + 25*x^2 + 25))*log(50*x + exp(2*exp(x))*(2*x + x^2 + 1) - exp(exp(x))*(20*x + 10*x^2 + 10) + 25*x^2 + 25)*(5*x - exp(exp(x)) *(x + 1) + 5)))/(log(50*x + exp(2*exp(x))*(2*x + x^2 + 1) - exp(exp(x))*(2 0*x + 10*x^2 + 10) + 25*x^2 + 25)*(5*x - exp(exp(x))*(x + 1) + 5)),x)