∫ 1−2x+(1−x) log(16)+(−1+x) log(−log(15) −1+x ) ____________________________________________________________________________ −3+4x−x2+(−3+4x−x2) log(16)+(−x+x2) log(−log(15) −1+x ) dx

Optimal. Leaf size=25 \[ \log \left (3-x+\frac {x \log \left (\frac {\log (15)}{1-x}\right )}{1+\log (16)}\right ) \]

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Rubi [A] time = 0.25, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6741, 6684} \begin {gather*} \log \left (x \log \left (\frac {\log (15)}{1-x}\right )-x (1+\log (16))+3+\log (4096)\right ) \end {gather*}

Int[(1 - 2*x + (1 - x)*Log[16] + (-1 + x)*Log[-(Log[15]/(-1 + x))])/(-3 + 4*x - x^2 + (-3 + 4*x - x^2)*Log[16]

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /; !Fa

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+2 x-(1-x) \log (16)-(-1+x) \log \left (-\frac {\log (15)}{-1+x}\right )}{(1-x) \left (3 (1+\log (16))-x (1+\log (16))+x \log \left (-\frac {\log (15)}{-1+x}\right )\right )} \, dx\\ &=\log \left (3-x (1+\log (16))+\log (4096)+x \log \left (\frac {\log (15)}{1-x}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.68, size = 41, normalized size = 1.64 \begin {gather*} \log \left (3-x+2 \log (16)-(-1+x) \log (16)+\log \left (-\frac {\log (15)}{-1+x}\right )+(-1+x) \log \left (-\frac {\log (15)}{-1+x}\right )\right ) \end {gather*}

Integrate[(1 - 2*x + (1 - x)*Log[16] + (-1 + x)*Log[-(Log[15]/(-1 + x))])/(-3 + 4*x - x^2 + (-3 + 4*x - x^2)*L

Log[3 - x + 2*Log[16] - (-1 + x)*Log[16] + Log[-(Log[15]/(-1 + x))] + (-1 + x)*Log[-(Log[15]/(-1 + x))]]

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fricas [A] time = 0.84, size = 32, normalized size = 1.28 \begin {gather*} \log \relax (x) + \log \left (-\frac {4 \, {\left (x - 3\right )} \log \relax (2) - x \log \left (-\frac {\log \left (15\right )}{x - 1}\right ) + x - 3}{x}\right ) \end {gather*}

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

log(x) + log(-(4*(x - 3)*log(2) - x*log(-log(15)/(x - 1)) + x - 3)/x)

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giac [B] time = 0.29, size = 86, normalized size = 3.44 \begin {gather*} \frac {\log \left (15\right ) \log \left (4 \, \log \left (15\right ) \log \relax (2) - \log \left (15\right ) \log \left (-\frac {\log \left (15\right )}{x - 1}\right ) - \frac {8 \, \log \left (15\right ) \log \relax (2)}{x - 1} - \frac {\log \left (15\right ) \log \left (-\frac {\log \left (15\right )}{x - 1}\right )}{x - 1} - \frac {2 \, \log \left (15\right )}{x - 1} + \log \left (15\right )\right ) - \log \left (15\right ) \log \left (-\frac {\log \left (15\right )}{x - 1}\right )}{\log \left (15\right )} \end {gather*}

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

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

)/(x - 1))/(x - 1) - 2*log(15)/(x - 1) + log(15)) - log(15)*log(-log(15)/(x - 1)))/log(15)

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method	result	size

norman	\(\ln \left (4 x \ln \relax (2)-\ln \left (-\frac {\ln \left (15\right )}{x -1}\right ) x -12 \ln \relax (2)+x -3\right )\)	\(27\)
risch	\(\ln \relax (x )+\ln \left (\ln \left (-\frac {\ln \relax (3)+\ln \relax (5)}{x -1}\right )-\frac {4 x \ln \relax (2)-12 \ln \relax (2)+x -3}{x}\right )\)	\(36\)

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

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maxima [A] time = 0.58, size = 38, normalized size = 1.52 \begin {gather*} \log \relax (x) + \log \left (\frac {x {\left (4 \, \log \relax (2) - \log \left (\log \relax (5) + \log \relax (3)\right ) + 1\right )} + x \log \left (-x + 1\right ) - 12 \, \log \relax (2) - 3}{x}\right ) \end {gather*}

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

log(x) + log((x*(4*log(2) - log(log(5) + log(3)) + 1) + x*log(-x + 1) - 12*log(2) - 3)/x)

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mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {2\,x+4\,\ln \relax (2)\,\left (x-1\right )-\ln \left (-\frac {\ln \left (15\right )}{x-1}\right )\,\left (x-1\right )-1}{x^2-4\,x+4\,\ln \relax (2)\,\left (x^2-4\,x+3\right )+\ln \left (-\frac {\ln \left (15\right )}{x-1}\right )\,\left (x-x^2\right )+3} \,d x \end {gather*}

int((2*x + 4*log(2)*(x - 1) - log(-log(15)/(x - 1))*(x - 1) - 1)/(x^2 - 4*x + 4*log(2)*(x^2 - 4*x + 3) + log(-

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sympy [A] time = 0.33, size = 31, normalized size = 1.24 \begin {gather*} \log {\relax (x )} + \log {\left (\log {\left (- \frac {\log {\left (15 \right )}}{x - 1} \right )} + \frac {- 4 x \log {\relax (2 )} - x + 3 + 12 \log {\relax (2 )}}{x} \right )} \end {gather*}

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

log(x) + log(log(-log(15)/(x - 1)) + (-4*x*log(2) - x + 3 + 12*log(2))/x)

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3.7.33 \(\int \frac {1-2 x+(1-x) \log (16)+(-1+x) \log (-\frac {\log (15)}{-1+x})}{-3+4 x-x^2+(-3+4 x-x^2) \log (16)+(-x+x^2) \log (-\frac {\log (15)}{-1+x})} \, dx\)