\(\int \frac {1}{x^5 (4+6 x)^2} \, dx\) [265]

Optimal result

Integrand size = 11, antiderivative size = 56 \[ \int \frac {1}{x^5 (4+6 x)^2} \, dx=-\frac {1}{64 x^4}+\frac {1}{16 x^3}-\frac {27}{128 x^2}+\frac {27}{32 x}+\frac {81}{128 (2+3 x)}+\frac {405 \log (x)}{256}-\frac {405}{256} \log (2+3 x) \]

[Out]

-1/64/x^4+1/16/x^3-27/128/x^2+27/32/x+81/128/(2+3*x)+405/256*ln(x)-405/256*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, 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} \[ \int \frac {1}{x^5 (4+6 x)^2} \, dx=-\frac {1}{64 x^4}+\frac {1}{16 x^3}-\frac {27}{128 x^2}+\frac {27}{32 x}+\frac {81}{128 (3 x+2)}+\frac {405 \log (x)}{256}-\frac {405}{256} \log (3 x+2) \]

[In]

Int[1/(x^5*(4 + 6*x)^2),x]

[Out]

-1/64*1/x^4 + 1/(16*x^3) - 27/(128*x^2) + 27/(32*x) + 81/(128*(2 + 3*x)) + (405*Log[x])/256 - (405*Log[2 + 3*x
])/256

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^5}-\frac {3}{16 x^4}+\frac {27}{64 x^3}-\frac {27}{32 x^2}+\frac {405}{256 x}-\frac {243}{128 (2+3 x)^2}-\frac {1215}{256 (2+3 x)}\right ) \, dx \\ & = -\frac {1}{64 x^4}+\frac {1}{16 x^3}-\frac {27}{128 x^2}+\frac {27}{32 x}+\frac {81}{128 (2+3 x)}+\frac {405 \log (x)}{256}-\frac {405}{256} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^5 (4+6 x)^2} \, dx=-\frac {1}{64 x^4}+\frac {1}{16 x^3}-\frac {27}{128 x^2}+\frac {27}{32 x}+\frac {81}{128 (2+3 x)}+\frac {405 \log (x)}{256}-\frac {405}{256} \log (2+3 x) \]

[In]

Integrate[1/(x^5*(4 + 6*x)^2),x]

[Out]

-1/64*1/x^4 + 1/(16*x^3) - 27/(128*x^2) + 27/(32*x) + 81/(128*(2 + 3*x)) + (405*Log[x])/256 - (405*Log[2 + 3*x
])/256

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.77

method	result	size
default	\(-\frac {1}{64 x^{4}}+\frac {1}{16 x^{3}}-\frac {27}{128 x^{2}}+\frac {27}{32 x}+\frac {81}{128 \left (2+3 x \right )}+\frac {405 \ln \left (x \right )}{256}-\frac {405 \ln \left (2+3 x \right )}{256}\)	\(43\)
norman	\(\frac {-\frac {1}{32}-\frac {1215}{256} x^{5}+\frac {5}{64} x -\frac {15}{64} x^{2}+\frac {135}{128} x^{3}}{x^{4} \left (2+3 x \right )}+\frac {405 \ln \left (x \right )}{256}-\frac {405 \ln \left (2+3 x \right )}{256}\)	\(45\)
risch	\(\frac {\frac {405}{128} x^{4}+\frac {135}{128} x^{3}-\frac {15}{64} x^{2}+\frac {5}{64} x -\frac {1}{32}}{x^{4} \left (2+3 x \right )}+\frac {405 \ln \left (x \right )}{256}-\frac {405 \ln \left (2+3 x \right )}{256}\)	\(46\)
meijerg	\(-\frac {1}{64 x^{4}}+\frac {1}{16 x^{3}}-\frac {27}{128 x^{2}}+\frac {27}{32 x}+\frac {81}{256}+\frac {405 \ln \left (x \right )}{256}+\frac {405 \ln \left (3\right )}{256}-\frac {405 \ln \left (2\right )}{256}-\frac {729 x}{256 \left (9 x +6\right )}-\frac {405 \ln \left (1+\frac {3 x}{2}\right )}{256}\)	\(53\)
parallelrisch	\(\frac {1215 \ln \left (x \right ) x^{5}-1215 \ln \left (\frac {2}{3}+x \right ) x^{5}-8+810 \ln \left (x \right ) x^{4}-810 \ln \left (\frac {2}{3}+x \right ) x^{4}-1215 x^{5}+270 x^{3}-60 x^{2}+20 x}{256 x^{4} \left (2+3 x \right )}\)	\(65\)

[In]

int(1/x^5/(4+6*x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/64/x^4+1/16/x^3-27/128/x^2+27/32/x+81/128/(2+3*x)+405/256*ln(x)-405/256*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^5 (4+6 x)^2} \, dx=\frac {810 \, x^{4} + 270 \, x^{3} - 60 \, x^{2} - 405 \, {\left (3 \, x^{5} + 2 \, x^{4}\right )} \log \left (3 \, x + 2\right ) + 405 \, {\left (3 \, x^{5} + 2 \, x^{4}\right )} \log \left (x\right ) + 20 \, x - 8}{256 \, {\left (3 \, x^{5} + 2 \, x^{4}\right )}} \]

[In]

integrate(1/x^5/(4+6*x)^2,x, algorithm="fricas")

[Out]

1/256*(810*x^4 + 270*x^3 - 60*x^2 - 405*(3*x^5 + 2*x^4)*log(3*x + 2) + 405*(3*x^5 + 2*x^4)*log(x) + 20*x - 8)/
(3*x^5 + 2*x^4)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^5 (4+6 x)^2} \, dx=\frac {405 \log {\left (x \right )}}{256} - \frac {405 \log {\left (x + \frac {2}{3} \right )}}{256} + \frac {405 x^{4} + 135 x^{3} - 30 x^{2} + 10 x - 4}{384 x^{5} + 256 x^{4}} \]

[In]

integrate(1/x**5/(4+6*x)**2,x)

[Out]

405*log(x)/256 - 405*log(x + 2/3)/256 + (405*x**4 + 135*x**3 - 30*x**2 + 10*x - 4)/(384*x**5 + 256*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^5 (4+6 x)^2} \, dx=\frac {405 \, x^{4} + 135 \, x^{3} - 30 \, x^{2} + 10 \, x - 4}{128 \, {\left (3 \, x^{5} + 2 \, x^{4}\right )}} - \frac {405}{256} \, \log \left (3 \, x + 2\right ) + \frac {405}{256} \, \log \left (x\right ) \]

[In]

integrate(1/x^5/(4+6*x)^2,x, algorithm="maxima")

[Out]

1/128*(405*x^4 + 135*x^3 - 30*x^2 + 10*x - 4)/(3*x^5 + 2*x^4) - 405/256*log(3*x + 2) + 405/256*log(x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^5 (4+6 x)^2} \, dx=\frac {81}{128 \, {\left (3 \, x + 2\right )}} - \frac {27 \, {\left (\frac {520}{3 \, x + 2} - \frac {1200}{{\left (3 \, x + 2\right )}^{2}} + \frac {960}{{\left (3 \, x + 2\right )}^{3}} - 77\right )}}{1024 \, {\left (\frac {2}{3 \, x + 2} - 1\right )}^{4}} + \frac {405}{256} \, \log \left ({\left | -\frac {2}{3 \, x + 2} + 1 \right |}\right ) \]

[In]

integrate(1/x^5/(4+6*x)^2,x, algorithm="giac")

[Out]

81/128/(3*x + 2) - 27/1024*(520/(3*x + 2) - 1200/(3*x + 2)^2 + 960/(3*x + 2)^3 - 77)/(2/(3*x + 2) - 1)^4 + 405
/256*log(abs(-2/(3*x + 2) + 1))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^5 (4+6 x)^2} \, dx=\frac {\frac {135\,x^4}{128}+\frac {45\,x^3}{128}-\frac {5\,x^2}{64}+\frac {5\,x}{192}-\frac {1}{96}}{x^5+\frac {2\,x^4}{3}}-\frac {405\,\mathrm {atanh}\left (3\,x+1\right )}{128} \]

[In]

int(1/(x^5*(6*x + 4)^2),x)

[Out]

((5*x)/192 - (5*x^2)/64 + (45*x^3)/128 + (135*x^4)/128 - 1/96)/((2*x^4)/3 + x^5) - (405*atanh(3*x + 1))/128