\(\int \frac {1}{5} (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)) \, dx\) [7670]

Optimal result

Integrand size = 29, antiderivative size = 20 \[ \int \frac {1}{5} \left (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)\right ) \, dx=3 \left (e^{4-\frac {x}{15}}-e^{-4+x} x\right ) \]

[Out]

3*exp(-1/15*x+4)-3*x*exp(x-4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 2225, 2207} \[ \int \frac {1}{5} \left (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)\right ) \, dx=-3 e^{x-4} (x+1)+3 e^{\frac {60-x}{15}}+3 e^{x-4} \]

[In]

Int[(-E^((60 - x)/15) + E^(-4 + x)*(-15 - 15*x))/5,x]

[Out]

3*E^((60 - x)/15) + 3*E^(-4 + x) - 3*E^(-4 + x)*(1 + x)

Rule 12

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

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int \left (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)\right ) \, dx \\ & = -\left (\frac {1}{5} \int e^{\frac {60-x}{15}} \, dx\right )+\frac {1}{5} \int e^{-4+x} (-15-15 x) \, dx \\ & = 3 e^{\frac {60-x}{15}}-3 e^{-4+x} (1+x)+3 \int e^{-4+x} \, dx \\ & = 3 e^{\frac {60-x}{15}}+3 e^{-4+x}-3 e^{-4+x} (1+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} \left (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)\right ) \, dx=3 e^{4-\frac {x}{15}}-3 e^{-4+x} x \]

[In]

Integrate[(-E^((60 - x)/15) + E^(-4 + x)*(-15 - 15*x))/5,x]

[Out]

3*E^(4 - x/15) - 3*E^(-4 + x)*x

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
risch	\(3 \,{\mathrm e}^{-\frac {x}{15}+4}-3 x \,{\mathrm e}^{x -4}\)	\(17\)
parallelrisch	\(3 \,{\mathrm e}^{-\frac {x}{15}+4}-3 x \,{\mathrm e}^{x -4}\)	\(17\)
default	\(-3 \,{\mathrm e}^{x -4} \left (x -4\right )-12 \,{\mathrm e}^{x -4}+3 \,{\mathrm e}^{-\frac {x}{15}+4}\)	\(25\)
parts	\(-3 \,{\mathrm e}^{x -4} \left (x -4\right )-12 \,{\mathrm e}^{x -4}+3 \,{\mathrm e}^{-\frac {x}{15}+4}\)	\(25\)
meijerg	\(3 \,{\mathrm e}^{-4} \left (1-{\mathrm e}^{x}\right )-3 \,{\mathrm e}^{-4} \left (1-\frac {\left (2-2 x \right ) {\mathrm e}^{x}}{2}\right )-3 \,{\mathrm e}^{4} \left (1-{\mathrm e}^{-\frac {x}{15}}\right )\)	\(39\)

[In]

int(1/5*(-15*x-15)*exp(x-4)-1/5*exp(-1/15*x+4),x,method=_RETURNVERBOSE)

[Out]

3*exp(-1/15*x+4)-3*x*exp(x-4)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {1}{5} \left (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)\right ) \, dx=-3 \, {\left (x e^{56} - e^{\left (-\frac {16}{15} \, x + 64\right )}\right )} e^{\left (x - 60\right )} \]

[In]

integrate(1/5*(-15*x-15)*exp(x-4)-1/5*exp(-1/15*x+4),x, algorithm="fricas")

[Out]

-3*(x*e^56 - e^(-16/15*x + 64))*e^(x - 60)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {1}{5} \left (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)\right ) \, dx=- 3 x e^{56} e^{x - 60} + 3 e^{4 - \frac {x}{15}} \]

[In]

integrate(1/5*(-15*x-15)*exp(x-4)-1/5*exp(-1/15*x+4),x)

[Out]

-3*x*exp(56)*exp(x - 60) + 3*exp(4 - x/15)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {1}{5} \left (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)\right ) \, dx=-3 \, {\left (x - 1\right )} e^{\left (x - 4\right )} - 3 \, e^{\left (x - 4\right )} + 3 \, e^{\left (-\frac {1}{15} \, x + 4\right )} \]

[In]

integrate(1/5*(-15*x-15)*exp(x-4)-1/5*exp(-1/15*x+4),x, algorithm="maxima")

[Out]

-3*(x - 1)*e^(x - 4) - 3*e^(x - 4) + 3*e^(-1/15*x + 4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{5} \left (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)\right ) \, dx=-3 \, x e^{\left (x - 4\right )} + 3 \, e^{\left (-\frac {1}{15} \, x + 4\right )} \]

[In]

integrate(1/5*(-15*x-15)*exp(x-4)-1/5*exp(-1/15*x+4),x, algorithm="giac")

[Out]

-3*x*e^(x - 4) + 3*e^(-1/15*x + 4)

Mupad [B] (verification not implemented)

Time = 12.42 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{5} \left (-e^{\frac {60-x}{15}}+e^{-4+x} (-15-15 x)\right ) \, dx=3\,{\mathrm {e}}^{4-\frac {x}{15}}-3\,x\,{\mathrm {e}}^{x-4} \]

[In]

int(- exp(4 - x/15)/5 - (exp(x - 4)*(15*x + 15))/5,x)

[Out]

3*exp(4 - x/15) - 3*x*exp(x - 4)