\(\int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log (5-\frac {e}{x}) \log (\log (5-\frac {e}{x}))}{(-5 x-5 x^2+\frac {e (x+x^2)}{x}) \log (5-\frac {e}{x})} \, dx\) [7745]

Optimal result

Integrand size = 95, antiderivative size = 24 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=\left (-e^{16}+(2+2 x)^5\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=\int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx \]

[In]

Int[((E^17*(1 + x))/x + (E*(-1 - x)*(2 + 2*x)^5)/x + (5*E - 25*x)*(2 + 2*x)^5*Log[5 - E/x]*Log[Log[5 - E/x]])/
((-5*x - 5*x^2 + (E*(x + x^2))/x)*Log[5 - E/x]),x]

[Out]

E*Defer[Int][(-32*(1 - E^16/32) - 160*x - 320*x^2 - 320*x^3 - 160*x^4 - 32*x^5)/((E - 5*x)*x*Log[(-E + 5*x)/x]
), x] + 160*Defer[Int][Log[Log[5 - E/x]], x] + 640*Defer[Int][x*Log[Log[5 - E/x]], x] + 960*Defer[Int][x^2*Log
[Log[5 - E/x]], x] + 640*Defer[Int][x^3*Log[Log[5 - E/x]], x] + 160*Defer[Int][x^4*Log[Log[5 - E/x]], x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{17}-32 e (1+x)^5+160 (e-5 x) x (1+x)^4 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{(e-5 x) x \log \left (5-\frac {e}{x}\right )} \, dx \\ & = \int \left (\frac {e \left (-32 \left (1-\frac {e^{16}}{32}\right )-160 x-320 x^2-320 x^3-160 x^4-32 x^5\right )}{(e-5 x) x \log \left (5-\frac {e}{x}\right )}+160 (1+x)^4 \log \left (\log \left (5-\frac {e}{x}\right )\right )\right ) \, dx \\ & = 160 \int (1+x)^4 \log \left (\log \left (5-\frac {e}{x}\right )\right ) \, dx+e \int \frac {-32 \left (1-\frac {e^{16}}{32}\right )-160 x-320 x^2-320 x^3-160 x^4-32 x^5}{(e-5 x) x \log \left (5-\frac {e}{x}\right )} \, dx \\ & = 160 \int \left (\log \left (\log \left (5-\frac {e}{x}\right )\right )+4 x \log \left (\log \left (5-\frac {e}{x}\right )\right )+6 x^2 \log \left (\log \left (5-\frac {e}{x}\right )\right )+4 x^3 \log \left (\log \left (5-\frac {e}{x}\right )\right )+x^4 \log \left (\log \left (5-\frac {e}{x}\right )\right )\right ) \, dx+e \int \frac {-32 \left (1-\frac {e^{16}}{32}\right )-160 x-320 x^2-320 x^3-160 x^4-32 x^5}{(e-5 x) x \log \left (\frac {-e+5 x}{x}\right )} \, dx \\ & = 160 \int \log \left (\log \left (5-\frac {e}{x}\right )\right ) \, dx+160 \int x^4 \log \left (\log \left (5-\frac {e}{x}\right )\right ) \, dx+640 \int x \log \left (\log \left (5-\frac {e}{x}\right )\right ) \, dx+640 \int x^3 \log \left (\log \left (5-\frac {e}{x}\right )\right ) \, dx+960 \int x^2 \log \left (\log \left (5-\frac {e}{x}\right )\right ) \, dx+e \int \frac {-32 \left (1-\frac {e^{16}}{32}\right )-160 x-320 x^2-320 x^3-160 x^4-32 x^5}{(e-5 x) x \log \left (\frac {-e+5 x}{x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=\left (-e^{16}+32 (1+x)^5\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right ) \]

[In]

Integrate[((E^17*(1 + x))/x + (E*(-1 - x)*(2 + 2*x)^5)/x + (5*E - 25*x)*(2 + 2*x)^5*Log[5 - E/x]*Log[Log[5 - E
/x]])/((-5*x - 5*x^2 + (E*(x + x^2))/x)*Log[5 - E/x]),x]

[Out]

(-E^16 + 32*(1 + x)^5)*Log[Log[5 - E/x]]

Maple [F]

\[\int \frac {\left (5 x \,{\mathrm e}^{1-\ln \left (x \right )}-25 x \right ) \left (2+2 x \right )^{5} \ln \left (-{\mathrm e}^{1-\ln \left (x \right )}+5\right ) \ln \left (\ln \left (-{\mathrm e}^{1-\ln \left (x \right )}+5\right )\right )+\left (-1-x \right ) {\mathrm e}^{1-\ln \left (x \right )} \left (2+2 x \right )^{5}+\left (1+x \right ) {\mathrm e}^{16} {\mathrm e}^{1-\ln \left (x \right )}}{\left (\left (x^{2}+x \right ) {\mathrm e}^{1-\ln \left (x \right )}-5 x^{2}-5 x \right ) \ln \left (-{\mathrm e}^{1-\ln \left (x \right )}+5\right )}d x\]

[In]

int(((5*x*exp(1-ln(x))-25*x)*(2+2*x)^5*ln(-exp(1-ln(x))+5)*ln(ln(-exp(1-ln(x))+5))+(-1-x)*exp(1-ln(x))*(2+2*x)
^5+(1+x)*exp(16)*exp(1-ln(x)))/((x^2+x)*exp(1-ln(x))-5*x^2-5*x)/ln(-exp(1-ln(x))+5),x)

[Out]

int(((5*x*exp(1-ln(x))-25*x)*(2+2*x)^5*ln(-exp(1-ln(x))+5)*ln(ln(-exp(1-ln(x))+5))+(-1-x)*exp(1-ln(x))*(2+2*x)
^5+(1+x)*exp(16)*exp(1-ln(x)))/((x^2+x)*exp(1-ln(x))-5*x^2-5*x)/ln(-exp(1-ln(x))+5),x)

Fricas [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx={\left (32 \, x^{5} + 160 \, x^{4} + 320 \, x^{3} + 320 \, x^{2} + 160 \, x - e^{16} + 32\right )} \log \left (\log \left (\frac {5 \, x - e}{x}\right )\right ) \]

[In]

integrate(((5*x*exp(1-log(x))-25*x)*(2+2*x)^5*log(-exp(1-log(x))+5)*log(log(-exp(1-log(x))+5))+(-1-x)*exp(1-lo
g(x))*(2+2*x)^5+(1+x)*exp(16)*exp(1-log(x)))/((x^2+x)*exp(1-log(x))-5*x^2-5*x)/log(-exp(1-log(x))+5),x, algori
thm="fricas")

[Out]

(32*x^5 + 160*x^4 + 320*x^3 + 320*x^2 + 160*x - e^16 + 32)*log(log((5*x - e)/x))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (19) = 38\).

Time = 0.70 (sec) , antiderivative size = 110, normalized size of antiderivative = 4.58 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=\left (32 x^{5} + 160 x^{4} + 320 x^{3} + 320 x^{2} + 160 x - \frac {32 e^{2}}{15} - \frac {16 e}{3} - \frac {48 e^{3}}{125} - \frac {64 e^{4}}{1875} - \frac {16 e^{5}}{13125}\right ) \log {\left (\log {\left (5 - \frac {e}{x} \right )} \right )} - \frac {\left (-420000 - 28000 e^{2} - 70000 e - 5040 e^{3} - 448 e^{4} - 16 e^{5} + 13125 e^{16}\right ) \log {\left (\log {\left (5 - \frac {e}{x} \right )} \right )}}{13125} \]

[In]

integrate(((5*x*exp(1-ln(x))-25*x)*(2+2*x)**5*ln(-exp(1-ln(x))+5)*ln(ln(-exp(1-ln(x))+5))+(-1-x)*exp(1-ln(x))*
(2+2*x)**5+(1+x)*exp(16)*exp(1-ln(x)))/((x**2+x)*exp(1-ln(x))-5*x**2-5*x)/ln(-exp(1-ln(x))+5),x)

[Out]

(32*x**5 + 160*x**4 + 320*x**3 + 320*x**2 + 160*x - 32*exp(2)/15 - 16*E/3 - 48*exp(3)/125 - 64*exp(4)/1875 - 1
6*exp(5)/13125)*log(log(5 - E/x)) - (-420000 - 28000*exp(2) - 70000*E - 5040*exp(3) - 448*exp(4) - 16*exp(5) +
13125*exp(16))*log(log(5 - E/x))/13125

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.88 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx={\left (32 \, x^{5} + 160 \, x^{4} + 320 \, x^{3} + 320 \, x^{2} + 160 \, x - e^{16} + 32\right )} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) \]

[In]

integrate(((5*x*exp(1-log(x))-25*x)*(2+2*x)^5*log(-exp(1-log(x))+5)*log(log(-exp(1-log(x))+5))+(-1-x)*exp(1-lo
g(x))*(2+2*x)^5+(1+x)*exp(16)*exp(1-log(x)))/((x^2+x)*exp(1-log(x))-5*x^2-5*x)/log(-exp(1-log(x))+5),x, algori
thm="maxima")

[Out]

(32*x^5 + 160*x^4 + 320*x^3 + 320*x^2 + 160*x - e^16 + 32)*log(log(5*x - e) - log(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (26) = 52\).

Time = 0.33 (sec) , antiderivative size = 135, normalized size of antiderivative = 5.62 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=32 \, x^{5} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) + 160 \, x^{4} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) + 320 \, x^{3} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) + 320 \, x^{2} \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) + 160 \, x \log \left (\log \left (5 \, x - e\right ) - \log \left (x\right )\right ) - e^{16} \log \left (-\log \left (5 \, x - e\right ) + \log \left (x\right )\right ) + 32 \, \log \left (-\log \left (5 \, x - e\right ) + \log \left (x\right )\right ) \]

[In]

integrate(((5*x*exp(1-log(x))-25*x)*(2+2*x)^5*log(-exp(1-log(x))+5)*log(log(-exp(1-log(x))+5))+(-1-x)*exp(1-lo
g(x))*(2+2*x)^5+(1+x)*exp(16)*exp(1-log(x)))/((x^2+x)*exp(1-log(x))-5*x^2-5*x)/log(-exp(1-log(x))+5),x, algori
thm="giac")

[Out]

32*x^5*log(log(5*x - e) - log(x)) + 160*x^4*log(log(5*x - e) - log(x)) + 320*x^3*log(log(5*x - e) - log(x)) +
320*x^2*log(log(5*x - e) - log(x)) + 160*x*log(log(5*x - e) - log(x)) - e^16*log(-log(5*x - e) + log(x)) + 32*
log(-log(5*x - e) + log(x))

Mupad [B] (verification not implemented)

Time = 12.87 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {\frac {e^{17} (1+x)}{x}+\frac {e (-1-x) (2+2 x)^5}{x}+(5 e-25 x) (2+2 x)^5 \log \left (5-\frac {e}{x}\right ) \log \left (\log \left (5-\frac {e}{x}\right )\right )}{\left (-5 x-5 x^2+\frac {e \left (x+x^2\right )}{x}\right ) \log \left (5-\frac {e}{x}\right )} \, dx=\ln \left (\ln \left (5-\frac {\mathrm {e}}{x}\right )\right )\,\left (32\,x^5+160\,x^4+320\,x^3+320\,x^2+160\,x-{\mathrm {e}}^{16}+32\right ) \]

[In]

int((exp(1 - log(x))*(2*x + 2)^5*(x + 1) - exp(1 - log(x))*exp(16)*(x + 1) + log(5 - exp(1 - log(x)))*log(log(
5 - exp(1 - log(x))))*(25*x - 5*x*exp(1 - log(x)))*(2*x + 2)^5)/(log(5 - exp(1 - log(x)))*(5*x - exp(1 - log(x
))*(x + x^2) + 5*x^2)),x)

[Out]

log(log(5 - exp(1)/x))*(160*x - exp(16) + 320*x^2 + 320*x^3 + 160*x^4 + 32*x^5 + 32)