\(\int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx\) [5013]

Optimal result

Integrand size = 13, antiderivative size = 12 \[ \int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx=5 x \left (-\frac {25}{6}-2 x+\log (4)\right ) \]

[Out]

x*(10*ln(2)-125/6-10*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.25, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {9} \[ \int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx=-\frac {5}{288} (24 x+25-6 \log (4))^2 \]

[In]

Int[(-125 - 120*x + 30*Log[4])/6,x]

[Out]

(-5*(25 + 24*x - 6*Log[4])^2)/288

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 {5}{288} (25+24 x-6 \log (4))^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.33 \[ \int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx=-\frac {125 x}{6}-10 x^2+5 x \log (4) \]

[In]

Integrate[(-125 - 120*x + 30*Log[4])/6,x]

[Out]

(-125*x)/6 - 10*x^2 + 5*x*Log[4]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08

method	result	size
gosper	\(\frac {5 x \left (-12 x +12 \ln \left (2\right )-25\right )}{6}\)	\(13\)
default	\(10 x \ln \left (2\right )-10 x^{2}-\frac {125 x}{6}\)	\(15\)
norman	\(\left (10 \ln \left (2\right )-\frac {125}{6}\right ) x -10 x^{2}\)	\(15\)
risch	\(10 x \ln \left (2\right )-10 x^{2}-\frac {125 x}{6}\)	\(15\)
parallelrisch	\(\left (10 \ln \left (2\right )-\frac {125}{6}\right ) x -10 x^{2}\)	\(15\)
parts	\(10 x \ln \left (2\right )-10 x^{2}-\frac {125 x}{6}\)	\(15\)

[In]

int(10*ln(2)-20*x-125/6,x,method=_RETURNVERBOSE)

[Out]

5/6*x*(-12*x+12*ln(2)-25)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx=-10 \, x^{2} + 10 \, x \log \left (2\right ) - \frac {125}{6} \, x \]

[In]

integrate(10*log(2)-20*x-125/6,x, algorithm="fricas")

[Out]

-10*x^2 + 10*x*log(2) - 125/6*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx=- 10 x^{2} + x \left (- \frac {125}{6} + 10 \log {\left (2 \right )}\right ) \]

[In]

integrate(10*ln(2)-20*x-125/6,x)

[Out]

-10*x**2 + x*(-125/6 + 10*log(2))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx=-10 \, x^{2} + 10 \, x \log \left (2\right ) - \frac {125}{6} \, x \]

[In]

integrate(10*log(2)-20*x-125/6,x, algorithm="maxima")

[Out]

-10*x^2 + 10*x*log(2) - 125/6*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx=-10 \, x^{2} + 10 \, x \log \left (2\right ) - \frac {125}{6} \, x \]

[In]

integrate(10*log(2)-20*x-125/6,x, algorithm="giac")

[Out]

-10*x^2 + 10*x*log(2) - 125/6*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {1}{6} (-125-120 x+30 \log (4)) \, dx=x\,\left (10\,\ln \left (2\right )-\frac {125}{6}\right )-10\,x^2 \]

[In]

int(10*log(2) - 20*x - 125/6,x)

[Out]

x*(10*log(2) - 125/6) - 10*x^2