\(\int \frac {1}{-x^3+b x^5} \, dx\) [320]

Optimal result

Integrand size = 13, antiderivative size = 27 \[ \int \frac {1}{-x^3+b x^5} \, dx=\frac {1}{2 x^2}-b \log (x)+\frac {1}{2} b \log \left (1-b x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1607, 272, 46} \[ \int \frac {1}{-x^3+b x^5} \, dx=\frac {1}{2} b \log \left (1-b x^2\right )-b \log (x)+\frac {1}{2 x^2} \]

[In]

Int[(-x^3 + b*x^5)^(-1),x]

[Out]

1/(2*x^2) - b*Log[x] + (b*Log[1 - b*x^2])/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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^3 \left (-1+b x^2\right )} \, dx \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^2 (-1+b x)} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (-\frac {1}{x^2}-\frac {b}{x}+\frac {b^2}{-1+b x}\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{2 x^2}-b \log (x)+\frac {1}{2} b \log \left (1-b x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{-x^3+b x^5} \, dx=\frac {1}{2 x^2}-b \log (x)+\frac {1}{2} b \log \left (1-b x^2\right ) \]

[In]

Integrate[(-x^3 + b*x^5)^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 1.72 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {1}{2 x^{2}}-b \ln \left (x \right )+\frac {b \ln \left (b \,x^{2}-1\right )}{2}\)	\(23\)
norman	\(\frac {1}{2 x^{2}}-b \ln \left (x \right )+\frac {b \ln \left (b \,x^{2}-1\right )}{2}\)	\(23\)
risch	\(\frac {1}{2 x^{2}}-b \ln \left (x \right )+\frac {b \ln \left (-b \,x^{2}+1\right )}{2}\)	\(24\)
parallelrisch	\(-\frac {2 b \ln \left (x \right ) x^{2}-b \ln \left (b \,x^{2}-1\right ) x^{2}-1}{2 x^{2}}\)	\(30\)
meijerg	\(\frac {b \left (\frac {1}{x^{2} b}-2 \ln \left (x \right )-\ln \left (-b \right )+\ln \left (-b \,x^{2}+1\right )\right )}{2}\)	\(31\)

[In]

int(1/(b*x^5-x^3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1}{-x^3+b x^5} \, dx=\frac {b x^{2} \log \left (b x^{2} - 1\right ) - 2 \, b x^{2} \log \left (x\right ) + 1}{2 \, x^{2}} \]

[In]

integrate(1/(b*x^5-x^3),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-x^3+b x^5} \, dx=- b \log {\left (x \right )} + \frac {b \log {\left (x^{2} - \frac {1}{b} \right )}}{2} + \frac {1}{2 x^{2}} \]

[In]

integrate(1/(b*x**5-x**3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-x^3+b x^5} \, dx=\frac {1}{2} \, b \log \left (b x^{2} - 1\right ) - b \log \left (x\right ) + \frac {1}{2 \, x^{2}} \]

[In]

integrate(1/(b*x^5-x^3),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {1}{-x^3+b x^5} \, dx=-\frac {1}{2} \, b \log \left (x^{2}\right ) + \frac {1}{2} \, b \log \left ({\left | b x^{2} - 1 \right |}\right ) + \frac {b x^{2} + 1}{2 \, x^{2}} \]

[In]

integrate(1/(b*x^5-x^3),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{-x^3+b x^5} \, dx=\frac {b\,\ln \left (b\,x^2-1\right )}{2}-b\,\ln \left (x\right )+\frac {1}{2\,x^2} \]

[In]

int(1/(b*x^5 - x^3),x)

[Out]

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