3.2621 \(\int \frac{x^{-1-n}}{2+b x^n} \, dx\)

Optimal. Leaf size=36 \[ \frac{b \log \left (b x^n+2\right )}{4 n}-\frac{1}{4} b \log (x)-\frac{x^{-n}}{2 n} \]

[Out]

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

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Rubi [A]  time = 0.0190837, antiderivative size = 36, normalized size of antiderivative = 1., 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 = {266, 44} \[ \frac{b \log \left (b x^n+2\right )}{4 n}-\frac{1}{4} b \log (x)-\frac{x^{-n}}{2 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - n)/(2 + b*x^n),x]

[Out]

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

Rule 266

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 44

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] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0270057, size = 31, normalized size = 0.86 \[ -\frac{-b \log \left (b x^n+2\right )+b n \log (x)+2 x^{-n}}{4 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - n)/(2 + b*x^n),x]

[Out]

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

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Maple [A]  time = 0.016, size = 42, normalized size = 1.2 \begin{align*}{\frac{1}{{{\rm e}^{n\ln \left ( x \right ) }}} \left ( -{\frac{b\ln \left ( x \right ){{\rm e}^{n\ln \left ( x \right ) }}}{4}}-{\frac{1}{2\,n}} \right ) }+{\frac{b\ln \left ( 2+b{{\rm e}^{n\ln \left ( x \right ) }} \right ) }{4\,n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.985828, size = 46, normalized size = 1.28 \begin{align*} -\frac{1}{4} \, b \log \left (x\right ) + \frac{b \log \left (\frac{b x^{n} + 2}{b}\right )}{4 \, n} - \frac{1}{2 \, n x^{n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.02583, size = 78, normalized size = 2.17 \begin{align*} -\frac{b n x^{n} \log \left (x\right ) - b x^{n} \log \left (b x^{n} + 2\right ) + 2}{4 \, n x^{n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 34.9272, size = 29, normalized size = 0.81 \begin{align*} \begin{cases} \frac{b \log{\left (\frac{b}{2} + x^{- n} \right )}}{4 n} - \frac{x^{- n}}{2 n} & \text{for}\: n \neq 0 \\\frac{\log{\left (x \right )}}{b + 2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((b*log(b/2 + x**(-n))/(4*n) - x**(-n)/(2*n), Ne(n, 0)), (log(x)/(b + 2), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{-n - 1}}{b x^{n} + 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(-n - 1)/(b*x^n + 2), x)