\(\int \frac {1}{51} (125-50 x-25 \log (3)) \, dx\) [6381]

Optimal result

Integrand size = 13, antiderivative size = 10 \[ \int \frac {1}{51} (125-50 x-25 \log (3)) \, dx=-\frac {25}{51} x (-5+x+\log (3)) \]

[Out]

-25/51*x*(ln(3)-5+x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.50, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {9} \[ \int \frac {1}{51} (125-50 x-25 \log (3)) \, dx=-\frac {25}{204} (-2 x+5-\log (3))^2 \]

[In]

Int[(125 - 50*x - 25*Log[3])/51,x]

[Out]

(-25*(5 - 2*x - Log[3])^2)/204

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 {25}{204} (5-2 x-\log (3))^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.50 \[ \int \frac {1}{51} (125-50 x-25 \log (3)) \, dx=-\frac {25}{51} \left (-5 x+x^2+x \log (3)\right ) \]

[In]

Integrate[(125 - 50*x - 25*Log[3])/51,x]

[Out]

(-25*(-5*x + x^2 + x*Log[3]))/51

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.90

method	result	size
gosper	\(-\frac {25 x \left (\ln \left (3\right )-5+x \right )}{51}\)	\(9\)
default	\(-\frac {25 x \ln \left (3\right )}{51}-\frac {25 x^{2}}{51}+\frac {125 x}{51}\)	\(15\)
norman	\(\left (-\frac {25 \ln \left (3\right )}{51}+\frac {125}{51}\right ) x -\frac {25 x^{2}}{51}\)	\(15\)
risch	\(-\frac {25 x \ln \left (3\right )}{51}-\frac {25 x^{2}}{51}+\frac {125 x}{51}\)	\(15\)
parallelrisch	\(\left (-\frac {25 \ln \left (3\right )}{51}+\frac {125}{51}\right ) x -\frac {25 x^{2}}{51}\)	\(15\)
parts	\(-\frac {25 x \ln \left (3\right )}{51}-\frac {25 x^{2}}{51}+\frac {125 x}{51}\)	\(15\)

[In]

int(-25/51*ln(3)-50/51*x+125/51,x,method=_RETURNVERBOSE)

[Out]

-25/51*x*(ln(3)-5+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.40 \[ \int \frac {1}{51} (125-50 x-25 \log (3)) \, dx=-\frac {25}{51} \, x^{2} - \frac {25}{51} \, x \log \left (3\right ) + \frac {125}{51} \, x \]

[In]

integrate(-25/51*log(3)-50/51*x+125/51,x, algorithm="fricas")

[Out]

-25/51*x^2 - 25/51*x*log(3) + 125/51*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.70 \[ \int \frac {1}{51} (125-50 x-25 \log (3)) \, dx=- \frac {25 x^{2}}{51} + x \left (\frac {125}{51} - \frac {25 \log {\left (3 \right )}}{51}\right ) \]

[In]

integrate(-25/51*ln(3)-50/51*x+125/51,x)

[Out]

-25*x**2/51 + x*(125/51 - 25*log(3)/51)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.40 \[ \int \frac {1}{51} (125-50 x-25 \log (3)) \, dx=-\frac {25}{51} \, x^{2} - \frac {25}{51} \, x \log \left (3\right ) + \frac {125}{51} \, x \]

[In]

integrate(-25/51*log(3)-50/51*x+125/51,x, algorithm="maxima")

[Out]

-25/51*x^2 - 25/51*x*log(3) + 125/51*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.40 \[ \int \frac {1}{51} (125-50 x-25 \log (3)) \, dx=-\frac {25}{51} \, x^{2} - \frac {25}{51} \, x \log \left (3\right ) + \frac {125}{51} \, x \]

[In]

integrate(-25/51*log(3)-50/51*x+125/51,x, algorithm="giac")

[Out]

-25/51*x^2 - 25/51*x*log(3) + 125/51*x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.80 \[ \int \frac {1}{51} (125-50 x-25 \log (3)) \, dx=-\frac {25\,x\,\left (x+\ln \left (3\right )-5\right )}{51} \]

[In]

int(125/51 - (25*log(3))/51 - (50*x)/51,x)

[Out]

-(25*x*(x + log(3) - 5))/51