\(\int \frac {x}{(a+\frac {b}{x})^2} \, dx\) [1623]

Optimal result

Integrand size = 11, antiderivative size = 46 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^2} \, dx=-\frac {2 b x}{a^3}+\frac {x^2}{2 a^2}+\frac {b^3}{a^4 (b+a x)}+\frac {3 b^2 \log (b+a x)}{a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {269, 45} \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {b^3}{a^4 (a x+b)}+\frac {3 b^2 \log (a x+b)}{a^4}-\frac {2 b x}{a^3}+\frac {x^2}{2 a^2} \]

[In]

Int[x/(a + b/x)^2,x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 {x^3}{(b+a x)^2} \, dx \\ & = \int \left (-\frac {2 b}{a^3}+\frac {x}{a^2}-\frac {b^3}{a^3 (b+a x)^2}+\frac {3 b^2}{a^3 (b+a x)}\right ) \, dx \\ & = -\frac {2 b x}{a^3}+\frac {x^2}{2 a^2}+\frac {b^3}{a^4 (b+a x)}+\frac {3 b^2 \log (b+a x)}{a^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.93 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {-4 a b x+a^2 x^2+\frac {2 b^3}{b+a x}+6 b^2 \log (b+a x)}{2 a^4} \]

[In]

Integrate[x/(a + b/x)^2,x]

[Out]

(-4*a*b*x + a^2*x^2 + (2*b^3)/(b + a*x) + 6*b^2*Log[b + a*x])/(2*a^4)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.98

method	result	size
default	\(\frac {\frac {1}{2} a \,x^{2}-2 b x}{a^{3}}+\frac {b^{3}}{a^{4} \left (a x +b \right )}+\frac {3 b^{2} \ln \left (a x +b \right )}{a^{4}}\)	\(45\)
risch	\(-\frac {2 b x}{a^{3}}+\frac {x^{2}}{2 a^{2}}+\frac {b^{3}}{a^{4} \left (a x +b \right )}+\frac {3 b^{2} \ln \left (a x +b \right )}{a^{4}}\)	\(45\)
norman	\(\frac {\frac {3 b^{3}}{a^{4}}+\frac {x^{3}}{2 a}-\frac {3 b \,x^{2}}{2 a^{2}}}{a x +b}+\frac {3 b^{2} \ln \left (a x +b \right )}{a^{4}}\)	\(50\)
parallelrisch	\(\frac {a^{3} x^{3}+6 \ln \left (a x +b \right ) x a \,b^{2}-3 a^{2} b \,x^{2}+6 b^{3} \ln \left (a x +b \right )+6 b^{3}}{2 a^{4} \left (a x +b \right )}\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.35 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {a^{3} x^{3} - 3 \, a^{2} b x^{2} - 4 \, a b^{2} x + 2 \, b^{3} + 6 \, {\left (a b^{2} x + b^{3}\right )} \log \left (a x + b\right )}{2 \, {\left (a^{5} x + a^{4} b\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {b^{3}}{a^{5} x + a^{4} b} + \frac {x^{2}}{2 a^{2}} - \frac {2 b x}{a^{3}} + \frac {3 b^{2} \log {\left (a x + b \right )}}{a^{4}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {b^{3}}{a^{5} x + a^{4} b} + \frac {3 \, b^{2} \log \left (a x + b\right )}{a^{4}} + \frac {a x^{2} - 4 \, b x}{2 \, a^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.04 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {3 \, b^{2} \log \left ({\left | a x + b \right |}\right )}{a^{4}} + \frac {b^{3}}{{\left (a x + b\right )} a^{4}} + \frac {a^{2} x^{2} - 4 \, a b x}{2 \, a^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.09 \[ \int \frac {x}{\left (a+\frac {b}{x}\right )^2} \, dx=\frac {x^2}{2\,a^2}+\frac {3\,b^2\,\ln \left (b+a\,x\right )}{a^4}+\frac {b^3}{a\,\left (x\,a^4+b\,a^3\right )}-\frac {2\,b\,x}{a^3} \]

[In]

int(x/(a + b/x)^2,x)

[Out]

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