Integrand size = 99, antiderivative size = 28 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{x+2 \left (x+2 \left (e^{8 x-x \log (x)}-x\right )^2 x\right )} \] Output:
exp(3*x+4*(exp(-x*ln(x)+8*x)-x)^2*x)
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{3 x+4 x^3+4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}} \] Input:
Integrate[E^(3*x + 4*E^(16*x - 2*x*Log[x])*x - 8*E^(8*x - x*Log[x])*x^2 + 4*x^3)*(3 + 12*x^2 + E^(16*x - 2*x*Log[x])*(4 + 56*x - 8*x*Log[x]) + E^(8* x - x*Log[x])*(-16*x - 56*x^2 + 8*x^2*Log[x])),x]
Output:
E^(3*x + 4*x^3 + 4*E^(16*x)*x^(1 - 2*x) - 8*E^(8*x)*x^(2 - x))
Time = 1.83 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, 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.
| \(\displaystyle \int \left (12 x^2+e^{8 x-x \log (x)} \left (-56 x^2+8 x^2 \log (x)-16 x\right )+e^{16 x-2 x \log (x)} (56 x-8 x \log (x)+4)+3\right ) \exp \left (4 x^3-8 x^2 e^{8 x-x \log (x)}+3 x+4 x e^{16 x-2 x \log (x)}\right ) \, dx\) |
| \(\Big \downarrow \) 7257 |
| \(\displaystyle \exp \left (4 e^{16 x} x^{1-2 x}-8 e^{8 x} x^{2-x}+4 x^3+3 x\right )\) |
Input:
Int[E^(3*x + 4*E^(16*x - 2*x*Log[x])*x - 8*E^(8*x - x*Log[x])*x^2 + 4*x^3) *(3 + 12*x^2 + E^(16*x - 2*x*Log[x])*(4 + 56*x - 8*x*Log[x]) + E^(8*x - x* Log[x])*(-16*x - 56*x^2 + 8*x^2*Log[x])),x]
Output:
E^(3*x + 4*x^3 + 4*E^(16*x)*x^(1 - 2*x) - 8*E^(8*x)*x^(2 - x))
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 = 1.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29
| method | result | size |
| risch | \({\mathrm e}^{x \left (-8 x^{-x} {\mathrm e}^{8 x} x +4 x^{-2 x} {\mathrm e}^{16 x}+4 x^{2}+3\right )}\) | \(36\) |
| parallelrisch | \({\mathrm e}^{4 x \,{\mathrm e}^{-2 x \ln \left (x \right )+16 x}-8 x^{2} {\mathrm e}^{-x \ln \left (x \right )+8 x}+4 x^{3}+3 x}\) | \(41\) |
Input:
int(((-8*x*ln(x)+56*x+4)*exp(-x*ln(x)+8*x)^2+(8*x^2*ln(x)-56*x^2-16*x)*exp (-x*ln(x)+8*x)+12*x^2+3)*exp(4*x*exp(-x*ln(x)+8*x)^2-8*x^2*exp(-x*ln(x)+8* x)+4*x^3+3*x),x,method=_RETURNVERBOSE)
Output:
exp(x*(4*(x^(-x))^2*exp(16*x)-8*x^(-x)*exp(8*x)*x+4*x^2+3))
Time = 0.11 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3 \, x\right )} \] Input:
integrate(((-8*x*log(x)+56*x+4)*exp(-x*log(x)+8*x)^2+(8*x^2*log(x)-56*x^2- 16*x)*exp(-x*log(x)+8*x)+12*x^2+3)*exp(4*x*exp(-x*log(x)+8*x)^2-8*x^2*exp( -x*log(x)+8*x)+4*x^3+3*x),x, algorithm="fricas")
Output:
e^(4*x^3 - 8*x^2*e^(-x*log(x) + 8*x) + 4*x*e^(-2*x*log(x) + 16*x) + 3*x)
Time = 0.43 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=e^{4 x^{3} - 8 x^{2} e^{- x \log {\left (x \right )} + 8 x} + 4 x e^{- 2 x \log {\left (x \right )} + 16 x} + 3 x} \] Input:
integrate(((-8*x*ln(x)+56*x+4)*exp(-x*ln(x)+8*x)**2+(8*x**2*ln(x)-56*x**2- 16*x)*exp(-x*ln(x)+8*x)+12*x**2+3)*exp(4*x*exp(-x*ln(x)+8*x)**2-8*x**2*exp (-x*ln(x)+8*x)+4*x**3+3*x),x)
Output:
exp(4*x**3 - 8*x**2*exp(-x*log(x) + 8*x) + 4*x*exp(-2*x*log(x) + 16*x) + 3 *x)
\[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=\int { {\left (12 \, x^{2} + 8 \, {\left (x^{2} \log \left (x\right ) - 7 \, x^{2} - 2 \, x\right )} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} - 4 \, {\left (2 \, x \log \left (x\right ) - 14 \, x - 1\right )} e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3\right )} e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3 \, x\right )} \,d x } \] Input:
integrate(((-8*x*log(x)+56*x+4)*exp(-x*log(x)+8*x)^2+(8*x^2*log(x)-56*x^2- 16*x)*exp(-x*log(x)+8*x)+12*x^2+3)*exp(4*x*exp(-x*log(x)+8*x)^2-8*x^2*exp( -x*log(x)+8*x)+4*x^3+3*x),x, algorithm="maxima")
Output:
integrate((12*x^2 + 8*(x^2*log(x) - 7*x^2 - 2*x)*e^(-x*log(x) + 8*x) - 4*( 2*x*log(x) - 14*x - 1)*e^(-2*x*log(x) + 16*x) + 3)*e^(4*x^3 - 8*x^2*e^(-x* log(x) + 8*x) + 4*x*e^(-2*x*log(x) + 16*x) + 3*x), x)
\[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=\int { {\left (12 \, x^{2} + 8 \, {\left (x^{2} \log \left (x\right ) - 7 \, x^{2} - 2 \, x\right )} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} - 4 \, {\left (2 \, x \log \left (x\right ) - 14 \, x - 1\right )} e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3\right )} e^{\left (4 \, x^{3} - 8 \, x^{2} e^{\left (-x \log \left (x\right ) + 8 \, x\right )} + 4 \, x e^{\left (-2 \, x \log \left (x\right ) + 16 \, x\right )} + 3 \, x\right )} \,d x } \] Input:
integrate(((-8*x*log(x)+56*x+4)*exp(-x*log(x)+8*x)^2+(8*x^2*log(x)-56*x^2- 16*x)*exp(-x*log(x)+8*x)+12*x^2+3)*exp(4*x*exp(-x*log(x)+8*x)^2-8*x^2*exp( -x*log(x)+8*x)+4*x^3+3*x),x, algorithm="giac")
Output:
integrate((12*x^2 + 8*(x^2*log(x) - 7*x^2 - 2*x)*e^(-x*log(x) + 8*x) - 4*( 2*x*log(x) - 14*x - 1)*e^(-2*x*log(x) + 16*x) + 3)*e^(4*x^3 - 8*x^2*e^(-x* log(x) + 8*x) + 4*x*e^(-2*x*log(x) + 16*x) + 3*x), x)
Time = 3.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx={\mathrm {e}}^{-8\,x^{2-x}\,{\mathrm {e}}^{8\,x}}\,{\mathrm {e}}^{4\,x^{1-2\,x}\,{\mathrm {e}}^{16\,x}}\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{4\,x^3} \] Input:
int(exp(3*x + 4*x*exp(16*x - 2*x*log(x)) - 8*x^2*exp(8*x - x*log(x)) + 4*x ^3)*(exp(16*x - 2*x*log(x))*(56*x - 8*x*log(x) + 4) - exp(8*x - x*log(x))* (16*x - 8*x^2*log(x) + 56*x^2) + 12*x^2 + 3),x)
Output:
exp(-8*x^(2 - x)*exp(8*x))*exp(4*x^(1 - 2*x)*exp(16*x))*exp(3*x)*exp(4*x^3 )
\[ \int e^{3 x+4 e^{16 x-2 x \log (x)} x-8 e^{8 x-x \log (x)} x^2+4 x^3} \left (3+12 x^2+e^{16 x-2 x \log (x)} (4+56 x-8 x \log (x))+e^{8 x-x \log (x)} \left (-16 x-56 x^2+8 x^2 \log (x)\right )\right ) \, dx=4 \left (\int \frac {e^{\frac {4 e^{16 x} x +4 x^{2 x} x^{3}+19 x^{2 x} x}{x^{2 x}}}}{x^{2 x} e^{\frac {8 e^{8 x} x^{2}}{x^{x}}}}d x \right )+3 \left (\int \frac {e^{\frac {4 e^{16 x} x +4 x^{2 x} x^{3}+3 x^{2 x} x}{x^{2 x}}}}{e^{\frac {8 e^{8 x} x^{2}}{x^{x}}}}d x \right )-8 \left (\int \frac {e^{\frac {4 e^{16 x} x +4 x^{2 x} x^{3}+19 x^{2 x} x}{x^{2 x}}} \mathrm {log}\left (x \right ) x}{x^{2 x} e^{\frac {8 e^{8 x} x^{2}}{x^{x}}}}d x \right )+56 \left (\int \frac {e^{\frac {4 e^{16 x} x +4 x^{2 x} x^{3}+19 x^{2 x} x}{x^{2 x}}} x}{x^{2 x} e^{\frac {8 e^{8 x} x^{2}}{x^{x}}}}d x \right )-56 \left (\int \frac {e^{\frac {4 e^{16 x} x +4 x^{2 x} x^{3}+11 x^{2 x} x}{x^{2 x}}} x^{2}}{x^{x} e^{\frac {8 e^{8 x} x^{2}}{x^{x}}}}d x \right )+8 \left (\int \frac {e^{\frac {4 e^{16 x} x +4 x^{2 x} x^{3}+11 x^{2 x} x}{x^{2 x}}} \mathrm {log}\left (x \right ) x^{2}}{x^{x} e^{\frac {8 e^{8 x} x^{2}}{x^{x}}}}d x \right )-16 \left (\int \frac {e^{\frac {4 e^{16 x} x +4 x^{2 x} x^{3}+11 x^{2 x} x}{x^{2 x}}} x}{x^{x} e^{\frac {8 e^{8 x} x^{2}}{x^{x}}}}d x \right )+12 \left (\int \frac {e^{\frac {4 e^{16 x} x +4 x^{2 x} x^{3}+3 x^{2 x} x}{x^{2 x}}} x^{2}}{e^{\frac {8 e^{8 x} x^{2}}{x^{x}}}}d x \right ) \] Input:
int(((-8*x*log(x)+56*x+4)*exp(-x*log(x)+8*x)^2+(8*x^2*log(x)-56*x^2-16*x)* exp(-x*log(x)+8*x)+12*x^2+3)*exp(4*x*exp(-x*log(x)+8*x)^2-8*x^2*exp(-x*log (x)+8*x)+4*x^3+3*x),x)
Output:
4*int(e**((4*e**(16*x)*x + 4*x**(2*x)*x**3 + 19*x**(2*x)*x)/x**(2*x))/(x** (2*x)*e**((8*e**(8*x)*x**2)/x**x)),x) + 3*int(e**((4*e**(16*x)*x + 4*x**(2 *x)*x**3 + 3*x**(2*x)*x)/x**(2*x))/e**((8*e**(8*x)*x**2)/x**x),x) - 8*int( (e**((4*e**(16*x)*x + 4*x**(2*x)*x**3 + 19*x**(2*x)*x)/x**(2*x))*log(x)*x) /(x**(2*x)*e**((8*e**(8*x)*x**2)/x**x)),x) + 56*int((e**((4*e**(16*x)*x + 4*x**(2*x)*x**3 + 19*x**(2*x)*x)/x**(2*x))*x)/(x**(2*x)*e**((8*e**(8*x)*x* *2)/x**x)),x) - 56*int((e**((4*e**(16*x)*x + 4*x**(2*x)*x**3 + 11*x**(2*x) *x)/x**(2*x))*x**2)/(x**x*e**((8*e**(8*x)*x**2)/x**x)),x) + 8*int((e**((4* e**(16*x)*x + 4*x**(2*x)*x**3 + 11*x**(2*x)*x)/x**(2*x))*log(x)*x**2)/(x** x*e**((8*e**(8*x)*x**2)/x**x)),x) - 16*int((e**((4*e**(16*x)*x + 4*x**(2*x )*x**3 + 11*x**(2*x)*x)/x**(2*x))*x)/(x**x*e**((8*e**(8*x)*x**2)/x**x)),x) + 12*int((e**((4*e**(16*x)*x + 4*x**(2*x)*x**3 + 3*x**(2*x)*x)/x**(2*x))* x**2)/e**((8*e**(8*x)*x**2)/x**x),x)