\(\int \frac {e^{-\frac {-3+x}{x}} (-30-19 x-8 e^{\frac {-3+x}{x}} x^6)}{2 x^3} \, dx\) [7614]

Optimal result

Integrand size = 36, antiderivative size = 29 \[ \int \frac {e^{-\frac {-3+x}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{2 x^3} \, dx=\frac {3 e^{-1+\frac {3}{x}} \left (\frac {5}{3}+\frac {x}{2}\right )}{x}-x^4 \]

[Out]

3*(5/3+1/2*x)/x/exp(1-3/x)-x^4

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {12, 6873, 6874, 2243, 2240} \[ \int \frac {e^{-\frac {-3+x}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{2 x^3} \, dx=-x^4+\frac {3}{2} e^{\frac {3}{x}-1}+\frac {5 e^{\frac {3}{x}-1}}{x} \]

[In]

Int[(-30 - 19*x - 8*E^((-3 + x)/x)*x^6)/(2*E^((-3 + x)/x)*x^3),x]

[Out]

(3*E^(-1 + 3/x))/2 + (5*E^(-1 + 3/x))/x - x^4

Rule 12

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

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{-\frac {-3+x}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{x^3} \, dx \\ & = \frac {1}{2} \int \frac {e^{-1+\frac {3}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{x^3} \, dx \\ & = \frac {1}{2} \int \left (\frac {e^{-1+\frac {3}{x}} (-30-19 x)}{x^3}-8 x^3\right ) \, dx \\ & = -x^4+\frac {1}{2} \int \frac {e^{-1+\frac {3}{x}} (-30-19 x)}{x^3} \, dx \\ & = -x^4+\frac {1}{2} \int \left (-\frac {30 e^{-1+\frac {3}{x}}}{x^3}-\frac {19 e^{-1+\frac {3}{x}}}{x^2}\right ) \, dx \\ & = -x^4-\frac {19}{2} \int \frac {e^{-1+\frac {3}{x}}}{x^2} \, dx-15 \int \frac {e^{-1+\frac {3}{x}}}{x^3} \, dx \\ & = \frac {19}{6} e^{-1+\frac {3}{x}}+\frac {5 e^{-1+\frac {3}{x}}}{x}-x^4+5 \int \frac {e^{-1+\frac {3}{x}}}{x^2} \, dx \\ & = \frac {3}{2} e^{-1+\frac {3}{x}}+\frac {5 e^{-1+\frac {3}{x}}}{x}-x^4 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-\frac {-3+x}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{2 x^3} \, dx=\frac {3}{2} e^{-1+\frac {3}{x}}+\frac {5 e^{-1+\frac {3}{x}}}{x}-x^4 \]

[In]

Integrate[(-30 - 19*x - 8*E^((-3 + x)/x)*x^6)/(2*E^((-3 + x)/x)*x^3),x]

[Out]

(3*E^(-1 + 3/x))/2 + (5*E^(-1 + 3/x))/x - x^4

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90

method	result	size
risch	\(-x^{4}+\frac {\left (3 x +10\right ) {\mathrm e}^{-\frac {-3+x}{x}}}{2 x}\)	\(26\)
parallelrisch	\(\frac {\left (-6 x^{5} {\mathrm e}^{\frac {-3+x}{x}}+30+9 x \right ) {\mathrm e}^{-\frac {-3+x}{x}}}{6 x}\)	\(34\)
derivativedivides	\(-x^{4}-\frac {5 \,{\mathrm e}^{\frac {3}{x}-1} \left (1-\frac {3}{x}\right )}{3}+\frac {19 \,{\mathrm e}^{\frac {3}{x}-1}}{6}\)	\(36\)
default	\(-x^{4}-\frac {5 \,{\mathrm e}^{\frac {3}{x}-1} \left (1-\frac {3}{x}\right )}{3}+\frac {19 \,{\mathrm e}^{\frac {3}{x}-1}}{6}\)	\(36\)
parts	\(-x^{4}-\frac {5 \,{\mathrm e}^{\frac {3}{x}-1} \left (1-\frac {3}{x}\right )}{3}+\frac {19 \,{\mathrm e}^{\frac {3}{x}-1}}{6}\)	\(36\)
norman	\(\frac {\left (5 x +\frac {3 x^{2}}{2}-x^{6} {\mathrm e}^{\frac {-3+x}{x}}\right ) {\mathrm e}^{-\frac {-3+x}{x}}}{x^{2}}\)	\(37\)
meijerg	\(\frac {5 \,{\mathrm e}^{\frac {3}{x}+1-\frac {3 \,{\mathrm e}^{-1}}{x}} \left (1-\frac {\left (2-\frac {6 \,{\mathrm e}^{-1}}{x}\right ) {\mathrm e}^{\frac {3 \,{\mathrm e}^{-1}}{x}}}{2}\right )}{3}+324 \,{\mathrm e}^{\frac {3}{x}-4-\frac {3 \,{\mathrm e}^{-1}}{x}} \left (1-{\mathrm e}\right )^{4} \left (-\frac {x^{4} {\mathrm e}^{4}}{324 \left (1-{\mathrm e}\right )^{4}}-\frac {x^{3} {\mathrm e}^{3}}{81 \left (1-{\mathrm e}\right )^{3}}-\frac {x^{2} {\mathrm e}^{2}}{36 \left (1-{\mathrm e}\right )^{2}}-\frac {x \,{\mathrm e}}{18 \left (1-{\mathrm e}\right )}-\frac {37}{288}-\frac {\ln \left (x \right )}{24}+\frac {\ln \left (3\right )}{24}+\frac {i \pi }{24}+\frac {\ln \left (1-{\mathrm e}\right )}{24}+\frac {x^{4} {\mathrm e}^{4} \left (\frac {10125 \,{\mathrm e}^{-4} \left (1-{\mathrm e}\right )^{4}}{x^{4}}+\frac {6480 \,{\mathrm e}^{-3} \left (1-{\mathrm e}\right )^{3}}{x^{3}}+\frac {3240 \,{\mathrm e}^{-2} \left (1-{\mathrm e}\right )^{2}}{x^{2}}+\frac {1440 \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )}{x}+360\right )}{116640 \left (1-{\mathrm e}\right )^{4}}-\frac {x^{4} {\mathrm e}^{4+\frac {3 \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )}{x}} \left (\frac {135 \,{\mathrm e}^{-3} \left (1-{\mathrm e}\right )^{3}}{x^{3}}+\frac {45 \,{\mathrm e}^{-2} \left (1-{\mathrm e}\right )^{2}}{x^{2}}+\frac {30 \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )}{x}+30\right )}{9720 \left (1-{\mathrm e}\right )^{4}}-\frac {\ln \left (-\frac {3 \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )}{x}\right )}{24}-\frac {\operatorname {Ei}_{1}\left (-\frac {3 \,{\mathrm e}^{-1} \left (1-{\mathrm e}\right )}{x}\right )}{24}\right )-\frac {19 \,{\mathrm e}^{\frac {3}{x}-\frac {3 \,{\mathrm e}^{-1}}{x}} \left (1-{\mathrm e}^{\frac {3 \,{\mathrm e}^{-1}}{x}}\right )}{6}\)	\(356\)

[In]

int(1/2*(-8*x^6*exp((-3+x)/x)-19*x-30)/x^3/exp((-3+x)/x),x,method=_RETURNVERBOSE)

[Out]

-x^4+1/2*(3*x+10)/x*exp(-(-3+x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-\frac {-3+x}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{2 x^3} \, dx=-\frac {{\left (2 \, x^{5} e^{\left (\frac {x - 3}{x}\right )} - 3 \, x - 10\right )} e^{\left (-\frac {x - 3}{x}\right )}}{2 \, x} \]

[In]

integrate(1/2*(-8*x^6*exp((-3+x)/x)-19*x-30)/x^3/exp((-3+x)/x),x, algorithm="fricas")

[Out]

-1/2*(2*x^5*e^((x - 3)/x) - 3*x - 10)*e^(-(x - 3)/x)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.59 \[ \int \frac {e^{-\frac {-3+x}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{2 x^3} \, dx=- x^{4} + \frac {\left (3 x + 10\right ) e^{- \frac {x - 3}{x}}}{2 x} \]

[In]

integrate(1/2*(-8*x**6*exp((-3+x)/x)-19*x-30)/x**3/exp((-3+x)/x),x)

[Out]

-x**4 + (3*x + 10)*exp(-(x - 3)/x)/(2*x)

Maxima [C] (verification not implemented)

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

Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-\frac {-3+x}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{2 x^3} \, dx=-x^{4} - \frac {5}{3} \, e^{\left (-1\right )} \Gamma \left (2, -\frac {3}{x}\right ) + \frac {19}{6} \, e^{\left (\frac {3}{x} - 1\right )} \]

[In]

integrate(1/2*(-8*x^6*exp((-3+x)/x)-19*x-30)/x^3/exp((-3+x)/x),x, algorithm="maxima")

[Out]

-x^4 - 5/3*e^(-1)*gamma(2, -3/x) + 19/6*e^(3/x - 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-\frac {-3+x}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{2 x^3} \, dx=\frac {1}{2} \, x^{4} {\left (\frac {3 \, e^{\frac {3}{x}}}{x^{4}} + \frac {10 \, e^{\frac {3}{x}}}{x^{5}} - 2 \, e\right )} e^{\left (-1\right )} \]

[In]

integrate(1/2*(-8*x^6*exp((-3+x)/x)-19*x-30)/x^3/exp((-3+x)/x),x, algorithm="giac")

[Out]

1/2*x^4*(3*e^(3/x)/x^4 + 10*e^(3/x)/x^5 - 2*e)*e^(-1)

Mupad [B] (verification not implemented)

Time = 12.45 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {-3+x}{x}} \left (-30-19 x-8 e^{\frac {-3+x}{x}} x^6\right )}{2 x^3} \, dx=\frac {3\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{3/x}}{2}-x^4+\frac {5\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{3/x}}{x} \]

[In]

int(-(exp(-(x - 3)/x)*((19*x)/2 + 4*x^6*exp((x - 3)/x) + 15))/x^3,x)

[Out]

(3*exp(-1)*exp(3/x))/2 - x^4 + (5*exp(-1)*exp(3/x))/x