\(\int \frac {1}{x^3 (4+6 x)^2} \, dx\) [263]

Optimal result

Integrand size = 11, antiderivative size = 42 \[ \int \frac {1}{x^3 (4+6 x)^2} \, dx=-\frac {1}{32 x^2}+\frac {3}{16 x}+\frac {9}{32 (2+3 x)}+\frac {27 \log (x)}{64}-\frac {27}{64} \log (2+3 x) \]

[Out]

-1/32/x^2+3/16/x+9/32/(2+3*x)+27/64*ln(x)-27/64*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \[ \int \frac {1}{x^3 (4+6 x)^2} \, dx=-\frac {1}{32 x^2}+\frac {3}{16 x}+\frac {9}{32 (3 x+2)}+\frac {27 \log (x)}{64}-\frac {27}{64} \log (3 x+2) \]

[In]

Int[1/(x^3*(4 + 6*x)^2),x]

[Out]

-1/32*1/x^2 + 3/(16*x) + 9/(32*(2 + 3*x)) + (27*Log[x])/64 - (27*Log[2 + 3*x])/64

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 {1}{16 x^3}-\frac {3}{16 x^2}+\frac {27}{64 x}-\frac {27}{32 (2+3 x)^2}-\frac {81}{64 (2+3 x)}\right ) \, dx \\ & = -\frac {1}{32 x^2}+\frac {3}{16 x}+\frac {9}{32 (2+3 x)}+\frac {27 \log (x)}{64}-\frac {27}{64} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^3 (4+6 x)^2} \, dx=\frac {1}{64} \left (-\frac {2}{x^2}+\frac {12}{x}+\frac {18}{2+3 x}+27 \log (x)-27 \log (2+3 x)\right ) \]

[In]

Integrate[1/(x^3*(4 + 6*x)^2),x]

[Out]

(-2/x^2 + 12/x + 18/(2 + 3*x) + 27*Log[x] - 27*Log[2 + 3*x])/64

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.79

method	result	size
default	\(-\frac {1}{32 x^{2}}+\frac {3}{16 x}+\frac {9}{32 \left (2+3 x \right )}+\frac {27 \ln \left (x \right )}{64}-\frac {27 \ln \left (2+3 x \right )}{64}\)	\(33\)
norman	\(\frac {-\frac {1}{16}-\frac {81}{64} x^{3}+\frac {9}{32} x}{x^{2} \left (2+3 x \right )}+\frac {27 \ln \left (x \right )}{64}-\frac {27 \ln \left (2+3 x \right )}{64}\)	\(35\)
risch	\(\frac {\frac {27}{32} x^{2}+\frac {9}{32} x -\frac {1}{16}}{x^{2} \left (2+3 x \right )}+\frac {27 \ln \left (x \right )}{64}-\frac {27 \ln \left (2+3 x \right )}{64}\)	\(36\)
meijerg	\(-\frac {1}{32 x^{2}}+\frac {3}{16 x}+\frac {9}{64}+\frac {27 \ln \left (x \right )}{64}+\frac {27 \ln \left (3\right )}{64}-\frac {27 \ln \left (2\right )}{64}-\frac {27 x}{32 \left (4+6 x \right )}-\frac {27 \ln \left (1+\frac {3 x}{2}\right )}{64}\)	\(43\)
parallelrisch	\(\frac {81 \ln \left (x \right ) x^{3}-81 \ln \left (\frac {2}{3}+x \right ) x^{3}-4+54 \ln \left (x \right ) x^{2}-54 \ln \left (\frac {2}{3}+x \right ) x^{2}-81 x^{3}+18 x}{64 x^{2} \left (2+3 x \right )}\)	\(55\)

[In]

int(1/x^3/(4+6*x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/32/x^2+3/16/x+9/32/(2+3*x)+27/64*ln(x)-27/64*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.40 \[ \int \frac {1}{x^3 (4+6 x)^2} \, dx=\frac {54 \, x^{2} - 27 \, {\left (3 \, x^{3} + 2 \, x^{2}\right )} \log \left (3 \, x + 2\right ) + 27 \, {\left (3 \, x^{3} + 2 \, x^{2}\right )} \log \left (x\right ) + 18 \, x - 4}{64 \, {\left (3 \, x^{3} + 2 \, x^{2}\right )}} \]

[In]

integrate(1/x^3/(4+6*x)^2,x, algorithm="fricas")

[Out]

1/64*(54*x^2 - 27*(3*x^3 + 2*x^2)*log(3*x + 2) + 27*(3*x^3 + 2*x^2)*log(x) + 18*x - 4)/(3*x^3 + 2*x^2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^3 (4+6 x)^2} \, dx=\frac {27 \log {\left (x \right )}}{64} - \frac {27 \log {\left (x + \frac {2}{3} \right )}}{64} + \frac {27 x^{2} + 9 x - 2}{96 x^{3} + 64 x^{2}} \]

[In]

integrate(1/x**3/(4+6*x)**2,x)

[Out]

27*log(x)/64 - 27*log(x + 2/3)/64 + (27*x**2 + 9*x - 2)/(96*x**3 + 64*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^3 (4+6 x)^2} \, dx=\frac {27 \, x^{2} + 9 \, x - 2}{32 \, {\left (3 \, x^{3} + 2 \, x^{2}\right )}} - \frac {27}{64} \, \log \left (3 \, x + 2\right ) + \frac {27}{64} \, \log \left (x\right ) \]

[In]

integrate(1/x^3/(4+6*x)^2,x, algorithm="maxima")

[Out]

1/32*(27*x^2 + 9*x - 2)/(3*x^3 + 2*x^2) - 27/64*log(3*x + 2) + 27/64*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^3 (4+6 x)^2} \, dx=\frac {9}{32 \, {\left (3 \, x + 2\right )}} - \frac {9 \, {\left (\frac {12}{3 \, x + 2} - 5\right )}}{128 \, {\left (\frac {2}{3 \, x + 2} - 1\right )}^{2}} + \frac {27}{64} \, \log \left ({\left | -\frac {2}{3 \, x + 2} + 1 \right |}\right ) \]

[In]

integrate(1/x^3/(4+6*x)^2,x, algorithm="giac")

[Out]

9/32/(3*x + 2) - 9/128*(12/(3*x + 2) - 5)/(2/(3*x + 2) - 1)^2 + 27/64*log(abs(-2/(3*x + 2) + 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.74 \[ \int \frac {1}{x^3 (4+6 x)^2} \, dx=\frac {\frac {9\,x^2}{32}+\frac {3\,x}{32}-\frac {1}{48}}{x^3+\frac {2\,x^2}{3}}-\frac {27\,\mathrm {atanh}\left (3\,x+1\right )}{32} \]

[In]

int(1/(x^3*(6*x + 4)^2),x)

[Out]

((3*x)/32 + (9*x^2)/32 - 1/48)/((2*x^2)/3 + x^3) - (27*atanh(3*x + 1))/32