\(\int \frac {-4 x+(-3+4 x) \log (\frac {1}{3} (-3+4 x))}{-3 x^2+4 x^3} \, dx\) [7405]

Optimal result

Integrand size = 34, antiderivative size = 13 \[ \int \frac {-4 x+(-3+4 x) \log \left (\frac {1}{3} (-3+4 x)\right )}{-3 x^2+4 x^3} \, dx=-\frac {\log \left (-1+\frac {4 x}{3}\right )}{x} \]

[Out]

-ln(4/3*x-1)/x

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {1607, 6820, 14, 36, 29, 31, 2442} \[ \int \frac {-4 x+(-3+4 x) \log \left (\frac {1}{3} (-3+4 x)\right )}{-3 x^2+4 x^3} \, dx=-\frac {\log \left (\frac {4 x}{3}-1\right )}{x} \]

[In]

Int[(-4*x + (-3 + 4*x)*Log[(-3 + 4*x)/3])/(-3*x^2 + 4*x^3),x]

[Out]

-(Log[-1 + (4*x)/3]/x)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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]

Rule 2442

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f + g*
x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/(g*(q + 1))), x] - Dist[b*e*(n/(g*(q + 1))), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 x+(-3+4 x) \log \left (\frac {1}{3} (-3+4 x)\right )}{x^2 (-3+4 x)} \, dx \\ & = \int \frac {\frac {4 x}{3-4 x}+\log \left (-1+\frac {4 x}{3}\right )}{x^2} \, dx \\ & = \int \left (-\frac {4}{x (-3+4 x)}+\frac {\log \left (-1+\frac {4 x}{3}\right )}{x^2}\right ) \, dx \\ & = -\left (4 \int \frac {1}{x (-3+4 x)} \, dx\right )+\int \frac {\log \left (-1+\frac {4 x}{3}\right )}{x^2} \, dx \\ & = -\frac {\log \left (-1+\frac {4 x}{3}\right )}{x}+\frac {4}{3} \int \frac {1}{x} \, dx+\frac {4}{3} \int \frac {1}{x \left (-1+\frac {4 x}{3}\right )} \, dx-\frac {16}{3} \int \frac {1}{-3+4 x} \, dx \\ & = -\frac {4}{3} \log (3-4 x)+\frac {4 \log (x)}{3}-\frac {\log \left (-1+\frac {4 x}{3}\right )}{x}-\frac {4}{3} \int \frac {1}{x} \, dx+\frac {16}{9} \int \frac {1}{-1+\frac {4 x}{3}} \, dx \\ & = -\frac {\log \left (-1+\frac {4 x}{3}\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x+(-3+4 x) \log \left (\frac {1}{3} (-3+4 x)\right )}{-3 x^2+4 x^3} \, dx=-\frac {\log \left (-1+\frac {4 x}{3}\right )}{x} \]

[In]

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

[Out]

-(Log[-1 + (4*x)/3]/x)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
norman	\(-\frac {\ln \left (\frac {4 x}{3}-1\right )}{x}\)	\(12\)
risch	\(-\frac {\ln \left (\frac {4 x}{3}-1\right )}{x}\)	\(12\)
parallelrisch	\(-\frac {\ln \left (\frac {4 x}{3}-1\right )}{x}\)	\(12\)
derivativedivides	\(-\frac {4 \ln \left (\frac {4 x}{3}-1\right )}{3}+\frac {\ln \left (\frac {4 x}{3}-1\right ) \left (\frac {4 x}{3}-1\right )}{x}\)	\(25\)
default	\(-\frac {4 \ln \left (\frac {4 x}{3}-1\right )}{3}+\frac {\ln \left (\frac {4 x}{3}-1\right ) \left (\frac {4 x}{3}-1\right )}{x}\)	\(25\)
parts	\(-\frac {4 \ln \left (\frac {4 x}{3}\right )}{3}+\frac {\ln \left (\frac {4 x}{3}-1\right ) \left (\frac {4 x}{3}-1\right )}{x}+\frac {4 \ln \left (x \right )}{3}-\frac {4 \ln \left (-3+4 x \right )}{3}\)	\(35\)

[In]

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

[Out]

-ln(4/3*x-1)/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {-4 x+(-3+4 x) \log \left (\frac {1}{3} (-3+4 x)\right )}{-3 x^2+4 x^3} \, dx=-\frac {\log \left (\frac {4}{3} \, x - 1\right )}{x} \]

[In]

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

[Out]

-log(4/3*x - 1)/x

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77 \[ \int \frac {-4 x+(-3+4 x) \log \left (\frac {1}{3} (-3+4 x)\right )}{-3 x^2+4 x^3} \, dx=- \frac {\log {\left (\frac {4 x}{3} - 1 \right )}}{x} \]

[In]

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

[Out]

-log(4*x/3 - 1)/x

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (11) = 22\).

Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.38 \[ \int \frac {-4 x+(-3+4 x) \log \left (\frac {1}{3} (-3+4 x)\right )}{-3 x^2+4 x^3} \, dx=\frac {{\left (4 \, x - 3\right )} \log \left (4 \, x - 3\right ) + 3 \, \log \left (3\right )}{3 \, x} - \frac {4}{3} \, \log \left (4 \, x - 3\right ) \]

[In]

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

[Out]

1/3*((4*x - 3)*log(4*x - 3) + 3*log(3))/x - 4/3*log(4*x - 3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {-4 x+(-3+4 x) \log \left (\frac {1}{3} (-3+4 x)\right )}{-3 x^2+4 x^3} \, dx=-\frac {\log \left (\frac {4}{3} \, x - 1\right )}{x} \]

[In]

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

[Out]

-log(4/3*x - 1)/x

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {-4 x+(-3+4 x) \log \left (\frac {1}{3} (-3+4 x)\right )}{-3 x^2+4 x^3} \, dx=-\frac {\ln \left (\frac {4\,x}{3}-1\right )}{x} \]

[In]

int((4*x - log((4*x)/3 - 1)*(4*x - 3))/(3*x^2 - 4*x^3),x)

[Out]

-log((4*x)/3 - 1)/x