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

Optimal. Leaf size=21 \[ x+\log \left (16 e^{-4+\log (5) \log (3-x)} x^2\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 0.71, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6, 1593, 893} \begin {gather*} x+\log (5) \log (3-x)+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[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 893

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 1593

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 15, normalized size = 0.71 \begin {gather*} x+\log (5) \log (3-x)+2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + Log[5]*Log[3 - x] + 2*Log[x]

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 13, normalized size = 0.62 \begin {gather*} \log \relax (5) \log \left (x - 3\right ) + x + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 15, normalized size = 0.71 \begin {gather*} \log \relax (5) \log \left ({\left | x - 3 \right |}\right ) + x + 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

maple [A]  time = 0.41, size = 14, normalized size = 0.67

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 13, normalized size = 0.62 \begin {gather*} \log \relax (5) \log \left (x - 3\right ) + x + 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 4.01, size = 13, normalized size = 0.62 \begin {gather*} x+2\,\ln \relax (x)+\ln \left (x-3\right )\,\ln \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 2*log(x) + log(x - 3)*log(5)

________________________________________________________________________________________

sympy [A]  time = 0.38, size = 24, normalized size = 1.14 \begin {gather*} x + 2 \log {\relax (x )} + \log {\relax (5 )} \log {\left (x + \frac {6 - 3 \log {\relax (5 )}}{-2 + \log {\relax (5 )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

________________________________________________________________________________________