3.2617 \(\int \frac{x^{-1+3 n}}{2+b x^n} \, dx\)

Optimal. Leaf size=43 \[ -\frac{2 x^n}{b^2 n}+\frac{4 \log \left (b x^n+2\right )}{b^3 n}+\frac{x^{2 n}}{2 b n} \]

[Out]

(-2*x^n)/(b^2*n) + x^(2*n)/(2*b*n) + (4*Log[2 + b*x^n])/(b^3*n)

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Rubi [A]  time = 0.0214277, antiderivative size = 43, 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} \[ -\frac{2 x^n}{b^2 n}+\frac{4 \log \left (b x^n+2\right )}{b^3 n}+\frac{x^{2 n}}{2 b n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 3*n)/(2 + b*x^n),x]

[Out]

(-2*x^n)/(b^2*n) + x^(2*n)/(2*b*n) + (4*Log[2 + b*x^n])/(b^3*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+3 n}}{2+b x^n} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{2+b x} \, dx,x,x^n\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{2}{b^2}+\frac{x}{b}+\frac{4}{b^2 (2+b x)}\right ) \, dx,x,x^n\right )}{n}\\ &=-\frac{2 x^n}{b^2 n}+\frac{x^{2 n}}{2 b n}+\frac{4 \log \left (2+b x^n\right )}{b^3 n}\\ \end{align*}

Mathematica [A]  time = 0.0168301, size = 33, normalized size = 0.77 \[ \frac{b x^n \left (b x^n-4\right )+8 \log \left (b x^n+2\right )}{2 b^3 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 3*n)/(2 + b*x^n),x]

[Out]

(b*x^n*(-4 + b*x^n) + 8*Log[2 + b*x^n])/(2*b^3*n)

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Maple [A]  time = 0.019, size = 48, normalized size = 1.1 \begin{align*} -2\,{\frac{{{\rm e}^{n\ln \left ( x \right ) }}}{{b}^{2}n}}+{\frac{ \left ({{\rm e}^{n\ln \left ( x \right ) }} \right ) ^{2}}{2\,bn}}+4\,{\frac{\ln \left ( 2+b{{\rm e}^{n\ln \left ( x \right ) }} \right ) }{{b}^{3}n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+3*n)/(2+b*x^n),x)

[Out]

-2/b^2/n*exp(n*ln(x))+1/2/b/n*exp(n*ln(x))^2+4/b^3/n*ln(2+b*exp(n*ln(x)))

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Maxima [A]  time = 0.980038, size = 57, normalized size = 1.33 \begin{align*} \frac{b x^{2 \, n} - 4 \, x^{n}}{2 \, b^{2} n} + \frac{4 \, \log \left (\frac{b x^{n} + 2}{b}\right )}{b^{3} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(2+b*x^n),x, algorithm="maxima")

[Out]

1/2*(b*x^(2*n) - 4*x^n)/(b^2*n) + 4*log((b*x^n + 2)/b)/(b^3*n)

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Fricas [A]  time = 1.06841, size = 76, normalized size = 1.77 \begin{align*} \frac{b^{2} x^{2 \, n} - 4 \, b x^{n} + 8 \, \log \left (b x^{n} + 2\right )}{2 \, b^{3} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(2+b*x^n),x, algorithm="fricas")

[Out]

1/2*(b^2*x^(2*n) - 4*b*x^n + 8*log(b*x^n + 2))/(b^3*n)

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Sympy [A]  time = 21.6661, size = 53, normalized size = 1.23 \begin{align*} \begin{cases} \frac{\log{\left (x \right )}}{2} & \text{for}\: b = 0 \wedge n = 0 \\\frac{x^{3 n}}{6 n} & \text{for}\: b = 0 \\\frac{\log{\left (x \right )}}{b + 2} & \text{for}\: n = 0 \\\frac{x^{2 n}}{2 b n} - \frac{2 x^{n}}{b^{2} n} + \frac{4 \log{\left (x^{n} + \frac{2}{b} \right )}}{b^{3} n} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+3*n)/(2+b*x**n),x)

[Out]

Piecewise((log(x)/2, Eq(b, 0) & Eq(n, 0)), (x**(3*n)/(6*n), Eq(b, 0)), (log(x)/(b + 2), Eq(n, 0)), (x**(2*n)/(
2*b*n) - 2*x**n/(b**2*n) + 4*log(x**n + 2/b)/(b**3*n), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3 \, n - 1}}{b x^{n} + 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+3*n)/(2+b*x^n),x, algorithm="giac")

[Out]

integrate(x^(3*n - 1)/(b*x^n + 2), x)