∫ 15+8e15−8 log(x) _______________ 4 log(x) x2 log2(x) ______________________________________________ 4x log2(x)+4e15−8 log(x) ______________

Optimal. Leaf size=19 \[ 25+\log \left (e^{2-\frac {15}{4 \log (x)}}+x^2\right ) \]

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Rubi [F] time = 1.85, 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 {15+8 e^{\frac {15-8 \log (x)}{4 \log (x)}} x^2 \log ^2(x)}{4 x \log ^2(x)+4 e^{\frac {15-8 \log (x)}{4 \log (x)}} x^3 \log ^2(x)} \, dx \end {gather*}

Int[(15 + 8*E^((15 - 8*Log[x])/(4*Log[x]))*x^2*Log[x]^2)/(4*x*Log[x]^2 + 4*E^((15 - 8*Log[x])/(4*Log[x]))*x^3*

2*Log[x] - 2*E^2*Defer[Int][1/(x*(E^2 + E^(15/(4*Log[x]))*x^2)), x] + (15*E^2*Defer[Int][1/(x*(E^2 + E^(15/(4*

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15 e^2+8 e^{\frac {15}{4 \log (x)}} x^2 \log ^2(x)}{4 x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)} \, dx\\ &=\frac {1}{4} \int \frac {15 e^2+8 e^{\frac {15}{4 \log (x)}} x^2 \log ^2(x)}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)} \, dx\\ &=\frac {1}{4} \int \left (\frac {8}{x}-\frac {e^2 \left (-15+8 \log ^2(x)\right )}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)}\right ) \, dx\\ &=2 \log (x)-\frac {1}{4} e^2 \int \frac {-15+8 \log ^2(x)}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)} \, dx\\ &=2 \log (x)-\frac {1}{4} e^2 \int \left (\frac {8}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right )}-\frac {15}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)}\right ) \, dx\\ &=2 \log (x)-\left (2 e^2\right ) \int \frac {1}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right )} \, dx+\frac {1}{4} \left (15 e^2\right ) \int \frac {1}{x \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \log ^2(x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.24, size = 28, normalized size = 1.47 \begin {gather*} -\frac {15}{4 \log (x)}+\log \left (e^2+e^{\frac {15}{4 \log (x)}} x^2\right ) \end {gather*}

Integrate[(15 + 8*E^((15 - 8*Log[x])/(4*Log[x]))*x^2*Log[x]^2)/(4*x*Log[x]^2 + 4*E^((15 - 8*Log[x])/(4*Log[x])

-15/(4*Log[x]) + Log[E^2 + E^(15/(4*Log[x]))*x^2]

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fricas [B] time = 0.96, size = 42, normalized size = 2.21 \begin {gather*} \frac {8 \, \log \relax (x)^{2} + 4 \, \log \relax (x) \log \left (\frac {x^{2} e^{\left (-\frac {8 \, \log \relax (x) - 15}{4 \, \log \relax (x)}\right )} + 1}{x^{2}}\right ) - 15}{4 \, \log \relax (x)} \end {gather*}

integrate((8*x^2*log(x)^2*exp(1/4*(-8*log(x)+15)/log(x))+15)/(4*x^3*log(x)^2*exp(1/4*(-8*log(x)+15)/log(x))+4*

1/4*(8*log(x)^2 + 4*log(x)*log((x^2*e^(-1/4*(8*log(x) - 15)/log(x)) + 1)/x^2) - 15)/log(x)

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giac [A] time = 0.20, size = 32, normalized size = 1.68 \begin {gather*} \frac {4 \, \log \left (x^{2} e^{\left (-\frac {8 \, \log \relax (x) - 15}{4 \, \log \relax (x)}\right )} + 1\right ) \log \relax (x) - 15}{4 \, \log \relax (x)} \end {gather*}

integrate((8*x^2*log(x)^2*exp(1/4*(-8*log(x)+15)/log(x))+15)/(4*x^3*log(x)^2*exp(1/4*(-8*log(x)+15)/log(x))+4*

1/4*(4*log(x^2*e^(-1/4*(8*log(x) - 15)/log(x)) + 1)*log(x) - 15)/log(x)

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method	result	size

risch	\(2 \ln \relax (x )-\frac {-8 \ln \relax (x )+15}{4 \ln \relax (x )}+\ln \left ({\mathrm e}^{-\frac {8 \ln \relax (x )-15}{4 \ln \relax (x )}}+\frac {1}{x^{2}}\right )\)	\(36\)

int((8*x^2*ln(x)^2*exp(1/4*(-8*ln(x)+15)/ln(x))+15)/(4*x^3*ln(x)^2*exp(1/4*(-8*ln(x)+15)/ln(x))+4*x*ln(x)^2),x

2*ln(x)-1/4*(-8*ln(x)+15)/ln(x)+ln(exp(-1/4*(8*ln(x)-15)/ln(x))+1/x^2)

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maxima [A] time = 0.47, size = 30, normalized size = 1.58 \begin {gather*} -\frac {15}{4 \, \log \relax (x)} + 2 \, \log \relax (x) + \log \left (\frac {x^{2} e^{\left (\frac {15}{4 \, \log \relax (x)}\right )} + e^{2}}{x^{2}}\right ) \end {gather*}

integrate((8*x^2*log(x)^2*exp(1/4*(-8*log(x)+15)/log(x))+15)/(4*x^3*log(x)^2*exp(1/4*(-8*log(x)+15)/log(x))+4*

-15/4/log(x) + 2*log(x) + log((x^2*e^(15/4/log(x)) + e^2)/x^2)

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mupad [B] time = 4.20, size = 31, normalized size = 1.63 \begin {gather*} \ln \left (\frac {1}{x^2}\right )+\ln \left (x^2\,{\mathrm {e}}^{\frac {15}{4\,\ln \relax (x)}}\,{\mathrm {e}}^{-2}+1\right )+2\,\ln \relax (x)-\frac {15}{4\,\ln \relax (x)} \end {gather*}

int((8*x^2*exp(-(2*log(x) - 15/4)/log(x))*log(x)^2 + 15)/(4*x*log(x)^2 + 4*x^3*exp(-(2*log(x) - 15/4)/log(x))*

log(1/x^2) + log(x^2*exp(15/(4*log(x)))*exp(-2) + 1) + 2*log(x) - 15/(4*log(x))

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sympy [B] time = 0.38, size = 31, normalized size = 1.63 \begin {gather*} 2 \log {\relax (x )} + \log {\left (e^{\frac {\frac {15}{4} - 2 \log {\relax (x )}}{\log {\relax (x )}}} + \frac {1}{x^{2}} \right )} - \frac {15}{4 \log {\relax (x )}} \end {gather*}

integrate((8*x**2*ln(x)**2*exp(1/4*(-8*ln(x)+15)/ln(x))+15)/(4*x**3*ln(x)**2*exp(1/4*(-8*ln(x)+15)/ln(x))+4*x*

2*log(x) + log(exp((15/4 - 2*log(x))/log(x)) + x**(-2)) - 15/(4*log(x))

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3.71.66 \(\int \frac {15+8 e^{\frac {15-8 \log (x)}{4 \log (x)}} x^2 \log ^2(x)}{4 x \log ^2(x)+4 e^{\frac {15-8 \log (x)}{4 \log (x)}} x^3 \log ^2(x)} \, dx\)