\(\int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx\) [183]

Optimal result

Integrand size = 15, antiderivative size = 43 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {1}{10 (1-4 x)^2}-\frac {3}{25 (1-4 x)}-\frac {9}{125} \log (1-4 x)+\frac {9}{125} \log (2-3 x) \]

[Out]

1/10/(-4*x+1)^2-3/25/(-4*x+1)-9/125*ln(-4*x+1)+9/125*ln(2-3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {46} \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=-\frac {3}{25 (1-4 x)}+\frac {1}{10 (1-4 x)^2}-\frac {9}{125} \log (1-4 x)+\frac {9}{125} \log (2-3 x) \]

[In]

Int[1/((1 - 4*x)^3*(2 - 3*x)),x]

[Out]

1/(10*(1 - 4*x)^2) - 3/(25*(1 - 4*x)) - (9*Log[1 - 4*x])/125 + (9*Log[2 - 3*x])/125

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {27}{125 (-2+3 x)}-\frac {4}{5 (-1+4 x)^3}-\frac {12}{25 (-1+4 x)^2}-\frac {36}{125 (-1+4 x)}\right ) \, dx \\ & = \frac {1}{10 (1-4 x)^2}-\frac {3}{25 (1-4 x)}-\frac {9}{125} \log (1-4 x)+\frac {9}{125} \log (2-3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {-5+120 x+18 (1-4 x)^2 \log (8-12 x)-18 (1-4 x)^2 \log (-1+4 x)}{250 (1-4 x)^2} \]

[In]

Integrate[1/((1 - 4*x)^3*(2 - 3*x)),x]

[Out]

(-5 + 120*x + 18*(1 - 4*x)^2*Log[8 - 12*x] - 18*(1 - 4*x)^2*Log[-1 + 4*x])/(250*(1 - 4*x)^2)

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\frac {12 x}{25}-\frac {1}{50}}{\left (-1+4 x \right )^{2}}+\frac {9 \ln \left (-2+3 x \right )}{125}-\frac {9 \ln \left (-1+4 x \right )}{125}\)	\(32\)
norman	\(\frac {\frac {8}{25} x +\frac {8}{25} x^{2}}{\left (-1+4 x \right )^{2}}+\frac {9 \ln \left (-2+3 x \right )}{125}-\frac {9 \ln \left (-1+4 x \right )}{125}\)	\(35\)
default	\(\frac {1}{10 \left (-1+4 x \right )^{2}}+\frac {3}{25 \left (-1+4 x \right )}-\frac {9 \ln \left (-1+4 x \right )}{125}+\frac {9 \ln \left (-2+3 x \right )}{125}\)	\(36\)
parallelrisch	\(\frac {144 \ln \left (x -\frac {2}{3}\right ) x^{2}-144 \ln \left (x -\frac {1}{4}\right ) x^{2}-72 \ln \left (x -\frac {2}{3}\right ) x +72 \ln \left (x -\frac {1}{4}\right ) x +40 x^{2}+9 \ln \left (x -\frac {2}{3}\right )-9 \ln \left (x -\frac {1}{4}\right )+40 x}{125 \left (-1+4 x \right )^{2}}\)	\(63\)

[In]

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

[Out]

16*(3/100*x-1/800)/(-1+4*x)^2+9/125*ln(-2+3*x)-9/125*ln(-1+4*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=-\frac {18 \, {\left (16 \, x^{2} - 8 \, x + 1\right )} \log \left (4 \, x - 1\right ) - 18 \, {\left (16 \, x^{2} - 8 \, x + 1\right )} \log \left (3 \, x - 2\right ) - 120 \, x + 5}{250 \, {\left (16 \, x^{2} - 8 \, x + 1\right )}} \]

[In]

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

[Out]

-1/250*(18*(16*x^2 - 8*x + 1)*log(4*x - 1) - 18*(16*x^2 - 8*x + 1)*log(3*x - 2) - 120*x + 5)/(16*x^2 - 8*x + 1
)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {24 x - 1}{800 x^{2} - 400 x + 50} + \frac {9 \log {\left (x - \frac {2}{3} \right )}}{125} - \frac {9 \log {\left (x - \frac {1}{4} \right )}}{125} \]

[In]

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

[Out]

(24*x - 1)/(800*x**2 - 400*x + 50) + 9*log(x - 2/3)/125 - 9*log(x - 1/4)/125

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {24 \, x - 1}{50 \, {\left (16 \, x^{2} - 8 \, x + 1\right )}} - \frac {9}{125} \, \log \left (4 \, x - 1\right ) + \frac {9}{125} \, \log \left (3 \, x - 2\right ) \]

[In]

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

[Out]

1/50*(24*x - 1)/(16*x^2 - 8*x + 1) - 9/125*log(4*x - 1) + 9/125*log(3*x - 2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {24 \, x - 1}{50 \, {\left (4 \, x - 1\right )}^{2}} - \frac {9}{125} \, \log \left ({\left | 4 \, x - 1 \right |}\right ) + \frac {9}{125} \, \log \left ({\left | 3 \, x - 2 \right |}\right ) \]

[In]

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

[Out]

1/50*(24*x - 1)/(4*x - 1)^2 - 9/125*log(abs(4*x - 1)) + 9/125*log(abs(3*x - 2))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.58 \[ \int \frac {1}{(1-4 x)^3 (2-3 x)} \, dx=\frac {\frac {3\,x}{100}-\frac {1}{800}}{x^2-\frac {x}{2}+\frac {1}{16}}-\frac {18\,\mathrm {atanh}\left (\frac {24\,x}{5}-\frac {11}{5}\right )}{125} \]

[In]

int(1/((3*x - 2)*(4*x - 1)^3),x)

[Out]

((3*x)/100 - 1/800)/(x^2 - x/2 + 1/16) - (18*atanh((24*x)/5 - 11/5))/125