3.27.19 \(\int \frac {x^{-1+n}}{2+b x^n} \, dx\) [2619]

Optimal. Leaf size=15 \[ \frac {\log \left (2+b x^n\right )}{b n} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {266} \begin {gather*} \frac {\log \left (b x^n+2\right )}{b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + n)/(2 + b*x^n),x]

[Out]

Log[2 + b*x^n]/(b*n)

Rule 266

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

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 18, normalized size = 1.20 \begin {gather*} \frac {\log \left (2 n+b n x^n\right )}{b n} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + n)/(2 + b*x^n),x]

[Out]

Log[2*n + b*n*x^n]/(b*n)

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Maple [A]
time = 0.27, size = 18, normalized size = 1.20

method	result	size
norman	\(\frac {\ln \left (2+b \,{\mathrm e}^{n \ln \left (x \right )}\right )}{b n}\)	\(18\)
risch	\(\frac {\ln \left (x^{n}+\frac {2}{b}\right )}{b n}\)	\(18\)
meijerg	\(-\frac {i \left (-1\right )^{-\frac {\mathrm {csgn}\left (i b \right )}{2}-\frac {\mathrm {csgn}\left (i x^{n}\right )}{2}+\frac {\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i b \right )}{2}-\frac {-1+n}{n}-\frac {1}{n}} \ln \left (1-\frac {i x^{n} b \left (-1\right )^{\frac {\mathrm {csgn}\left (i b \right )}{2}+\frac {\mathrm {csgn}\left (i x^{n}\right )}{2}-\frac {\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i b \right )}{2}}}{2}\right )}{b n}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 15, normalized size = 1.00 \begin {gather*} \frac {\log \left (b x^{n} + 2\right )}{b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b*x^n + 2)/(b*n)

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Fricas [A]
time = 0.40, size = 15, normalized size = 1.00 \begin {gather*} \frac {\log \left (b x^{n} + 2\right )}{b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b*x^n + 2)/(b*n)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (10) = 20\).
time = 0.70, size = 27, normalized size = 1.80 \begin {gather*} \begin {cases} \frac {\log {\left (x \right )}}{2} & \text {for}\: b = 0 \wedge n = 0 \\\frac {x^{n}}{2 n} & \text {for}\: b = 0 \\\frac {\log {\left (x \right )}}{b + 2} & \text {for}\: n = 0 \\\frac {\log {\left (x^{n} + \frac {2}{b} \right )}}{b n} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.30, size = 16, normalized size = 1.07 \begin {gather*} \frac {\log \left ({\left | b x^{n} + 2 \right |}\right )}{b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(b*x^n + 2))/(b*n)

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Mupad [B]
time = 1.21, size = 15, normalized size = 1.00 \begin {gather*} \frac {\ln \left (b\,x^n+2\right )}{b\,n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(n - 1)/(b*x^n + 2),x)

[Out]

log(b*x^n + 2)/(b*n)

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