\(\int \frac {-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)}{10 \log (2)} \, dx\) [7619]

Optimal result

Integrand size = 27, antiderivative size = 21 \[ \int \frac {-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)}{10 \log (2)} \, dx=-\frac {x \log (4) (x-\log (x)+\log (4 x))}{10 \log (2)} \]

[Out]

-1/5*x*(x-ln(x)+ln(4*x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {12, 2332} \[ \int \frac {-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)}{10 \log (2)} \, dx=-\frac {x^2 \log (4)}{10 \log (2)}+\frac {x \log (4) \log (x)}{10 \log (2)}-\frac {x \log (4) \log (4 x)}{10 \log (2)} \]

[In]

Int[(-2*x*Log[4] + Log[4]*Log[x] - Log[4]*Log[4*x])/(10*Log[2]),x]

[Out]

-1/10*(x^2*Log[4])/Log[2] + (x*Log[4]*Log[x])/(10*Log[2]) - (x*Log[4]*Log[4*x])/(10*Log[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)) \, dx}{10 \log (2)} \\ & = -\frac {x^2 \log (4)}{10 \log (2)}+\frac {\log (4) \int \log (x) \, dx}{10 \log (2)}-\frac {\log (4) \int \log (4 x) \, dx}{10 \log (2)} \\ & = -\frac {x^2 \log (4)}{10 \log (2)}+\frac {x \log (4) \log (x)}{10 \log (2)}-\frac {x \log (4) \log (4 x)}{10 \log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)}{10 \log (2)} \, dx=-\frac {\log (4) \left (x^2+x \log (4)\right )}{\log (1024)} \]

[In]

Integrate[(-2*x*Log[4] + Log[4]*Log[x] - Log[4]*Log[4*x])/(10*Log[2]),x]

[Out]

-((Log[4]*(x^2 + x*Log[4]))/Log[1024])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57

method	result	size
risch	\(-\frac {2 x \ln \left (2\right )}{5}-\frac {x^{2}}{5}\)	\(12\)
norman	\(-\frac {x^{2}}{5}+\frac {x \ln \left (x \right )}{5}-\frac {x \ln \left (4 x \right )}{5}\)	\(19\)
parts	\(-\frac {x^{2}}{5}+\frac {x \ln \left (x \right )}{5}-\frac {x \ln \left (4 x \right )}{5}\)	\(19\)
parallelrisch	\(\frac {-2 x^{2} \ln \left (2\right )+2 x \ln \left (2\right ) \ln \left (x \right )-2 \ln \left (2\right ) x \ln \left (4 x \right )}{10 \ln \left (2\right )}\)	\(31\)
default	\(\frac {\ln \left (2\right ) \left (x \ln \left (x \right )-x \right )-x^{2} \ln \left (2\right )-\ln \left (2\right ) \left (x \ln \left (4 x \right )-x \right )}{5 \ln \left (2\right )}\)	\(40\)

[In]

int(1/10*(-2*ln(2)*ln(4*x)+2*ln(2)*ln(x)-4*x*ln(2))/ln(2),x,method=_RETURNVERBOSE)

[Out]

-2/5*x*ln(2)-1/5*x^2

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52 \[ \int \frac {-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)}{10 \log (2)} \, dx=-\frac {1}{5} \, x^{2} - \frac {2}{5} \, x \log \left (2\right ) \]

[In]

integrate(1/10*(-2*log(2)*log(4*x)+2*log(2)*log(x)-4*x*log(2))/log(2),x, algorithm="fricas")

[Out]

-1/5*x^2 - 2/5*x*log(2)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67 \[ \int \frac {-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)}{10 \log (2)} \, dx=- \frac {x^{2}}{5} - \frac {2 x \log {\left (2 \right )}}{5} \]

[In]

integrate(1/10*(-2*ln(2)*ln(4*x)+2*ln(2)*ln(x)-4*x*ln(2))/ln(2),x)

[Out]

-x**2/5 - 2*x*log(2)/5

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (13) = 26\).

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81 \[ \int \frac {-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)}{10 \log (2)} \, dx=-\frac {x^{2} \log \left (2\right ) + {\left (x \log \left (4 \, x\right ) - x\right )} \log \left (2\right ) - {\left (x \log \left (x\right ) - x\right )} \log \left (2\right )}{5 \, \log \left (2\right )} \]

[In]

integrate(1/10*(-2*log(2)*log(4*x)+2*log(2)*log(x)-4*x*log(2))/log(2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (13) = 26\).

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.81 \[ \int \frac {-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)}{10 \log (2)} \, dx=-\frac {x^{2} \log \left (2\right ) + {\left (x \log \left (4 \, x\right ) - x\right )} \log \left (2\right ) - {\left (x \log \left (x\right ) - x\right )} \log \left (2\right )}{5 \, \log \left (2\right )} \]

[In]

integrate(1/10*(-2*log(2)*log(4*x)+2*log(2)*log(x)-4*x*log(2))/log(2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.08 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.33 \[ \int \frac {-2 x \log (4)+\log (4) \log (x)-\log (4) \log (4 x)}{10 \log (2)} \, dx=-\frac {x\,\left (x+\ln \left (4\right )\right )}{5} \]

[In]

int(-((log(4*x)*log(2))/5 + (2*x*log(2))/5 - (log(2)*log(x))/5)/log(2),x)

[Out]

-(x*(x + log(4)))/5