\(\int \frac {-4+3 x+x^2}{(-1+2 x)^2 (3+2 x)} \, dx\) [164]

Optimal result

Integrand size = 23, antiderivative size = 32 \[ \int \frac {-4+3 x+x^2}{(-1+2 x)^2 (3+2 x)} \, dx=-\frac {9}{32 (1-2 x)}+\frac {41}{128} \log (1-2 x)-\frac {25}{128} \log (3+2 x) \]

[Out]

-9/32/(1-2*x)+41/128*ln(1-2*x)-25/128*ln(3+2*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {907} \[ \int \frac {-4+3 x+x^2}{(-1+2 x)^2 (3+2 x)} \, dx=-\frac {9}{32 (1-2 x)}+\frac {41}{128} \log (1-2 x)-\frac {25}{128} \log (2 x+3) \]

[In]

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

[Out]

-9/(32*(1 - 2*x)) + (41*Log[1 - 2*x])/128 - (25*Log[3 + 2*x])/128

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]))

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {9}{16 (-1+2 x)^2}+\frac {41}{64 (-1+2 x)}-\frac {25}{64 (3+2 x)}\right ) \, dx \\ & = -\frac {9}{32 (1-2 x)}+\frac {41}{128} \log (1-2 x)-\frac {25}{128} \log (3+2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {-4+3 x+x^2}{(-1+2 x)^2 (3+2 x)} \, dx=\frac {9}{32 (-1+2 x)}+\frac {41}{128} \log (1-2 x)-\frac {25}{128} \log (3+2 x) \]

[In]

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

[Out]

9/(32*(-1 + 2*x)) + (41*Log[1 - 2*x])/128 - (25*Log[3 + 2*x])/128

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {9}{64 \left (x -\frac {1}{2}\right )}-\frac {25 \ln \left (3+2 x \right )}{128}+\frac {41 \ln \left (2 x -1\right )}{128}\)	\(25\)
default	\(-\frac {25 \ln \left (3+2 x \right )}{128}+\frac {9}{32 \left (2 x -1\right )}+\frac {41 \ln \left (2 x -1\right )}{128}\)	\(27\)
norman	\(\frac {9 x}{16 \left (2 x -1\right )}-\frac {25 \ln \left (3+2 x \right )}{128}+\frac {41 \ln \left (2 x -1\right )}{128}\)	\(28\)
parallelrisch	\(\frac {82 \ln \left (x -\frac {1}{2}\right ) x -50 \ln \left (\frac {3}{2}+x \right ) x -41 \ln \left (x -\frac {1}{2}\right )+25 \ln \left (\frac {3}{2}+x \right )+72 x}{256 x -128}\)	\(40\)

[In]

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

[Out]

9/64/(x-1/2)-25/128*ln(3+2*x)+41/128*ln(2*x-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16 \[ \int \frac {-4+3 x+x^2}{(-1+2 x)^2 (3+2 x)} \, dx=-\frac {25 \, {\left (2 \, x - 1\right )} \log \left (2 \, x + 3\right ) - 41 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 36}{128 \, {\left (2 \, x - 1\right )}} \]

[In]

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

[Out]

-1/128*(25*(2*x - 1)*log(2*x + 3) - 41*(2*x - 1)*log(2*x - 1) - 36)/(2*x - 1)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-4+3 x+x^2}{(-1+2 x)^2 (3+2 x)} \, dx=\frac {41 \log {\left (x - \frac {1}{2} \right )}}{128} - \frac {25 \log {\left (x + \frac {3}{2} \right )}}{128} + \frac {9}{64 x - 32} \]

[In]

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

[Out]

41*log(x - 1/2)/128 - 25*log(x + 3/2)/128 + 9/(64*x - 32)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {-4+3 x+x^2}{(-1+2 x)^2 (3+2 x)} \, dx=\frac {9}{32 \, {\left (2 \, x - 1\right )}} - \frac {25}{128} \, \log \left (2 \, x + 3\right ) + \frac {41}{128} \, \log \left (2 \, x - 1\right ) \]

[In]

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

[Out]

9/32/(2*x - 1) - 25/128*log(2*x + 3) + 41/128*log(2*x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {-4+3 x+x^2}{(-1+2 x)^2 (3+2 x)} \, dx=\frac {9}{32 \, {\left (2 \, x - 1\right )}} - \frac {1}{8} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) - \frac {25}{128} \, \log \left ({\left | -\frac {4}{2 \, x - 1} - 1 \right |}\right ) \]

[In]

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

[Out]

9/32/(2*x - 1) - 1/8*log(1/2*abs(2*x - 1)/(2*x - 1)^2) - 25/128*log(abs(-4/(2*x - 1) - 1))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-4+3 x+x^2}{(-1+2 x)^2 (3+2 x)} \, dx=\frac {41\,\ln \left (x-\frac {1}{2}\right )}{128}-\frac {25\,\ln \left (x+\frac {3}{2}\right )}{128}+\frac {9}{64\,\left (x-\frac {1}{2}\right )} \]

[In]

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

[Out]

(41*log(x - 1/2))/128 - (25*log(x + 3/2))/128 + 9/(64*(x - 1/2))