\(\int \frac {1}{(a+\frac {b}{x}) x^3} \, dx\) [1614]

Optimal result

Integrand size = 13, antiderivative size = 28 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^3} \, dx=-\frac {1}{b x}-\frac {a \log (x)}{b^2}+\frac {a \log (b+a x)}{b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, 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} \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^3} \, dx=-\frac {a \log (x)}{b^2}+\frac {a \log (a x+b)}{b^2}-\frac {1}{b x} \]

[In]

Int[1/((a + b/x)*x^3),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^3} \, dx=-\frac {1}{b x}-\frac {a \log (x)}{b^2}+\frac {a \log (b+a x)}{b^2} \]

[In]

Integrate[1/((a + b/x)*x^3),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93

method	result	size
parallelrisch	\(-\frac {a \ln \left (x \right ) x -a \ln \left (a x +b \right ) x +b}{x \,b^{2}}\)	\(26\)
default	\(-\frac {1}{b x}-\frac {a \ln \left (x \right )}{b^{2}}+\frac {a \ln \left (a x +b \right )}{b^{2}}\)	\(29\)
norman	\(-\frac {1}{b x}-\frac {a \ln \left (x \right )}{b^{2}}+\frac {a \ln \left (a x +b \right )}{b^{2}}\)	\(29\)
risch	\(-\frac {1}{b x}-\frac {a \ln \left (x \right )}{b^{2}}+\frac {a \ln \left (-a x -b \right )}{b^{2}}\)	\(32\)

[In]

int(1/(a+b/x)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^3} \, dx=\frac {a x \log \left (a x + b\right ) - a x \log \left (x\right ) - b}{b^{2} x} \]

[In]

integrate(1/(a+b/x)/x^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^3} \, dx=\frac {a \left (- \log {\left (x \right )} + \log {\left (x + \frac {b}{a} \right )}\right )}{b^{2}} - \frac {1}{b x} \]

[In]

integrate(1/(a+b/x)/x**3,x)

[Out]

a*(-log(x) + log(x + b/a))/b**2 - 1/(b*x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^3} \, dx=\frac {a \log \left (a x + b\right )}{b^{2}} - \frac {a \log \left (x\right )}{b^{2}} - \frac {1}{b x} \]

[In]

integrate(1/(a+b/x)/x^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^3} \, dx=\frac {a \log \left ({\left | a x + b \right |}\right )}{b^{2}} - \frac {a \log \left ({\left | x \right |}\right )}{b^{2}} - \frac {1}{b x} \]

[In]

integrate(1/(a+b/x)/x^3,x, algorithm="giac")

[Out]

a*log(abs(a*x + b))/b^2 - a*log(abs(x))/b^2 - 1/(b*x)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^3} \, dx=\frac {2\,a\,\mathrm {atanh}\left (\frac {2\,a\,x}{b}+1\right )}{b^2}-\frac {1}{b\,x} \]

[In]

int(1/(x^3*(a + b/x)),x)

[Out]

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