\(\int \frac {-1+(6+2 x-3 x^2) \log (5)}{\log (5)} \, dx\) [7739]

   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 = 20, antiderivative size = 22 \[ \int \frac {-1+\left (6+2 x-3 x^2\right ) \log (5)}{\log (5)} \, dx=-1+2 x-x^3+(2+x)^2-\frac {x}{\log (5)} \]

[Out]

2*x-1-x^3-x/ln(5)+(2+x)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {12} \[ \int \frac {-1+\left (6+2 x-3 x^2\right ) \log (5)}{\log (5)} \, dx=-x^3+x^2+6 x-\frac {x}{\log (5)} \]

[In]

Int[(-1 + (6 + 2*x - 3*x^2)*Log[5])/Log[5],x]

[Out]

6*x + x^2 - x^3 - x/Log[5]

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 {\int \left (-1+\left (6+2 x-3 x^2\right ) \log (5)\right ) \, dx}{\log (5)} \\ & = -\frac {x}{\log (5)}+\int \left (6+2 x-3 x^2\right ) \, dx \\ & = 6 x+x^2-x^3-\frac {x}{\log (5)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-1+\left (6+2 x-3 x^2\right ) \log (5)}{\log (5)} \, dx=6 x+x^2-x^3-\frac {x}{\log (5)} \]

[In]

Integrate[(-1 + (6 + 2*x - 3*x^2)*Log[5])/Log[5],x]

[Out]

6*x + x^2 - x^3 - x/Log[5]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method result size
risch \(-x^{3}+x^{2}+6 x -\frac {x}{\ln \left (5\right )}\) \(20\)
norman \(x^{2}+\frac {\left (6 \ln \left (5\right )-1\right ) x}{\ln \left (5\right )}-x^{3}\) \(22\)
gosper \(-\frac {x \left (x^{2} \ln \left (5\right )-x \ln \left (5\right )-6 \ln \left (5\right )+1\right )}{\ln \left (5\right )}\) \(25\)
parallelrisch \(\frac {\ln \left (5\right ) \left (-x^{3}+x^{2}+6 x \right )-x}{\ln \left (5\right )}\) \(25\)
default \(\frac {-x^{3} \ln \left (5\right )+x^{2} \ln \left (5\right )+6 x \ln \left (5\right )-x}{\ln \left (5\right )}\) \(28\)

[In]

int(((-3*x^2+2*x+6)*ln(5)-1)/ln(5),x,method=_RETURNVERBOSE)

[Out]

-x^3+x^2+6*x-x/ln(5)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-1+\left (6+2 x-3 x^2\right ) \log (5)}{\log (5)} \, dx=-\frac {{\left (x^{3} - x^{2} - 6 \, x\right )} \log \left (5\right ) + x}{\log \left (5\right )} \]

[In]

integrate(((-3*x^2+2*x+6)*log(5)-1)/log(5),x, algorithm="fricas")

[Out]

-((x^3 - x^2 - 6*x)*log(5) + x)/log(5)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-1+\left (6+2 x-3 x^2\right ) \log (5)}{\log (5)} \, dx=- x^{3} + x^{2} + \frac {x \left (-1 + 6 \log {\left (5 \right )}\right )}{\log {\left (5 \right )}} \]

[In]

integrate(((-3*x**2+2*x+6)*ln(5)-1)/ln(5),x)

[Out]

-x**3 + x**2 + x*(-1 + 6*log(5))/log(5)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-1+\left (6+2 x-3 x^2\right ) \log (5)}{\log (5)} \, dx=-\frac {{\left (x^{3} - x^{2} - 6 \, x\right )} \log \left (5\right ) + x}{\log \left (5\right )} \]

[In]

integrate(((-3*x^2+2*x+6)*log(5)-1)/log(5),x, algorithm="maxima")

[Out]

-((x^3 - x^2 - 6*x)*log(5) + x)/log(5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-1+\left (6+2 x-3 x^2\right ) \log (5)}{\log (5)} \, dx=-\frac {{\left (x^{3} - x^{2} - 6 \, x\right )} \log \left (5\right ) + x}{\log \left (5\right )} \]

[In]

integrate(((-3*x^2+2*x+6)*log(5)-1)/log(5),x, algorithm="giac")

[Out]

-((x^3 - x^2 - 6*x)*log(5) + x)/log(5)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-1+\left (6+2 x-3 x^2\right ) \log (5)}{\log (5)} \, dx=-x^3+x^2+\frac {\left (6\,\ln \left (5\right )-1\right )\,x}{\ln \left (5\right )} \]

[In]

int((log(5)*(2*x - 3*x^2 + 6) - 1)/log(5),x)

[Out]

x^2 - x^3 + (x*(6*log(5) - 1))/log(5)