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

Optimal. Leaf size=18 \[ e^{-6+\frac {10}{x}} x (5+10 x \log (3)) \]

________________________________________________________________________________________

Rubi [C]  time = 0.42, antiderivative size = 90, normalized size of antiderivative = 5.00, number of steps used = 9, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6741, 6742, 2210, 2206, 2214} \begin {gather*} \frac {50 \text {Ei}\left (\frac {10}{x}\right )}{e^6}-\frac {1000 \log (3) \text {Ei}\left (\frac {10}{x}\right )}{e^6}-\frac {50 (1-20 \log (3)) \text {Ei}\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)) \end {gather*}

Antiderivative was successfully verified.

[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 2206

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 2210

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 2214

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 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 \text {Ei}\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 \text {Ei}\left (\frac {10}{x}\right )}{e^6}+5 e^{-6+\frac {10}{x}} x (1-20 \log (3))-\frac {50 \text {Ei}\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 \text {Ei}\left (\frac {10}{x}\right )}{e^6}+5 e^{-6+\frac {10}{x}} x (1-20 \log (3))-\frac {50 \text {Ei}\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 \text {Ei}\left (\frac {10}{x}\right ) \log (3)}{e^6}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 20, normalized size = 1.11 \begin {gather*} \frac {5}{2} e^{-6+\frac {10}{x}} x (2+x \log (81)) \end {gather*}

Antiderivative was successfully verified.

[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*(2 + x*Log[81]))/2

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 22, normalized size = 1.22 \begin {gather*} 5 \, {\left (2 \, x^{2} \log \relax (3) + x\right )} e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [B]  time = 0.31, size = 77, normalized size = 4.28 \begin {gather*} \frac {25 \, {\left (10 \, e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )} \log \relax (3) - \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.12, size = 23, normalized size = 1.28

method	result	size
gosper	\(5 \left (2 x \ln \relax (3)+1\right ) x \,{\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}\)	\(23\)
risch	\(\left (5 x +10 x^{2} \ln \relax (3)\right ) {\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}\)	\(24\)
norman	\(\left (5 x +10 x^{2} \ln \relax (3)\right ) {\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}\)	\(25\)
derivativedivides	\(5 \,{\mathrm e}^{-6+\frac {10}{x}} x -20 \ln \relax (3) {\mathrm e}^{-6+\frac {10}{x}} x^{2} \left (3-\frac {5}{x}\right )+70 \ln \relax (3) {\mathrm e}^{-6+\frac {10}{x}} x^{2}-100 \ln \relax (3) {\mathrm e}^{-6+\frac {10}{x}} x\)	\(63\)
default	\(5 \,{\mathrm e}^{-6+\frac {10}{x}} x -20 \ln \relax (3) {\mathrm e}^{-6+\frac {10}{x}} x^{2} \left (3-\frac {5}{x}\right )+70 \ln \relax (3) {\mathrm e}^{-6+\frac {10}{x}} x^{2}-100 \ln \relax (3) {\mathrm e}^{-6+\frac {10}{x}} x\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maxima [C]  time = 0.51, size = 48, normalized size = 2.67 \begin {gather*} 1000 \, e^{\left (-6\right )} \Gamma \left (-1, -\frac {10}{x}\right ) \log \relax (3) + 2000 \, e^{\left (-6\right )} \Gamma \left (-2, -\frac {10}{x}\right ) \log \relax (3) + 50 \, {\rm Ei}\left (\frac {10}{x}\right ) e^{\left (-6\right )} - 50 \, e^{\left (-6\right )} \Gamma \left (-1, -\frac {10}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 0.46, size = 18, normalized size = 1.00 \begin {gather*} 5\,x\,{\mathrm {e}}^{\frac {10}{x}-6}\,\left (2\,x\,\ln \relax (3)+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 0.24, size = 19, normalized size = 1.06 \begin {gather*} \left (10 x^{2} \log {\relax (3 )} + 5 x\right ) e^{- \frac {2 \left (3 x - 5\right )}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________