Integrand size = 138, antiderivative size = 24 \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=e^{7-e^{\frac {x}{x-\log (10+5 x)}}} x \] Output:
x/exp(exp(x/(x-ln(5*x+10)))-7)
Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=e^{7-e^{-\frac {x}{-x+\log (5 (2+x))}}} x \] Input:
Integrate[(E^(7 - E^(-(x/(-x + Log[10 + 5*x]))))*(2*x^2 + x^3 + (-4*x - 2* x^2)*Log[10 + 5*x] + (2 + x)*Log[10 + 5*x]^2 + (-x^2 + (2*x + x^2)*Log[10 + 5*x])/E^(x/(-x + Log[10 + 5*x]))))/(2*x^2 + x^3 + (-4*x - 2*x^2)*Log[10 + 5*x] + (2 + x)*Log[10 + 5*x]^2),x]
Output:
E^(7 - E^(-(x/(-x + Log[5*(2 + x)]))))*x
Leaf count is larger than twice the leaf count of optimal. \(121\) vs. \(2(24)=48\).
Time = 0.63 (sec) , antiderivative size = 121, normalized size of antiderivative = 5.04, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.007, Rules used = {2726}
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.
\(\displaystyle \int \frac {e^{7-e^{-\frac {x}{\log (5 x+10)-x}}} \left (x^3+2 x^2+\left (-2 x^2-4 x\right ) \log (5 x+10)+e^{-\frac {x}{\log (5 x+10)-x}} \left (\left (x^2+2 x\right ) \log (5 x+10)-x^2\right )+(x+2) \log ^2(5 x+10)\right )}{x^3+2 x^2+\left (-2 x^2-4 x\right ) \log (5 x+10)+(x+2) \log ^2(5 x+10)} \, dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle -\frac {e^{7-e^{\frac {x}{x-\log (5 x+10)}}} \left (x^2-\left (x^2+2 x\right ) \log (5 x+10)\right )}{\left (\frac {x \left (1-\frac {1}{x+2}\right )}{(x-\log (5 x+10))^2}-\frac {1}{x-\log (5 x+10)}\right ) \left (x^3+2 x^2-2 \left (x^2+2 x\right ) \log (5 x+10)+(x+2) \log ^2(5 x+10)\right )}\) |
Input:
Int[(E^(7 - E^(-(x/(-x + Log[10 + 5*x]))))*(2*x^2 + x^3 + (-4*x - 2*x^2)*L og[10 + 5*x] + (2 + x)*Log[10 + 5*x]^2 + (-x^2 + (2*x + x^2)*Log[10 + 5*x] )/E^(x/(-x + Log[10 + 5*x]))))/(2*x^2 + x^3 + (-4*x - 2*x^2)*Log[10 + 5*x] + (2 + x)*Log[10 + 5*x]^2),x]
Output:
-((E^(7 - E^(x/(x - Log[10 + 5*x])))*(x^2 - (2*x + x^2)*Log[10 + 5*x]))/(( (x*(1 - (2 + x)^(-1)))/(x - Log[10 + 5*x])^2 - (x - Log[10 + 5*x])^(-1))*( 2*x^2 + x^3 - 2*(2*x + x^2)*Log[10 + 5*x] + (2 + x)*Log[10 + 5*x]^2)))
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 2.70 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
risch | \(x \,{\mathrm e}^{-{\mathrm e}^{\frac {x}{x -\ln \left (5 x +10\right )}}+7}\) | \(23\) |
parallelrisch | \(x \,{\mathrm e}^{-{\mathrm e}^{-\frac {x}{\ln \left (5 x +10\right )-x}}+7}\) | \(24\) |
Input:
int((((x^2+2*x)*ln(5*x+10)-x^2)*exp(-x/(ln(5*x+10)-x))+(2+x)*ln(5*x+10)^2+ (-2*x^2-4*x)*ln(5*x+10)+x^3+2*x^2)/((2+x)*ln(5*x+10)^2+(-2*x^2-4*x)*ln(5*x +10)+x^3+2*x^2)/exp(exp(-x/(ln(5*x+10)-x))-7),x,method=_RETURNVERBOSE)
Output:
x*exp(-exp(x/(x-ln(5*x+10)))+7)
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=x e^{\left (-e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} + 7\right )} \] Input:
integrate((((x^2+2*x)*log(5*x+10)-x^2)*exp(-x/(log(5*x+10)-x))+(2+x)*log(5 *x+10)^2+(-2*x^2-4*x)*log(5*x+10)+x^3+2*x^2)/((2+x)*log(5*x+10)^2+(-2*x^2- 4*x)*log(5*x+10)+x^3+2*x^2)/exp(exp(-x/(log(5*x+10)-x))-7),x, algorithm="f ricas")
Output:
x*e^(-e^(x/(x - log(5*x + 10))) + 7)
Exception generated. \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((((x**2+2*x)*ln(5*x+10)-x**2)*exp(-x/(ln(5*x+10)-x))+(2+x)*ln(5* x+10)**2+(-2*x**2-4*x)*ln(5*x+10)+x**3+2*x**2)/((2+x)*ln(5*x+10)**2+(-2*x* *2-4*x)*ln(5*x+10)+x**3+2*x**2)/exp(exp(-x/(ln(5*x+10)-x))-7),x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
\[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=\int { \frac {{\left (x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - {\left (x^{2} - {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (-e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} + 7\right )}}{x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )} \,d x } \] Input:
integrate((((x^2+2*x)*log(5*x+10)-x^2)*exp(-x/(log(5*x+10)-x))+(2+x)*log(5 *x+10)^2+(-2*x^2-4*x)*log(5*x+10)+x^3+2*x^2)/((2+x)*log(5*x+10)^2+(-2*x^2- 4*x)*log(5*x+10)+x^3+2*x^2)/exp(exp(-x/(log(5*x+10)-x))-7),x, algorithm="m axima")
Output:
integrate((x^3 + (x + 2)*log(5*x + 10)^2 + 2*x^2 - (x^2 - (x^2 + 2*x)*log( 5*x + 10))*e^(x/(x - log(5*x + 10))) - 2*(x^2 + 2*x)*log(5*x + 10))*e^(-e^ (x/(x - log(5*x + 10))) + 7)/(x^3 + (x + 2)*log(5*x + 10)^2 + 2*x^2 - 2*(x ^2 + 2*x)*log(5*x + 10)), x)
\[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=\int { \frac {{\left (x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - {\left (x^{2} - {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )\right )} e^{\left (-e^{\left (\frac {x}{x - \log \left (5 \, x + 10\right )}\right )} + 7\right )}}{x^{3} + {\left (x + 2\right )} \log \left (5 \, x + 10\right )^{2} + 2 \, x^{2} - 2 \, {\left (x^{2} + 2 \, x\right )} \log \left (5 \, x + 10\right )} \,d x } \] Input:
integrate((((x^2+2*x)*log(5*x+10)-x^2)*exp(-x/(log(5*x+10)-x))+(2+x)*log(5 *x+10)^2+(-2*x^2-4*x)*log(5*x+10)+x^3+2*x^2)/((2+x)*log(5*x+10)^2+(-2*x^2- 4*x)*log(5*x+10)+x^3+2*x^2)/exp(exp(-x/(log(5*x+10)-x))-7),x, algorithm="g iac")
Output:
integrate((x^3 + (x + 2)*log(5*x + 10)^2 + 2*x^2 - (x^2 - (x^2 + 2*x)*log( 5*x + 10))*e^(x/(x - log(5*x + 10))) - 2*(x^2 + 2*x)*log(5*x + 10))*e^(-e^ (x/(x - log(5*x + 10))) + 7)/(x^3 + (x + 2)*log(5*x + 10)^2 + 2*x^2 - 2*(x ^2 + 2*x)*log(5*x + 10)), x)
Time = 3.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=x\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {x}{x-\ln \left (5\,x+10\right )}}}\,{\mathrm {e}}^7 \] Input:
int((exp(7 - exp(x/(x - log(5*x + 10))))*(log(5*x + 10)^2*(x + 2) - log(5* x + 10)*(4*x + 2*x^2) + exp(x/(x - log(5*x + 10)))*(log(5*x + 10)*(2*x + x ^2) - x^2) + 2*x^2 + x^3))/(log(5*x + 10)^2*(x + 2) - log(5*x + 10)*(4*x + 2*x^2) + 2*x^2 + x^3),x)
Output:
x*exp(-exp(x/(x - log(5*x + 10))))*exp(7)
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{7-e^{-\frac {x}{-x+\log (10+5 x)}}} \left (2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)+e^{-\frac {x}{-x+\log (10+5 x)}} \left (-x^2+\left (2 x+x^2\right ) \log (10+5 x)\right )\right )}{2 x^2+x^3+\left (-4 x-2 x^2\right ) \log (10+5 x)+(2+x) \log ^2(10+5 x)} \, dx=\frac {e^{7} x}{e^{\frac {1}{e^{\frac {x}{\mathrm {log}\left (5 x +10\right )-x}}}}} \] Input:
int((((x^2+2*x)*log(5*x+10)-x^2)*exp(-x/(log(5*x+10)-x))+(2+x)*log(5*x+10) ^2+(-2*x^2-4*x)*log(5*x+10)+x^3+2*x^2)/((2+x)*log(5*x+10)^2+(-2*x^2-4*x)*l og(5*x+10)+x^3+2*x^2)/exp(exp(-x/(log(5*x+10)-x))-7),x)
Output:
(e**7*x)/e**(1/e**(x/(log(5*x + 10) - x)))