3.2600 \(\int \frac {x^{-1+2 n}}{a+b x^n} \, dx\)

Optimal. Leaf size=28 \[ \frac {x^n}{b n}-\frac {a \log \left (a+b x^n\right )}{b^2 n} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 28, 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} \[ \frac {x^n}{b n}-\frac {a \log \left (a+b x^n\right )}{b^2 n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)/(a + b*x^n),x]

[Out]

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

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Mathematica [A]  time = 0.02, size = 26, normalized size = 0.93 \[ \frac {\frac {x^n}{b}-\frac {a \log \left (a+b x^n\right )}{b^2}}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 2*n)/(a + b*x^n),x]

[Out]

(x^n/b - (a*Log[a + b*x^n])/b^2)/n

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fricas [A]  time = 0.66, size = 24, normalized size = 0.86 \[ \frac {b x^{n} - a \log \left (b x^{n} + a\right )}{b^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x^n - a*log(b*x^n + a))/(b^2*n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(2*n - 1)/(b*x^n + a), x)

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maple [A]  time = 0.02, size = 33, normalized size = 1.18 \[ -\frac {a \ln \left (b \,{\mathrm e}^{n \ln \relax (x )}+a \right )}{b^{2} n}+\frac {{\mathrm e}^{n \ln \relax (x )}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.60, size = 32, normalized size = 1.14 \[ \frac {x^{n}}{b n} - \frac {a \log \left (\frac {b x^{n} + a}{b}\right )}{b^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {x^{2\,n-1}}{a+b\,x^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(2*n - 1)/(a + b*x^n),x)

[Out]

int(x^(2*n - 1)/(a + b*x^n), x)

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sympy [A]  time = 12.23, size = 41, normalized size = 1.46 \[ \begin {cases} \frac {\log {\relax (x )}}{a} & \text {for}\: b = 0 \wedge n = 0 \\\frac {\log {\relax (x )}}{a + b} & \text {for}\: n = 0 \\\frac {x^{2 n}}{2 a n} & \text {for}\: b = 0 \\- \frac {a \log {\left (\frac {a}{b} + x^{n} \right )}}{b^{2} n} + \frac {x^{n}}{b n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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