3.2615 \(\int \frac {x^{-1+5 n}}{2+b x^n} \, dx\)

Optimal. Leaf size=71 \[ \frac {16 \log \left (b x^n+2\right )}{b^5 n}-\frac {8 x^n}{b^4 n}+\frac {2 x^{2 n}}{b^3 n}-\frac {2 x^{3 n}}{3 b^2 n}+\frac {x^{4 n}}{4 b n} \]

[Out]

-8*x^n/b^4/n+2*x^(2*n)/b^3/n-2/3*x^(3*n)/b^2/n+1/4*x^(4*n)/b/n+16*ln(2+b*x^n)/b^5/n

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Rubi [A]  time = 0.03, antiderivative size = 71, 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 = {266, 43} \[ -\frac {8 x^n}{b^4 n}+\frac {2 x^{2 n}}{b^3 n}-\frac {2 x^{3 n}}{3 b^2 n}+\frac {16 \log \left (b x^n+2\right )}{b^5 n}+\frac {x^{4 n}}{4 b n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 5*n)/(2 + b*x^n),x]

[Out]

(-8*x^n)/(b^4*n) + (2*x^(2*n))/(b^3*n) - (2*x^(3*n))/(3*b^2*n) + x^(4*n)/(4*b*n) + (16*Log[2 + b*x^n])/(b^5*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])

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]]

Rubi steps

\begin {align*} \int \frac {x^{-1+5 n}}{2+b x^n} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^4}{2+b x} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (-\frac {8}{b^4}+\frac {4 x}{b^3}-\frac {2 x^2}{b^2}+\frac {x^3}{b}+\frac {16}{b^4 (2+b x)}\right ) \, dx,x,x^n\right )}{n}\\ &=-\frac {8 x^n}{b^4 n}+\frac {2 x^{2 n}}{b^3 n}-\frac {2 x^{3 n}}{3 b^2 n}+\frac {x^{4 n}}{4 b n}+\frac {16 \log \left (2+b x^n\right )}{b^5 n}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 54, normalized size = 0.76 \[ \frac {b x^n \left (3 b^3 x^{3 n}-8 b^2 x^{2 n}+24 b x^n-96\right )+192 \log \left (b x^n+2\right )}{12 b^5 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 5*n)/(2 + b*x^n),x]

[Out]

(b*x^n*(-96 + 24*b*x^n - 8*b^2*x^(2*n) + 3*b^3*x^(3*n)) + 192*Log[2 + b*x^n])/(12*b^5*n)

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fricas [A]  time = 0.47, size = 55, normalized size = 0.77 \[ \frac {3 \, b^{4} x^{4 \, n} - 8 \, b^{3} x^{3 \, n} + 24 \, b^{2} x^{2 \, n} - 96 \, b x^{n} + 192 \, \log \left (b x^{n} + 2\right )}{12 \, b^{5} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+5*n)/(2+b*x^n),x, algorithm="fricas")

[Out]

1/12*(3*b^4*x^(4*n) - 8*b^3*x^(3*n) + 24*b^2*x^(2*n) - 96*b*x^n + 192*log(b*x^n + 2))/(b^5*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5 \, n - 1}}{b x^{n} + 2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+5*n)/(2+b*x^n),x, algorithm="giac")

[Out]

integrate(x^(5*n - 1)/(b*x^n + 2), x)

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maple [A]  time = 0.03, size = 78, normalized size = 1.10 \[ \frac {{\mathrm e}^{4 n \ln \relax (x )}}{4 b n}-\frac {2 \,{\mathrm e}^{3 n \ln \relax (x )}}{3 b^{2} n}+\frac {2 \,{\mathrm e}^{2 n \ln \relax (x )}}{b^{3} n}-\frac {8 \,{\mathrm e}^{n \ln \relax (x )}}{b^{4} n}+\frac {16 \ln \left (b \,{\mathrm e}^{n \ln \relax (x )}+2\right )}{b^{5} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(-1+5*n)/(2+b*x^n),x)

[Out]

-8/b^4/n*exp(n*ln(x))+2/b^3/n*exp(n*ln(x))^2-2/3/b^2/n*exp(n*ln(x))^3+1/4/b/n*exp(n*ln(x))^4+16/b^5/n*ln(2+b*e
xp(n*ln(x)))

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maxima [A]  time = 0.49, size = 63, normalized size = 0.89 \[ \frac {3 \, b^{3} x^{4 \, n} - 8 \, b^{2} x^{3 \, n} + 24 \, b x^{2 \, n} - 96 \, x^{n}}{12 \, b^{4} n} + \frac {16 \, \log \left (\frac {b x^{n} + 2}{b}\right )}{b^{5} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+5*n)/(2+b*x^n),x, algorithm="maxima")

[Out]

1/12*(3*b^3*x^(4*n) - 8*b^2*x^(3*n) + 24*b*x^(2*n) - 96*x^n)/(b^4*n) + 16*log((b*x^n + 2)/b)/(b^5*n)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{5\,n-1}}{b\,x^n+2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5*n - 1)/(b*x^n + 2),x)

[Out]

int(x^(5*n - 1)/(b*x^n + 2), x)

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sympy [A]  time = 76.52, size = 78, normalized size = 1.10 \[ \begin {cases} \frac {\log {\relax (x )}}{2} & \text {for}\: b = 0 \wedge n = 0 \\\frac {\log {\relax (x )}}{b + 2} & \text {for}\: n = 0 \\\frac {x^{5 n}}{10 n} & \text {for}\: b = 0 \\\frac {x^{4 n}}{4 b n} - \frac {2 x^{3 n}}{3 b^{2} n} + \frac {2 x^{2 n}}{b^{3} n} - \frac {8 x^{n}}{b^{4} n} + \frac {16 \log {\left (x^{n} + \frac {2}{b} \right )}}{b^{5} n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(-1+5*n)/(2+b*x**n),x)

[Out]

Piecewise((log(x)/2, Eq(b, 0) & Eq(n, 0)), (log(x)/(b + 2), Eq(n, 0)), (x**(5*n)/(10*n), Eq(b, 0)), (x**(4*n)/
(4*b*n) - 2*x**(3*n)/(3*b**2*n) + 2*x**(2*n)/(b**3*n) - 8*x**n/(b**4*n) + 16*log(x**n + 2/b)/(b**5*n), True))

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