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

Optimal. Leaf size=26 \[ \frac {\log (x)}{\log \left (\frac {5 e^5 \left (x+\log \left (e^{32 x^2}\right )\right )}{9 x}\right )} \]

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Rubi [F]  time = 1.48, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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+\operatorname {Subst}\left (\int \frac {1}{-5-\log \left (\frac {5 x}{9}\right )} \, dx,x,1+32 x\right )+\operatorname {Subst}\left (\int \frac {1}{5+\log \left (\frac {5 x}{9}\right )} \, dx,x,1+32 x\right )\\ &=\frac {9}{5} \operatorname {Subst}\left (\int \frac {e^x}{-5-x} \, dx,x,\log \left (\frac {5}{9} (1+32 x)\right )\right )+\frac {9}{5} \operatorname {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 \operatorname {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-\operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 1.23, size = 17, normalized size = 0.65 \begin {gather*} \frac {\log (x)}{5+\log \left (\frac {5}{9} (1+32 x)\right )} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.78, size = 15, normalized size = 0.58 \begin {gather*} \frac {\log \relax (x)}{\log \left (\frac {5}{9} \, {\left (32 \, x + 1\right )} e^{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.23, size = 20, normalized size = 0.77 \begin {gather*} -\frac {\log \relax (x)}{2 \, \log \relax (3) - \log \left (160 \, x + 5\right ) - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.35, size = 16, normalized size = 0.62

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.91, size = 19, normalized size = 0.73 \begin {gather*} \frac {\log \relax (x)}{\log \relax (5) - 2 \, \log \relax (3) + \log \left (32 \, x + 1\right ) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.57, size = 16, normalized size = 0.62 \begin {gather*} \frac {\ln \relax (x)}{\ln \left (\frac {5\,{\mathrm {e}}^5}{9}+\frac {160\,x\,{\mathrm {e}}^5}{9}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.24, size = 15, normalized size = 0.58 \begin {gather*} \frac {\log {\relax (x )}}{\log {\left (\left (\frac {160 x}{9} + \frac {5}{9}\right ) e^{5} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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