\(\int (e^x+e^{240-240 x-60 x^2+60 x^3+15 x^4} (768+384 x-576 x^2-192 x^3)) \, dx\) [8360]

Optimal result

Integrand size = 42, antiderivative size = 22 \[ \int \left (e^x+e^{240-240 x-60 x^2+60 x^3+15 x^4} \left (768+384 x-576 x^2-192 x^3\right )\right ) \, dx=3+e^x-\frac {16}{5} e^{15 (-4+x (2+x))^2} \]

[Out]

exp(x)-16/5*exp(15*(x*(2+x)-4)^2)+3

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2225, 6838} \[ \int \left (e^x+e^{240-240 x-60 x^2+60 x^3+15 x^4} \left (768+384 x-576 x^2-192 x^3\right )\right ) \, dx=e^x-\frac {16}{5} e^{15 x^4+60 x^3-60 x^2-240 x+240} \]

[In]

Int[E^x + E^(240 - 240*x - 60*x^2 + 60*x^3 + 15*x^4)*(768 + 384*x - 576*x^2 - 192*x^3),x]

[Out]

E^x - (16*E^(240 - 240*x - 60*x^2 + 60*x^3 + 15*x^4))/5

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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}& = \int e^x \, dx+\int e^{240-240 x-60 x^2+60 x^3+15 x^4} \left (768+384 x-576 x^2-192 x^3\right ) \, dx \\ & = e^x-\frac {16}{5} e^{240-240 x-60 x^2+60 x^3+15 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \left (e^x+e^{240-240 x-60 x^2+60 x^3+15 x^4} \left (768+384 x-576 x^2-192 x^3\right )\right ) \, dx=e^x-\frac {16}{5} e^{15 \left (-4+2 x+x^2\right )^2} \]

[In]

Integrate[E^x + E^(240 - 240*x - 60*x^2 + 60*x^3 + 15*x^4)*(768 + 384*x - 576*x^2 - 192*x^3),x]

[Out]

E^x - (16*E^(15*(-4 + 2*x + x^2)^2))/5

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
risch	\(-\frac {16 \,{\mathrm e}^{15 \left (x^{2}+2 x -4\right )^{2}}}{5}+{\mathrm e}^{x}\)	\(19\)
default	\(-\frac {16 \,{\mathrm e}^{15 x^{4}+60 x^{3}-60 x^{2}-240 x +240}}{5}+{\mathrm e}^{x}\)	\(27\)
norman	\(-\frac {16 \,{\mathrm e}^{15 x^{4}+60 x^{3}-60 x^{2}-240 x +240}}{5}+{\mathrm e}^{x}\)	\(27\)
parallelrisch	\(-\frac {16 \,{\mathrm e}^{15 x^{4}+60 x^{3}-60 x^{2}-240 x +240}}{5}+{\mathrm e}^{x}\)	\(27\)
parts	\(-\frac {16 \,{\mathrm e}^{15 x^{4}+60 x^{3}-60 x^{2}-240 x +240}}{5}+{\mathrm e}^{x}\)	\(27\)

[In]

int((-192*x^3-576*x^2+384*x+768)*exp(15*x^4+60*x^3-60*x^2-240*x+240)+exp(x),x,method=_RETURNVERBOSE)

[Out]

-16/5*exp(15*(x^2+2*x-4)^2)+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \left (e^x+e^{240-240 x-60 x^2+60 x^3+15 x^4} \left (768+384 x-576 x^2-192 x^3\right )\right ) \, dx=-\frac {16}{5} \, e^{\left (15 \, x^{4} + 60 \, x^{3} - 60 \, x^{2} - 240 \, x + 240\right )} + e^{x} \]

[In]

integrate((-192*x^3-576*x^2+384*x+768)*exp(15*x^4+60*x^3-60*x^2-240*x+240)+exp(x),x, algorithm="fricas")

[Out]

-16/5*e^(15*x^4 + 60*x^3 - 60*x^2 - 240*x + 240) + e^x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \left (e^x+e^{240-240 x-60 x^2+60 x^3+15 x^4} \left (768+384 x-576 x^2-192 x^3\right )\right ) \, dx=e^{x} - \frac {16 e^{15 x^{4} + 60 x^{3} - 60 x^{2} - 240 x + 240}}{5} \]

[In]

integrate((-192*x**3-576*x**2+384*x+768)*exp(15*x**4+60*x**3-60*x**2-240*x+240)+exp(x),x)

[Out]

exp(x) - 16*exp(15*x**4 + 60*x**3 - 60*x**2 - 240*x + 240)/5

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \left (e^x+e^{240-240 x-60 x^2+60 x^3+15 x^4} \left (768+384 x-576 x^2-192 x^3\right )\right ) \, dx=-\frac {16}{5} \, e^{\left (15 \, x^{4} + 60 \, x^{3} - 60 \, x^{2} - 240 \, x + 240\right )} + e^{x} \]

[In]

integrate((-192*x^3-576*x^2+384*x+768)*exp(15*x^4+60*x^3-60*x^2-240*x+240)+exp(x),x, algorithm="maxima")

[Out]

-16/5*e^(15*x^4 + 60*x^3 - 60*x^2 - 240*x + 240) + e^x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \left (e^x+e^{240-240 x-60 x^2+60 x^3+15 x^4} \left (768+384 x-576 x^2-192 x^3\right )\right ) \, dx=-\frac {16}{5} \, e^{\left (15 \, x^{4} + 60 \, x^{3} - 60 \, x^{2} - 240 \, x + 240\right )} + e^{x} \]

[In]

integrate((-192*x^3-576*x^2+384*x+768)*exp(15*x^4+60*x^3-60*x^2-240*x+240)+exp(x),x, algorithm="giac")

[Out]

-16/5*e^(15*x^4 + 60*x^3 - 60*x^2 - 240*x + 240) + e^x

Mupad [B] (verification not implemented)

Time = 14.75 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \left (e^x+e^{240-240 x-60 x^2+60 x^3+15 x^4} \left (768+384 x-576 x^2-192 x^3\right )\right ) \, dx={\mathrm {e}}^x-\frac {16\,{\mathrm {e}}^{-240\,x}\,{\mathrm {e}}^{240}\,{\mathrm {e}}^{15\,x^4}\,{\mathrm {e}}^{-60\,x^2}\,{\mathrm {e}}^{60\,x^3}}{5} \]

[In]

int(exp(x) + exp(60*x^3 - 60*x^2 - 240*x + 15*x^4 + 240)*(384*x - 576*x^2 - 192*x^3 + 768),x)

[Out]

exp(x) - (16*exp(-240*x)*exp(240)*exp(15*x^4)*exp(-60*x^2)*exp(60*x^3))/5