3.27.22 \(\int \frac {x^{-1-2 n}}{2+b x^n} \, dx\) [2622]

Optimal. Leaf size=53 \[ -\frac {x^{-2 n}}{4 n}+\frac {b x^{-n}}{4 n}+\frac {1}{8} b^2 \log (x)-\frac {b^2 \log \left (2+b x^n\right )}{8 n} \]

[Out]

-1/4/n/(x^(2*n))+1/4*b/n/(x^n)+1/8*b^2*ln(x)-1/8*b^2*ln(2+b*x^n)/n

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Rubi [A]
time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {272, 46} \begin {gather*} -\frac {b^2 \log \left (b x^n+2\right )}{8 n}+\frac {1}{8} b^2 \log (x)+\frac {b x^{-n}}{4 n}-\frac {x^{-2 n}}{4 n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 - 2*n)/(2 + b*x^n),x]

[Out]

-1/4*1/(n*x^(2*n)) + b/(4*n*x^n) + (b^2*Log[x])/8 - (b^2*Log[2 + b*x^n])/(8*n)

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]]

Rubi steps

\begin {align*} \int \frac {x^{-1-2 n}}{2+b x^n} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x^3 (2+b x)} \, dx,x,x^n\right )}{n}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{2 x^3}-\frac {b}{4 x^2}+\frac {b^2}{8 x}-\frac {b^3}{8 (2+b x)}\right ) \, dx,x,x^n\right )}{n}\\ &=-\frac {x^{-2 n}}{4 n}+\frac {b x^{-n}}{4 n}+\frac {1}{8} b^2 \log (x)-\frac {b^2 \log \left (2+b x^n\right )}{8 n}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 45, normalized size = 0.85 \begin {gather*} -\frac {x^{-2 n} \left (2-2 b x^n\right )-b^2 \log \left (x^n\right )+b^2 \log \left (n \left (2+b x^n\right )\right )}{8 n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - 2*n)/(2 + b*x^n),x]

[Out]

-1/8*((2 - 2*b*x^n)/x^(2*n) - b^2*Log[x^n] + b^2*Log[n*(2 + b*x^n)])/n

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Maple [A]
time = 0.33, size = 48, normalized size = 0.91

method	result	size
risch	\(\frac {b \,x^{-n}}{4 n}-\frac {x^{-2 n}}{4 n}+\frac {b^{2} \ln \left (x \right )}{8}-\frac {b^{2} \ln \left (x^{n}+\frac {2}{b}\right )}{8 n}\)	\(48\)
norman	\(\left (\frac {b^{2} \ln \left (x \right ) {\mathrm e}^{2 n \ln \left (x \right )}}{8}-\frac {1}{4 n}+\frac {b \,{\mathrm e}^{n \ln \left (x \right )}}{4 n}\right ) {\mathrm e}^{-2 n \ln \left (x \right )}-\frac {b^{2} \ln \left (2+b \,{\mathrm e}^{n \ln \left (x \right )}\right )}{8 n}\)	\(59\)
meijerg	\(-\frac {\left (-1\right )^{\mathrm {csgn}\left (i b \right )+\mathrm {csgn}\left (i x^{n}\right )+\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i b \right )} b^{2} \left (-\left (-1\right )^{-\frac {-1-2 n}{n}-\frac {1}{n}} \ln \left (1-\frac {i x^{n} b \left (-1\right )^{\frac {\mathrm {csgn}\left (i b \right )}{2}+\frac {\mathrm {csgn}\left (i x^{n}\right )}{2}-\frac {\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i b \right )}{2}}}{2}\right )+\left (-1\right )^{-\frac {-1-2 n}{n}-\frac {1}{n}} \left (n \ln \left (x \right )-\ln \left (2\right )+\ln \left (b \right )+i \left (\frac {\mathrm {csgn}\left (i b \right )}{2}+\frac {\mathrm {csgn}\left (i x^{n}\right )}{2}-\frac {\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i b \right )}{2}-\frac {1}{2}\right ) \pi \right )-\frac {\left (-1\right )^{\frac {\left (\frac {-1-2 n}{n}+\frac {1}{n}\right ) \mathrm {csgn}\left (i b \right )}{2}+\frac {\left (\frac {-1-2 n}{n}+\frac {1}{n}\right ) \mathrm {csgn}\left (i x^{n}\right )}{2}-\frac {\left (\frac {-1-2 n}{n}+\frac {1}{n}\right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i b \right )}{2}-\frac {-1-2 n}{2 n}-\frac {1}{2 n}} x^{\left (\frac {-1-2 n}{n}+\frac {1}{n}\right ) n} 2^{-\frac {-1-2 n}{n}-\frac {1}{n}} b^{\frac {-1-2 n}{n}+\frac {1}{n}}}{2}-\left (-1\right )^{\frac {\left (1+\frac {-1-2 n}{n}+\frac {1}{n}\right ) \mathrm {csgn}\left (i b \right )}{2}+\frac {\left (1+\frac {-1-2 n}{n}+\frac {1}{n}\right ) \mathrm {csgn}\left (i x^{n}\right )}{2}-\frac {\left (1+\frac {-1-2 n}{n}+\frac {1}{n}\right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i b \right )}{2}-\frac {-1-2 n}{2 n}-\frac {1}{2 n}+\frac {1}{2}} x^{\left (1+\frac {-1-2 n}{n}+\frac {1}{n}\right ) n} 2^{-1-\frac {-1-2 n}{n}-\frac {1}{n}} b^{1+\frac {-1-2 n}{n}+\frac {1}{n}}\right )}{8 n}\)	\(452\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1-2*n)/(2+b*x^n),x,method=_RETURNVERBOSE)

[Out]

1/4*b/n/(x^n)-1/4/n/(x^n)^2+1/8*b^2*ln(x)-1/8*b^2/n*ln(x^n+2/b)

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Maxima [A]
time = 0.30, size = 47, normalized size = 0.89 \begin {gather*} \frac {1}{8} \, b^{2} \log \left (x\right ) - \frac {b^{2} \log \left (\frac {b x^{n} + 2}{b}\right )}{8 \, n} + \frac {b x^{n} - 1}{4 \, n x^{2 \, n}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-2*n)/(2+b*x^n),x, algorithm="maxima")

[Out]

1/8*b^2*log(x) - 1/8*b^2*log((b*x^n + 2)/b)/n + 1/4*(b*x^n - 1)/(n*x^(2*n))

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Fricas [A]
time = 0.50, size = 50, normalized size = 0.94 \begin {gather*} \frac {b^{2} n x^{2 \, n} \log \left (x\right ) - b^{2} x^{2 \, n} \log \left (b x^{n} + 2\right ) + 2 \, b x^{n} - 2}{8 \, n x^{2 \, n}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-2*n)/(2+b*x^n),x, algorithm="fricas")

[Out]

1/8*(b^2*n*x^(2*n)*log(x) - b^2*x^(2*n)*log(b*x^n + 2) + 2*b*x^n - 2)/(n*x^(2*n))

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Sympy [A]
time = 20.64, size = 66, normalized size = 1.25 \begin {gather*} \begin {cases} \frac {\log {\left (x \right )}}{2} & \text {for}\: b = 0 \wedge n = 0 \\\frac {\log {\left (x \right )}}{b + 2} & \text {for}\: n = 0 \\- \frac {x^{- 2 n}}{4 n} & \text {for}\: b = 0 \\\frac {b^{2} \log {\left (x^{n} \right )}}{8 n} - \frac {b^{2} \log {\left (x^{n} + \frac {2}{b} \right )}}{8 n} + \frac {b x^{- n}}{4 n} - \frac {x^{- 2 n}}{4 n} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1-2*n)/(2+b*x**n),x)

[Out]

Piecewise((log(x)/2, Eq(b, 0) & Eq(n, 0)), (log(x)/(b + 2), Eq(n, 0)), (-1/(4*n*x**(2*n)), Eq(b, 0)), (b**2*lo
g(x**n)/(8*n) - b**2*log(x**n + 2/b)/(8*n) + b/(4*n*x**n) - 1/(4*n*x**(2*n)), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1-2*n)/(2+b*x^n),x, algorithm="giac")

[Out]

integrate(x^(-2*n - 1)/(b*x^n + 2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{x^{2\,n+1}\,\left (b\,x^n+2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(2*n + 1)*(b*x^n + 2)),x)

[Out]

int(1/(x^(2*n + 1)*(b*x^n + 2)), x)

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