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

3.5.2.1 Optimal result

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

output

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

3.5.2.2 Mathematica [A] (verified)

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

input

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

output

5*E^(-6 + 10/x)*x*(1 + x*Log[9])
 

3.5.2.3 Rubi [C] (verified)

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

Time = 0.57 (sec) , antiderivative size = 90, normalized size of antiderivative = 5.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {7292, 7293, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {10}{x}-6} \left (20 x^2 \log (3)+5 x (1-20 \log (3))-50\right )}{x}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {50 e^{\frac {10}{x}-6}}{x}+20 e^{\frac {10}{x}-6} x \log (3)+5 e^{\frac {10}{x}-6} (1-20 \log (3))\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}-\frac {1000 \log (3) \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}-\frac {50 (1-20 \log (3)) \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}+10 e^{\frac {10}{x}-6} x^2 \log (3)+100 e^{\frac {10}{x}-6} x \log (3)+5 e^{\frac {10}{x}-6} x (1-20 \log (3))\)

input

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

output

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

3.5.2.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

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

rule 7293

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

3.5.2.4 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	\(5 \,{\mathrm e}^{-6+\frac {10}{x}} x +500 \ln \left (3\right ) \left (-\frac {{\mathrm e}^{-6+\frac {10}{x}} \left (-1-\frac {10}{x}\right ) x^{2}}{50}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )-500 \ln \left (3\right ) \left (\frac {{\mathrm e}^{-6+\frac {10}{x}} x}{5}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )\)	\(76\)

input

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

output

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

3.5.2.5 Fricas [A] (verification not implemented)

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

input

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

output

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

3.5.2.6 Sympy [A] (verification not implemented)

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

input

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

output

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

3.5.2.7 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 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=1000 \, e^{\left (-6\right )} \Gamma \left (-1, -\frac {10}{x}\right ) \log \left (3\right ) + 2000 \, e^{\left (-6\right )} \Gamma \left (-2, -\frac {10}{x}\right ) \log \left (3\right ) + 50 \, {\rm Ei}\left (\frac {10}{x}\right ) e^{\left (-6\right )} - 50 \, e^{\left (-6\right )} \Gamma \left (-1, -\frac {10}{x}\right ) \]

input

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

output

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)*gamma(-1, -10/x)
 

3.5.2.8 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 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=\frac {25 \, {\left (10 \, e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )} \log \left (3\right ) - \frac {{\left (3 \, x - 5\right )} e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )}}{x} + 3 \, e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )}\right )}}{\frac {{\left (3 \, x - 5\right )}^{2}}{x^{2}} - \frac {6 \, {\left (3 \, x - 5\right )}}{x} + 9} \]

input

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

output

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)
 

3.5.2.9 Mupad [B] (verification not implemented)

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

input

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

output

5*x*exp(10/x - 6)*(2*x*log(3) + 1)