3.12.74 \(\int \frac {-160+(-4000+e^{7-x} (-1600-800 x)+e^{14-2 x} (-160-160 x)) \log (x)+160 \log (x) \log (4 \log ^2(x))}{x^3 \log (x)} \, dx\) [1174]

3.12.74.1 Optimal result
3.12.74.2 Mathematica [A] (verified)
3.12.74.3 Rubi [C] (verified)
3.12.74.4 Maple [A] (verified)
3.12.74.5 Fricas [A] (verification not implemented)
3.12.74.6 Sympy [A] (verification not implemented)
3.12.74.7 Maxima [C] (verification not implemented)
3.12.74.8 Giac [A] (verification not implemented)
3.12.74.9 Mupad [B] (verification not implemented)

3.12.74.1 Optimal result

Integrand size = 52, antiderivative size = 26 \[ \int \frac {-160+\left (-4000+e^{7-x} (-1600-800 x)+e^{14-2 x} (-160-160 x)\right ) \log (x)+160 \log (x) \log \left (4 \log ^2(x)\right )}{x^3 \log (x)} \, dx=\frac {80 \left (\left (5+e^{7-x}\right )^2-\log \left (4 \log ^2(x)\right )\right )}{x^2} \]

output
80*((exp(-x+7)+5)^2-ln(4*ln(x)^2))/x^2
 
3.12.74.2 Mathematica [A] (verified)

Time = 2.34 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {-160+\left (-4000+e^{7-x} (-1600-800 x)+e^{14-2 x} (-160-160 x)\right ) \log (x)+160 \log (x) \log \left (4 \log ^2(x)\right )}{x^3 \log (x)} \, dx=\frac {80 \left (e^{-2 x} \left (e^7+5 e^x\right )^2-\log \left (4 \log ^2(x)\right )\right )}{x^2} \]

input
Integrate[(-160 + (-4000 + E^(7 - x)*(-1600 - 800*x) + E^(14 - 2*x)*(-160 
- 160*x))*Log[x] + 160*Log[x]*Log[4*Log[x]^2])/(x^3*Log[x]),x]
 
output
(80*((E^7 + 5*E^x)^2/E^(2*x) - Log[4*Log[x]^2]))/x^2
 
3.12.74.3 Rubi [C] (verified)

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

Time = 0.87 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {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.

\(\displaystyle \int \frac {160 \log \left (4 \log ^2(x)\right ) \log (x)+\left (e^{7-x} (-800 x-1600)+e^{14-2 x} (-160 x-160)-4000\right ) \log (x)-160}{x^3 \log (x)} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {160 e^{14-2 x} (x+1)}{x^3}-\frac {800 e^{7-x} (x+2)}{x^3}+\frac {160 \left (\log \left (4 \log ^2(x)\right ) \log (x)-25 \log (x)-1\right )}{x^3 \log (x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 4000 \log (x) \operatorname {ExpIntegralEi}(-2 \log (x))-160 (25 \log (x)+1) \operatorname {ExpIntegralEi}(-2 \log (x))+160 \operatorname {ExpIntegralEi}(-2 \log (x))+\frac {80 e^{14-2 x}}{x^2}+\frac {800 e^{7-x}}{x^2}+\frac {2000}{x^2}-\frac {80 \log \left (4 \log ^2(x)\right )}{x^2}\)

input
Int[(-160 + (-4000 + E^(7 - x)*(-1600 - 800*x) + E^(14 - 2*x)*(-160 - 160* 
x))*Log[x] + 160*Log[x]*Log[4*Log[x]^2])/(x^3*Log[x]),x]
 
output
2000/x^2 + (80*E^(14 - 2*x))/x^2 + (800*E^(7 - x))/x^2 + 160*ExpIntegralEi 
[-2*Log[x]] + 4000*ExpIntegralEi[-2*Log[x]]*Log[x] - 160*ExpIntegralEi[-2* 
Log[x]]*(1 + 25*Log[x]) - (80*Log[4*Log[x]^2])/x^2
 

3.12.74.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 
3.12.74.4 Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35

method result size
parallelrisch \(-\frac {-2000-80 \,{\mathrm e}^{-2 x +14}-800 \,{\mathrm e}^{-x +7}+80 \ln \left (4 \ln \left (x \right )^{2}\right )}{x^{2}}\) \(35\)
risch \(-\frac {160 \ln \left (\ln \left (x \right )\right )}{x^{2}}-\frac {40 \left (-50-i \pi \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )+2 i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )-i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}-2 \,{\mathrm e}^{-2 x +14}+4 \ln \left (2\right )-20 \,{\mathrm e}^{-x +7}\right )}{x^{2}}\) \(91\)
default \(\frac {2000}{x^{2}}-\frac {160 \ln \left (\ln \left (x \right )\right )}{x^{2}}+\frac {40 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \left (\operatorname {csgn}\left (i \ln \left (x \right )\right )^{2}-2 \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right )+\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}\right )}{x^{2}}-\frac {160 \ln \left (2\right )}{x^{2}}+\frac {80 \,{\mathrm e}^{-2 x +14}}{x^{2}}+\frac {800 \,{\mathrm e}^{-x +7}}{x^{2}}\) \(96\)
parts \(\frac {2000}{x^{2}}-\frac {160 \ln \left (\ln \left (x \right )\right )}{x^{2}}+\frac {40 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \left (\operatorname {csgn}\left (i \ln \left (x \right )\right )^{2}-2 \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right )+\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}\right )}{x^{2}}-\frac {160 \ln \left (2\right )}{x^{2}}+\frac {80 \,{\mathrm e}^{-2 x +14}}{x^{2}}+\frac {800 \,{\mathrm e}^{-x +7}}{x^{2}}\) \(96\)

input
int((160*ln(x)*ln(4*ln(x)^2)+((-160*x-160)*exp(-x+7)^2+(-800*x-1600)*exp(- 
x+7)-4000)*ln(x)-160)/x^3/ln(x),x,method=_RETURNVERBOSE)
 
output
-1/x^2*(-2000-80*exp(-x+7)^2-800*exp(-x+7)+80*ln(4*ln(x)^2))
 
3.12.74.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-160+\left (-4000+e^{7-x} (-1600-800 x)+e^{14-2 x} (-160-160 x)\right ) \log (x)+160 \log (x) \log \left (4 \log ^2(x)\right )}{x^3 \log (x)} \, dx=\frac {80 \, {\left (10 \, e^{\left (-x + 7\right )} + e^{\left (-2 \, x + 14\right )} - \log \left (4 \, \log \left (x\right )^{2}\right ) + 25\right )}}{x^{2}} \]

input
integrate((160*log(x)*log(4*log(x)^2)+((-160*x-160)*exp(-x+7)^2+(-800*x-16 
00)*exp(-x+7)-4000)*log(x)-160)/x^3/log(x),x, algorithm=\
 
output
80*(10*e^(-x + 7) + e^(-2*x + 14) - log(4*log(x)^2) + 25)/x^2
 
3.12.74.6 Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {-160+\left (-4000+e^{7-x} (-1600-800 x)+e^{14-2 x} (-160-160 x)\right ) \log (x)+160 \log (x) \log \left (4 \log ^2(x)\right )}{x^3 \log (x)} \, dx=- \frac {80 \log {\left (4 \log {\left (x \right )}^{2} \right )}}{x^{2}} + \frac {2000}{x^{2}} + \frac {800 x^{2} e^{7 - x} + 80 x^{2} e^{14 - 2 x}}{x^{4}} \]

input
integrate((160*ln(x)*ln(4*ln(x)**2)+((-160*x-160)*exp(-x+7)**2+(-800*x-160 
0)*exp(-x+7)-4000)*ln(x)-160)/x**3/ln(x),x)
 
output
-80*log(4*log(x)**2)/x**2 + 2000/x**2 + (800*x**2*exp(7 - x) + 80*x**2*exp 
(14 - 2*x))/x**4
 
3.12.74.7 Maxima [C] (verification not implemented)

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

Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92 \[ \int \frac {-160+\left (-4000+e^{7-x} (-1600-800 x)+e^{14-2 x} (-160-160 x)\right ) \log (x)+160 \log (x) \log \left (4 \log ^2(x)\right )}{x^3 \log (x)} \, dx=320 \, e^{14} \Gamma \left (-1, 2 \, x\right ) + 800 \, e^{7} \Gamma \left (-1, x\right ) + 640 \, e^{14} \Gamma \left (-2, 2 \, x\right ) + 1600 \, e^{7} \Gamma \left (-2, x\right ) - \frac {80 \, \log \left (4 \, \log \left (x\right )^{2}\right )}{x^{2}} + \frac {2000}{x^{2}} \]

input
integrate((160*log(x)*log(4*log(x)^2)+((-160*x-160)*exp(-x+7)^2+(-800*x-16 
00)*exp(-x+7)-4000)*log(x)-160)/x^3/log(x),x, algorithm=\
 
output
320*e^14*gamma(-1, 2*x) + 800*e^7*gamma(-1, x) + 640*e^14*gamma(-2, 2*x) + 
 1600*e^7*gamma(-2, x) - 80*log(4*log(x)^2)/x^2 + 2000/x^2
 
3.12.74.8 Giac [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {-160+\left (-4000+e^{7-x} (-1600-800 x)+e^{14-2 x} (-160-160 x)\right ) \log (x)+160 \log (x) \log \left (4 \log ^2(x)\right )}{x^3 \log (x)} \, dx=\frac {80 \, {\left (10 \, e^{\left (-x + 7\right )} + e^{\left (-2 \, x + 14\right )} - \log \left (4 \, \log \left (x\right )^{2}\right ) + 25\right )}}{x^{2}} \]

input
integrate((160*log(x)*log(4*log(x)^2)+((-160*x-160)*exp(-x+7)^2+(-800*x-16 
00)*exp(-x+7)-4000)*log(x)-160)/x^3/log(x),x, algorithm=\
 
output
80*(10*e^(-x + 7) + e^(-2*x + 14) - log(4*log(x)^2) + 25)/x^2
 
3.12.74.9 Mupad [B] (verification not implemented)

Time = 13.42 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {-160+\left (-4000+e^{7-x} (-1600-800 x)+e^{14-2 x} (-160-160 x)\right ) \log (x)+160 \log (x) \log \left (4 \log ^2(x)\right )}{x^3 \log (x)} \, dx=\frac {800\,{\mathrm {e}}^{7-x}}{x^2}-\frac {80\,\ln \left (4\,{\ln \left (x\right )}^2\right )}{x^2}+\frac {80\,{\mathrm {e}}^{14-2\,x}}{x^2}+\frac {2000}{x^2} \]

input
int(-(log(x)*(exp(14 - 2*x)*(160*x + 160) + exp(7 - x)*(800*x + 1600) + 40 
00) - 160*log(4*log(x)^2)*log(x) + 160)/(x^3*log(x)),x)
 
output
(800*exp(7 - x))/x^2 - (80*log(4*log(x)^2))/x^2 + (80*exp(14 - 2*x))/x^2 + 
 2000/x^2