Integrand size = 43, antiderivative size = 19 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{-2-\frac {1}{5} e^3 x \left (e^{43/16}+x\right )} \] Output:
exp(-2-1/15*exp(3)*(3*x+3*exp(43/16))*x)
Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{-2-\frac {1}{5} e^{91/16} x-\frac {e^3 x^2}{5}} \] Input:
Integrate[(E^(3 + (-10 + E^3*(-(E^(43/16)*x) - x^2))/5)*(-E^(43/16) - 2*x) )/5,x]
Output:
E^(-2 - (E^(91/16)*x)/5 - (E^3*x^2)/5)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {27, 25, 2674, 2666}
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.
\(\displaystyle \int \frac {1}{5} e^{\frac {1}{5} \left (e^3 \left (-x^2-e^{43/16} x\right )-10\right )+3} \left (-2 x-e^{43/16}\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -e^{\frac {1}{5} \left (-e^3 \left (x^2+e^{43/16} x\right )-10\right )+3} \left (2 x+e^{43/16}\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int e^{\frac {1}{5} \left (-e^3 \left (x^2+e^{43/16} x\right )-10\right )+3} \left (2 x+e^{43/16}\right )dx\) |
\(\Big \downarrow \) 2674 |
\(\displaystyle -\frac {1}{5} \int e^{-\frac {1}{5} e^3 x^2-\frac {1}{5} e^{91/16} x+1} \left (2 x+e^{43/16}\right )dx\) |
\(\Big \downarrow \) 2666 |
\(\displaystyle e^{-\frac {1}{5} e^3 x^2-\frac {1}{5} e^{91/16} x-2}\) |
Input:
Int[(E^(3 + (-10 + E^3*(-(E^(43/16)*x) - x^2))/5)*(-E^(43/16) - 2*x))/5,x]
Output:
E^(-2 - (E^(91/16)*x)/5 - (E^3*x^2)/5)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol ] :> Simp[e*(F^(a + b*x + c*x^2)/(2*c*Log[F])), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]
Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSu m[v, x], x] /; FreeQ[{F, m}, x] && LinearQ[u, x] && QuadraticQ[v, x] && !( LinearMatchQ[u, x] && QuadraticMatchQ[v, x])
Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84
method | result | size |
risch | \({\mathrm e}^{-\frac {x \,{\mathrm e}^{\frac {91}{16}}}{5}-\frac {x^{2} {\mathrm e}^{3}}{5}-2}\) | \(16\) |
gosper | \({\mathrm e}^{-\frac {{\mathrm e}^{\frac {43}{16}} {\mathrm e}^{3} x}{5}-\frac {x^{2} {\mathrm e}^{3}}{5}-2}\) | \(18\) |
derivativedivides | \({\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}\) | \(19\) |
default | \({\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}\) | \(19\) |
norman | \({\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}\) | \(19\) |
parallelrisch | \({\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}\) | \(19\) |
orering | \(-\frac {\left (-{\mathrm e}^{\frac {43}{16}}-2 x \right ) {\mathrm e}^{\frac {\left (-x \,{\mathrm e}^{\frac {43}{16}}-x^{2}\right ) {\mathrm e}^{3}}{5}-2}}{{\mathrm e}^{\frac {43}{16}}+2 x}\) | \(37\) |
parts | \(-\frac {\sqrt {\pi }\, {\mathrm e}^{-2+\frac {{\mathrm e}^{\frac {43}{8}} {\mathrm e}^{3}}{20}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}} \operatorname {erf}\left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right ) {\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}}}{10}-\frac {\sqrt {\pi }\, {\mathrm e}^{-2+\frac {{\mathrm e}^{\frac {43}{8}} {\mathrm e}^{3}}{20}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}} \operatorname {erf}\left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right ) {\mathrm e}^{3} x}{5}+{\mathrm e}^{-2+\frac {{\mathrm e}^{\frac {43}{8}} {\mathrm e}^{3}}{20}} {\mathrm e}^{-3} {\mathrm e}^{3} \left (\left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right ) \operatorname {erf}\left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right ) \sqrt {\pi }+{\mathrm e}^{-\left (\frac {\sqrt {5}\, {\mathrm e}^{\frac {3}{2}} x}{5}+\frac {{\mathrm e}^{3} {\mathrm e}^{\frac {43}{16}} \sqrt {5}\, {\mathrm e}^{-\frac {3}{2}}}{10}\right )^{2}}\right )\) | \(196\) |
Input:
int(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x^2)*exp(3)-2),x,m ethod=_RETURNVERBOSE)
Output:
exp(-1/5*x*exp(91/16)-1/5*x^2*exp(3)-2)
Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{\left (-\frac {1}{5} \, x^{2} e^{3} - \frac {1}{5} \, x e^{\frac {91}{16}} - 2\right )} \] Input:
integrate(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x^2)*exp(3)- 2),x, algorithm="fricas")
Output:
e^(-1/5*x^2*e^3 - 1/5*x*e^(91/16) - 2)
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{\left (- \frac {x^{2}}{5} - \frac {x e^{\frac {43}{16}}}{5}\right ) e^{3} - 2} \] Input:
integrate(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x**2)*exp(3) -2),x)
Output:
exp((-x**2/5 - x*exp(43/16)/5)*exp(3) - 2)
Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{\left (-\frac {1}{5} \, {\left (x^{2} + x e^{\frac {43}{16}}\right )} e^{3} - 2\right )} \] Input:
integrate(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x^2)*exp(3)- 2),x, algorithm="maxima")
Output:
e^(-1/5*(x^2 + x*e^(43/16))*e^3 - 2)
Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=e^{\left (-\frac {1}{5} \, x^{2} e^{3} - \frac {1}{5} \, x e^{\frac {91}{16}} - 2\right )} \] Input:
integrate(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x^2)*exp(3)- 2),x, algorithm="giac")
Output:
e^(-1/5*x^2*e^3 - 1/5*x*e^(91/16) - 2)
Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx={\mathrm {e}}^{-\frac {x^2\,{\mathrm {e}}^3}{5}}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{91/16}}{5}} \] Input:
int(-(exp(3)*exp(- (exp(3)*(x*exp(43/16) + x^2))/5 - 2)*(2*x + exp(43/16)) )/5,x)
Output:
exp(-(x^2*exp(3))/5)*exp(-2)*exp(-(x*exp(91/16))/5)
Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {1}{5} e^{3+\frac {1}{5} \left (-10+e^3 \left (-e^{43/16} x-x^2\right )\right )} \left (-e^{43/16}-2 x\right ) \, dx=\frac {1}{e^{\frac {e^{\frac {91}{16}} x}{5}+\frac {e^{3} x^{2}}{5}} e^{2}} \] Input:
int(1/5*(-exp(43/16)-2*x)*exp(3)*exp(1/5*(-x*exp(43/16)-x^2)*exp(3)-2),x)
Output:
1/(e**((e**(11/16)*e**5*x + e**3*x**2)/5)*e**2)