\(\int \frac {1}{(a+\frac {b}{x^2}) x^3} \, dx\) [1852]

Optimal result

Integrand size = 13, antiderivative size = 15 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right ) x^3} \, dx=-\frac {\log \left (a+\frac {b}{x^2}\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {266} \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right ) x^3} \, dx=-\frac {\log \left (a+\frac {b}{x^2}\right )}{2 b} \]

[In]

Int[1/((a + b/x^2)*x^3),x]

[Out]

-1/2*Log[a + b/x^2]/b

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\log \left (a+\frac {b}{x^2}\right )}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right ) x^3} \, dx=\frac {\log (x)}{b}-\frac {\log \left (b+a x^2\right )}{2 b} \]

[In]

Integrate[1/((a + b/x^2)*x^3),x]

[Out]

Log[x]/b - Log[b + a*x^2]/(2*b)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(-\frac {\ln \left (a +\frac {b}{x^{2}}\right )}{2 b}\)	\(14\)
default	\(\frac {\ln \left (x \right )}{b}-\frac {\ln \left (a \,x^{2}+b \right )}{2 b}\)	\(21\)
norman	\(\frac {\ln \left (x \right )}{b}-\frac {\ln \left (a \,x^{2}+b \right )}{2 b}\)	\(21\)
risch	\(\frac {\ln \left (x \right )}{b}-\frac {\ln \left (a \,x^{2}+b \right )}{2 b}\)	\(21\)
parallelrisch	\(\frac {2 \ln \left (x \right )-\ln \left (a \,x^{2}+b \right )}{2 b}\)	\(21\)

[In]

int(1/(a+b/x^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right ) x^3} \, dx=-\frac {\log \left (a x^{2} + b\right ) - 2 \, \log \left (x\right )}{2 \, b} \]

[In]

integrate(1/(a+b/x^2)/x^3,x, algorithm="fricas")

[Out]

-1/2*(log(a*x^2 + b) - 2*log(x))/b

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right ) x^3} \, dx=\frac {\log {\left (x \right )}}{b} - \frac {\log {\left (x^{2} + \frac {b}{a} \right )}}{2 b} \]

[In]

integrate(1/(a+b/x**2)/x**3,x)

[Out]

log(x)/b - log(x**2 + b/a)/(2*b)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right ) x^3} \, dx=-\frac {\log \left (a + \frac {b}{x^{2}}\right )}{2 \, b} \]

[In]

integrate(1/(a+b/x^2)/x^3,x, algorithm="maxima")

[Out]

-1/2*log(a + b/x^2)/b

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right ) x^3} \, dx=\frac {\log \left (x^{2}\right )}{2 \, b} - \frac {\log \left ({\left | a x^{2} + b \right |}\right )}{2 \, b} \]

[In]

integrate(1/(a+b/x^2)/x^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.85 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right ) x^3} \, dx=-\frac {\ln \left (a\,x^2+b\right )-2\,\ln \left (x\right )}{2\,b} \]

[In]

int(1/(x^3*(a + b/x^2)),x)

[Out]

-(log(b + a*x^2) - 2*log(x))/(2*b)