\(\int \frac {(16-x)^{\frac {1}{2-4 x-8 x^2+2 \log (5)}} (1-2 x-4 x^2+\log (5)+(-32-126 x+8 x^2) \log (16-x))}{-32+130 x+120 x^2-520 x^3-480 x^4+32 x^5+(-64+132 x+248 x^2-16 x^3) \log (5)+(-32+2 x) \log ^2(5)} \, dx\) [688]

Optimal result

Integrand size = 107, antiderivative size = 24 \[ \int \frac {(16-x)^{\frac {1}{2-4 x-8 x^2+2 \log (5)}} \left (1-2 x-4 x^2+\log (5)+\left (-32-126 x+8 x^2\right ) \log (16-x)\right )}{-32+130 x+120 x^2-520 x^3-480 x^4+32 x^5+\left (-64+132 x+248 x^2-16 x^3\right ) \log (5)+(-32+2 x) \log ^2(5)} \, dx=(16-x)^{\frac {1}{2 \left (1-2 x-4 x^2+\log (5)\right )}} \] Output:

exp(ln(16-x)/(2*ln(5)-8*x^2-4*x+2))
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(52\) vs. \(2(24)=48\).

Time = 0.79 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \[ \int \frac {(16-x)^{\frac {1}{2-4 x-8 x^2+2 \log (5)}} \left (1-2 x-4 x^2+\log (5)+\left (-32-126 x+8 x^2\right ) \log (16-x)\right )}{-32+130 x+120 x^2-520 x^3-480 x^4+32 x^5+\left (-64+132 x+248 x^2-16 x^3\right ) \log (5)+(-32+2 x) \log ^2(5)} \, dx=\frac {(16-x)^{\frac {1}{2-4 x-8 x^2+\log (25)}} \left (2-4 x-8 x^2+\log (25)\right )^2}{4 \left (1-2 x-4 x^2+\log (5)\right )^2} \] Input:

Integrate[((16 - x)^(2 - 4*x - 8*x^2 + 2*Log[5])^(-1)*(1 - 2*x - 4*x^2 + L 
og[5] + (-32 - 126*x + 8*x^2)*Log[16 - x]))/(-32 + 130*x + 120*x^2 - 520*x 
^3 - 480*x^4 + 32*x^5 + (-64 + 132*x + 248*x^2 - 16*x^3)*Log[5] + (-32 + 2 
*x)*Log[5]^2),x]
 

Output:

((16 - x)^(2 - 4*x - 8*x^2 + Log[25])^(-1)*(2 - 4*x - 8*x^2 + Log[25])^2)/ 
(4*(1 - 2*x - 4*x^2 + Log[5])^2)
 

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.

\(\displaystyle \int \frac {(16-x)^{\frac {1}{-8 x^2-4 x+2+2 \log (5)}} \left (-4 x^2+\left (8 x^2-126 x-32\right ) \log (16-x)-2 x+1+\log (5)\right )}{32 x^5-480 x^4-520 x^3+120 x^2+\left (-16 x^3+248 x^2+132 x-64\right ) \log (5)+130 x+(2 x-32) \log ^2(5)-32} \, dx\)

\(\Big \downarrow \) 2463

\(\displaystyle \int \left (-\frac {\left (-4 x^2+\left (8 x^2-126 x-32\right ) \log (16-x)-2 x+1+\log (5)\right ) (16-x)^{\frac {1}{-8 x^2-4 x+2+2 \log (5)}-1}}{2 (\log (5)-1055)^2}+\frac {(2 x+33) \left (-4 x^2+\left (8 x^2-126 x-32\right ) \log (16-x)-2 x+1+\log (5)\right ) (16-x)^{\frac {1}{-8 x^2-4 x+2+2 \log (5)}}}{(\log (5)-1055) \left (4 x^2+2 x-1-\log (5)\right )^2}+\frac {(-2 x-33) \left (-4 x^2+\left (8 x^2-126 x-32\right ) \log (16-x)-2 x+1+\log (5)\right ) (16-x)^{\frac {1}{-8 x^2-4 x+2+2 \log (5)}}}{(\log (5)-1055)^2 \left (4 x^2+2 x-1-\log (5)\right )}\right )dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {(16-x)^{\frac {1}{-8 x^2-4 x+2+\log (25)}-1} \left (4 x^2-2 \left (4 x^2-63 x-16\right ) \log (16-x)+2 x-1-\log (5)\right )}{2 \left (-4 x^2-2 x+1+\log (5)\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int -\frac {(16-x)^{\frac {1}{-8 x^2-4 x+\log (25)+2}-1} \left (-4 x^2-2 x-2 \left (-4 x^2+63 x+16\right ) \log (16-x)+\log (5)+1\right )}{\left (-4 x^2-2 x+\log (5)+1\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{2} \int \frac {(16-x)^{\frac {1}{-8 x^2-4 x+\log (25)+2}-1} \left (-4 x^2-2 x-2 \left (-4 x^2+63 x+16\right ) \log (16-x)+\log (5)+1\right )}{\left (-4 x^2-2 x+\log (5)+1\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{2} \int \left (-\frac {4 x^2 (16-x)^{\frac {1}{-8 x^2-4 x+\log (25)+2}-1}}{\left (4 x^2+2 x-\log (5)-1\right )^2}-\frac {2 x (16-x)^{\frac {1}{-8 x^2-4 x+\log (25)+2}-1}}{\left (4 x^2+2 x-\log (5)-1\right )^2}+\frac {(1+\log (5)) (16-x)^{\frac {1}{-8 x^2-4 x+\log (25)+2}-1}}{\left (4 x^2+2 x-\log (5)-1\right )^2}-\frac {2 (4 x+1) \log (16-x) (16-x)^{\frac {1}{-8 x^2-4 x+\log (25)+2}}}{\left (4 x^2+2 x-\log (5)-1\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (2 \int \frac {(16-x)^{\frac {1}{-8 x^2-4 x+\log (25)+2}} \log (16-x)}{\left (4 x^2+2 x-\log (5)-1\right )^2}dx+8 \int \frac {(16-x)^{\frac {1}{-8 x^2-4 x+\log (25)+2}} x \log (16-x)}{\left (4 x^2+2 x-\log (5)-1\right )^2}dx+\frac {2 \left (-8 x^2-4 x+2+\log (25)\right ) (16-x)^{\frac {1}{-8 x^2-4 x+2+\log (25)}} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{-8 x^2-4 x+\log (25)+2},1+\frac {1}{-8 x^2-4 x+\log (25)+2},\frac {4 (16-x)}{65+\sqrt {5+\log (625)}}\right )}{5+\log (625)+65 \sqrt {5+\log (625)}}+\frac {2 \left (-8 x^2-4 x+2+\log (25)\right ) (16-x)^{\frac {1}{-8 x^2-4 x+2+\log (25)}} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{-8 x^2-4 x+\log (25)+2},1+\frac {1}{-8 x^2-4 x+\log (25)+2},\frac {4 (16-x)}{65-\sqrt {5+\log (625)}}\right )}{5+\log (625)-65 \sqrt {5+\log (625)}}\right )\)

Input:

Int[((16 - x)^(2 - 4*x - 8*x^2 + 2*Log[5])^(-1)*(1 - 2*x - 4*x^2 + Log[5] 
+ (-32 - 126*x + 8*x^2)*Log[16 - x]))/(-32 + 130*x + 120*x^2 - 520*x^3 - 4 
80*x^4 + 32*x^5 + (-64 + 132*x + 248*x^2 - 16*x^3)*Log[5] + (-32 + 2*x)*Lo 
g[5]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 5.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
risch	\(\left (16-x \right )^{\frac {1}{2 \ln \left (5\right )-8 x^{2}-4 x +2}}\)	\(23\)
parallelrisch	\(\frac {512 \ln \left (5\right )^{4} {\mathrm e}^{\frac {\ln \left (16-x \right )}{2 \ln \left (5\right )-8 x^{2}-4 x +2}}+2048 \ln \left (5\right )^{3} {\mathrm e}^{\frac {\ln \left (16-x \right )}{2 \ln \left (5\right )-8 x^{2}-4 x +2}}+3072 \ln \left (5\right )^{2} {\mathrm e}^{\frac {\ln \left (16-x \right )}{2 \ln \left (5\right )-8 x^{2}-4 x +2}}+2048 \ln \left (5\right ) {\mathrm e}^{\frac {\ln \left (16-x \right )}{2 \ln \left (5\right )-8 x^{2}-4 x +2}}+512 \,{\mathrm e}^{\frac {\ln \left (16-x \right )}{2 \ln \left (5\right )-8 x^{2}-4 x +2}}}{512 \left (\ln \left (5\right )^{2}+2 \ln \left (5\right )+1\right ) \left (\ln \left (5\right )+1\right )^{2}}\)	\(161\)

Input:

int(((8*x^2-126*x-32)*ln(16-x)+ln(5)-4*x^2-2*x+1)*exp(ln(16-x)/(2*ln(5)-8* 
x^2-4*x+2))/((2*x-32)*ln(5)^2+(-16*x^3+248*x^2+132*x-64)*ln(5)+32*x^5-480* 
x^4-520*x^3+120*x^2+130*x-32),x,method=_RETURNVERBOSE)
 

Output:

(16-x)^(1/2/(-4*x^2+ln(5)-2*x+1))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {(16-x)^{\frac {1}{2-4 x-8 x^2+2 \log (5)}} \left (1-2 x-4 x^2+\log (5)+\left (-32-126 x+8 x^2\right ) \log (16-x)\right )}{-32+130 x+120 x^2-520 x^3-480 x^4+32 x^5+\left (-64+132 x+248 x^2-16 x^3\right ) \log (5)+(-32+2 x) \log ^2(5)} \, dx=\frac {1}{{\left (-x + 16\right )}^{\frac {1}{2 \, {\left (4 \, x^{2} + 2 \, x - \log \left (5\right ) - 1\right )}}}} \] Input:

integrate(((8*x^2-126*x-32)*log(16-x)+log(5)-4*x^2-2*x+1)*exp(log(16-x)/(2 
*log(5)-8*x^2-4*x+2))/((2*x-32)*log(5)^2+(-16*x^3+248*x^2+132*x-64)*log(5) 
+32*x^5-480*x^4-520*x^3+120*x^2+130*x-32),x, algorithm="fricas")
 

Output:

1/((-x + 16)^(1/2/(4*x^2 + 2*x - log(5) - 1)))
 

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {(16-x)^{\frac {1}{2-4 x-8 x^2+2 \log (5)}} \left (1-2 x-4 x^2+\log (5)+\left (-32-126 x+8 x^2\right ) \log (16-x)\right )}{-32+130 x+120 x^2-520 x^3-480 x^4+32 x^5+\left (-64+132 x+248 x^2-16 x^3\right ) \log (5)+(-32+2 x) \log ^2(5)} \, dx=e^{\frac {\log {\left (16 - x \right )}}{- 8 x^{2} - 4 x + 2 + 2 \log {\left (5 \right )}}} \] Input:

integrate(((8*x**2-126*x-32)*ln(16-x)+ln(5)-4*x**2-2*x+1)*exp(ln(16-x)/(2* 
ln(5)-8*x**2-4*x+2))/((2*x-32)*ln(5)**2+(-16*x**3+248*x**2+132*x-64)*ln(5) 
+32*x**5-480*x**4-520*x**3+120*x**2+130*x-32),x)
 

Output:

exp(log(16 - x)/(-8*x**2 - 4*x + 2 + 2*log(5)))
 

Maxima [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {(16-x)^{\frac {1}{2-4 x-8 x^2+2 \log (5)}} \left (1-2 x-4 x^2+\log (5)+\left (-32-126 x+8 x^2\right ) \log (16-x)\right )}{-32+130 x+120 x^2-520 x^3-480 x^4+32 x^5+\left (-64+132 x+248 x^2-16 x^3\right ) \log (5)+(-32+2 x) \log ^2(5)} \, dx=\frac {1}{{\left (-x + 16\right )}^{\frac {1}{2 \, {\left (4 \, x^{2} + 2 \, x - \log \left (5\right ) - 1\right )}}}} \] Input:

integrate(((8*x^2-126*x-32)*log(16-x)+log(5)-4*x^2-2*x+1)*exp(log(16-x)/(2 
*log(5)-8*x^2-4*x+2))/((2*x-32)*log(5)^2+(-16*x^3+248*x^2+132*x-64)*log(5) 
+32*x^5-480*x^4-520*x^3+120*x^2+130*x-32),x, algorithm="maxima")
 

Output:

1/((-x + 16)^(1/2/(4*x^2 + 2*x - log(5) - 1)))
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {(16-x)^{\frac {1}{2-4 x-8 x^2+2 \log (5)}} \left (1-2 x-4 x^2+\log (5)+\left (-32-126 x+8 x^2\right ) \log (16-x)\right )}{-32+130 x+120 x^2-520 x^3-480 x^4+32 x^5+\left (-64+132 x+248 x^2-16 x^3\right ) \log (5)+(-32+2 x) \log ^2(5)} \, dx=\frac {1}{{\left (-x + 16\right )}^{\frac {1}{2 \, {\left (4 \, {\left (x - 16\right )}^{2} + 130 \, x - \log \left (5\right ) - 1025\right )}}}} \] Input:

integrate(((8*x^2-126*x-32)*log(16-x)+log(5)-4*x^2-2*x+1)*exp(log(16-x)/(2 
*log(5)-8*x^2-4*x+2))/((2*x-32)*log(5)^2+(-16*x^3+248*x^2+132*x-64)*log(5) 
+32*x^5-480*x^4-520*x^3+120*x^2+130*x-32),x, algorithm="giac")
 

Output:

1/((-x + 16)^(1/2/(4*(x - 16)^2 + 130*x - log(5) - 1025)))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(16-x)^{\frac {1}{2-4 x-8 x^2+2 \log (5)}} \left (1-2 x-4 x^2+\log (5)+\left (-32-126 x+8 x^2\right ) \log (16-x)\right )}{-32+130 x+120 x^2-520 x^3-480 x^4+32 x^5+\left (-64+132 x+248 x^2-16 x^3\right ) \log (5)+(-32+2 x) \log ^2(5)} \, dx=-\int \frac {{\mathrm {e}}^{-\frac {\ln \left (16-x\right )}{8\,x^2+4\,x-2\,\ln \left (5\right )-2}}\,\left (2\,x-\ln \left (5\right )+\ln \left (16-x\right )\,\left (-8\,x^2+126\,x+32\right )+4\,x^2-1\right )}{130\,x+{\ln \left (5\right )}^2\,\left (2\,x-32\right )+\ln \left (5\right )\,\left (-16\,x^3+248\,x^2+132\,x-64\right )+120\,x^2-520\,x^3-480\,x^4+32\,x^5-32} \,d x \] Input:

int(-(exp(-log(16 - x)/(4*x - 2*log(5) + 8*x^2 - 2))*(2*x - log(5) + log(1 
6 - x)*(126*x - 8*x^2 + 32) + 4*x^2 - 1))/(130*x + log(5)^2*(2*x - 32) + l 
og(5)*(132*x + 248*x^2 - 16*x^3 - 64) + 120*x^2 - 520*x^3 - 480*x^4 + 32*x 
^5 - 32),x)
 

Output:

-int((exp(-log(16 - x)/(4*x - 2*log(5) + 8*x^2 - 2))*(2*x - log(5) + log(1 
6 - x)*(126*x - 8*x^2 + 32) + 4*x^2 - 1))/(130*x + log(5)^2*(2*x - 32) + l 
og(5)*(132*x + 248*x^2 - 16*x^3 - 64) + 120*x^2 - 520*x^3 - 480*x^4 + 32*x 
^5 - 32), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {(16-x)^{\frac {1}{2-4 x-8 x^2+2 \log (5)}} \left (1-2 x-4 x^2+\log (5)+\left (-32-126 x+8 x^2\right ) \log (16-x)\right )}{-32+130 x+120 x^2-520 x^3-480 x^4+32 x^5+\left (-64+132 x+248 x^2-16 x^3\right ) \log (5)+(-32+2 x) \log ^2(5)} \, dx=e^{\frac {\mathrm {log}\left (-x +16\right )}{2 \,\mathrm {log}\left (5\right )-8 x^{2}-4 x +2}} \] Input:

int(((8*x^2-126*x-32)*log(16-x)+log(5)-4*x^2-2*x+1)*exp(log(16-x)/(2*log(5 
)-8*x^2-4*x+2))/((2*x-32)*log(5)^2+(-16*x^3+248*x^2+132*x-64)*log(5)+32*x^ 
5-480*x^4-520*x^3+120*x^2+130*x-32),x)
 

Output:

e**(log( - x + 16)/(2*log(5) - 8*x**2 - 4*x + 2))