\(\int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx\) [9589]

Optimal result

Integrand size = 45, antiderivative size = 28 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 \left (-4-e^{-5+\frac {9 (5-x)}{x^2}}+e^{2-2 x}+x\right ) \]

[Out]

3*exp(2-2*x)+3*x-3*exp(9*(5-x)/x^2)/exp(5)-12

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 14, 2225, 6820, 6838} \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=-3 e^{\frac {9 (5-x)}{x^2}-5}+3 x+3 e^{2-2 x} \]

[In]

Int[(E^((45 - 9*x)/x^2)*(270 - 27*x) + 3*E^5*x^3 - 6*E^(7 - 2*x)*x^3)/(E^5*x^3),x]

[Out]

-3*E^(-5 + (9*(5 - x))/x^2) + 3*E^(2 - 2*x) + 3*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 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 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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}& = \frac {\int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{x^3} \, dx}{e^5} \\ & = \frac {\int \left (-6 e^{7-2 x}+\frac {3 e^{-9/x} \left (90 e^{\frac {45}{x^2}}-9 e^{\frac {45}{x^2}} x+e^{5+\frac {9}{x}} x^3\right )}{x^3}\right ) \, dx}{e^5} \\ & = \frac {3 \int \frac {e^{-9/x} \left (90 e^{\frac {45}{x^2}}-9 e^{\frac {45}{x^2}} x+e^{5+\frac {9}{x}} x^3\right )}{x^3} \, dx}{e^5}-\frac {6 \int e^{7-2 x} \, dx}{e^5} \\ & = 3 e^{2-2 x}+\frac {3 \int \left (e^5-\frac {9 e^{-\frac {9 (-5+x)}{x^2}} (-10+x)}{x^3}\right ) \, dx}{e^5} \\ & = 3 e^{2-2 x}+3 x-\frac {27 \int \frac {e^{-\frac {9 (-5+x)}{x^2}} (-10+x)}{x^3} \, dx}{e^5} \\ & = -3 e^{-5+\frac {9 (5-x)}{x^2}}+3 e^{2-2 x}+3 x \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=-3 e^{-5+\frac {45}{x^2}-\frac {9}{x}}+3 e^{2-2 x}+3 x \]

[In]

Integrate[(E^((45 - 9*x)/x^2)*(270 - 27*x) + 3*E^5*x^3 - 6*E^(7 - 2*x)*x^3)/(E^5*x^3),x]

[Out]

-3*E^(-5 + 45/x^2 - 9/x) + 3*E^(2 - 2*x) + 3*x

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
parts	\(3 x +3 \,{\mathrm e}^{2-2 x}-3 \,{\mathrm e}^{-5} {\mathrm e}^{\frac {-9 x +45}{x^{2}}}\)	\(29\)
risch	\(3 x +3 \,{\mathrm e}^{2-2 x}-3 \,{\mathrm e}^{-\frac {5 x^{2}+9 x -45}{x^{2}}}\)	\(31\)
parallelrisch	\({\mathrm e}^{-5} \left (3 x \,{\mathrm e}^{5}+3 \,{\mathrm e}^{5} {\mathrm e}^{2-2 x}-3 \,{\mathrm e}^{-\frac {9 \left (-5+x \right )}{x^{2}}}\right )\)	\(33\)
default	\({\mathrm e}^{-5} \left (-3 \,{\mathrm e}^{-\frac {9}{x}+\frac {45}{x^{2}}}+3 \,{\mathrm e}^{5} {\mathrm e}^{2} {\mathrm e}^{-2 x}+3 x \,{\mathrm e}^{5}\right )\)	\(36\)
norman	\(\frac {3 x^{3}+3 x^{2} {\mathrm e}^{2-2 x}-3 x^{2} {\mathrm e}^{-5} {\mathrm e}^{\frac {-9 x +45}{x^{2}}}}{x^{2}}\)	\(41\)

[In]

int(((-27*x+270)*exp((-9*x+45)/x^2)-6*x^3*exp(5)*exp(2-2*x)+3*x^3*exp(5))/x^3/exp(5),x,method=_RETURNVERBOSE)

[Out]

3*x+3*exp(2-2*x)-3/exp(5)*exp((-9*x+45)/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 \, {\left (x e^{5} + e^{\left (-2 \, x + 7\right )} - e^{\left (-\frac {9 \, {\left (x - 5\right )}}{x^{2}}\right )}\right )} e^{\left (-5\right )} \]

[In]

integrate(((-27*x+270)*exp((-9*x+45)/x^2)-6*x^3*exp(5)*exp(2-2*x)+3*x^3*exp(5))/x^3/exp(5),x, algorithm="frica
s")

[Out]

3*(x*e^5 + e^(-2*x + 7) - e^(-9*(x - 5)/x^2))*e^(-5)

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 x - \frac {3 e^{\frac {45 - 9 x}{x^{2}}}}{e^{5}} + 3 e^{2 - 2 x} \]

[In]

integrate(((-27*x+270)*exp((-9*x+45)/x**2)-6*x**3*exp(5)*exp(2-2*x)+3*x**3*exp(5))/x**3/exp(5),x)

[Out]

3*x - 3*exp(-5)*exp((45 - 9*x)/x**2) + 3*exp(2 - 2*x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 \, {\left (x e^{5} - {\left (e^{\left (2 \, x + \frac {45}{x^{2}}\right )} - e^{\left (\frac {9}{x} + 7\right )}\right )} e^{\left (-2 \, x - \frac {9}{x}\right )}\right )} e^{\left (-5\right )} \]

[In]

integrate(((-27*x+270)*exp((-9*x+45)/x^2)-6*x^3*exp(5)*exp(2-2*x)+3*x^3*exp(5))/x^3/exp(5),x, algorithm="maxim
a")

[Out]

3*(x*e^5 - (e^(2*x + 45/x^2) - e^(9/x + 7))*e^(-2*x - 9/x))*e^(-5)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3 \, {\left (x e^{5} + e^{\left (-2 \, x + 7\right )} - e^{\left (-\frac {9}{x} + \frac {45}{x^{2}}\right )}\right )} e^{\left (-5\right )} \]

[In]

integrate(((-27*x+270)*exp((-9*x+45)/x^2)-6*x^3*exp(5)*exp(2-2*x)+3*x^3*exp(5))/x^3/exp(5),x, algorithm="giac"
)

[Out]

3*(x*e^5 + e^(-2*x + 7) - e^(-9/x + 45/x^2))*e^(-5)

Mupad [B] (verification not implemented)

Time = 14.45 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {45-9 x}{x^2}} (270-27 x)+3 e^5 x^3-6 e^{7-2 x} x^3}{e^5 x^3} \, dx=3\,x+3\,{\mathrm {e}}^{-2\,x}\,{\mathrm {e}}^2-3\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{-\frac {9}{x}}\,{\mathrm {e}}^{\frac {45}{x^2}} \]

[In]

int(-(exp(-5)*(exp(-(9*x - 45)/x^2)*(27*x - 270) - 3*x^3*exp(5) + 6*x^3*exp(5)*exp(2 - 2*x)))/x^3,x)

[Out]

3*x + 3*exp(-2*x)*exp(2) - 3*exp(-5)*exp(-9/x)*exp(45/x^2)