\(\int \frac {-4+8 x \log (2)}{\log (2)} \, dx\) [8506]

Optimal result

Integrand size = 12, antiderivative size = 24 \[ \int \frac {-4+8 x \log (2)}{\log (2)} \, dx=4 \left (-\frac {1}{9}+x^2+\frac {x-x^2}{x \log (2)}\right ) \]

[Out]

4*(-x^2+x)/x/ln(2)+4*x^2-4/9

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {9} \[ \int \frac {-4+8 x \log (2)}{\log (2)} \, dx=\frac {(1-2 x \log (2))^2}{\log ^2(2)} \]

[In]

Int[(-4 + 8*x*Log[2])/Log[2],x]

[Out]

(1 - 2*x*Log[2])^2/Log[2]^2

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[a*((b + c*x)^2/(2*c)), x] /; FreeQ[{a, b, c}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {(1-2 x \log (2))^2}{\log ^2(2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.54 \[ \int \frac {-4+8 x \log (2)}{\log (2)} \, dx=4 x^2-\frac {4 x}{\log (2)} \]

[In]

Integrate[(-4 + 8*x*Log[2])/Log[2],x]

[Out]

4*x^2 - (4*x)/Log[2]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58

method	result	size
gosper	\(\frac {4 x \left (x \ln \left (2\right )-1\right )}{\ln \left (2\right )}\)	\(14\)
norman	\(4 x^{2}-\frac {4 x}{\ln \left (2\right )}\)	\(14\)
risch	\(4 x^{2}-\frac {4 x}{\ln \left (2\right )}\)	\(14\)
default	\(\frac {4 x^{2} \ln \left (2\right )-4 x}{\ln \left (2\right )}\)	\(17\)
parallelrisch	\(\frac {4 x^{2} \ln \left (2\right )-4 x}{\ln \left (2\right )}\)	\(17\)

[In]

int((8*x*ln(2)-4)/ln(2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-4+8 x \log (2)}{\log (2)} \, dx=\frac {4 \, {\left (x^{2} \log \left (2\right ) - x\right )}}{\log \left (2\right )} \]

[In]

integrate((8*x*log(2)-4)/log(2),x, algorithm="fricas")

[Out]

4*(x^2*log(2) - x)/log(2)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.42 \[ \int \frac {-4+8 x \log (2)}{\log (2)} \, dx=4 x^{2} - \frac {4 x}{\log {\left (2 \right )}} \]

[In]

integrate((8*x*ln(2)-4)/ln(2),x)

[Out]

4*x**2 - 4*x/log(2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-4+8 x \log (2)}{\log (2)} \, dx=\frac {4 \, {\left (x^{2} \log \left (2\right ) - x\right )}}{\log \left (2\right )} \]

[In]

integrate((8*x*log(2)-4)/log(2),x, algorithm="maxima")

[Out]

4*(x^2*log(2) - x)/log(2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-4+8 x \log (2)}{\log (2)} \, dx=\frac {4 \, {\left (x^{2} \log \left (2\right ) - x\right )}}{\log \left (2\right )} \]

[In]

integrate((8*x*log(2)-4)/log(2),x, algorithm="giac")

[Out]

4*(x^2*log(2) - x)/log(2)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {-4+8 x \log (2)}{\log (2)} \, dx=\frac {{\left (8\,x\,\ln \left (2\right )-4\right )}^2}{16\,{\ln \left (2\right )}^2} \]

[In]

int((8*x*log(2) - 4)/log(2),x)

[Out]

(8*x*log(2) - 4)^2/(16*log(2)^2)