\(\int \frac {1}{(a+\frac {b}{x})^2 x^6} \, dx\) [1630]

Optimal result

Integrand size = 13, antiderivative size = 69 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^6} \, dx=-\frac {1}{3 b^2 x^3}+\frac {a}{b^3 x^2}-\frac {3 a^2}{b^4 x}-\frac {a^3}{b^4 (b+a x)}-\frac {4 a^3 \log (x)}{b^5}+\frac {4 a^3 \log (b+a x)}{b^5} \]

[Out]

-1/3/b^2/x^3+a/b^3/x^2-3*a^2/b^4/x-a^3/b^4/(a*x+b)-4*a^3*ln(x)/b^5+4*a^3*ln(a*x+b)/b^5

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {269, 46} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^6} \, dx=-\frac {4 a^3 \log (x)}{b^5}+\frac {4 a^3 \log (a x+b)}{b^5}-\frac {a^3}{b^4 (a x+b)}-\frac {3 a^2}{b^4 x}+\frac {a}{b^3 x^2}-\frac {1}{3 b^2 x^3} \]

[In]

Int[1/((a + b/x)^2*x^6),x]

[Out]

-1/3*1/(b^2*x^3) + a/(b^3*x^2) - (3*a^2)/(b^4*x) - a^3/(b^4*(b + a*x)) - (4*a^3*Log[x])/b^5 + (4*a^3*Log[b + a
*x])/b^5

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])

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^4 (b+a x)^2} \, dx \\ & = \int \left (\frac {1}{b^2 x^4}-\frac {2 a}{b^3 x^3}+\frac {3 a^2}{b^4 x^2}-\frac {4 a^3}{b^5 x}+\frac {a^4}{b^4 (b+a x)^2}+\frac {4 a^4}{b^5 (b+a x)}\right ) \, dx \\ & = -\frac {1}{3 b^2 x^3}+\frac {a}{b^3 x^2}-\frac {3 a^2}{b^4 x}-\frac {a^3}{b^4 (b+a x)}-\frac {4 a^3 \log (x)}{b^5}+\frac {4 a^3 \log (b+a x)}{b^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^6} \, dx=-\frac {\frac {b \left (b^3-2 a b^2 x+6 a^2 b x^2+12 a^3 x^3\right )}{x^3 (b+a x)}+12 a^3 \log (x)-12 a^3 \log (b+a x)}{3 b^5} \]

[In]

Integrate[1/((a + b/x)^2*x^6),x]

[Out]

-1/3*((b*(b^3 - 2*a*b^2*x + 6*a^2*b*x^2 + 12*a^3*x^3))/(x^3*(b + a*x)) + 12*a^3*Log[x] - 12*a^3*Log[b + a*x])/
b^5

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99

method	result	size
default	\(-\frac {1}{3 b^{2} x^{3}}+\frac {a}{b^{3} x^{2}}-\frac {3 a^{2}}{b^{4} x}-\frac {a^{3}}{b^{4} \left (a x +b \right )}-\frac {4 a^{3} \ln \left (x \right )}{b^{5}}+\frac {4 a^{3} \ln \left (a x +b \right )}{b^{5}}\)	\(68\)
risch	\(\frac {-\frac {4 a^{3} x^{3}}{b^{4}}-\frac {2 a^{2} x^{2}}{b^{3}}+\frac {2 a x}{3 b^{2}}-\frac {1}{3 b}}{\left (a x +b \right ) x^{3}}+\frac {4 a^{3} \ln \left (-a x -b \right )}{b^{5}}-\frac {4 a^{3} \ln \left (x \right )}{b^{5}}\)	\(75\)
norman	\(\frac {\frac {4 a^{4} x^{6}}{b^{5}}-\frac {x^{2}}{3 b}+\frac {2 a \,x^{3}}{3 b^{2}}-\frac {2 a^{2} x^{4}}{b^{3}}}{\left (a x +b \right ) x^{5}}-\frac {4 a^{3} \ln \left (x \right )}{b^{5}}+\frac {4 a^{3} \ln \left (a x +b \right )}{b^{5}}\)	\(77\)
parallelrisch	\(-\frac {12 a^{4} \ln \left (x \right ) x^{4}-12 a^{4} \ln \left (a x +b \right ) x^{4}+12 \ln \left (x \right ) x^{3} a^{3} b -12 \ln \left (a x +b \right ) x^{3} a^{3} b -12 a^{4} x^{4}+6 a^{2} b^{2} x^{2}-2 a \,b^{3} x +b^{4}}{3 b^{5} x^{3} \left (a x +b \right )}\)	\(96\)

[In]

int(1/(a+b/x)^2/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/3/b^2/x^3+a/b^3/x^2-3*a^2/b^4/x-a^3/b^4/(a*x+b)-4*a^3*ln(x)/b^5+4*a^3*ln(a*x+b)/b^5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^6} \, dx=-\frac {12 \, a^{3} b x^{3} + 6 \, a^{2} b^{2} x^{2} - 2 \, a b^{3} x + b^{4} - 12 \, {\left (a^{4} x^{4} + a^{3} b x^{3}\right )} \log \left (a x + b\right ) + 12 \, {\left (a^{4} x^{4} + a^{3} b x^{3}\right )} \log \left (x\right )}{3 \, {\left (a b^{5} x^{4} + b^{6} x^{3}\right )}} \]

[In]

integrate(1/(a+b/x)^2/x^6,x, algorithm="fricas")

[Out]

-1/3*(12*a^3*b*x^3 + 6*a^2*b^2*x^2 - 2*a*b^3*x + b^4 - 12*(a^4*x^4 + a^3*b*x^3)*log(a*x + b) + 12*(a^4*x^4 + a
^3*b*x^3)*log(x))/(a*b^5*x^4 + b^6*x^3)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^6} \, dx=\frac {4 a^{3} \left (- \log {\left (x \right )} + \log {\left (x + \frac {b}{a} \right )}\right )}{b^{5}} + \frac {- 12 a^{3} x^{3} - 6 a^{2} b x^{2} + 2 a b^{2} x - b^{3}}{3 a b^{4} x^{4} + 3 b^{5} x^{3}} \]

[In]

integrate(1/(a+b/x)**2/x**6,x)

[Out]

4*a**3*(-log(x) + log(x + b/a))/b**5 + (-12*a**3*x**3 - 6*a**2*b*x**2 + 2*a*b**2*x - b**3)/(3*a*b**4*x**4 + 3*
b**5*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^6} \, dx=-\frac {12 \, a^{3} x^{3} + 6 \, a^{2} b x^{2} - 2 \, a b^{2} x + b^{3}}{3 \, {\left (a b^{4} x^{4} + b^{5} x^{3}\right )}} + \frac {4 \, a^{3} \log \left (a x + b\right )}{b^{5}} - \frac {4 \, a^{3} \log \left (x\right )}{b^{5}} \]

[In]

integrate(1/(a+b/x)^2/x^6,x, algorithm="maxima")

[Out]

-1/3*(12*a^3*x^3 + 6*a^2*b*x^2 - 2*a*b^2*x + b^3)/(a*b^4*x^4 + b^5*x^3) + 4*a^3*log(a*x + b)/b^5 - 4*a^3*log(x
)/b^5

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^6} \, dx=\frac {4 \, a^{3} \log \left ({\left | a x + b \right |}\right )}{b^{5}} - \frac {4 \, a^{3} \log \left ({\left | x \right |}\right )}{b^{5}} - \frac {12 \, a^{3} b x^{3} + 6 \, a^{2} b^{2} x^{2} - 2 \, a b^{3} x + b^{4}}{3 \, {\left (a x + b\right )} b^{5} x^{3}} \]

[In]

integrate(1/(a+b/x)^2/x^6,x, algorithm="giac")

[Out]

4*a^3*log(abs(a*x + b))/b^5 - 4*a^3*log(abs(x))/b^5 - 1/3*(12*a^3*b*x^3 + 6*a^2*b^2*x^2 - 2*a*b^3*x + b^4)/((a
*x + b)*b^5*x^3)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^6} \, dx=\frac {8\,a^3\,\mathrm {atanh}\left (\frac {2\,a\,x}{b}+1\right )}{b^5}-\frac {\frac {1}{3\,b}+\frac {2\,a^2\,x^2}{b^3}+\frac {4\,a^3\,x^3}{b^4}-\frac {2\,a\,x}{3\,b^2}}{a\,x^4+b\,x^3} \]

[In]

int(1/(x^6*(a + b/x)^2),x)

[Out]

(8*a^3*atanh((2*a*x)/b + 1))/b^5 - (1/(3*b) + (2*a^2*x^2)/b^3 + (4*a^3*x^3)/b^4 - (2*a*x)/(3*b^2))/(a*x^4 + b*
x^3)