3.11.51 \(\int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} (-4 x^2-x^2 \log (169)+(-400 x+8 x^2+(-100 x+2 x^2) \log (169)) \log (50-x))}{(-1000+20 x) \log ^2(50-x)} \, dx\) [1051]

3.11.51.1 Optimal result

Integrand size = 83, antiderivative size = 21 \[ \int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (-4 x^2-x^2 \log (169)+\left (-400 x+8 x^2+\left (-100 x+2 x^2\right ) \log (169)\right ) \log (50-x)\right )}{(-1000+20 x) \log ^2(50-x)} \, dx=e^{\frac {x^2 (4+\log (169))}{20 \log (50-x)}} \]

output

exp(1/4*x^2/ln(-x+50)*(4/5+2/5*ln(13)))
 

3.11.51.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(48\) vs. \(2(21)=42\).

Time = 0.70 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29 \[ \int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (-4 x^2-x^2 \log (169)+\left (-400 x+8 x^2+\left (-100 x+2 x^2\right ) \log (169)\right ) \log (50-x)\right )}{(-1000+20 x) \log ^2(50-x)} \, dx=\frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}} (4+\log (169))}{2 (2+\log (13))} \]

input

Integrate[(E^((4*x^2 + x^2*Log[169])/(20*Log[50 - x]))*(-4*x^2 - x^2*Log[1 
69] + (-400*x + 8*x^2 + (-100*x + 2*x^2)*Log[169])*Log[50 - x]))/((-1000 + 
 20*x)*Log[50 - x]^2),x]
 

output

(13^(x^2/(10*Log[50 - x]))*E^(x^2/(5*Log[50 - x]))*(4 + Log[169]))/(2*(2 + 
 Log[13]))
 

3.11.51.3 Rubi [F]

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.

\(\displaystyle \int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (-4 x^2+x^2 (-\log (169))+\left (8 x^2+\left (2 x^2-100 x\right ) \log (169)-400 x\right ) \log (50-x)\right )}{(20 x-1000) \log ^2(50-x)} \, dx\)

\(\Big \downarrow \) 6

\(\displaystyle \int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (x^2 (-4-\log (169))+\left (8 x^2+\left (2 x^2-100 x\right ) \log (169)-400 x\right ) \log (50-x)\right )}{(20 x-1000) \log ^2(50-x)}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {x (4+\log (169)) e^{\frac {x^2 (4+\log (169))}{20 \log (50-x)}} (x-2 x \log (50-x)+100 \log (50-x))}{(1000-20 x) \log ^2(50-x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle (4+\log (169)) \int \frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}} x (-2 \log (50-x) x+x+100 \log (50-x))}{20 (50-x) \log ^2(50-x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{20} (4+\log (169)) \int \frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}} x (-2 \log (50-x) x+x+100 \log (50-x))}{(50-x) \log ^2(50-x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{20} (4+\log (169)) \int \left (\frac {2\ 13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}} x}{\log (50-x)}-\frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}} x^2}{(x-50) \log ^2(50-x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{20} (4+\log (169)) \left (-100 \int \frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}}}{\log ^2(50-x)}dx+\int \frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}} (50-x)}{\log ^2(50-x)}dx-2500 \int \frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}}}{(x-50) \log ^2(50-x)}dx+100 \int \frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}}}{\log (50-x)}dx-2 \int \frac {13^{\frac {x^2}{10 \log (50-x)}} e^{\frac {x^2}{5 \log (50-x)}} (50-x)}{\log (50-x)}dx\right )\)

input

Int[(E^((4*x^2 + x^2*Log[169])/(20*Log[50 - x]))*(-4*x^2 - x^2*Log[169] + 
(-400*x + 8*x^2 + (-100*x + 2*x^2)*Log[169])*Log[50 - x]))/((-1000 + 20*x) 
*Log[50 - x]^2),x]
 

output

$Aborted
 

3.11.51.3.1 Defintions of rubi rules used

rule 6

Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v 
+ (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] &&  !FreeQ[Fx, x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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] 
]
 

3.11.51.4 Maple [A] (verified)

Time = 3.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
risch	\({\mathrm e}^{\frac {x^{2} \left (\ln \left (13\right )+2\right )}{10 \ln \left (-x +50\right )}}\)	\(19\)
parallelrisch	\({\mathrm e}^{\frac {x^{2} \left (\ln \left (13\right )+2\right )}{10 \ln \left (-x +50\right )}}\)	\(19\)
norman	\({\mathrm e}^{\frac {2 x^{2} \ln \left (13\right )+4 x^{2}}{20 \ln \left (-x +50\right )}}\)	\(25\)

input

int(((2*(2*x^2-100*x)*ln(13)+8*x^2-400*x)*ln(-x+50)-2*x^2*ln(13)-4*x^2)*ex 
p(1/20*(2*x^2*ln(13)+4*x^2)/ln(-x+50))/(20*x-1000)/ln(-x+50)^2,x,method=_R 
ETURNVERBOSE)
 

output

exp(1/10*x^2*(ln(13)+2)/ln(-x+50))
 

3.11.51.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (-4 x^2-x^2 \log (169)+\left (-400 x+8 x^2+\left (-100 x+2 x^2\right ) \log (169)\right ) \log (50-x)\right )}{(-1000+20 x) \log ^2(50-x)} \, dx=e^{\left (\frac {x^{2} \log \left (13\right ) + 2 \, x^{2}}{10 \, \log \left (-x + 50\right )}\right )} \]

input

integrate(((2*(2*x^2-100*x)*log(13)+8*x^2-400*x)*log(-x+50)-2*x^2*log(13)- 
4*x^2)*exp(1/20*(2*x^2*log(13)+4*x^2)/log(-x+50))/(20*x-1000)/log(-x+50)^2 
,x, algorithm=\
 

output

e^(1/10*(x^2*log(13) + 2*x^2)/log(-x + 50))
 

3.11.51.6 Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (-4 x^2-x^2 \log (169)+\left (-400 x+8 x^2+\left (-100 x+2 x^2\right ) \log (169)\right ) \log (50-x)\right )}{(-1000+20 x) \log ^2(50-x)} \, dx=\text {Exception raised: TypeError} \]

input

integrate(((2*(2*x**2-100*x)*ln(13)+8*x**2-400*x)*ln(-x+50)-2*x**2*ln(13)- 
4*x**2)*exp(1/20*(2*x**2*ln(13)+4*x**2)/ln(-x+50))/(20*x-1000)/ln(-x+50)** 
2,x)
 

output

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

3.11.51.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (-4 x^2-x^2 \log (169)+\left (-400 x+8 x^2+\left (-100 x+2 x^2\right ) \log (169)\right ) \log (50-x)\right )}{(-1000+20 x) \log ^2(50-x)} \, dx=\text {Exception raised: RuntimeError} \]

input

integrate(((2*(2*x^2-100*x)*log(13)+8*x^2-400*x)*log(-x+50)-2*x^2*log(13)- 
4*x^2)*exp(1/20*(2*x^2*log(13)+4*x^2)/log(-x+50))/(20*x-1000)/log(-x+50)^2 
,x, algorithm=\
 

output

Exception raised: RuntimeError >> ECL says: In function CAR, the value of 
the first argument is  0which is not of the expected type LIST
 

3.11.51.8 Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (-4 x^2-x^2 \log (169)+\left (-400 x+8 x^2+\left (-100 x+2 x^2\right ) \log (169)\right ) \log (50-x)\right )}{(-1000+20 x) \log ^2(50-x)} \, dx=e^{\left (\frac {x^{2} \log \left (13\right )}{10 \, \log \left (-x + 50\right )} + \frac {x^{2}}{5 \, \log \left (-x + 50\right )}\right )} \]

input

integrate(((2*(2*x^2-100*x)*log(13)+8*x^2-400*x)*log(-x+50)-2*x^2*log(13)- 
4*x^2)*exp(1/20*(2*x^2*log(13)+4*x^2)/log(-x+50))/(20*x-1000)/log(-x+50)^2 
,x, algorithm=\
 

output

e^(1/10*x^2*log(13)/log(-x + 50) + 1/5*x^2/log(-x + 50))
 

3.11.51.9 Mupad [B] (verification not implemented)

Time = 10.90 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {4 x^2+x^2 \log (169)}{20 \log (50-x)}} \left (-4 x^2-x^2 \log (169)+\left (-400 x+8 x^2+\left (-100 x+2 x^2\right ) \log (169)\right ) \log (50-x)\right )}{(-1000+20 x) \log ^2(50-x)} \, dx={\mathrm {e}}^{\frac {x^2\,\ln \left (13\right )+2\,x^2}{10\,\ln \left (50-x\right )}} \]

input

int(-(exp(((x^2*log(13))/10 + x^2/5)/log(50 - x))*(2*x^2*log(13) + log(50 
- x)*(400*x + 2*log(13)*(100*x - 2*x^2) - 8*x^2) + 4*x^2))/(log(50 - x)^2* 
(20*x - 1000)),x)
 

output

exp((x^2*log(13) + 2*x^2)/(10*log(50 - x)))