Integrand size = 86, antiderivative size = 32 \[ \int \frac {-10 x-4 e^5 x+x^2+\left (-20 x-8 e^5 x+4 x^2\right ) \log \left (4 e^x\right )+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^4+8 x^3 \log ^2\left (4 e^x\right )+8 x^2 \log ^4\left (4 e^x\right )} \, dx=\frac {e^5-x+\frac {5+x}{2}}{x \left (x+2 \log ^2\left (4 e^x\right )\right )} \]
Time = 0.41 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-10 x-4 e^5 x+x^2+\left (-20 x-8 e^5 x+4 x^2\right ) \log \left (4 e^x\right )+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^4+8 x^3 \log ^2\left (4 e^x\right )+8 x^2 \log ^4\left (4 e^x\right )} \, dx=\frac {5+2 e^5-x}{2 x \left (x+2 \log ^2\left (4 e^x\right )\right )} \]
Integrate[(-10*x - 4*E^5*x + x^2 + (-20*x - 8*E^5*x + 4*x^2)*Log[4*E^x] + (-10 - 4*E^5)*Log[4*E^x]^2)/(2*x^4 + 8*x^3*Log[4*E^x]^2 + 8*x^2*Log[4*E^x] ^4),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.
\(\displaystyle \int \frac {x^2+\left (4 x^2-8 e^5 x-20 x\right ) \log \left (4 e^x\right )-4 e^5 x-10 x+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^4+8 x^3 \log ^2\left (4 e^x\right )+8 x^2 \log ^4\left (4 e^x\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x^2+\left (4 x^2-8 e^5 x-20 x\right ) \log \left (4 e^x\right )+\left (-10-4 e^5\right ) x+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^4+8 x^3 \log ^2\left (4 e^x\right )+8 x^2 \log ^4\left (4 e^x\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {x^2+\left (4 x^2-8 e^5 x-20 x\right ) \log \left (4 e^x\right )+\left (-10-4 e^5\right ) x+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^2 \left (x+2 \log ^2\left (4 e^x\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {-x^2+2 \left (5+2 e^5\right ) x+2 \left (5+2 e^5\right ) \log ^2\left (4 e^x\right )+4 \left (-x^2+2 e^5 x+5 x\right ) \log \left (4 e^x\right )}{x^2 \left (2 \log ^2\left (4 e^x\right )+x\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {-x^2+2 \left (5+2 e^5\right ) x+2 \left (5+2 e^5\right ) \log ^2\left (4 e^x\right )+4 \left (-x^2+2 e^5 x+5 x\right ) \log \left (4 e^x\right )}{x^2 \left (2 \log ^2\left (4 e^x\right )+x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {\left (-x+2 e^5+5\right ) \left (4 \log \left (4 e^x\right )+1\right )}{x \left (2 \log ^2\left (4 e^x\right )+x\right )^2}+\frac {5+2 e^5}{x^2 \left (2 \log ^2\left (4 e^x\right )+x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\left (5+2 e^5\right ) \int \frac {1}{x^2 \left (2 \log ^2\left (4 e^x\right )+x\right )}dx-\left (\left (5+2 e^5\right ) \int \frac {1}{x \left (2 \log ^2\left (4 e^x\right )+x\right )^2}dx\right )-4 \left (5+2 e^5\right ) \int \frac {\log \left (4 e^x\right )}{x \left (2 \log ^2\left (4 e^x\right )+x\right )^2}dx-\frac {1}{x+2 \log ^2\left (4 e^x\right )}\right )\) |
Int[(-10*x - 4*E^5*x + x^2 + (-20*x - 8*E^5*x + 4*x^2)*Log[4*E^x] + (-10 - 4*E^5)*Log[4*E^x]^2)/(2*x^4 + 8*x^3*Log[4*E^x]^2 + 8*x^2*Log[4*E^x]^4),x]
3.10.92.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.74 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {10-2 x +4 \,{\mathrm e}^{5}}{4 x \left (x +2 \ln \left (4 \,{\mathrm e}^{x}\right )^{2}\right )}\) | \(28\) |
risch | \(\frac {2 \,{\mathrm e}^{5}-x +5}{x \left (16 \ln \left (2\right )^{2}+16 \ln \left (2\right ) \ln \left ({\mathrm e}^{x}\right )+4 \ln \left ({\mathrm e}^{x}\right )^{2}+2 x \right )}\) | \(40\) |
default | \(\frac {\left (-2 \,{\mathrm e}^{5}-5\right ) x -4 \,{\mathrm e}^{5} \left (\ln \left (4 \,{\mathrm e}^{x}\right )-x \right )-{\left (\ln \left (4 \,{\mathrm e}^{x}\right )-x \right )}^{2}-{\mathrm e}^{5}-\frac {5}{2}-10 \ln \left (4 \,{\mathrm e}^{x}\right )+10 x}{4 {\left (\ln \left (4 \,{\mathrm e}^{x}\right )-x \right )}^{2} \left (x^{2}+2 x \left (\ln \left (4 \,{\mathrm e}^{x}\right )-x \right )+{\left (\ln \left (4 \,{\mathrm e}^{x}\right )-x \right )}^{2}+\frac {x}{2}\right )}-\frac {-2 \,{\mathrm e}^{5}-5}{4 {\left (\ln \left (4 \,{\mathrm e}^{x}\right )-x \right )}^{2} x}\) | \(119\) |
int(((-4*exp(5)-10)*ln(4*exp(x))^2+(-8*x*exp(5)+4*x^2-20*x)*ln(4*exp(x))-4 *x*exp(5)+x^2-10*x)/(8*x^2*ln(4*exp(x))^4+8*x^3*ln(4*exp(x))^2+2*x^4),x,me thod=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-10 x-4 e^5 x+x^2+\left (-20 x-8 e^5 x+4 x^2\right ) \log \left (4 e^x\right )+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^4+8 x^3 \log ^2\left (4 e^x\right )+8 x^2 \log ^4\left (4 e^x\right )} \, dx=-\frac {x - 2 \, e^{5} - 5}{2 \, {\left (2 \, x^{3} + 8 \, x^{2} \log \left (2\right ) + 8 \, x \log \left (2\right )^{2} + x^{2}\right )}} \]
integrate(((-4*exp(5)-10)*log(4*exp(x))^2+(-8*x*exp(5)+4*x^2-20*x)*log(4*e xp(x))-4*x*exp(5)+x^2-10*x)/(8*x^2*log(4*exp(x))^4+8*x^3*log(4*exp(x))^2+2 *x^4),x, algorithm=\
Time = 5.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-10 x-4 e^5 x+x^2+\left (-20 x-8 e^5 x+4 x^2\right ) \log \left (4 e^x\right )+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^4+8 x^3 \log ^2\left (4 e^x\right )+8 x^2 \log ^4\left (4 e^x\right )} \, dx=\frac {- x + 5 + 2 e^{5}}{4 x^{3} + x^{2} \cdot \left (2 + 16 \log {\left (2 \right )}\right ) + 16 x \log {\left (2 \right )}^{2}} \]
integrate(((-4*exp(5)-10)*ln(4*exp(x))**2+(-8*x*exp(5)+4*x**2-20*x)*ln(4*e xp(x))-4*x*exp(5)+x**2-10*x)/(8*x**2*ln(4*exp(x))**4+8*x**3*ln(4*exp(x))** 2+2*x**4),x)
Leaf count of result is larger than twice the leaf count of optimal. 5288 vs. \(2 (25) = 50\).
Time = 0.37 (sec) , antiderivative size = 5288, normalized size of antiderivative = 165.25 \[ \int \frac {-10 x-4 e^5 x+x^2+\left (-20 x-8 e^5 x+4 x^2\right ) \log \left (4 e^x\right )+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^4+8 x^3 \log ^2\left (4 e^x\right )+8 x^2 \log ^4\left (4 e^x\right )} \, dx=\text {Too large to display} \]
integrate(((-4*exp(5)-10)*log(4*exp(x))^2+(-8*x*exp(5)+4*x^2-20*x)*log(4*e xp(x))-4*x*exp(5)+x^2-10*x)/(8*x^2*log(4*exp(x))^4+8*x^3*log(4*exp(x))^2+2 *x^4),x, algorithm=\
1/256*((512*log(2)^4 - 512*log(2)^3 - 288*log(2)^2 - 32*log(2) - 1)*log((4 *x - sqrt(16*log(2) + 1) + 8*log(2) + 1)/(4*x + sqrt(16*log(2) + 1) + 8*lo g(2) + 1))/((16*log(2)^7 + log(2)^6)*sqrt(16*log(2) + 1)) + 16*(2*(16*log( 2)^2 + 16*log(2) + 1)*x^2 + 64*log(2)^3 + (64*log(2)^3 + 136*log(2)^2 + 24 *log(2) + 1)*x + 4*log(2)^2)/(2*(16*log(2)^5 + log(2)^4)*x^3 + (128*log(2) ^6 + 24*log(2)^5 + log(2)^4)*x^2 + 8*(16*log(2)^7 + log(2)^6)*x) - (8*log( 2) + 1)*log(2*x^2 + x*(8*log(2) + 1) + 8*log(2)^2)/log(2)^6 + 2*(8*log(2) + 1)*log(x)/log(2)^6)*e^5*log(4*e^x)^2 - 1/32*((256*log(2)^3 - 96*log(2)^2 - 24*log(2) - 1)*log((4*x - sqrt(16*log(2) + 1) + 8*log(2) + 1)/(4*x + sq rt(16*log(2) + 1) + 8*log(2) + 1))/((16*log(2)^5 + log(2)^4)*sqrt(16*log(2 ) + 1)) + 16*(2*x*(8*log(2) + 1) + 32*log(2)^2 + 16*log(2) + 1)/(128*log(2 )^5 + 8*log(2)^4 + 2*(16*log(2)^3 + log(2)^2)*x^2 + (128*log(2)^4 + 24*log (2)^3 + log(2)^2)*x) - log(2*x^2 + x*(8*log(2) + 1) + 8*log(2)^2)/log(2)^4 + 2*log(x)/log(2)^4)*e^5*log(4*e^x) + 5/512*((512*log(2)^4 - 512*log(2)^3 - 288*log(2)^2 - 32*log(2) - 1)*log((4*x - sqrt(16*log(2) + 1) + 8*log(2) + 1)/(4*x + sqrt(16*log(2) + 1) + 8*log(2) + 1))/((16*log(2)^7 + log(2)^6 )*sqrt(16*log(2) + 1)) + 16*(2*(16*log(2)^2 + 16*log(2) + 1)*x^2 + 64*log( 2)^3 + (64*log(2)^3 + 136*log(2)^2 + 24*log(2) + 1)*x + 4*log(2)^2)/(2*(16 *log(2)^5 + log(2)^4)*x^3 + (128*log(2)^6 + 24*log(2)^5 + log(2)^4)*x^2 + 8*(16*log(2)^7 + log(2)^6)*x) - (8*log(2) + 1)*log(2*x^2 + x*(8*log(2) ...
Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-10 x-4 e^5 x+x^2+\left (-20 x-8 e^5 x+4 x^2\right ) \log \left (4 e^x\right )+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^4+8 x^3 \log ^2\left (4 e^x\right )+8 x^2 \log ^4\left (4 e^x\right )} \, dx=-\frac {x - 2 \, e^{5} - 5}{2 \, {\left (2 \, x^{3} + 8 \, x^{2} \log \left (2\right ) + 8 \, x \log \left (2\right )^{2} + x^{2}\right )}} \]
integrate(((-4*exp(5)-10)*log(4*exp(x))^2+(-8*x*exp(5)+4*x^2-20*x)*log(4*e xp(x))-4*x*exp(5)+x^2-10*x)/(8*x^2*log(4*exp(x))^4+8*x^3*log(4*exp(x))^2+2 *x^4),x, algorithm=\
Time = 16.88 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {-10 x-4 e^5 x+x^2+\left (-20 x-8 e^5 x+4 x^2\right ) \log \left (4 e^x\right )+\left (-10-4 e^5\right ) \log ^2\left (4 e^x\right )}{2 x^4+8 x^3 \log ^2\left (4 e^x\right )+8 x^2 \log ^4\left (4 e^x\right )} \, dx=\frac {2\,{\mathrm {e}}^5-x+5}{4\,x^3+\left (8\,\ln \left (4\right )+2\right )\,x^2+4\,{\ln \left (4\right )}^2\,x} \]