3.2642 \(\int \frac {x^{-1-n}}{a+b x^n} \, dx\)

Optimal. Leaf size=38 \[ \frac {b \log \left (a+b x^n\right )}{a^2 n}-\frac {b \log (x)}{a^2}-\frac {x^{-n}}{a n} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, 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, 44} \[ \frac {b \log \left (a+b x^n\right )}{a^2 n}-\frac {b \log (x)}{a^2}-\frac {x^{-n}}{a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - n)/(a + b*x^n),x]

[Out]

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

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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-n}}{a+b x^n} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^2 (a+b x)} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a x^2}-\frac {b}{a^2 x}+\frac {b^2}{a^2 (a+b x)}\right ) \, dx,x,x^n\right )}{n}\\ &=-\frac {x^{-n}}{a n}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^n\right )}{a^2 n}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 32, normalized size = 0.84 \[ -\frac {-b \log \left (a+b x^n\right )+a x^{-n}+b n \log (x)}{a^2 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - n)/(a + b*x^n),x]

[Out]

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

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fricas [A]  time = 0.63, size = 37, normalized size = 0.97 \[ -\frac {b n x^{n} \log \relax (x) - b x^{n} \log \left (b x^{n} + a\right ) + a}{a^{2} n x^{n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(-n - 1)/(b*x^n + a), x)

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maple [A]  time = 0.00, size = 50, normalized size = 1.32 \[ \left (-\frac {b \,{\mathrm e}^{n \ln \relax (x )} \ln \relax (x )}{a^{2}}-\frac {1}{a n}\right ) {\mathrm e}^{-n \ln \relax (x )}+\frac {b \ln \left (b \,{\mathrm e}^{n \ln \relax (x )}+a \right )}{a^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 42, normalized size = 1.11 \[ -\frac {b \log \relax (x)}{a^{2}} + \frac {b \log \left (\frac {b x^{n} + a}{b}\right )}{a^{2} n} - \frac {1}{a n x^{n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {1}{x^{n+1}\,\left (a+b\,x^n\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(n + 1)*(a + b*x^n)),x)

[Out]

int(1/(x^(n + 1)*(a + b*x^n)), x)

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sympy [A]  time = 25.36, size = 48, normalized size = 1.26 \[ \begin {cases} \frac {\log {\relax (x )}}{b} & \text {for}\: a = 0 \wedge n = 0 \\- \frac {x^{- 2 n}}{2 b n} & \text {for}\: a = 0 \\\frac {\log {\relax (x )}}{a + b} & \text {for}\: n = 0 \\- \frac {x^{- n}}{a n} + \frac {b \log {\left (x^{- n} + \frac {b}{a} \right )}}{a^{2} n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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