\(\int \frac {e^x (2-x)-14 x^3+e^{5-x} (14 x^3-14 x^4)}{14 x^3} \, dx\) [9590]

Optimal result

Integrand size = 41, antiderivative size = 23 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=-\frac {e^x}{14 x^2}-x+e^{5-x} x \]

[Out]

x*exp(5-x)-x-1/14*exp(x)/x^2

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {12, 14, 2207, 2225, 2228} \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=-\frac {e^x}{14 x^2}-e^{5-x} (1-x)+e^{5-x}-x \]

[In]

Int[(E^x*(2 - x) - 14*x^3 + E^(5 - x)*(14*x^3 - 14*x^4))/(14*x^3),x]

[Out]

E^(5 - x) - E^(5 - x)*(1 - x) - E^x/(14*x^2) - 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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]

Rule 2228

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[g*u^(m + 1)*(F^(c*v)/(b*c*
e*Log[F])), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{14} \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{x^3} \, dx \\ & = \frac {1}{14} \int \left (-14-14 e^{5-x} (-1+x)-\frac {e^x (-2+x)}{x^3}\right ) \, dx \\ & = -x-\frac {1}{14} \int \frac {e^x (-2+x)}{x^3} \, dx-\int e^{5-x} (-1+x) \, dx \\ & = -e^{5-x} (1-x)-\frac {e^x}{14 x^2}-x-\int e^{5-x} \, dx \\ & = e^{5-x}-e^{5-x} (1-x)-\frac {e^x}{14 x^2}-x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=-\frac {e^x}{14 x^2}-x+e^{5-x} x \]

[In]

Integrate[(E^x*(2 - x) - 14*x^3 + E^(5 - x)*(14*x^3 - 14*x^4))/(14*x^3),x]

[Out]

-1/14*E^x/x^2 - x + E^(5 - x)*x

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
risch	\(x \,{\mathrm e}^{5-x}-x -\frac {{\mathrm e}^{x}}{14 x^{2}}\)	\(20\)
parallelrisch	\(\frac {14 \,{\mathrm e}^{5-x} x^{3}-14 x^{3}-{\mathrm e}^{x}}{14 x^{2}}\)	\(27\)
norman	\(\frac {\left (x^{3} {\mathrm e}^{5}-\frac {{\mathrm e}^{2 x}}{14}-{\mathrm e}^{x} x^{3}\right ) {\mathrm e}^{-x}}{x^{2}}\)	\(29\)
parts	\(-x -{\mathrm e}^{5-x} \left (5-x \right )+5 \,{\mathrm e}^{5-x}-\frac {{\mathrm e}^{x}}{14 x^{2}}\)	\(33\)
default	\(-x -{\mathrm e}^{5} {\mathrm e}^{-x}-\frac {{\mathrm e}^{x}}{14 x^{2}}-{\mathrm e}^{5} \left (-x \,{\mathrm e}^{-x}-{\mathrm e}^{-x}\right )\)	\(38\)

[In]

int(1/14*((2-x)*exp(x)+(-14*x^4+14*x^3)*exp(5-x)-14*x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

x*exp(5-x)-x-1/14*exp(x)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=\frac {{\left (14 \, x^{3} e^{5} - 14 \, x^{3} e^{x} - e^{\left (2 \, x\right )}\right )} e^{\left (-x\right )}}{14 \, x^{2}} \]

[In]

integrate(1/14*((2-x)*exp(x)+(-14*x^4+14*x^3)*exp(5-x)-14*x^3)/x^3,x, algorithm="fricas")

[Out]

1/14*(14*x^3*e^5 - 14*x^3*e^x - e^(2*x))*e^(-x)/x^2

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=- x + \frac {14 x^{3} e^{5} e^{- x} - e^{x}}{14 x^{2}} \]

[In]

integrate(1/14*((2-x)*exp(x)+(-14*x**4+14*x**3)*exp(5-x)-14*x**3)/x**3,x)

[Out]

-x + (14*x**3*exp(5)*exp(-x) - exp(x))/(14*x**2)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx={\left (x e^{5} + e^{5}\right )} e^{\left (-x\right )} - x - e^{\left (-x + 5\right )} - \frac {1}{14} \, \Gamma \left (-1, -x\right ) - \frac {1}{7} \, \Gamma \left (-2, -x\right ) \]

[In]

integrate(1/14*((2-x)*exp(x)+(-14*x^4+14*x^3)*exp(5-x)-14*x^3)/x^3,x, algorithm="maxima")

[Out]

(x*e^5 + e^5)*e^(-x) - x - e^(-x + 5) - 1/14*gamma(-1, -x) - 1/7*gamma(-2, -x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=\frac {14 \, x^{3} e^{\left (-x + 5\right )} - 14 \, x^{3} - e^{x}}{14 \, x^{2}} \]

[In]

integrate(1/14*((2-x)*exp(x)+(-14*x^4+14*x^3)*exp(5-x)-14*x^3)/x^3,x, algorithm="giac")

[Out]

1/14*(14*x^3*e^(-x + 5) - 14*x^3 - e^x)/x^2

Mupad [B] (verification not implemented)

Time = 14.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {e^x (2-x)-14 x^3+e^{5-x} \left (14 x^3-14 x^4\right )}{14 x^3} \, dx=x\,\left ({\mathrm {e}}^{5-x}-1\right )-\frac {{\mathrm {e}}^x}{14\,x^2} \]

[In]

int(-((exp(x)*(x - 2))/14 - (exp(5 - x)*(14*x^3 - 14*x^4))/14 + x^3)/x^3,x)

[Out]

x*(exp(5 - x) - 1) - exp(x)/(14*x^2)