\(\int \frac {-6-x+x^2+x \log (5)}{-3 x+x^2} \, dx\) [6205]

Optimal result

Integrand size = 22, antiderivative size = 21 \[ \int \frac {-6-x+x^2+x \log (5)}{-3 x+x^2} \, dx=x+\log \left (16 e^{-4+\log (5) \log (3-x)} x^2\right ) \]

[Out]

x+ln(16*exp(ln(5)*ln(-x+3)-4)*x^2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6, 1607, 907} \[ \int \frac {-6-x+x^2+x \log (5)}{-3 x+x^2} \, dx=x+\log (5) \log (3-x)+2 \log (x) \]

[In]

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

[Out]

x + Log[5]*Log[3 - x] + 2*Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-6-x+x^2+x \log (5)}{-3 x+x^2} \, dx=x+\log (5) \log (3-x)+2 \log (x) \]

[In]

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

[Out]

x + Log[5]*Log[3 - x] + 2*Log[x]

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
default	\(x +\ln \left (5\right ) \ln \left (-3+x \right )+2 \ln \left (x \right )\)	\(14\)
norman	\(x +\ln \left (5\right ) \ln \left (-3+x \right )+2 \ln \left (x \right )\)	\(14\)
parallelrisch	\(x +\ln \left (5\right ) \ln \left (-3+x \right )+2 \ln \left (x \right )\)	\(14\)
risch	\(x +2 \ln \left (x \right )+\ln \left (5\right ) \ln \left (-x +3\right )\)	\(16\)
meijerg	\(2 \ln \left (x \right )-2 \ln \left (3\right )+2 i \pi +\ln \left (1-\frac {x}{3}\right )-3 \left (-\frac {\ln \left (5\right )}{3}+\frac {1}{3}\right ) \ln \left (1-\frac {x}{3}\right )+x\)	\(35\)

[In]

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

[Out]

x+ln(5)*ln(-3+x)+2*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {-6-x+x^2+x \log (5)}{-3 x+x^2} \, dx=\log \left (5\right ) \log \left (x - 3\right ) + x + 2 \, \log \left (x\right ) \]

[In]

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

[Out]

log(5)*log(x - 3) + x + 2*log(x)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {-6-x+x^2+x \log (5)}{-3 x+x^2} \, dx=x + 2 \log {\left (x \right )} + \log {\left (5 \right )} \log {\left (x + \frac {6 - 3 \log {\left (5 \right )}}{-2 + \log {\left (5 \right )}} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {-6-x+x^2+x \log (5)}{-3 x+x^2} \, dx=\log \left (5\right ) \log \left (x - 3\right ) + x + 2 \, \log \left (x\right ) \]

[In]

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

[Out]

log(5)*log(x - 3) + x + 2*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-6-x+x^2+x \log (5)}{-3 x+x^2} \, dx=\log \left (5\right ) \log \left ({\left | x - 3 \right |}\right ) + x + 2 \, \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

log(5)*log(abs(x - 3)) + x + 2*log(abs(x))

Mupad [B] (verification not implemented)

Time = 12.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {-6-x+x^2+x \log (5)}{-3 x+x^2} \, dx=x+2\,\ln \left (x\right )+\ln \left (x-3\right )\,\ln \left (5\right ) \]

[In]

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

[Out]

x + 2*log(x) + log(x - 3)*log(5)