\(\int \frac {\log (2)-\log (2) \log (x)+(1-\log (2)) \log ^2(x)}{\log ^2(x)} \, dx\) [6529]

Optimal result

Integrand size = 25, antiderivative size = 16 \[ \int \frac {\log (2)-\log (2) \log (x)+(1-\log (2)) \log ^2(x)}{\log ^2(x)} \, dx=x+\frac {x \log (2) (-1-\log (x))}{\log (x)} \]

[Out]

x+x/ln(x)*(-ln(x)-1)*ln(2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6874, 2334, 2335} \[ \int \frac {\log (2)-\log (2) \log (x)+(1-\log (2)) \log ^2(x)}{\log ^2(x)} \, dx=x (1-\log (2))-\frac {x \log (2)}{\log (x)} \]

[In]

Int[(Log[2] - Log[2]*Log[x] + (1 - Log[2])*Log[x]^2)/Log[x]^2,x]

[Out]

x*(1 - Log[2]) - (x*Log[2])/Log[x]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \left (1-\log (2)+\frac {\log (2)}{\log ^2(x)}-\frac {\log (2)}{\log (x)}\right ) \, dx \\ & = x (1-\log (2))+\log (2) \int \frac {1}{\log ^2(x)} \, dx-\log (2) \int \frac {1}{\log (x)} \, dx \\ & = x (1-\log (2))-\frac {x \log (2)}{\log (x)}-\log (2) \text {li}(x)+\log (2) \int \frac {1}{\log (x)} \, dx \\ & = x (1-\log (2))-\frac {x \log (2)}{\log (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\log (2)-\log (2) \log (x)+(1-\log (2)) \log ^2(x)}{\log ^2(x)} \, dx=x-x \log (2)-\frac {x \log (2)}{\log (x)} \]

[In]

Integrate[(Log[2] - Log[2]*Log[x] + (1 - Log[2])*Log[x]^2)/Log[x]^2,x]

[Out]

x - x*Log[2] - (x*Log[2])/Log[x]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06

method	result	size
risch	\(-x \ln \left (2\right )+x -\frac {x \ln \left (2\right )}{\ln \left (x \right )}\)	\(17\)
norman	\(\frac {x \left (1-\ln \left (2\right )\right ) \ln \left (x \right )-x \ln \left (2\right )}{\ln \left (x \right )}\)	\(22\)
parallelrisch	\(-\frac {x \ln \left (2\right ) \ln \left (x \right )+x \ln \left (2\right )-x \ln \left (x \right )}{\ln \left (x \right )}\)	\(23\)
default	\(-x \ln \left (2\right )+\ln \left (2\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )+x +\ln \left (2\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )\)	\(36\)
parts	\(-x \ln \left (2\right )+\ln \left (2\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )+x +\ln \left (2\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )\)	\(36\)

[In]

int(((1-ln(2))*ln(x)^2-ln(2)*ln(x)+ln(2))/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-x*ln(2)+x-x*ln(2)/ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38 \[ \int \frac {\log (2)-\log (2) \log (x)+(1-\log (2)) \log ^2(x)}{\log ^2(x)} \, dx=-\frac {x \log \left (2\right ) + {\left (x \log \left (2\right ) - x\right )} \log \left (x\right )}{\log \left (x\right )} \]

[In]

integrate(((1-log(2))*log(x)^2-log(2)*log(x)+log(2))/log(x)^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {\log (2)-\log (2) \log (x)+(1-\log (2)) \log ^2(x)}{\log ^2(x)} \, dx=x \left (1 - \log {\left (2 \right )}\right ) - \frac {x \log {\left (2 \right )}}{\log {\left (x \right )}} \]

[In]

integrate(((1-ln(2))*ln(x)**2-ln(2)*ln(x)+ln(2))/ln(x)**2,x)

[Out]

x*(1 - log(2)) - x*log(2)/log(x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int \frac {\log (2)-\log (2) \log (x)+(1-\log (2)) \log ^2(x)}{\log ^2(x)} \, dx=-x \log \left (2\right ) - {\rm Ei}\left (\log \left (x\right )\right ) \log \left (2\right ) + \Gamma \left (-1, -\log \left (x\right )\right ) \log \left (2\right ) + x \]

[In]

integrate(((1-log(2))*log(x)^2-log(2)*log(x)+log(2))/log(x)^2,x, algorithm="maxima")

[Out]

-x*log(2) - Ei(log(x))*log(2) + gamma(-1, -log(x))*log(2) + x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\log (2)-\log (2) \log (x)+(1-\log (2)) \log ^2(x)}{\log ^2(x)} \, dx=-x \log \left (2\right ) + x - \frac {x \log \left (2\right )}{\log \left (x\right )} \]

[In]

integrate(((1-log(2))*log(x)^2-log(2)*log(x)+log(2))/log(x)^2,x, algorithm="giac")

[Out]

-x*log(2) + x - x*log(2)/log(x)

Mupad [B] (verification not implemented)

Time = 12.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int \frac {\log (2)-\log (2) \log (x)+(1-\log (2)) \log ^2(x)}{\log ^2(x)} \, dx=-x\,\left (\ln \left (2\right )-1\right )-\frac {x\,\ln \left (2\right )}{\ln \left (x\right )} \]

[In]

int(-(log(x)^2*(log(2) - 1) - log(2) + log(2)*log(x))/log(x)^2,x)

[Out]

- x*(log(2) - 1) - (x*log(2))/log(x)