\(\int \frac {e^{-\frac {2 (-5+3 x)}{x}} (-50+5 x+(-100 x+20 x^2) \log (3))}{x} \, dx\) [402]

Optimal result

Integrand size = 33, antiderivative size = 18 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=e^{-6+\frac {10}{x}} x (5+10 x \log (3)) \]

[Out]

(10*x*ln(3)+5)/exp(3-5/x)^2*x

Rubi [C] (verified)

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

Time = 0.39 (sec) , antiderivative size = 90, normalized size of antiderivative = 5.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6873, 6874, 2241, 2237, 2245} \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=\frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}-\frac {1000 \log (3) \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}-\frac {50 (1-20 \log (3)) \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}+10 e^{\frac {10}{x}-6} x^2 \log (3)+100 e^{\frac {10}{x}-6} x \log (3)+5 e^{\frac {10}{x}-6} x (1-20 \log (3)) \]

[In]

Int[(-50 + 5*x + (-100*x + 20*x^2)*Log[3])/(E^((2*(-5 + 3*x))/x)*x),x]

[Out]

(50*ExpIntegralEi[10/x])/E^6 + 5*E^(-6 + 10/x)*x*(1 - 20*Log[3]) - (50*ExpIntegralEi[10/x]*(1 - 20*Log[3]))/E^
6 + 100*E^(-6 + 10/x)*x*Log[3] + 10*E^(-6 + 10/x)*x^2*Log[3] - (1000*ExpIntegralEi[10/x]*Log[3])/E^6

Rule 2237

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

Rule 2241

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

Rule 2245

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*(F^(a + b*(c + d*x)^n)/(d*(m + 1))), x] - Dist[b*n*(Log[F]/(m + 1)), 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[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

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}& = \int \frac {e^{-6+\frac {10}{x}} \left (-50+5 x (1-20 \log (3))+20 x^2 \log (3)\right )}{x} \, dx \\ & = \int \left (-\frac {50 e^{-6+\frac {10}{x}}}{x}+5 e^{-6+\frac {10}{x}} (1-20 \log (3))+20 e^{-6+\frac {10}{x}} x \log (3)\right ) \, dx \\ & = -\left (50 \int \frac {e^{-6+\frac {10}{x}}}{x} \, dx\right )+(5 (1-20 \log (3))) \int e^{-6+\frac {10}{x}} \, dx+(20 \log (3)) \int e^{-6+\frac {10}{x}} x \, dx \\ & = \frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}+5 e^{-6+\frac {10}{x}} x (1-20 \log (3))+10 e^{-6+\frac {10}{x}} x^2 \log (3)+(50 (1-20 \log (3))) \int \frac {e^{-6+\frac {10}{x}}}{x} \, dx+(100 \log (3)) \int e^{-6+\frac {10}{x}} \, dx \\ & = \frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}+5 e^{-6+\frac {10}{x}} x (1-20 \log (3))-\frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right ) (1-20 \log (3))}{e^6}+100 e^{-6+\frac {10}{x}} x \log (3)+10 e^{-6+\frac {10}{x}} x^2 \log (3)+(1000 \log (3)) \int \frac {e^{-6+\frac {10}{x}}}{x} \, dx \\ & = \frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}+5 e^{-6+\frac {10}{x}} x (1-20 \log (3))-\frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right ) (1-20 \log (3))}{e^6}+100 e^{-6+\frac {10}{x}} x \log (3)+10 e^{-6+\frac {10}{x}} x^2 \log (3)-\frac {1000 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right ) \log (3)}{e^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=5 e^{-6+\frac {10}{x}} x (1+x \log (9)) \]

[In]

Integrate[(-50 + 5*x + (-100*x + 20*x^2)*Log[3])/(E^((2*(-5 + 3*x))/x)*x),x]

[Out]

5*E^(-6 + 10/x)*x*(1 + x*Log[9])

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28

method	result	size
gosper	\(5 \left (2 x \ln \left (3\right )+1\right ) x \,{\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}\)	\(23\)
risch	\(\left (5 x +10 x^{2} \ln \left (3\right )\right ) {\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}\)	\(24\)
norman	\(\left (5 x +10 x^{2} \ln \left (3\right )\right ) {\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}\)	\(25\)
parallelrisch	\(\frac {\left (10 x^{3} \ln \left (3\right )+5 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}}{x}\)	\(30\)
derivativedivides	\(5 \,{\mathrm e}^{-6+\frac {10}{x}} x +500 \ln \left (3\right ) \left (-\frac {{\mathrm e}^{-6+\frac {10}{x}} \left (-1-\frac {10}{x}\right ) x^{2}}{50}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )-500 \ln \left (3\right ) \left (\frac {{\mathrm e}^{-6+\frac {10}{x}} x}{5}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )\)	\(76\)
default	\(5 \,{\mathrm e}^{-6+\frac {10}{x}} x +500 \ln \left (3\right ) \left (-\frac {{\mathrm e}^{-6+\frac {10}{x}} \left (-1-\frac {10}{x}\right ) x^{2}}{50}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )-500 \ln \left (3\right ) \left (\frac {{\mathrm e}^{-6+\frac {10}{x}} x}{5}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )\)	\(76\)

[In]

int(((20*x^2-100*x)*ln(3)+5*x-50)/x/exp((3*x-5)/x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=5 \, {\left (2 \, x^{2} \log \left (3\right ) + x\right )} e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )} \]

[In]

integrate(((20*x^2-100*x)*log(3)+5*x-50)/x/exp((3*x-5)/x)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=\left (10 x^{2} \log {\left (3 \right )} + 5 x\right ) e^{- \frac {2 \cdot \left (3 x - 5\right )}{x}} \]

[In]

integrate(((20*x**2-100*x)*ln(3)+5*x-50)/x/exp((3*x-5)/x)**2,x)

[Out]

(10*x**2*log(3) + 5*x)*exp(-2*(3*x - 5)/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 = 48, normalized size of antiderivative = 2.67 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=1000 \, e^{\left (-6\right )} \Gamma \left (-1, -\frac {10}{x}\right ) \log \left (3\right ) + 2000 \, e^{\left (-6\right )} \Gamma \left (-2, -\frac {10}{x}\right ) \log \left (3\right ) + 50 \, {\rm Ei}\left (\frac {10}{x}\right ) e^{\left (-6\right )} - 50 \, e^{\left (-6\right )} \Gamma \left (-1, -\frac {10}{x}\right ) \]

[In]

integrate(((20*x^2-100*x)*log(3)+5*x-50)/x/exp((3*x-5)/x)^2,x, algorithm="maxima")

[Out]

1000*e^(-6)*gamma(-1, -10/x)*log(3) + 2000*e^(-6)*gamma(-2, -10/x)*log(3) + 50*Ei(10/x)*e^(-6) - 50*e^(-6)*gam
ma(-1, -10/x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (18) = 36\).

Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 4.28 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=\frac {25 \, {\left (10 \, e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )} \log \left (3\right ) - \frac {{\left (3 \, x - 5\right )} e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )}}{x} + 3 \, e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )}\right )}}{\frac {{\left (3 \, x - 5\right )}^{2}}{x^{2}} - \frac {6 \, {\left (3 \, x - 5\right )}}{x} + 9} \]

[In]

integrate(((20*x^2-100*x)*log(3)+5*x-50)/x/exp((3*x-5)/x)^2,x, algorithm="giac")

[Out]

25*(10*e^(-2*(3*x - 5)/x)*log(3) - (3*x - 5)*e^(-2*(3*x - 5)/x)/x + 3*e^(-2*(3*x - 5)/x))/((3*x - 5)^2/x^2 - 6
*(3*x - 5)/x + 9)

Mupad [B] (verification not implemented)

Time = 7.61 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=5\,x\,{\mathrm {e}}^{\frac {10}{x}-6}\,\left (2\,x\,\ln \left (3\right )+1\right ) \]

[In]

int(-(exp(-(2*(3*x - 5))/x)*(log(3)*(100*x - 20*x^2) - 5*x + 50))/x,x)

[Out]

5*x*exp(10/x - 6)*(2*x*log(3) + 1)