Integrand size = 105, antiderivative size = 25 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=-1+\left (e^{2+2 x}-\log \left (-2-e^{5 x}+x\right )\right )^2 \] Output:
(exp(1+x)^2-ln(-exp(5*x)+x-2))^2-1
Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=\left (e^{2+2 x}-\log \left (-2-e^{5 x}+x\right )\right )^2 \] Input:
Integrate[(2*E^(2 + 2*x) + E^(5*x)*(-10*E^(2 + 2*x) + 4*E^(4 + 4*x)) + E^( 4 + 4*x)*(8 - 4*x) + (-2 + E^(5*x)*(10 - 4*E^(2 + 2*x)) + E^(2 + 2*x)*(-8 + 4*x))*Log[-2 - E^(5*x) + x])/(2 + E^(5*x) - x),x]
Output:
(E^(2 + 2*x) - Log[-2 - E^(5*x) + x])^2
Time = 0.95 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {7239, 27, 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.
\(\displaystyle \int \frac {e^{5 x} \left (4 e^{4 x+4}-10 e^{2 x+2}\right )+2 e^{2 x+2}+e^{4 x+4} (8-4 x)+\left (e^{5 x} \left (10-4 e^{2 x+2}\right )+e^{2 x+2} (4 x-8)-2\right ) \log \left (x-e^{5 x}-2\right )}{-x+e^{5 x}+2} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (-2 e^{2 x+2} (x-2)-5 e^{5 x}+2 e^{7 x+2}+1\right ) \left (e^{2 x+2}-\log \left (x-e^{5 x}-2\right )\right )}{-x+e^{5 x}+2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {\left (2 e^{2 x+2} (2-x)-5 e^{5 x}+2 e^{7 x+2}+1\right ) \left (e^{2 x+2}-\log \left (x-e^{5 x}-2\right )\right )}{-x+e^{5 x}+2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \left (e^{2 x+2}-\log \left (x-e^{5 x}-2\right )\right )^2\) |
Input:
Int[(2*E^(2 + 2*x) + E^(5*x)*(-10*E^(2 + 2*x) + 4*E^(4 + 4*x)) + E^(4 + 4* x)*(8 - 4*x) + (-2 + E^(5*x)*(10 - 4*E^(2 + 2*x)) + E^(2 + 2*x)*(-8 + 4*x) )*Log[-2 - E^(5*x) + x])/(2 + E^(5*x) - x),x]
Output:
(E^(2 + 2*x) - Log[-2 - E^(5*x) + x])^2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.50 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52
method | result | size |
risch | \({\mathrm e}^{4+4 x}-2 \,{\mathrm e}^{2+2 x} \ln \left (-{\mathrm e}^{5 x}+x -2\right )+\ln \left (-{\mathrm e}^{5 x}+x -2\right )^{2}\) | \(38\) |
parallelrisch | \({\mathrm e}^{4+4 x}-2 \,{\mathrm e}^{2+2 x} \ln \left (-{\mathrm e}^{5 x}+x -2\right )+\ln \left (-{\mathrm e}^{5 x}+x -2\right )^{2}\) | \(38\) |
Input:
int((((-4*exp(1+x)^2+10)*exp(5*x)+(4*x-8)*exp(1+x)^2-2)*ln(-exp(5*x)+x-2)+ (4*exp(1+x)^4-10*exp(1+x)^2)*exp(5*x)+(-4*x+8)*exp(1+x)^4+2*exp(1+x)^2)/(e xp(5*x)+2-x),x,method=_RETURNVERBOSE)
Output:
exp(4+4*x)-2*exp(2+2*x)*ln(-exp(5*x)+x-2)+ln(-exp(5*x)+x-2)^2
Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (23) = 46\).
Time = 0.09 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=-2 \, e^{\left (2 \, x + 2\right )} \log \left ({\left ({\left (x - 2\right )} e^{5} - e^{\left (5 \, x + 5\right )}\right )} e^{\left (-5\right )}\right ) + \log \left ({\left ({\left (x - 2\right )} e^{5} - e^{\left (5 \, x + 5\right )}\right )} e^{\left (-5\right )}\right )^{2} + e^{\left (4 \, x + 4\right )} \] Input:
integrate((((-4*exp(1+x)^2+10)*exp(5*x)+(4*x-8)*exp(1+x)^2-2)*log(-exp(5*x )+x-2)+(4*exp(1+x)^4-10*exp(1+x)^2)*exp(5*x)+(-4*x+8)*exp(1+x)^4+2*exp(1+x )^2)/(exp(5*x)+2-x),x, algorithm="fricas")
Output:
-2*e^(2*x + 2)*log(((x - 2)*e^5 - e^(5*x + 5))*e^(-5)) + log(((x - 2)*e^5 - e^(5*x + 5))*e^(-5))^2 + e^(4*x + 4)
Timed out. \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=\text {Timed out} \] Input:
integrate((((-4*exp(1+x)**2+10)*exp(5*x)+(4*x-8)*exp(1+x)**2-2)*ln(-exp(5* x)+x-2)+(4*exp(1+x)**4-10*exp(1+x)**2)*exp(5*x)+(-4*x+8)*exp(1+x)**4+2*exp (1+x)**2)/(exp(5*x)+2-x),x)
Output:
Timed out
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=-2 \, e^{\left (2 \, x + 2\right )} \log \left (x - e^{\left (5 \, x\right )} - 2\right ) + \log \left (x - e^{\left (5 \, x\right )} - 2\right )^{2} + e^{\left (4 \, x + 4\right )} \] Input:
integrate((((-4*exp(1+x)^2+10)*exp(5*x)+(4*x-8)*exp(1+x)^2-2)*log(-exp(5*x )+x-2)+(4*exp(1+x)^4-10*exp(1+x)^2)*exp(5*x)+(-4*x+8)*exp(1+x)^4+2*exp(1+x )^2)/(exp(5*x)+2-x),x, algorithm="maxima")
Output:
-2*e^(2*x + 2)*log(x - e^(5*x) - 2) + log(x - e^(5*x) - 2)^2 + e^(4*x + 4)
\[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=\int { -\frac {2 \, {\left ({\left (2 \, e^{\left (4 \, x + 4\right )} - 5 \, e^{\left (2 \, x + 2\right )}\right )} e^{\left (5 \, x\right )} - 2 \, {\left (x - 2\right )} e^{\left (4 \, x + 4\right )} - {\left ({\left (2 \, e^{\left (2 \, x + 2\right )} - 5\right )} e^{\left (5 \, x\right )} - 2 \, {\left (x - 2\right )} e^{\left (2 \, x + 2\right )} + 1\right )} \log \left (x - e^{\left (5 \, x\right )} - 2\right ) + e^{\left (2 \, x + 2\right )}\right )}}{x - e^{\left (5 \, x\right )} - 2} \,d x } \] Input:
integrate((((-4*exp(1+x)^2+10)*exp(5*x)+(4*x-8)*exp(1+x)^2-2)*log(-exp(5*x )+x-2)+(4*exp(1+x)^4-10*exp(1+x)^2)*exp(5*x)+(-4*x+8)*exp(1+x)^4+2*exp(1+x )^2)/(exp(5*x)+2-x),x, algorithm="giac")
Output:
integrate(-2*((2*e^(4*x + 4) - 5*e^(2*x + 2))*e^(5*x) - 2*(x - 2)*e^(4*x + 4) - ((2*e^(2*x + 2) - 5)*e^(5*x) - 2*(x - 2)*e^(2*x + 2) + 1)*log(x - e^ (5*x) - 2) + e^(2*x + 2))/(x - e^(5*x) - 2), x)
Time = 2.72 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx={\ln \left (x-{\mathrm {e}}^{5\,x}-2\right )}^2-2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^2\,\ln \left (x-{\mathrm {e}}^{5\,x}-2\right )+{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^4 \] Input:
int(-(exp(4*x + 4)*(4*x - 8) - 2*exp(2*x + 2) + log(x - exp(5*x) - 2)*(exp (5*x)*(4*exp(2*x + 2) - 10) - exp(2*x + 2)*(4*x - 8) + 2) + exp(5*x)*(10*e xp(2*x + 2) - 4*exp(4*x + 4)))/(exp(5*x) - x + 2),x)
Output:
exp(4*x)*exp(4) + log(x - exp(5*x) - 2)^2 - 2*exp(2*x)*exp(2)*log(x - exp( 5*x) - 2)
Time = 0.24 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {2 e^{2+2 x}+e^{5 x} \left (-10 e^{2+2 x}+4 e^{4+4 x}\right )+e^{4+4 x} (8-4 x)+\left (-2+e^{5 x} \left (10-4 e^{2+2 x}\right )+e^{2+2 x} (-8+4 x)\right ) \log \left (-2-e^{5 x}+x\right )}{2+e^{5 x}-x} \, dx=e^{4 x} e^{4}-2 e^{2 x} \mathrm {log}\left (-e^{5 x}+x -2\right ) e^{2}+\mathrm {log}\left (-e^{5 x}+x -2\right )^{2} \] Input:
int((((-4*exp(1+x)^2+10)*exp(5*x)+(4*x-8)*exp(1+x)^2-2)*log(-exp(5*x)+x-2) +(4*exp(1+x)^4-10*exp(1+x)^2)*exp(5*x)+(-4*x+8)*exp(1+x)^4+2*exp(1+x)^2)/( exp(5*x)+2-x),x)
Output:
e**(4*x)*e**4 - 2*e**(2*x)*log( - e**(5*x) + x - 2)*e**2 + log( - e**(5*x) + x - 2)**2