\(\int \frac {1}{5} e^{\frac {1}{5} (60+15 x+e^{-2+x} x+100 x^2)} (15+200 x+e^{-2+x} (1+x)) \, dx\) [5094]

Optimal result

Integrand size = 41, antiderivative size = 29 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{2-x+\left (4+\frac {e^{-2+x}}{5}\right ) x+5 \left (2+4 x^2\right )} \]

[Out]

exp(12+x*(1/5*exp(-2+x)+4)-x+20*x^2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {12, 6838} \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{\frac {1}{5} \left (100 x^2+e^{x-2} x+15 x+60\right )} \]

[In]

Int[(E^((60 + 15*x + E^(-2 + x)*x + 100*x^2)/5)*(15 + 200*x + E^(-2 + x)*(1 + x)))/5,x]

[Out]

E^((60 + 15*x + E^(-2 + x)*x + 100*x^2)/5)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx \\ & = e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{12+3 x+\frac {1}{5} e^{-2+x} x+20 x^2} \]

[In]

Integrate[(E^((60 + 15*x + E^(-2 + x)*x + 100*x^2)/5)*(15 + 200*x + E^(-2 + x)*(1 + x)))/5,x]

[Out]

E^(12 + 3*x + (E^(-2 + x)*x)/5 + 20*x^2)

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66

method	result	size
norman	\({\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{5}+20 x^{2}+3 x +12}\)	\(19\)
risch	\({\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{5}+20 x^{2}+3 x +12}\)	\(19\)
parallelrisch	\({\mathrm e}^{\frac {x \,{\mathrm e}^{-2+x}}{5}+20 x^{2}+3 x +12}\)	\(19\)

[In]

int(1/5*((1+x)*exp(-2+x)+200*x+15)*exp(1/5*x*exp(-2+x)+20*x^2+3*x+12),x,method=_RETURNVERBOSE)

[Out]

exp(1/5*x*exp(-2+x)+20*x^2+3*x+12)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{\left (20 \, x^{2} + \frac {1}{5} \, x e^{\left (x - 2\right )} + 3 \, x + 12\right )} \]

[In]

integrate(1/5*((1+x)*exp(-2+x)+200*x+15)*exp(1/5*x*exp(-2+x)+20*x^2+3*x+12),x, algorithm="fricas")

[Out]

e^(20*x^2 + 1/5*x*e^(x - 2) + 3*x + 12)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{20 x^{2} + \frac {x e^{x - 2}}{5} + 3 x + 12} \]

[In]

integrate(1/5*((1+x)*exp(-2+x)+200*x+15)*exp(1/5*x*exp(-2+x)+20*x**2+3*x+12),x)

[Out]

exp(20*x**2 + x*exp(x - 2)/5 + 3*x + 12)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{\left (20 \, x^{2} + \frac {1}{5} \, x e^{\left (x - 2\right )} + 3 \, x + 12\right )} \]

[In]

integrate(1/5*((1+x)*exp(-2+x)+200*x+15)*exp(1/5*x*exp(-2+x)+20*x^2+3*x+12),x, algorithm="maxima")

[Out]

e^(20*x^2 + 1/5*x*e^(x - 2) + 3*x + 12)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx=e^{\left (20 \, x^{2} + \frac {1}{5} \, x e^{\left (x - 2\right )} + 3 \, x + 12\right )} \]

[In]

integrate(1/5*((1+x)*exp(-2+x)+200*x+15)*exp(1/5*x*exp(-2+x)+20*x^2+3*x+12),x, algorithm="giac")

[Out]

e^(20*x^2 + 1/5*x*e^(x - 2) + 3*x + 12)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (60+15 x+e^{-2+x} x+100 x^2\right )} \left (15+200 x+e^{-2+x} (1+x)\right ) \, dx={\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{12}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^{20\,x^2} \]

[In]

int((exp(3*x + (x*exp(x - 2))/5 + 20*x^2 + 12)*(200*x + exp(x - 2)*(x + 1) + 15))/5,x)

[Out]

exp(3*x)*exp(12)*exp((x*exp(-2)*exp(x))/5)*exp(20*x^2)