\(\int (-24 x+(48+24 x) \log (2)) \, dx\) [4670]

Optimal result

Integrand size = 12, antiderivative size = 16 \[ \int (-24 x+(48+24 x) \log (2)) \, dx=12 (4+x) (4+3 x+x (-4+\log (2))) \]

[Out]

12*(3*x+x*(ln(2)-4)+4)*(4+x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94, number of steps used = 1, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int (-24 x+(48+24 x) \log (2)) \, dx=12 (x+2)^2 \log (2)-12 x^2 \]

[In]

Int[-24*x + (48 + 24*x)*Log[2],x]

[Out]

-12*x^2 + 12*(2 + x)^2*Log[2]

Rubi steps \begin{align*} \text {integral}& = -12 x^2+12 (2+x)^2 \log (2) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.44 \[ \int (-24 x+(48+24 x) \log (2)) \, dx=24 \left (-\frac {x^2}{2}+\frac {1}{2} x^2 \log (2)+x \log (4)\right ) \]

[In]

Integrate[-24*x + (48 + 24*x)*Log[2],x]

[Out]

24*(-1/2*x^2 + (x^2*Log[2])/2 + x*Log[4])

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00

method	result	size
gosper	\(12 x \left (x \ln \left (2\right )+4 \ln \left (2\right )-x \right )\)	\(16\)
norman	\(\left (12 \ln \left (2\right )-12\right ) x^{2}+48 x \ln \left (2\right )\)	\(17\)
default	\(12 x^{2} \ln \left (2\right )+48 x \ln \left (2\right )-12 x^{2}\)	\(19\)
risch	\(12 x^{2} \ln \left (2\right )+48 x \ln \left (2\right )-12 x^{2}\)	\(19\)
parallelrisch	\(12 x^{2} \ln \left (2\right )+48 x \ln \left (2\right )-12 x^{2}\)	\(19\)
parts	\(12 x^{2} \ln \left (2\right )+48 x \ln \left (2\right )-12 x^{2}\)	\(19\)

[In]

int((24*x+48)*ln(2)-24*x,x,method=_RETURNVERBOSE)

[Out]

12*x*(x*ln(2)+4*ln(2)-x)

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int (-24 x+(48+24 x) \log (2)) \, dx=-12 \, x^{2} + 12 \, {\left (x^{2} + 4 \, x\right )} \log \left (2\right ) \]

[In]

integrate((24*x+48)*log(2)-24*x,x, algorithm="fricas")

[Out]

-12*x^2 + 12*(x^2 + 4*x)*log(2)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int (-24 x+(48+24 x) \log (2)) \, dx=x^{2} \left (-12 + 12 \log {\left (2 \right )}\right ) + 48 x \log {\left (2 \right )} \]

[In]

integrate((24*x+48)*ln(2)-24*x,x)

[Out]

x**2*(-12 + 12*log(2)) + 48*x*log(2)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int (-24 x+(48+24 x) \log (2)) \, dx=-12 \, x^{2} + 12 \, {\left (x^{2} + 4 \, x\right )} \log \left (2\right ) \]

[In]

integrate((24*x+48)*log(2)-24*x,x, algorithm="maxima")

[Out]

-12*x^2 + 12*(x^2 + 4*x)*log(2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06 \[ \int (-24 x+(48+24 x) \log (2)) \, dx=-12 \, x^{2} + 12 \, {\left (x^{2} + 4 \, x\right )} \log \left (2\right ) \]

[In]

integrate((24*x+48)*log(2)-24*x,x, algorithm="giac")

[Out]

-12*x^2 + 12*(x^2 + 4*x)*log(2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int (-24 x+(48+24 x) \log (2)) \, dx=\left (12\,\ln \left (2\right )-12\right )\,x^2+48\,\ln \left (2\right )\,x \]

[In]

int(log(2)*(24*x + 48) - 24*x,x)

[Out]

48*x*log(2) + x^2*(12*log(2) - 12)