3.2618 \(\int \frac{x^{-1+2 n}}{2+b x^n} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0155853, antiderivative size = 27, 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, 43} \[ \frac{x^n}{b n}-\frac{2 \log \left (b x^n+2\right )}{b^2 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)/(2 + b*x^n),x]

[Out]

x^n/(b*n) - (2*Log[2 + b*x^n])/(b^2*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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0105969, size = 25, normalized size = 0.93 \[ \frac{\frac{x^n}{b}-\frac{2 \log \left (b x^n+2\right )}{b^2}}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 2*n)/(2 + b*x^n),x]

[Out]

(x^n/b - (2*Log[2 + b*x^n])/b^2)/n

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Maple [A]  time = 0.013, size = 32, normalized size = 1.2 \begin{align*}{\frac{{{\rm e}^{n\ln \left ( x \right ) }}}{bn}}-2\,{\frac{\ln \left ( 2+b{{\rm e}^{n\ln \left ( x \right ) }} \right ) }{{b}^{2}n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.972354, size = 42, normalized size = 1.56 \begin{align*} \frac{x^{n}}{b n} - \frac{2 \, \log \left (\frac{b x^{n} + 2}{b}\right )}{b^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.04919, size = 49, normalized size = 1.81 \begin{align*} \frac{b x^{n} - 2 \, \log \left (b x^{n} + 2\right )}{b^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x^n - 2*log(b*x^n + 2))/(b^2*n)

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Sympy [A]  time = 14.5323, size = 39, normalized size = 1.44 \begin{align*} \begin{cases} \frac{\log{\left (x \right )}}{2} & \text{for}\: b = 0 \wedge n = 0 \\\frac{x^{2 n}}{4 n} & \text{for}\: b = 0 \\\frac{\log{\left (x \right )}}{b + 2} & \text{for}\: n = 0 \\\frac{x^{n}}{b n} - \frac{2 \log{\left (x^{n} + \frac{2}{b} \right )}}{b^{2} n} & \text{otherwise} \end{cases} \end{align*}

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

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

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)