3.4.93 \(\int \frac {6 \log (5)+e^x (-27-18 x \log (5)-3 x^2 \log ^2(5))}{(96+62 x \log (5)+10 x^2 \log ^2(5)+e^x (9+6 x \log (5)+x^2 \log ^2(5))) \log (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}) \log ^2(\log (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}))} \, dx\) [393]

3.4.93.1 Optimal result

Integrand size = 122, antiderivative size = 30 \begin {dmath*} \int \frac {6 \log (5)+e^x \left (-27-18 x \log (5)-3 x^2 \log ^2(5)\right )}{\left (96+62 x \log (5)+10 x^2 \log ^2(5)+e^x \left (9+6 x \log (5)+x^2 \log ^2(5)\right )\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx=e^4+\frac {3}{\log \left (\log \left (10+e^x+\frac {2}{x \left (\frac {3}{x}+\log (5)\right )}\right )\right )} \end {dmath*}

output

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

3.4.93.2 Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \begin {dmath*} \int \frac {6 \log (5)+e^x \left (-27-18 x \log (5)-3 x^2 \log ^2(5)\right )}{\left (96+62 x \log (5)+10 x^2 \log ^2(5)+e^x \left (9+6 x \log (5)+x^2 \log ^2(5)\right )\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx=\frac {3}{\log \left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \end {dmath*}

input

Integrate[(6*Log[5] + E^x*(-27 - 18*x*Log[5] - 3*x^2*Log[5]^2))/((96 + 62* 
x*Log[5] + 10*x^2*Log[5]^2 + E^x*(9 + 6*x*Log[5] + x^2*Log[5]^2))*Log[(32 
+ 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]*Log[Log[(32 + 10*x*Log 
[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]^2),x]
 

output

3/Log[Log[(32 + 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]
 

3.4.93.3 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.

\(\displaystyle \int \frac {e^x \left (-3 x^2 \log ^2(5)-18 x \log (5)-27\right )+6 \log (5)}{\left (10 x^2 \log ^2(5)+e^x \left (x^2 \log ^2(5)+6 x \log (5)+9\right )+62 x \log (5)+96\right ) \log \left (\frac {10 x \log (5)+e^x (x \log (5)+3)+32}{x \log (5)+3}\right ) \log ^2\left (\log \left (\frac {10 x \log (5)+e^x (x \log (5)+3)+32}{x \log (5)+3}\right )\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^x \left (-3 x^2 \log ^2(5)-18 x \log (5)-27\right )+6 \log (5)}{(x \log (5)+3) \left (3 e^x+e^x x \log (5)+10 x \log (5)+32\right ) \log \left (\frac {10 x \log (5)+e^x (x \log (5)+3)+32}{x \log (5)+3}\right ) \log ^2\left (\log \left (\frac {10 x \log (5)+e^x (x \log (5)+3)+32}{x \log (5)+3}\right )\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {6 \left (5 x^2 \log ^2(5)+31 x \log (5)+48+\log (5)\right )}{(x \log (5)+3) \left (3 e^x+e^x x \log (5)+10 x \log (5)+32\right ) \log \left (\frac {10 x \log (5)+e^x (x \log (5)+3)+32}{x \log (5)+3}\right ) \log ^2\left (\log \left (\frac {10 x \log (5)+e^x (x \log (5)+3)+32}{x \log (5)+3}\right )\right )}-\frac {3}{\log \left (\frac {10 x \log (5)+e^x (x \log (5)+3)+32}{x \log (5)+3}\right ) \log ^2\left (\log \left (\frac {10 x \log (5)+e^x (x \log (5)+3)+32}{x \log (5)+3}\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -3 \int \frac {1}{\log \left (\frac {10 \log (5) x+e^x (\log (5) x+3)+32}{\log (5) x+3}\right ) \log ^2\left (\log \left (\frac {10 \log (5) x+e^x (\log (5) x+3)+32}{\log (5) x+3}\right )\right )}dx+96 \int \frac {1}{\left (e^x \log (5) x+10 \log (5) x+3 e^x+32\right ) \log \left (\frac {10 \log (5) x+e^x (\log (5) x+3)+32}{\log (5) x+3}\right ) \log ^2\left (\log \left (\frac {10 \log (5) x+e^x (\log (5) x+3)+32}{\log (5) x+3}\right )\right )}dx+30 \log (5) \int \frac {x}{\left (e^x \log (5) x+10 \log (5) x+3 e^x+32\right ) \log \left (\frac {10 \log (5) x+e^x (\log (5) x+3)+32}{\log (5) x+3}\right ) \log ^2\left (\log \left (\frac {10 \log (5) x+e^x (\log (5) x+3)+32}{\log (5) x+3}\right )\right )}dx+6 \log (5) \int \frac {1}{(\log (5) x+3) \left (e^x \log (5) x+10 \log (5) x+3 e^x+32\right ) \log \left (\frac {10 \log (5) x+e^x (\log (5) x+3)+32}{\log (5) x+3}\right ) \log ^2\left (\log \left (\frac {10 \log (5) x+e^x (\log (5) x+3)+32}{\log (5) x+3}\right )\right )}dx\)

input

Int[(6*Log[5] + E^x*(-27 - 18*x*Log[5] - 3*x^2*Log[5]^2))/((96 + 62*x*Log[ 
5] + 10*x^2*Log[5]^2 + E^x*(9 + 6*x*Log[5] + x^2*Log[5]^2))*Log[(32 + 10*x 
*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]*Log[Log[(32 + 10*x*Log[5] + 
E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]^2),x]
 

output

$Aborted
 

3.4.93.3.1 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] 
]
 

3.4.93.4 Maple [A] (verified)

Time = 97.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10

method	result	size
parallelrisch	\(\frac {3}{\ln \left (\ln \left (\frac {x \,{\mathrm e}^{x} \ln \left (5\right )+10 x \ln \left (5\right )+3 \,{\mathrm e}^{x}+32}{x \ln \left (5\right )+3}\right )\right )}\)	\(33\)
risch	\(\frac {3}{\ln \left (-\ln \left (x \ln \left (5\right )+3\right )+\ln \left (\ln \left (5\right ) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )}{x \ln \left (5\right )+3}\right ) \left (-\operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )}{x \ln \left (5\right )+3}\right )+\operatorname {csgn}\left (\frac {i}{x \ln \left (5\right )+3}\right )\right ) \left (-\operatorname {csgn}\left (\frac {i \left (\ln \left (5\right ) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )}{x \ln \left (5\right )+3}\right )+\operatorname {csgn}\left (i \left (\ln \left (5\right ) \left ({\mathrm e}^{x}+10\right ) x +3 \,{\mathrm e}^{x}+32\right )\right )\right )}{2}\right )}\)	\(149\)

input

int(((-3*x^2*ln(5)^2-18*x*ln(5)-27)*exp(x)+6*ln(5))/((x^2*ln(5)^2+6*x*ln(5 
)+9)*exp(x)+10*x^2*ln(5)^2+62*x*ln(5)+96)/ln(((x*ln(5)+3)*exp(x)+10*x*ln(5 
)+32)/(x*ln(5)+3))/ln(ln(((x*ln(5)+3)*exp(x)+10*x*ln(5)+32)/(x*ln(5)+3)))^ 
2,x,method=_RETURNVERBOSE)
 

output

3/ln(ln((x*exp(x)*ln(5)+10*x*ln(5)+3*exp(x)+32)/(x*ln(5)+3)))
 

3.4.93.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \begin {dmath*} \int \frac {6 \log (5)+e^x \left (-27-18 x \log (5)-3 x^2 \log ^2(5)\right )}{\left (96+62 x \log (5)+10 x^2 \log ^2(5)+e^x \left (9+6 x \log (5)+x^2 \log ^2(5)\right )\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx=\frac {3}{\log \left (\log \left (\frac {{\left (x \log \left (5\right ) + 3\right )} e^{x} + 10 \, x \log \left (5\right ) + 32}{x \log \left (5\right ) + 3}\right )\right )} \end {dmath*}

input

integrate(((-3*x^2*log(5)^2-18*x*log(5)-27)*exp(x)+6*log(5))/((x^2*log(5)^ 
2+6*x*log(5)+9)*exp(x)+10*x^2*log(5)^2+62*x*log(5)+96)/log(((x*log(5)+3)*e 
xp(x)+10*x*log(5)+32)/(x*log(5)+3))/log(log(((x*log(5)+3)*exp(x)+10*x*log( 
5)+32)/(x*log(5)+3)))^2,x, algorithm=\
 

output

3/log(log(((x*log(5) + 3)*e^x + 10*x*log(5) + 32)/(x*log(5) + 3)))
 

3.4.93.6 Sympy [A] (verification not implemented)

Time = 3.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \begin {dmath*} \int \frac {6 \log (5)+e^x \left (-27-18 x \log (5)-3 x^2 \log ^2(5)\right )}{\left (96+62 x \log (5)+10 x^2 \log ^2(5)+e^x \left (9+6 x \log (5)+x^2 \log ^2(5)\right )\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx=\frac {3}{\log {\left (\log {\left (\frac {10 x \log {\left (5 \right )} + \left (x \log {\left (5 \right )} + 3\right ) e^{x} + 32}{x \log {\left (5 \right )} + 3} \right )} \right )}} \end {dmath*}

input

integrate(((-3*x**2*ln(5)**2-18*x*ln(5)-27)*exp(x)+6*ln(5))/((x**2*ln(5)** 
2+6*x*ln(5)+9)*exp(x)+10*x**2*ln(5)**2+62*x*ln(5)+96)/ln(((x*ln(5)+3)*exp( 
x)+10*x*ln(5)+32)/(x*ln(5)+3))/ln(ln(((x*ln(5)+3)*exp(x)+10*x*ln(5)+32)/(x 
*ln(5)+3)))**2,x)
 

output

3/log(log((10*x*log(5) + (x*log(5) + 3)*exp(x) + 32)/(x*log(5) + 3)))
 

3.4.93.7 Maxima [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \begin {dmath*} \int \frac {6 \log (5)+e^x \left (-27-18 x \log (5)-3 x^2 \log ^2(5)\right )}{\left (96+62 x \log (5)+10 x^2 \log ^2(5)+e^x \left (9+6 x \log (5)+x^2 \log ^2(5)\right )\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx=\frac {3}{\log \left (\log \left ({\left (x \log \left (5\right ) + 3\right )} e^{x} + 10 \, x \log \left (5\right ) + 32\right ) - \log \left (x \log \left (5\right ) + 3\right )\right )} \end {dmath*}

input

integrate(((-3*x^2*log(5)^2-18*x*log(5)-27)*exp(x)+6*log(5))/((x^2*log(5)^ 
2+6*x*log(5)+9)*exp(x)+10*x^2*log(5)^2+62*x*log(5)+96)/log(((x*log(5)+3)*e 
xp(x)+10*x*log(5)+32)/(x*log(5)+3))/log(log(((x*log(5)+3)*exp(x)+10*x*log( 
5)+32)/(x*log(5)+3)))^2,x, algorithm=\
 

output

3/log(log((x*log(5) + 3)*e^x + 10*x*log(5) + 32) - log(x*log(5) + 3))
 

3.4.93.8 Giac [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \begin {dmath*} \int \frac {6 \log (5)+e^x \left (-27-18 x \log (5)-3 x^2 \log ^2(5)\right )}{\left (96+62 x \log (5)+10 x^2 \log ^2(5)+e^x \left (9+6 x \log (5)+x^2 \log ^2(5)\right )\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx=\frac {3}{\log \left (\log \left (x e^{x} \log \left (5\right ) + 10 \, x \log \left (5\right ) + 3 \, e^{x} + 32\right ) - \log \left (x \log \left (5\right ) + 3\right )\right )} \end {dmath*}

input

integrate(((-3*x^2*log(5)^2-18*x*log(5)-27)*exp(x)+6*log(5))/((x^2*log(5)^ 
2+6*x*log(5)+9)*exp(x)+10*x^2*log(5)^2+62*x*log(5)+96)/log(((x*log(5)+3)*e 
xp(x)+10*x*log(5)+32)/(x*log(5)+3))/log(log(((x*log(5)+3)*exp(x)+10*x*log( 
5)+32)/(x*log(5)+3)))^2,x, algorithm=\
 

output

3/log(log(x*e^x*log(5) + 10*x*log(5) + 3*e^x + 32) - log(x*log(5) + 3))
 

3.4.93.9 Mupad [B] (verification not implemented)

Time = 15.61 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \begin {dmath*} \int \frac {6 \log (5)+e^x \left (-27-18 x \log (5)-3 x^2 \log ^2(5)\right )}{\left (96+62 x \log (5)+10 x^2 \log ^2(5)+e^x \left (9+6 x \log (5)+x^2 \log ^2(5)\right )\right ) \log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right ) \log ^2\left (\log \left (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}\right )\right )} \, dx=\frac {3}{\ln \left (\ln \left (\frac {10\,x\,\ln \left (5\right )+{\mathrm {e}}^x\,\left (x\,\ln \left (5\right )+3\right )+32}{x\,\ln \left (5\right )+3}\right )\right )} \end {dmath*}

input

int((6*log(5) - exp(x)*(3*x^2*log(5)^2 + 18*x*log(5) + 27))/(log((10*x*log 
(5) + exp(x)*(x*log(5) + 3) + 32)/(x*log(5) + 3))*log(log((10*x*log(5) + e 
xp(x)*(x*log(5) + 3) + 32)/(x*log(5) + 3)))^2*(10*x^2*log(5)^2 + 62*x*log( 
5) + exp(x)*(x^2*log(5)^2 + 6*x*log(5) + 9) + 96)),x)
 

output

3/log(log((10*x*log(5) + exp(x)*(x*log(5) + 3) + 32)/(x*log(5) + 3)))