Integrand size = 114, antiderivative size = 27 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=\log \left (e^{-e^{\frac {x}{\log \left (1+e^5-2 x\right )}}} \log \left (\frac {3 x}{2}\right )\right ) \]
Time = 0.94 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=-e^{\frac {x}{\log \left (1+e^5-2 x\right )}}+\log \left (\log \left (\frac {3 x}{2}\right )\right ) \]
Integrate[((1 + E^5 - 2*x)*Log[1 + E^5 - 2*x]^2 + E^(x/Log[1 + E^5 - 2*x]) *(-2*x^2*Log[(3*x)/2] + (-x - E^5*x + 2*x^2)*Log[1 + E^5 - 2*x]*Log[(3*x)/ 2]))/((x + E^5*x - 2*x^2)*Log[1 + E^5 - 2*x]^2*Log[(3*x)/2]),x]
Time = 1.99 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6, 2026, 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.
\(\displaystyle \int \frac {e^{\frac {x}{\log \left (-2 x+e^5+1\right )}} \left (\left (2 x^2-e^5 x-x\right ) \log \left (-2 x+e^5+1\right ) \log \left (\frac {3 x}{2}\right )-2 x^2 \log \left (\frac {3 x}{2}\right )\right )+\left (-2 x+e^5+1\right ) \log ^2\left (-2 x+e^5+1\right )}{\left (-2 x^2+e^5 x+x\right ) \log ^2\left (-2 x+e^5+1\right ) \log \left (\frac {3 x}{2}\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{\frac {x}{\log \left (-2 x+e^5+1\right )}} \left (\left (2 x^2-e^5 x-x\right ) \log \left (-2 x+e^5+1\right ) \log \left (\frac {3 x}{2}\right )-2 x^2 \log \left (\frac {3 x}{2}\right )\right )+\left (-2 x+e^5+1\right ) \log ^2\left (-2 x+e^5+1\right )}{\left (\left (1+e^5\right ) x-2 x^2\right ) \log ^2\left (-2 x+e^5+1\right ) \log \left (\frac {3 x}{2}\right )}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {x}{\log \left (-2 x+e^5+1\right )}} \left (\left (2 x^2-e^5 x-x\right ) \log \left (-2 x+e^5+1\right ) \log \left (\frac {3 x}{2}\right )-2 x^2 \log \left (\frac {3 x}{2}\right )\right )+\left (-2 x+e^5+1\right ) \log ^2\left (-2 x+e^5+1\right )}{\left (-2 x+e^5+1\right ) x \log ^2\left (-2 x+e^5+1\right ) \log \left (\frac {3 x}{2}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{\frac {x}{\log \left (-2 x+e^5+1\right )}} \left (-2 x+2 x \log \left (-2 x+e^5+1\right )-\left (1+e^5\right ) \log \left (-2 x+e^5+1\right )\right )}{\left (-2 x+e^5+1\right ) \log ^2\left (-2 x+e^5+1\right )}+\frac {1}{x \log \left (\frac {3 x}{2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log \left (\log \left (\frac {3 x}{2}\right )\right )-e^{\frac {x}{\log \left (-2 x+e^5+1\right )}}\) |
Int[((1 + E^5 - 2*x)*Log[1 + E^5 - 2*x]^2 + E^(x/Log[1 + E^5 - 2*x])*(-2*x ^2*Log[(3*x)/2] + (-x - E^5*x + 2*x^2)*Log[1 + E^5 - 2*x]*Log[(3*x)/2]))/( (x + E^5*x - 2*x^2)*Log[1 + E^5 - 2*x]^2*Log[(3*x)/2]),x]
3.10.22.3.1 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[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 35.66 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81
method | result | size |
default | \(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\) | \(22\) |
risch | \(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\) | \(22\) |
parallelrisch | \(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\) | \(22\) |
parts | \(\ln \left (\ln \left (\frac {3 x}{2}\right )\right )-{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{5}+1-2 x \right )}}\) | \(22\) |
int((((-x*exp(5)+2*x^2-x)*ln(3/2*x)*ln(exp(5)+1-2*x)-2*x^2*ln(3/2*x))*exp( x/ln(exp(5)+1-2*x))+(exp(5)+1-2*x)*ln(exp(5)+1-2*x)^2)/(x*exp(5)-2*x^2+x)/ ln(3/2*x)/ln(exp(5)+1-2*x)^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=-e^{\left (\frac {x}{\log \left (-2 \, x + e^{5} + 1\right )}\right )} + \log \left (\log \left (\frac {3}{2} \, x\right )\right ) \]
integrate((((-x*exp(5)+2*x^2-x)*log(3/2*x)*log(exp(5)+1-2*x)-2*x^2*log(3/2 *x))*exp(x/log(exp(5)+1-2*x))+(exp(5)+1-2*x)*log(exp(5)+1-2*x)^2)/(x*exp(5 )-2*x^2+x)/log(3/2*x)/log(exp(5)+1-2*x)^2,x, algorithm=\
Exception generated. \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=\text {Exception raised: TypeError} \]
integrate((((-x*exp(5)+2*x**2-x)*ln(3/2*x)*ln(exp(5)+1-2*x)-2*x**2*ln(3/2* x))*exp(x/ln(exp(5)+1-2*x))+(exp(5)+1-2*x)*ln(exp(5)+1-2*x)**2)/(x*exp(5)- 2*x**2+x)/ln(3/2*x)/ln(exp(5)+1-2*x)**2,x)
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=-e^{\left (\frac {x}{\log \left (-2 \, x + e^{5} + 1\right )}\right )} + \log \left (\log \left (3\right ) - \log \left (2\right ) + \log \left (x\right )\right ) \]
integrate((((-x*exp(5)+2*x^2-x)*log(3/2*x)*log(exp(5)+1-2*x)-2*x^2*log(3/2 *x))*exp(x/log(exp(5)+1-2*x))+(exp(5)+1-2*x)*log(exp(5)+1-2*x)^2)/(x*exp(5 )-2*x^2+x)/log(3/2*x)/log(exp(5)+1-2*x)^2,x, algorithm=\
\[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=\int { \frac {{\left (2 \, x - e^{5} - 1\right )} \log \left (-2 \, x + e^{5} + 1\right )^{2} + {\left (2 \, x^{2} \log \left (\frac {3}{2} \, x\right ) - {\left (2 \, x^{2} - x e^{5} - x\right )} \log \left (\frac {3}{2} \, x\right ) \log \left (-2 \, x + e^{5} + 1\right )\right )} e^{\left (\frac {x}{\log \left (-2 \, x + e^{5} + 1\right )}\right )}}{{\left (2 \, x^{2} - x e^{5} - x\right )} \log \left (\frac {3}{2} \, x\right ) \log \left (-2 \, x + e^{5} + 1\right )^{2}} \,d x } \]
integrate((((-x*exp(5)+2*x^2-x)*log(3/2*x)*log(exp(5)+1-2*x)-2*x^2*log(3/2 *x))*exp(x/log(exp(5)+1-2*x))+(exp(5)+1-2*x)*log(exp(5)+1-2*x)^2)/(x*exp(5 )-2*x^2+x)/log(3/2*x)/log(exp(5)+1-2*x)^2,x, algorithm=\
integrate(((2*x - e^5 - 1)*log(-2*x + e^5 + 1)^2 + (2*x^2*log(3/2*x) - (2* x^2 - x*e^5 - x)*log(3/2*x)*log(-2*x + e^5 + 1))*e^(x/log(-2*x + e^5 + 1)) )/((2*x^2 - x*e^5 - x)*log(3/2*x)*log(-2*x + e^5 + 1)^2), x)
Time = 9.60 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {\left (1+e^5-2 x\right ) \log ^2\left (1+e^5-2 x\right )+e^{\frac {x}{\log \left (1+e^5-2 x\right )}} \left (-2 x^2 \log \left (\frac {3 x}{2}\right )+\left (-x-e^5 x+2 x^2\right ) \log \left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )\right )}{\left (x+e^5 x-2 x^2\right ) \log ^2\left (1+e^5-2 x\right ) \log \left (\frac {3 x}{2}\right )} \, dx=\ln \left (\ln \left (\frac {3\,x}{2}\right )\right )-{\mathrm {e}}^{\frac {x}{\ln \left ({\mathrm {e}}^5-2\,x+1\right )}} \]