\(\int \frac {1}{(a+\frac {b}{x})^2 x^4} \, dx\) [1628]

Optimal result

Integrand size = 13, antiderivative size = 42 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^4} \, dx=-\frac {1}{b^2 x}-\frac {a}{b^2 (b+a x)}-\frac {2 a \log (x)}{b^3}+\frac {2 a \log (b+a x)}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 42, 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} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^4} \, dx=-\frac {2 a \log (x)}{b^3}+\frac {2 a \log (a x+b)}{b^3}-\frac {a}{b^2 (a x+b)}-\frac {1}{b^2 x} \]

[In]

Int[1/((a + b/x)^2*x^4),x]

[Out]

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

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^2 (b+a x)^2} \, dx \\ & = \int \left (\frac {1}{b^2 x^2}-\frac {2 a}{b^3 x}+\frac {a^2}{b^2 (b+a x)^2}+\frac {2 a^2}{b^3 (b+a x)}\right ) \, dx \\ & = -\frac {1}{b^2 x}-\frac {a}{b^2 (b+a x)}-\frac {2 a \log (x)}{b^3}+\frac {2 a \log (b+a x)}{b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^4} \, dx=-\frac {b \left (\frac {1}{x}+\frac {a}{b+a x}\right )+2 a \log (x)-2 a \log (b+a x)}{b^3} \]

[In]

Integrate[1/((a + b/x)^2*x^4),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {1}{b^{2} x}-\frac {a}{b^{2} \left (a x +b \right )}-\frac {2 a \ln \left (x \right )}{b^{3}}+\frac {2 a \ln \left (a x +b \right )}{b^{3}}\)	\(43\)
risch	\(\frac {-\frac {2 a x}{b^{2}}-\frac {1}{b}}{x \left (a x +b \right )}-\frac {2 a \ln \left (x \right )}{b^{3}}+\frac {2 a \ln \left (-a x -b \right )}{b^{3}}\)	\(49\)
norman	\(\frac {\frac {2 a^{2} x^{4}}{b^{3}}-\frac {x^{2}}{b}}{\left (a x +b \right ) x^{3}}-\frac {2 a \ln \left (x \right )}{b^{3}}+\frac {2 a \ln \left (a x +b \right )}{b^{3}}\)	\(53\)
parallelrisch	\(-\frac {2 a^{2} \ln \left (x \right ) x^{2}-2 a^{2} \ln \left (a x +b \right ) x^{2}+2 a b \ln \left (x \right ) x -2 \ln \left (a x +b \right ) x a b -2 a^{2} x^{2}+b^{2}}{b^{3} x \left (a x +b \right )}\)	\(70\)

[In]

int(1/(a+b/x)^2/x^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^4} \, dx=-\frac {2 \, a b x + b^{2} - 2 \, {\left (a^{2} x^{2} + a b x\right )} \log \left (a x + b\right ) + 2 \, {\left (a^{2} x^{2} + a b x\right )} \log \left (x\right )}{a b^{3} x^{2} + b^{4} x} \]

[In]

integrate(1/(a+b/x)^2/x^4,x, algorithm="fricas")

[Out]

-(2*a*b*x + b^2 - 2*(a^2*x^2 + a*b*x)*log(a*x + b) + 2*(a^2*x^2 + a*b*x)*log(x))/(a*b^3*x^2 + b^4*x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^4} \, dx=\frac {2 a \left (- \log {\left (x \right )} + \log {\left (x + \frac {b}{a} \right )}\right )}{b^{3}} + \frac {- 2 a x - b}{a b^{2} x^{2} + b^{3} x} \]

[In]

integrate(1/(a+b/x)**2/x**4,x)

[Out]

2*a*(-log(x) + log(x + b/a))/b**3 + (-2*a*x - b)/(a*b**2*x**2 + b**3*x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^4} \, dx=-\frac {2 \, a x + b}{a b^{2} x^{2} + b^{3} x} + \frac {2 \, a \log \left (a x + b\right )}{b^{3}} - \frac {2 \, a \log \left (x\right )}{b^{3}} \]

[In]

integrate(1/(a+b/x)^2/x^4,x, algorithm="maxima")

[Out]

-(2*a*x + b)/(a*b^2*x^2 + b^3*x) + 2*a*log(a*x + b)/b^3 - 2*a*log(x)/b^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^4} \, dx=\frac {2 \, a \log \left ({\left | a x + b \right |}\right )}{b^{3}} - \frac {2 \, a \log \left ({\left | x \right |}\right )}{b^{3}} - \frac {2 \, a x + b}{{\left (a x^{2} + b x\right )} b^{2}} \]

[In]

integrate(1/(a+b/x)^2/x^4,x, algorithm="giac")

[Out]

2*a*log(abs(a*x + b))/b^3 - 2*a*log(abs(x))/b^3 - (2*a*x + b)/((a*x^2 + b*x)*b^2)

Mupad [B] (verification not implemented)

Time = 5.76 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^2 x^4} \, dx=\frac {4\,a\,\mathrm {atanh}\left (\frac {2\,a\,x}{b}+1\right )}{b^3}-\frac {\frac {1}{b}+\frac {2\,a\,x}{b^2}}{a\,x^2+b\,x} \]

[In]

int(1/(x^4*(a + b/x)^2),x)

[Out]

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