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

Optimal. Leaf size=21 \[ \frac {23 (x+\log (x))}{4 x \log (5+4 (5+x))} \]

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Rubi [F]  time = 0.90, 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 {-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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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} \operatorname {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} \operatorname {Subst}\left (\int \frac {-\frac {25}{4}+\frac {x}{4}}{x \log ^2(x)} \, dx,x,25+4 x\right )+\frac {23}{25} \operatorname {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} \operatorname {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} \operatorname {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} \operatorname {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 \text {li}(25+4 x)}{100}+\frac {23}{100} \operatorname {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} \operatorname {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} \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,25+4 x\right )\\ &=-\frac {23}{100} \text {li}(25+4 x)+\frac {23}{100} \operatorname {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} \operatorname {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} \operatorname {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} \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 19, normalized size = 0.90 \begin {gather*} \frac {23 (x+\log (x))}{4 x \log (25+4 x)} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.77, size = 17, normalized size = 0.81 \begin {gather*} \frac {23 \, {\left (x + \log \relax (x)\right )}}{4 \, x \log \left (4 \, x + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.44, size = 17, normalized size = 0.81 \begin {gather*} \frac {23 \, {\left (x + \log \relax (x)\right )}}{4 \, x \log \left (4 \, x + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.04, size = 18, normalized size = 0.86

method	result	size
risch	\(\frac {\frac {23 x}{4}+\frac {23 \ln \relax (x )}{4}}{\ln \left (4 x +25\right ) x}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [A]  time = 0.55, size = 26, normalized size = 1.24 \begin {gather*} \frac {23}{4 \, \log \left (4 \, x + 25\right )} + \frac {23 \, \log \relax (x)}{4 \, x \log \left (4 \, x + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 0.47, size = 17, normalized size = 0.81 \begin {gather*} \frac {23\,\left (x+\ln \relax (x)\right )}{4\,x\,\ln \left (4\,x+25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.24, size = 17, normalized size = 0.81 \begin {gather*} \frac {23 x + 23 \log {\relax (x )}}{4 x \log {\left (4 x + 25 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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