\(\int \frac {-32 x \log (x)+(1+32 x) \log (\frac {1}{9} e^5 (5+160 x))}{(x+32 x^2) \log ^2(\frac {1}{9} e^5 (5+160 x))} \, dx\) [1771]

Optimal result

Integrand size = 50, antiderivative size = 26 \[ \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{\left (x+32 x^2\right ) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx=\frac {\log (x)}{\log \left (\frac {5 e^5 \left (x+\log \left (e^{32 x^2}\right )\right )}{9 x}\right )} \]

[Out]

ln(x)/ln(5/9/x*(ln(exp(16*x^2)^2)+x)*exp(5))

Rubi [F]

\[ \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{\left (x+32 x^2\right ) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx=\int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{\left (x+32 x^2\right ) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx \]

[In]

Int[(-32*x*Log[x] + (1 + 32*x)*Log[(E^5*(5 + 160*x))/9])/((x + 32*x^2)*Log[(E^5*(5 + 160*x))/9]^2),x]

[Out]

Defer[Int][1/(x*(5 + Log[5/9 + (160*x)/9])), x] - Defer[Subst][Defer[Int][Log[-1/32 + (9*x)/(160*E^5)]/(x*Log[
x]^2), x], x, (5*E^5)/9 + (160*E^5*x)/9]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{x (1+32 x) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx \\ & = \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{x (1+32 x) \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )} \, dx \\ & = \int \left (\frac {32 \left (32 x \log (x)-\log \left (\frac {5}{9} e^5 (1+32 x)\right )-32 x \log \left (\frac {5}{9} e^5 (1+32 x)\right )\right )}{(1+32 x) \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )}+\frac {-32 x \log (x)+\log \left (\frac {5}{9} e^5 (1+32 x)\right )+32 x \log \left (\frac {5}{9} e^5 (1+32 x)\right )}{x \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )}\right ) \, dx \\ & = 32 \int \frac {32 x \log (x)-\log \left (\frac {5}{9} e^5 (1+32 x)\right )-32 x \log \left (\frac {5}{9} e^5 (1+32 x)\right )}{(1+32 x) \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )} \, dx+\int \frac {-32 x \log (x)+\log \left (\frac {5}{9} e^5 (1+32 x)\right )+32 x \log \left (\frac {5}{9} e^5 (1+32 x)\right )}{x \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )} \, dx \\ & = 32 \int \frac {\frac {32 x \log (x)}{1+32 x}-\log \left (\frac {5}{9} e^5 (1+32 x)\right )}{\log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )} \, dx+\int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {5}{9} e^5 (1+32 x)\right )}{x \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )} \, dx \\ & = 32 \int \left (\frac {1}{-5-\log \left (\frac {5}{9} (1+32 x)\right )}+\frac {32 x \log (x)}{(1+32 x) \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )}\right ) \, dx+\int \left (\frac {1}{x \left (5+\log \left (\frac {5}{9}+\frac {160 x}{9}\right )\right )}+\frac {32}{5+\log \left (\frac {5}{9} (1+32 x)\right )}-\frac {32 \log (x)}{\log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )}\right ) \, dx \\ & = 32 \int \frac {1}{-5-\log \left (\frac {5}{9} (1+32 x)\right )} \, dx+32 \int \frac {1}{5+\log \left (\frac {5}{9} (1+32 x)\right )} \, dx-32 \int \frac {\log (x)}{\log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )} \, dx+1024 \int \frac {x \log (x)}{(1+32 x) \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )} \, dx+\int \frac {1}{x \left (5+\log \left (\frac {5}{9}+\frac {160 x}{9}\right )\right )} \, dx \\ & = -\left (32 \int \frac {\log (x)}{\log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )} \, dx\right )+1024 \int \left (\frac {\log (x)}{32 \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )}+\frac {\log (x)}{32 (-1-32 x) \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )}\right ) \, dx+\int \frac {1}{x \left (5+\log \left (\frac {5}{9}+\frac {160 x}{9}\right )\right )} \, dx+\text {Subst}\left (\int \frac {1}{-5-\log \left (\frac {5 x}{9}\right )} \, dx,x,1+32 x\right )+\text {Subst}\left (\int \frac {1}{5+\log \left (\frac {5 x}{9}\right )} \, dx,x,1+32 x\right ) \\ & = \frac {9}{5} \text {Subst}\left (\int \frac {e^x}{-5-x} \, dx,x,\log \left (\frac {5}{9} (1+32 x)\right )\right )+\frac {9}{5} \text {Subst}\left (\int \frac {e^x}{5+x} \, dx,x,\log \left (\frac {5}{9} (1+32 x)\right )\right )+32 \int \frac {\log (x)}{(-1-32 x) \log ^2\left (\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )} \, dx+\int \frac {1}{x \left (5+\log \left (\frac {5}{9}+\frac {160 x}{9}\right )\right )} \, dx \\ & = \frac {9 \text {Subst}\left (\int -\frac {5 e^5 \log \left (-\frac {1}{32}+\frac {9 x}{160 e^5}\right )}{9 x \log ^2(x)} \, dx,x,\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right )}{5 e^5}+\int \frac {1}{x \left (5+\log \left (\frac {5}{9}+\frac {160 x}{9}\right )\right )} \, dx \\ & = \int \frac {1}{x \left (5+\log \left (\frac {5}{9}+\frac {160 x}{9}\right )\right )} \, dx-\text {Subst}\left (\int \frac {\log \left (-\frac {1}{32}+\frac {9 x}{160 e^5}\right )}{x \log ^2(x)} \, dx,x,\frac {5 e^5}{9}+\frac {160 e^5 x}{9}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{\left (x+32 x^2\right ) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx=\frac {\log (x)}{5+\log \left (\frac {5}{9} (1+32 x)\right )} \]

[In]

Integrate[(-32*x*Log[x] + (1 + 32*x)*Log[(E^5*(5 + 160*x))/9])/((x + 32*x^2)*Log[(E^5*(5 + 160*x))/9]^2),x]

[Out]

Log[x]/(5 + Log[(5*(1 + 32*x))/9])

Maple [A] (verified)

Time = 1.60 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62

method	result	size
risch	\(\frac {\ln \left (x \right )}{\ln \left (\frac {\left (160 x +5\right ) {\mathrm e}^{5}}{9}\right )}\)	\(16\)

[In]

int(((32*x+1)*ln(1/9*(160*x+5)*exp(5))-32*x*ln(x))/(32*x^2+x)/ln(1/9*(160*x+5)*exp(5))^2,x,method=_RETURNVERBO
SE)

[Out]

ln(x)/ln(1/9*(160*x+5)*exp(5))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{\left (x+32 x^2\right ) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx=\frac {\log \left (x\right )}{\log \left (\frac {5}{9} \, {\left (32 \, x + 1\right )} e^{5}\right )} \]

[In]

integrate(((32*x+1)*log(1/9*(160*x+5)*exp(5))-32*x*log(x))/(32*x^2+x)/log(1/9*(160*x+5)*exp(5))^2,x, algorithm
="fricas")

[Out]

log(x)/log(5/9*(32*x + 1)*e^5)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.58 \[ \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{\left (x+32 x^2\right ) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx=\frac {\log {\left (x \right )}}{\log {\left (\left (\frac {160 x}{9} + \frac {5}{9}\right ) e^{5} \right )}} \]

[In]

integrate(((32*x+1)*ln(1/9*(160*x+5)*exp(5))-32*x*ln(x))/(32*x**2+x)/ln(1/9*(160*x+5)*exp(5))**2,x)

[Out]

log(x)/log((160*x/9 + 5/9)*exp(5))

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{\left (x+32 x^2\right ) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx=\frac {\log \left (x\right )}{\log \left (5\right ) - 2 \, \log \left (3\right ) + \log \left (32 \, x + 1\right ) + 5} \]

[In]

integrate(((32*x+1)*log(1/9*(160*x+5)*exp(5))-32*x*log(x))/(32*x^2+x)/log(1/9*(160*x+5)*exp(5))^2,x, algorithm
="maxima")

[Out]

log(x)/(log(5) - 2*log(3) + log(32*x + 1) + 5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{\left (x+32 x^2\right ) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx=-\frac {\log \left (x\right )}{2 \, \log \left (3\right ) - \log \left (160 \, x + 5\right ) - 5} \]

[In]

integrate(((32*x+1)*log(1/9*(160*x+5)*exp(5))-32*x*log(x))/(32*x^2+x)/log(1/9*(160*x+5)*exp(5))^2,x, algorithm
="giac")

[Out]

-log(x)/(2*log(3) - log(160*x + 5) - 5)

Mupad [B] (verification not implemented)

Time = 9.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int \frac {-32 x \log (x)+(1+32 x) \log \left (\frac {1}{9} e^5 (5+160 x)\right )}{\left (x+32 x^2\right ) \log ^2\left (\frac {1}{9} e^5 (5+160 x)\right )} \, dx=\frac {\ln \left (x\right )}{\ln \left (\frac {5\,{\mathrm {e}}^5}{9}+\frac {160\,x\,{\mathrm {e}}^5}{9}\right )} \]

[In]

int(-(32*x*log(x) - log((exp(5)*(160*x + 5))/9)*(32*x + 1))/(log((exp(5)*(160*x + 5))/9)^2*(x + 32*x^2)),x)

[Out]

log(x)/log((5*exp(5))/9 + (160*x*exp(5))/9)