\(\int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{(100 x^2+16 x^3) \log ^2(25+4 x)} \, dx\) [40]

Optimal result

Integrand size = 53, antiderivative size = 21 \[ \int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{\left (100 x^2+16 x^3\right ) \log ^2(25+4 x)} \, dx=\frac {23 (x+\log (x))}{4 x \log (5+4 (5+x))} \]

[Out]

23/4*(x+ln(x))/ln(4*x+25)/x

Rubi [F]

\[ \int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{\left (100 x^2+16 x^3\right ) \log ^2(25+4 x)} \, dx=\int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{\left (100 x^2+16 x^3\right ) \log ^2(25+4 x)} \, dx \]

[In]

Int[(-92*x^2 - 92*x*Log[x] + (575 + 92*x + (-575 - 92*x)*Log[x])*Log[25 + 4*x])/((100*x^2 + 16*x^3)*Log[25 + 4
*x]^2),x]

[Out]

23/(4*Log[25 + 4*x]) - (23*Defer[Int][Log[x]/(x*Log[25 + 4*x]^2), x])/25 - (23*Defer[Int][(-1 + Log[x])/(x^2*L
og[25 + 4*x]), x])/4 + (23*Defer[Subst][Defer[Int][Log[-25/4 + x/4]/(x*Log[x]^2), x], x, 25 + 4*x])/25

Rubi steps \begin{align*} \text {integral}& = \int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{x^2 (100+16 x) \log ^2(25+4 x)} \, dx \\ & = \int \left (-\frac {23 (x+\log (x))}{x (25+4 x) \log ^2(25+4 x)}-\frac {23 (-1+\log (x))}{4 x^2 \log (25+4 x)}\right ) \, dx \\ & = -\left (\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx\right )-23 \int \frac {x+\log (x)}{x (25+4 x) \log ^2(25+4 x)} \, dx \\ & = -\left (\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx\right )-23 \int \left (\frac {x+\log (x)}{25 x \log ^2(25+4 x)}-\frac {4 (x+\log (x))}{25 (25+4 x) \log ^2(25+4 x)}\right ) \, dx \\ & = -\left (\frac {23}{25} \int \frac {x+\log (x)}{x \log ^2(25+4 x)} \, dx\right )+\frac {92}{25} \int \frac {x+\log (x)}{(25+4 x) \log ^2(25+4 x)} \, dx-\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx \\ & = -\left (\frac {23}{25} \int \left (\frac {1}{\log ^2(25+4 x)}+\frac {\log (x)}{x \log ^2(25+4 x)}\right ) \, dx\right )+\frac {92}{25} \int \left (\frac {x}{(25+4 x) \log ^2(25+4 x)}+\frac {\log (x)}{(25+4 x) \log ^2(25+4 x)}\right ) \, dx-\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx \\ & = -\left (\frac {23}{25} \int \frac {1}{\log ^2(25+4 x)} \, dx\right )-\frac {23}{25} \int \frac {\log (x)}{x \log ^2(25+4 x)} \, dx+\frac {92}{25} \int \frac {x}{(25+4 x) \log ^2(25+4 x)} \, dx+\frac {92}{25} \int \frac {\log (x)}{(25+4 x) \log ^2(25+4 x)} \, dx-\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx \\ & = -\left (\frac {23}{100} \text {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,25+4 x\right )\right )-\frac {23}{25} \int \frac {\log (x)}{x \log ^2(25+4 x)} \, dx+\frac {23}{25} \text {Subst}\left (\int \frac {-\frac {25}{4}+\frac {x}{4}}{x \log ^2(x)} \, dx,x,25+4 x\right )+\frac {23}{25} \text {Subst}\left (\int \frac {\log \left (-\frac {25}{4}+\frac {x}{4}\right )}{x \log ^2(x)} \, dx,x,25+4 x\right )-\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx \\ & = \frac {23 (25+4 x)}{100 \log (25+4 x)}-\frac {23}{100} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,25+4 x\right )-\frac {23}{25} \int \frac {\log (x)}{x \log ^2(25+4 x)} \, dx+\frac {23}{25} \text {Subst}\left (\int \left (\frac {1}{4 \log ^2(x)}-\frac {25}{4 x \log ^2(x)}\right ) \, dx,x,25+4 x\right )+\frac {23}{25} \text {Subst}\left (\int \frac {\log \left (-\frac {25}{4}+\frac {x}{4}\right )}{x \log ^2(x)} \, dx,x,25+4 x\right )-\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx \\ & = \frac {23 (25+4 x)}{100 \log (25+4 x)}-\frac {23 \operatorname {LogIntegral}(25+4 x)}{100}+\frac {23}{100} \text {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,25+4 x\right )-\frac {23}{25} \int \frac {\log (x)}{x \log ^2(25+4 x)} \, dx+\frac {23}{25} \text {Subst}\left (\int \frac {\log \left (-\frac {25}{4}+\frac {x}{4}\right )}{x \log ^2(x)} \, dx,x,25+4 x\right )-\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx-\frac {23}{4} \text {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,25+4 x\right ) \\ & = -\frac {23}{100} \operatorname {LogIntegral}(25+4 x)+\frac {23}{100} \text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,25+4 x\right )-\frac {23}{25} \int \frac {\log (x)}{x \log ^2(25+4 x)} \, dx+\frac {23}{25} \text {Subst}\left (\int \frac {\log \left (-\frac {25}{4}+\frac {x}{4}\right )}{x \log ^2(x)} \, dx,x,25+4 x\right )-\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx-\frac {23}{4} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (25+4 x)\right ) \\ & = \frac {23}{4 \log (25+4 x)}-\frac {23}{25} \int \frac {\log (x)}{x \log ^2(25+4 x)} \, dx+\frac {23}{25} \text {Subst}\left (\int \frac {\log \left (-\frac {25}{4}+\frac {x}{4}\right )}{x \log ^2(x)} \, dx,x,25+4 x\right )-\frac {23}{4} \int \frac {-1+\log (x)}{x^2 \log (25+4 x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{\left (100 x^2+16 x^3\right ) \log ^2(25+4 x)} \, dx=\frac {23 (x+\log (x))}{4 x \log (25+4 x)} \]

[In]

Integrate[(-92*x^2 - 92*x*Log[x] + (575 + 92*x + (-575 - 92*x)*Log[x])*Log[25 + 4*x])/((100*x^2 + 16*x^3)*Log[
25 + 4*x]^2),x]

[Out]

(23*(x + Log[x]))/(4*x*Log[25 + 4*x])

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {\frac {23 x}{4}+\frac {23 \ln \left (x \right )}{4}}{\ln \left (4 x +25\right ) x}\)	\(18\)
parallelrisch	\(-\frac {-9200 x -9200 \ln \left (x \right )}{1600 x \ln \left (4 x +25\right )}\)	\(22\)

[In]

int((((-92*x-575)*ln(x)+92*x+575)*ln(4*x+25)-92*x*ln(x)-92*x^2)/(16*x^3+100*x^2)/ln(4*x+25)^2,x,method=_RETURN
VERBOSE)

[Out]

23/4*(x+ln(x))/ln(4*x+25)/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{\left (100 x^2+16 x^3\right ) \log ^2(25+4 x)} \, dx=\frac {23 \, {\left (x + \log \left (x\right )\right )}}{4 \, x \log \left (4 \, x + 25\right )} \]

[In]

integrate((((-92*x-575)*log(x)+92*x+575)*log(4*x+25)-92*x*log(x)-92*x^2)/(16*x^3+100*x^2)/log(4*x+25)^2,x, alg
orithm="fricas")

[Out]

23/4*(x + log(x))/(x*log(4*x + 25))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{\left (100 x^2+16 x^3\right ) \log ^2(25+4 x)} \, dx=\frac {23 x + 23 \log {\left (x \right )}}{4 x \log {\left (4 x + 25 \right )}} \]

[In]

integrate((((-92*x-575)*ln(x)+92*x+575)*ln(4*x+25)-92*x*ln(x)-92*x**2)/(16*x**3+100*x**2)/ln(4*x+25)**2,x)

[Out]

(23*x + 23*log(x))/(4*x*log(4*x + 25))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{\left (100 x^2+16 x^3\right ) \log ^2(25+4 x)} \, dx=\frac {23}{4 \, \log \left (4 \, x + 25\right )} + \frac {23 \, \log \left (x\right )}{4 \, x \log \left (4 \, x + 25\right )} \]

[In]

integrate((((-92*x-575)*log(x)+92*x+575)*log(4*x+25)-92*x*log(x)-92*x^2)/(16*x^3+100*x^2)/log(4*x+25)^2,x, alg
orithm="maxima")

[Out]

23/4/log(4*x + 25) + 23/4*log(x)/(x*log(4*x + 25))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{\left (100 x^2+16 x^3\right ) \log ^2(25+4 x)} \, dx=\frac {23 \, {\left (x + \log \left (x\right )\right )}}{4 \, x \log \left (4 \, x + 25\right )} \]

[In]

integrate((((-92*x-575)*log(x)+92*x+575)*log(4*x+25)-92*x*log(x)-92*x^2)/(16*x^3+100*x^2)/log(4*x+25)^2,x, alg
orithm="giac")

[Out]

23/4*(x + log(x))/(x*log(4*x + 25))

Mupad [B] (verification not implemented)

Time = 8.46 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-92 x^2-92 x \log (x)+(575+92 x+(-575-92 x) \log (x)) \log (25+4 x)}{\left (100 x^2+16 x^3\right ) \log ^2(25+4 x)} \, dx=\frac {23\,\left (x+\ln \left (x\right )\right )}{4\,x\,\ln \left (4\,x+25\right )} \]

[In]

int(-(92*x*log(x) - log(4*x + 25)*(92*x - log(x)*(92*x + 575) + 575) + 92*x^2)/(log(4*x + 25)^2*(100*x^2 + 16*
x^3)),x)

[Out]

(23*(x + log(x)))/(4*x*log(4*x + 25))