\(\int \frac {6 \log (5)+e^x (-27-18 x \log (5)-3 x^2 \log ^2(5))}{(96+62 x \log (5)+10 x^2 \log ^2(5)+e^x (9+6 x \log (5)+x^2 \log ^2(5))) \log (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}) \log ^2(\log (\frac {32+10 x \log (5)+e^x (3+x \log (5))}{3+x \log (5)}))} \, dx\) [9393]

Optimal result

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

[Out]

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

Rubi [F]

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

[In]

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]

[Out]

-3*Defer[Int][1/(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] + 96*Defer[Int][1/((32 + 3*E^x + 10*x*Log[5] + E^x*x*Log[5])*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] + 30*Log[5]*Defer[Int][x/((32 + 3*E^x + 10*x*Log[5] + E^x*x*Log[5])*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] + 6*Log[
5]*Defer[Int][1/((3 + x*Log[5])*(32 + 3*E^x + 10*x*Log[5] + E^x*x*Log[5])*Log[(32 + 10*x*Log[5] + E^x*(3 + x*L
og[5]))/(3 + x*Log[5])]*Log[Log[(32 + 10*x*Log[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {3 \left (2 \log (5)-e^x (3+x \log (5))^2\right )}{(3+x \log (5)) \left (32+10 x \log (5)+e^x (3+x \log (5))\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 \\ & = 3 \int \frac {2 \log (5)-e^x (3+x \log (5))^2}{(3+x \log (5)) \left (32+10 x \log (5)+e^x (3+x \log (5))\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 \\ & = 3 \int \left (-\frac {1}{\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 )}+\frac {2 \left (48+\log (5)+31 x \log (5)+5 x^2 \log ^2(5)\right )}{(3+x \log (5)) \left (32+3 e^x+10 x \log (5)+e^x x \log (5)\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 )}\right ) \, dx \\ & = -\left (3 \int \frac {1}{\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\right )+6 \int \frac {48+\log (5)+31 x \log (5)+5 x^2 \log ^2(5)}{(3+x \log (5)) \left (32+3 e^x+10 x \log (5)+e^x x \log (5)\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 \\ & = -\left (3 \int \frac {1}{\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\right )+6 \int \left (\frac {16}{\left (32+3 e^x+10 x \log (5)+e^x x \log (5)\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 )}+\frac {5 x \log (5)}{\left (32+3 e^x+10 x \log (5)+e^x x \log (5)\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 )}+\frac {\log (5)}{(3+x \log (5)) \left (32+3 e^x+10 x \log (5)+e^x x \log (5)\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 )}\right ) \, dx \\ & = -\left (3 \int \frac {1}{\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\right )+96 \int \frac {1}{\left (32+3 e^x+10 x \log (5)+e^x x \log (5)\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+(6 \log (5)) \int \frac {1}{(3+x \log (5)) \left (32+3 e^x+10 x \log (5)+e^x x \log (5)\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+(30 \log (5)) \int \frac {x}{\left (32+3 e^x+10 x \log (5)+e^x x \log (5)\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 \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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*L
og[5] + E^x*(3 + x*Log[5]))/(3 + x*Log[5])]]^2),x]

[Out]

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

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

[In]

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)

[Out]

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

Fricas [A] (verification not implemented)

none

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

[In]

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)*exp(x)+10*x*log(5)+32)/(x*log(5)+3))/log(log(((x*log(5)+3)*exp(x)+10*x*lo
g(5)+32)/(x*log(5)+3)))^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

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

[In]

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)

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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)*exp(x)+10*x*log(5)+32)/(x*log(5)+3))/log(log(((x*log(5)+3)*exp(x)+10*x*lo
g(5)+32)/(x*log(5)+3)))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

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

[In]

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)*exp(x)+10*x*log(5)+32)/(x*log(5)+3))/log(log(((x*log(5)+3)*exp(x)+10*x*lo
g(5)+32)/(x*log(5)+3)))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

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

[In]

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) + exp(x)*(x*log(5) + 3) + 32)/(x*log(5) + 3)))^2*(10*x^2*log(5)^2 + 62*x*lo
g(5) + exp(x)*(x^2*log(5)^2 + 6*x*log(5) + 9) + 96)),x)

[Out]

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