Integrand size = 81, antiderivative size = 28 \[ \int e^{-e^{16}+e^{-e^{16}} \left (-4 x^2+e^{e^{16}} \left (-4 x+e^x x\right )+\left (x^2-x^3\right ) \log (5)\right )} \left (-8 x+e^{e^{16}} \left (-4+e^x (1+x)\right )+\left (2 x-3 x^2\right ) \log (5)\right ) \, dx=e^{x \left (-4+e^x+e^{-e^{16}} x (-4+(1-x) \log (5))\right )} \] Output:
exp(x*(exp(x)-4+x/exp(exp(16))*((1-x)*ln(5)-4)))
Time = 1.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int e^{-e^{16}+e^{-e^{16}} \left (-4 x^2+e^{e^{16}} \left (-4 x+e^x x\right )+\left (x^2-x^3\right ) \log (5)\right )} \left (-8 x+e^{e^{16}} \left (-4+e^x (1+x)\right )+\left (2 x-3 x^2\right ) \log (5)\right ) \, dx=5^{-e^{-e^{16}} (-1+x) x^2} e^{x \left (-4+e^x-4 e^{-e^{16}} x\right )} \] Input:
Integrate[E^(-E^16 + (-4*x^2 + E^E^16*(-4*x + E^x*x) + (x^2 - x^3)*Log[5]) /E^E^16)*(-8*x + E^E^16*(-4 + E^x*(1 + x)) + (2*x - 3*x^2)*Log[5]),x]
Output:
E^(x*(-4 + E^x - (4*x)/E^E^16))/5^(((-1 + x)*x^2)/E^E^16)
Time = 0.61 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {7257}
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.
\(\displaystyle \int \left (\left (2 x-3 x^2\right ) \log (5)-8 x+e^{e^{16}} \left (e^x (x+1)-4\right )\right ) \exp \left (e^{-e^{16}} \left (-4 x^2+\left (x^2-x^3\right ) \log (5)+e^{e^{16}} \left (e^x x-4 x\right )\right )-e^{16}\right ) \, dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle 5^{e^{-e^{16}} \left (x^2-x^3\right )} \exp \left (-e^{-e^{16}} \left (4 x^2+e^{e^{16}} \left (4 x-e^x x\right )\right )\right )\) |
Input:
Int[E^(-E^16 + (-4*x^2 + E^E^16*(-4*x + E^x*x) + (x^2 - x^3)*Log[5])/E^E^1 6)*(-8*x + E^E^16*(-4 + E^x*(1 + x)) + (2*x - 3*x^2)*Log[5]),x]
Output:
5^((x^2 - x^3)/E^E^16)/E^((4*x^2 + E^E^16*(4*x - E^x*x))/E^E^16)
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 0.46 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32
method | result | size |
risch | \({\mathrm e}^{-x \left (x^{2} \ln \left (5\right )-x \ln \left (5\right )+4 \,{\mathrm e}^{{\mathrm e}^{16}}-{\mathrm e}^{{\mathrm e}^{16}+x}+4 x \right ) {\mathrm e}^{-{\mathrm e}^{16}}}\) | \(37\) |
norman | \({\mathrm e}^{\left (\left ({\mathrm e}^{x} x -4 x \right ) {\mathrm e}^{{\mathrm e}^{16}}+\left (-x^{3}+x^{2}\right ) \ln \left (5\right )-4 x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{16}}}\) | \(38\) |
parallelrisch | \({\mathrm e}^{\left (\left ({\mathrm e}^{x} x -4 x \right ) {\mathrm e}^{{\mathrm e}^{16}}+\left (-x^{3}+x^{2}\right ) \ln \left (5\right )-4 x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{16}}}\) | \(38\) |
Input:
int((((1+x)*exp(x)-4)*exp(exp(16))+(-3*x^2+2*x)*ln(5)-8*x)*exp(((exp(x)*x- 4*x)*exp(exp(16))+(-x^3+x^2)*ln(5)-4*x^2)/exp(exp(16)))/exp(exp(16)),x,met hod=_RETURNVERBOSE)
Output:
exp(-x*(x^2*ln(5)-x*ln(5)+4*exp(exp(16))-exp(exp(16)+x)+4*x)*exp(-exp(16)) )
Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int e^{-e^{16}+e^{-e^{16}} \left (-4 x^2+e^{e^{16}} \left (-4 x+e^x x\right )+\left (x^2-x^3\right ) \log (5)\right )} \left (-8 x+e^{e^{16}} \left (-4+e^x (1+x)\right )+\left (2 x-3 x^2\right ) \log (5)\right ) \, dx=e^{\left (-{\left (4 \, x^{2} - {\left (x e^{x} - 4 \, x - e^{16}\right )} e^{\left (e^{16}\right )} + {\left (x^{3} - x^{2}\right )} \log \left (5\right )\right )} e^{\left (-e^{16}\right )} + e^{16}\right )} \] Input:
integrate((((1+x)*exp(x)-4)*exp(exp(16))+(-3*x^2+2*x)*log(5)-8*x)*exp(((ex p(x)*x-4*x)*exp(exp(16))+(-x^3+x^2)*log(5)-4*x^2)/exp(exp(16)))/exp(exp(16 )),x, algorithm="fricas")
Output:
e^(-(4*x^2 - (x*e^x - 4*x - e^16)*e^(e^16) + (x^3 - x^2)*log(5))*e^(-e^16) + e^16)
Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int e^{-e^{16}+e^{-e^{16}} \left (-4 x^2+e^{e^{16}} \left (-4 x+e^x x\right )+\left (x^2-x^3\right ) \log (5)\right )} \left (-8 x+e^{e^{16}} \left (-4+e^x (1+x)\right )+\left (2 x-3 x^2\right ) \log (5)\right ) \, dx=e^{\frac {- 4 x^{2} + \left (- x^{3} + x^{2}\right ) \log {\left (5 \right )} + \left (x e^{x} - 4 x\right ) e^{e^{16}}}{e^{e^{16}}}} \] Input:
integrate((((1+x)*exp(x)-4)*exp(exp(16))+(-3*x**2+2*x)*ln(5)-8*x)*exp(((ex p(x)*x-4*x)*exp(exp(16))+(-x**3+x**2)*ln(5)-4*x**2)/exp(exp(16)))/exp(exp( 16)),x)
Output:
exp((-4*x**2 + (-x**3 + x**2)*log(5) + (x*exp(x) - 4*x)*exp(exp(16)))*exp( -exp(16)))
Time = 0.30 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int e^{-e^{16}+e^{-e^{16}} \left (-4 x^2+e^{e^{16}} \left (-4 x+e^x x\right )+\left (x^2-x^3\right ) \log (5)\right )} \left (-8 x+e^{e^{16}} \left (-4+e^x (1+x)\right )+\left (2 x-3 x^2\right ) \log (5)\right ) \, dx=e^{\left (-x^{3} e^{\left (-e^{16}\right )} \log \left (5\right ) + x^{2} e^{\left (-e^{16}\right )} \log \left (5\right ) - 4 \, x^{2} e^{\left (-e^{16}\right )} + x e^{x} - 4 \, x\right )} \] Input:
integrate((((1+x)*exp(x)-4)*exp(exp(16))+(-3*x^2+2*x)*log(5)-8*x)*exp(((ex p(x)*x-4*x)*exp(exp(16))+(-x^3+x^2)*log(5)-4*x^2)/exp(exp(16)))/exp(exp(16 )),x, algorithm="maxima")
Output:
e^(-x^3*e^(-e^16)*log(5) + x^2*e^(-e^16)*log(5) - 4*x^2*e^(-e^16) + x*e^x - 4*x)
\[ \int e^{-e^{16}+e^{-e^{16}} \left (-4 x^2+e^{e^{16}} \left (-4 x+e^x x\right )+\left (x^2-x^3\right ) \log (5)\right )} \left (-8 x+e^{e^{16}} \left (-4+e^x (1+x)\right )+\left (2 x-3 x^2\right ) \log (5)\right ) \, dx=\int { {\left ({\left ({\left (x + 1\right )} e^{x} - 4\right )} e^{\left (e^{16}\right )} - {\left (3 \, x^{2} - 2 \, x\right )} \log \left (5\right ) - 8 \, x\right )} e^{\left (-{\left (4 \, x^{2} - {\left (x e^{x} - 4 \, x\right )} e^{\left (e^{16}\right )} + {\left (x^{3} - x^{2}\right )} \log \left (5\right )\right )} e^{\left (-e^{16}\right )} - e^{16}\right )} \,d x } \] Input:
integrate((((1+x)*exp(x)-4)*exp(exp(16))+(-3*x^2+2*x)*log(5)-8*x)*exp(((ex p(x)*x-4*x)*exp(exp(16))+(-x^3+x^2)*log(5)-4*x^2)/exp(exp(16)))/exp(exp(16 )),x, algorithm="giac")
Output:
integrate((((x + 1)*e^x - 4)*e^(e^16) - (3*x^2 - 2*x)*log(5) - 8*x)*e^(-(4 *x^2 - (x*e^x - 4*x)*e^(e^16) + (x^3 - x^2)*log(5))*e^(-e^16) - e^16), x)
Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int e^{-e^{16}+e^{-e^{16}} \left (-4 x^2+e^{e^{16}} \left (-4 x+e^x x\right )+\left (x^2-x^3\right ) \log (5)\right )} \left (-8 x+e^{e^{16}} \left (-4+e^x (1+x)\right )+\left (2 x-3 x^2\right ) \log (5)\right ) \, dx=5^{{\mathrm {e}}^{-{\mathrm {e}}^{16}}\,\left (x^2-x^3\right )}\,{\mathrm {e}}^{x\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-4\,x^2\,{\mathrm {e}}^{-{\mathrm {e}}^{16}}} \] Input:
int(exp(-exp(16))*exp(-exp(-exp(16))*(exp(exp(16))*(4*x - x*exp(x)) - log( 5)*(x^2 - x^3) + 4*x^2))*(log(5)*(2*x - 3*x^2) - 8*x + exp(exp(16))*(exp(x )*(x + 1) - 4)),x)
Output:
5^(exp(-exp(16))*(x^2 - x^3))*exp(x*exp(x))*exp(-4*x)*exp(-4*x^2*exp(-exp( 16)))
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.11 \[ \int e^{-e^{16}+e^{-e^{16}} \left (-4 x^2+e^{e^{16}} \left (-4 x+e^x x\right )+\left (x^2-x^3\right ) \log (5)\right )} \left (-8 x+e^{e^{16}} \left (-4+e^x (1+x)\right )+\left (2 x-3 x^2\right ) \log (5)\right ) \, dx=\frac {e^{\frac {e^{e^{16}+x} x +\mathrm {log}\left (5\right ) x^{2}}{e^{e^{16}}}}}{e^{\frac {4 e^{e^{16}} x +\mathrm {log}\left (5\right ) x^{3}+4 x^{2}}{e^{e^{16}}}}} \] Input:
int((((1+x)*exp(x)-4)*exp(exp(16))+(-3*x^2+2*x)*log(5)-8*x)*exp(((exp(x)*x -4*x)*exp(exp(16))+(-x^3+x^2)*log(5)-4*x^2)/exp(exp(16)))/exp(exp(16)),x)
Output:
e**((e**(e**16 + x)*x + log(5)*x**2)/e**(e**16))/e**((4*e**(e**16)*x + log (5)*x**3 + 4*x**2)/e**(e**16))