\(\int \frac {1}{x^4 (4+6 x)^2} \, dx\) [264]

Optimal result

Integrand size = 11, antiderivative size = 49 \[ \int \frac {1}{x^4 (4+6 x)^2} \, dx=-\frac {1}{48 x^3}+\frac {3}{32 x^2}-\frac {27}{64 x}-\frac {27}{64 (2+3 x)}-\frac {27 \log (x)}{32}+\frac {27}{32} \log (2+3 x) \]

[Out]

-1/48/x^3+3/32/x^2-27/64/x-27/64/(2+3*x)-27/32*ln(x)+27/32*ln(2+3*x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, 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} \[ \int \frac {1}{x^4 (4+6 x)^2} \, dx=-\frac {1}{48 x^3}+\frac {3}{32 x^2}-\frac {27}{64 x}-\frac {27}{64 (3 x+2)}-\frac {27 \log (x)}{32}+\frac {27}{32} \log (3 x+2) \]

[In]

Int[1/(x^4*(4 + 6*x)^2),x]

[Out]

-1/48*1/x^3 + 3/(32*x^2) - 27/(64*x) - 27/(64*(2 + 3*x)) - (27*Log[x])/32 + (27*Log[2 + 3*x])/32

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^4}-\frac {3}{16 x^3}+\frac {27}{64 x^2}-\frac {27}{32 x}+\frac {81}{64 (2+3 x)^2}+\frac {81}{32 (2+3 x)}\right ) \, dx \\ & = -\frac {1}{48 x^3}+\frac {3}{32 x^2}-\frac {27}{64 x}-\frac {27}{64 (2+3 x)}-\frac {27 \log (x)}{32}+\frac {27}{32} \log (2+3 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^4 (4+6 x)^2} \, dx=\frac {1}{192} \left (-\frac {4 \left (2-6 x+27 x^2+81 x^3\right )}{x^3 (2+3 x)}-162 \log (x)+162 \log (2+3 x)\right ) \]

[In]

Integrate[1/(x^4*(4 + 6*x)^2),x]

[Out]

((-4*(2 - 6*x + 27*x^2 + 81*x^3))/(x^3*(2 + 3*x)) - 162*Log[x] + 162*Log[2 + 3*x])/192

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {1}{48 x^{3}}+\frac {3}{32 x^{2}}-\frac {27}{64 x}-\frac {27}{64 \left (2+3 x \right )}-\frac {27 \ln \left (x \right )}{32}+\frac {27 \ln \left (2+3 x \right )}{32}\)	\(38\)
norman	\(\frac {-\frac {1}{24}+\frac {81}{32} x^{4}+\frac {1}{8} x -\frac {9}{16} x^{2}}{x^{3} \left (2+3 x \right )}-\frac {27 \ln \left (x \right )}{32}+\frac {27 \ln \left (2+3 x \right )}{32}\)	\(40\)
risch	\(\frac {-\frac {27}{16} x^{3}-\frac {9}{16} x^{2}+\frac {1}{8} x -\frac {1}{24}}{x^{3} \left (2+3 x \right )}-\frac {27 \ln \left (x \right )}{32}+\frac {27 \ln \left (2+3 x \right )}{32}\)	\(41\)
meijerg	\(-\frac {1}{48 x^{3}}+\frac {3}{32 x^{2}}-\frac {27}{64 x}-\frac {27}{128}-\frac {27 \ln \left (x \right )}{32}-\frac {27 \ln \left (3\right )}{32}+\frac {27 \ln \left (2\right )}{32}+\frac {405 x}{256 \left (5+\frac {15 x}{2}\right )}+\frac {27 \ln \left (1+\frac {3 x}{2}\right )}{32}\)	\(48\)
parallelrisch	\(-\frac {243 \ln \left (x \right ) x^{4}-243 \ln \left (\frac {2}{3}+x \right ) x^{4}+4+162 \ln \left (x \right ) x^{3}-162 \ln \left (\frac {2}{3}+x \right ) x^{3}-243 x^{4}+54 x^{2}-12 x}{96 x^{3} \left (2+3 x \right )}\)	\(60\)

[In]

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

[Out]

-1/48/x^3+3/32/x^2-27/64/x-27/64/(2+3*x)-27/32*ln(x)+27/32*ln(2+3*x)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.31 \[ \int \frac {1}{x^4 (4+6 x)^2} \, dx=-\frac {162 \, x^{3} + 54 \, x^{2} - 81 \, {\left (3 \, x^{4} + 2 \, x^{3}\right )} \log \left (3 \, x + 2\right ) + 81 \, {\left (3 \, x^{4} + 2 \, x^{3}\right )} \log \left (x\right ) - 12 \, x + 4}{96 \, {\left (3 \, x^{4} + 2 \, x^{3}\right )}} \]

[In]

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

[Out]

-1/96*(162*x^3 + 54*x^2 - 81*(3*x^4 + 2*x^3)*log(3*x + 2) + 81*(3*x^4 + 2*x^3)*log(x) - 12*x + 4)/(3*x^4 + 2*x
^3)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^4 (4+6 x)^2} \, dx=- \frac {27 \log {\left (x \right )}}{32} + \frac {27 \log {\left (x + \frac {2}{3} \right )}}{32} + \frac {- 81 x^{3} - 27 x^{2} + 6 x - 2}{144 x^{4} + 96 x^{3}} \]

[In]

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

[Out]

-27*log(x)/32 + 27*log(x + 2/3)/32 + (-81*x**3 - 27*x**2 + 6*x - 2)/(144*x**4 + 96*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^4 (4+6 x)^2} \, dx=-\frac {81 \, x^{3} + 27 \, x^{2} - 6 \, x + 2}{48 \, {\left (3 \, x^{4} + 2 \, x^{3}\right )}} + \frac {27}{32} \, \log \left (3 \, x + 2\right ) - \frac {27}{32} \, \log \left (x\right ) \]

[In]

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

[Out]

-1/48*(81*x^3 + 27*x^2 - 6*x + 2)/(3*x^4 + 2*x^3) + 27/32*log(3*x + 2) - 27/32*log(x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^4 (4+6 x)^2} \, dx=-\frac {27}{64 \, {\left (3 \, x + 2\right )}} - \frac {9 \, {\left (\frac {60}{3 \, x + 2} - \frac {72}{{\left (3 \, x + 2\right )}^{2}} - 13\right )}}{128 \, {\left (\frac {2}{3 \, x + 2} - 1\right )}^{3}} - \frac {27}{32} \, \log \left ({\left | -\frac {2}{3 \, x + 2} + 1 \right |}\right ) \]

[In]

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

[Out]

-27/64/(3*x + 2) - 9/128*(60/(3*x + 2) - 72/(3*x + 2)^2 - 13)/(2/(3*x + 2) - 1)^3 - 27/32*log(abs(-2/(3*x + 2)
+ 1))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^4 (4+6 x)^2} \, dx=\frac {27\,\mathrm {atanh}\left (3\,x+1\right )}{16}-\frac {\frac {9\,x^3}{16}+\frac {3\,x^2}{16}-\frac {x}{24}+\frac {1}{72}}{x^4+\frac {2\,x^3}{3}} \]

[In]

int(1/(x^4*(6*x + 4)^2),x)

[Out]

(27*atanh(3*x + 1))/16 - ((3*x^2)/16 - x/24 + (9*x^3)/16 + 1/72)/((2*x^3)/3 + x^4)