\(\int \frac {1}{4} (45-86 x+6 x^2) \log (3) \, dx\) [2195]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 28 \[ \int \frac {1}{4} \left (45-86 x+6 x^2\right ) \log (3) \, dx=\left (x+(1-x) \left (x+\frac {1}{4} x (1+3 (12-x)+x)\right )\right ) \log (3) \]

[Out]

ln(3)*((x*(-1/2*x+37/4)+x)*(1-x)+x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12} \[ \int \frac {1}{4} \left (45-86 x+6 x^2\right ) \log (3) \, dx=\frac {1}{2} x^3 \log (3)-\frac {43}{4} x^2 \log (3)+\frac {45}{4} x \log (3) \]

[In]

Int[((45 - 86*x + 6*x^2)*Log[3])/4,x]

[Out]

(45*x*Log[3])/4 - (43*x^2*Log[3])/4 + (x^3*Log[3])/2

Rule 12

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \log (3) \int \left (45-86 x+6 x^2\right ) \, dx \\ & = \frac {45}{4} x \log (3)-\frac {43}{4} x^2 \log (3)+\frac {1}{2} x^3 \log (3) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {1}{4} \left (45-86 x+6 x^2\right ) \log (3) \, dx=\frac {1}{4} \left (45 x-43 x^2+2 x^3\right ) \log (3) \]

[In]

Integrate[((45 - 86*x + 6*x^2)*Log[3])/4,x]

[Out]

((45*x - 43*x^2 + 2*x^3)*Log[3])/4

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.57

method result size
gosper \(\frac {x \left (2 x^{2}-43 x +45\right ) \ln \left (3\right )}{4}\) \(16\)
default \(\frac {\left (2 x^{3}-43 x^{2}+45 x \right ) \ln \left (3\right )}{4}\) \(19\)
parallelrisch \(\frac {\left (2 x^{3}-43 x^{2}+45 x \right ) \ln \left (3\right )}{4}\) \(19\)
norman \(\frac {45 x \ln \left (3\right )}{4}-\frac {43 x^{2} \ln \left (3\right )}{4}+\frac {x^{3} \ln \left (3\right )}{2}\) \(21\)
risch \(\frac {45 x \ln \left (3\right )}{4}-\frac {43 x^{2} \ln \left (3\right )}{4}+\frac {x^{3} \ln \left (3\right )}{2}\) \(21\)

[In]

int(1/4*(6*x^2-86*x+45)*ln(3),x,method=_RETURNVERBOSE)

[Out]

1/4*x*(2*x^2-43*x+45)*ln(3)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {1}{4} \left (45-86 x+6 x^2\right ) \log (3) \, dx=\frac {1}{4} \, {\left (2 \, x^{3} - 43 \, x^{2} + 45 \, x\right )} \log \left (3\right ) \]

[In]

integrate(1/4*(6*x^2-86*x+45)*log(3),x, algorithm="fricas")

[Out]

1/4*(2*x^3 - 43*x^2 + 45*x)*log(3)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1}{4} \left (45-86 x+6 x^2\right ) \log (3) \, dx=\frac {x^{3} \log {\left (3 \right )}}{2} - \frac {43 x^{2} \log {\left (3 \right )}}{4} + \frac {45 x \log {\left (3 \right )}}{4} \]

[In]

integrate(1/4*(6*x**2-86*x+45)*ln(3),x)

[Out]

x**3*log(3)/2 - 43*x**2*log(3)/4 + 45*x*log(3)/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {1}{4} \left (45-86 x+6 x^2\right ) \log (3) \, dx=\frac {1}{4} \, {\left (2 \, x^{3} - 43 \, x^{2} + 45 \, x\right )} \log \left (3\right ) \]

[In]

integrate(1/4*(6*x^2-86*x+45)*log(3),x, algorithm="maxima")

[Out]

1/4*(2*x^3 - 43*x^2 + 45*x)*log(3)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {1}{4} \left (45-86 x+6 x^2\right ) \log (3) \, dx=\frac {1}{4} \, {\left (2 \, x^{3} - 43 \, x^{2} + 45 \, x\right )} \log \left (3\right ) \]

[In]

integrate(1/4*(6*x^2-86*x+45)*log(3),x, algorithm="giac")

[Out]

1/4*(2*x^3 - 43*x^2 + 45*x)*log(3)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54 \[ \int \frac {1}{4} \left (45-86 x+6 x^2\right ) \log (3) \, dx=\frac {x\,\ln \left (3\right )\,\left (2\,x^2-43\,x+45\right )}{4} \]

[In]

int((log(3)*(6*x^2 - 86*x + 45))/4,x)

[Out]

(x*log(3)*(2*x^2 - 43*x + 45))/4