\(\int e^{162+80 x} (400+e^{-162-80 x}) \, dx\) [409]

Optimal result

Integrand size = 17, antiderivative size = 16 \[ \int e^{162+80 x} \left (400+e^{-162-80 x}\right ) \, dx=2+5 e^{-x+81 (2+x)}+x \]

[Out]

5/exp(-80*x-162)+x+2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2280, 45} \[ \int e^{162+80 x} \left (400+e^{-162-80 x}\right ) \, dx=x+5 e^{80 x+162} \]

[In]

Int[E^(162 + 80*x)*(400 + E^(-162 - 80*x)),x]

[Out]

5*E^(162 + 80*x) + x

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2280

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[g*h*(Log[G]/(d*e*Log[F]))]}, Dist[Denominator[m]*(G^(f*h - c*g*(h/d))/(d*e*Log[F])), Subst
[Int[x^(Numerator[m] - 1)*(a + b*x^Denominator[m])^p, x], x, F^(e*((c + d*x)/Denominator[m]))], x] /; LeQ[m, -
1] || GeQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{80} \text {Subst}\left (\int \frac {400+x}{x^2} \, dx,x,e^{-162-80 x}\right )\right ) \\ & = -\left (\frac {1}{80} \text {Subst}\left (\int \left (\frac {400}{x^2}+\frac {1}{x}\right ) \, dx,x,e^{-162-80 x}\right )\right ) \\ & = 5 e^{162+80 x}+x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int e^{162+80 x} \left (400+e^{-162-80 x}\right ) \, dx=5 e^{162+80 x}+x \]

[In]

Integrate[E^(162 + 80*x)*(400 + E^(-162 - 80*x)),x]

[Out]

5*E^(162 + 80*x) + x

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69

method	result	size
risch	\(x +5 \,{\mathrm e}^{80 x +162}\)	\(11\)
parts	\(x +5 \,{\mathrm e}^{80 x +162}\)	\(13\)
norman	\(\left (5+x \,{\mathrm e}^{-80 x -162}\right ) {\mathrm e}^{80 x +162}\)	\(20\)
parallelrisch	\(\left (5+x \,{\mathrm e}^{-80 x -162}\right ) {\mathrm e}^{80 x +162}\)	\(20\)
derivativedivides	\(5 \,{\mathrm e}^{80 x +162}-\frac {\ln \left ({\mathrm e}^{-80 x -162}\right )}{80}\)	\(21\)
default	\(5 \,{\mathrm e}^{80 x +162}-\frac {\ln \left ({\mathrm e}^{-80 x -162}\right )}{80}\)	\(21\)
meijerg	\(-\frac {{\mathrm e}^{80 x -80 x \,{\mathrm e}^{162}-162} \left (1-{\mathrm e}^{80 x \,{\mathrm e}^{162} \left (1-{\mathrm e}^{-162}\right )}\right )}{80 \left (1-{\mathrm e}^{-162}\right )}-5 \,{\mathrm e}^{80 x -80 x \,{\mathrm e}^{162}} \left (1-{\mathrm e}^{80 x \,{\mathrm e}^{162}}\right )\)	\(61\)

[In]

int((exp(-80*x-162)+400)/exp(-80*x-162),x,method=_RETURNVERBOSE)

[Out]

x+5*exp(80*x+162)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int e^{162+80 x} \left (400+e^{-162-80 x}\right ) \, dx=x + 5 \, e^{\left (80 \, x + 162\right )} \]

[In]

integrate((exp(-80*x-162)+400)/exp(-80*x-162),x, algorithm="fricas")

[Out]

x + 5*e^(80*x + 162)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.50 \[ \int e^{162+80 x} \left (400+e^{-162-80 x}\right ) \, dx=x + 5 e^{80 x + 162} \]

[In]

integrate((exp(-80*x-162)+400)/exp(-80*x-162),x)

[Out]

x + 5*exp(80*x + 162)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.69 \[ \int e^{162+80 x} \left (400+e^{-162-80 x}\right ) \, dx=x + 5 \, e^{\left (80 \, x + 162\right )} + \frac {81}{40} \]

[In]

integrate((exp(-80*x-162)+400)/exp(-80*x-162),x, algorithm="maxima")

[Out]

x + 5*e^(80*x + 162) + 81/40

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int e^{162+80 x} \left (400+e^{-162-80 x}\right ) \, dx=x + 5 \, e^{\left (80 \, x + 162\right )} \]

[In]

integrate((exp(-80*x-162)+400)/exp(-80*x-162),x, algorithm="giac")

[Out]

x + 5*e^(80*x + 162)

Mupad [B] (verification not implemented)

Time = 7.63 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int e^{162+80 x} \left (400+e^{-162-80 x}\right ) \, dx=x+5\,{\mathrm {e}}^{80\,x}\,{\mathrm {e}}^{162} \]

[In]

int(exp(80*x + 162)*(exp(- 80*x - 162) + 400),x)

[Out]

x + 5*exp(80*x)*exp(162)