3.1628 \(\int \frac{1}{(a+\frac{b}{x})^2 x^4} \, dx\)

Optimal. Leaf size=42 \[ -\frac{a}{b^2 (a x+b)}-\frac{2 a \log (x)}{b^3}+\frac{2 a \log (a x+b)}{b^3}-\frac{1}{b^2 x} \]

[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

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Rubi [A]  time = 0.0240158, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {263, 44} \[ -\frac{a}{b^2 (a x+b)}-\frac{2 a \log (x)}{b^3}+\frac{2 a \log (a x+b)}{b^3}-\frac{1}{b^2 x} \]

Antiderivative was successfully verified.

[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 263

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]

Rule 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x}\right )^2 x^4} \, dx &=\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]  time = 0.0386083, size = 35, normalized size = 0.83 \[ -\frac{b \left (\frac{a}{a x+b}+\frac{1}{x}\right )-2 a \log (a x+b)+2 a \log (x)}{b^3} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.009, size = 43, normalized size = 1. \begin{align*} -{\frac{1}{{b}^{2}x}}-{\frac{a}{{b}^{2} \left ( ax+b \right ) }}-2\,{\frac{a\ln \left ( x \right ) }{{b}^{3}}}+2\,{\frac{a\ln \left ( ax+b \right ) }{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Maxima [A]  time = 0.988525, size = 61, normalized size = 1.45 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.4455, size = 138, normalized size = 3.29 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.388862, size = 36, normalized size = 0.86 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.11049, size = 61, normalized size = 1.45 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)