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

Optimal. Leaf size=27 \[ \frac {1}{2 x^2}-b \log (x)+\frac {1}{2} b \log \left (1-b x^2\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1607, 272, 46} \begin {gather*} \frac {1}{2} b \log \left (1-b x^2\right )-b \log (x)+\frac {1}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {1}{-x^3+b x^5} \, dx &=\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]
time = 0.00, size = 27, normalized size = 1.00 \begin {gather*} \frac {1}{2 x^2}-b \log (x)+\frac {1}{2} b \log \left (1-b x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[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]
time = 0.38, size = 23, normalized size = 0.85

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 22, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, b \log \left (b x^{2} - 1\right ) - b \log \left (x\right ) + \frac {1}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]
time = 1.91, size = 28, normalized size = 1.04 \begin {gather*} \frac {b x^{2} \log \left (b x^{2} - 1\right ) - 2 \, b x^{2} \log \left (x\right ) + 1}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]
time = 0.09, size = 22, normalized size = 0.81 \begin {gather*} - b \log {\left (x \right )} + \frac {b \log {\left (x^{2} - \frac {1}{b} \right )}}{2} + \frac {1}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 0.58, size = 32, normalized size = 1.19 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]
time = 5.21, size = 22, normalized size = 0.81 \begin {gather*} \frac {b\,\ln \left (b\,x^2-1\right )}{2}-b\,\ln \left (x\right )+\frac {1}{2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^5 - x^3),x)

[Out]

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

________________________________________________________________________________________