3.2620 \(\int \frac{1}{x (2+b x^n)} \, dx\)

Optimal. Leaf size=22 \[ \frac{\log (x)}{2}-\frac{\log \left (b x^n+2\right )}{2 n} \]

[Out]

Log[x]/2 - Log[2 + b*x^n]/(2*n)

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Rubi [A]  time = 0.00948, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {266, 36, 29, 31} \[ \frac{\log (x)}{2}-\frac{\log \left (b x^n+2\right )}{2 n} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(2 + b*x^n)),x]

[Out]

Log[x]/2 - Log[2 + b*x^n]/(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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0051662, size = 22, normalized size = 1. \[ \frac{n \log (x)-\log \left (b x^n+2\right )}{2 n} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(2 + b*x^n)),x]

[Out]

(n*Log[x] - Log[2 + b*x^n])/(2*n)

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Maple [A]  time = 0.002, size = 24, normalized size = 1.1 \begin{align*}{\frac{\ln \left ({x}^{n} \right ) }{2\,n}}-{\frac{\ln \left ( 2+b{x}^{n} \right ) }{2\,n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.953897, size = 31, normalized size = 1.41 \begin{align*} -\frac{\log \left (b x^{n} + 2\right )}{2 \, n} + \frac{\log \left (x^{n}\right )}{2 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.03745, size = 47, normalized size = 2.14 \begin{align*} \frac{n \log \left (x\right ) - \log \left (b x^{n} + 2\right )}{2 \, n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(n*log(x) - log(b*x^n + 2))/n

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Sympy [A]  time = 0.550214, size = 31, normalized size = 1.41 \begin{align*} \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{\log{\left (x \right )}}{2} & \text{for}\: b = 0 \\\frac{\log{\left (x \right )}}{2} - \frac{\log{\left (x^{n} + \frac{2}{b} \right )}}{2 n} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^n + 2)*x), x)