3.2613 \(\int \frac {x^{-3-2 (-1+n)}}{a+b x^n} \, dx\)

Optimal. Leaf size=57 \[ -\frac {b^2 \log \left (a+b x^n\right )}{a^3 n}+\frac {b^2 \log (x)}{a^3}+\frac {b x^{-n}}{a^2 n}-\frac {x^{-2 n}}{2 a n} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {266, 44} \[ -\frac {b^2 \log \left (a+b x^n\right )}{a^3 n}+\frac {b^2 \log (x)}{a^3}+\frac {b x^{-n}}{a^2 n}-\frac {x^{-2 n}}{2 a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-3 - 2*(-1 + n))/(a + b*x^n),x]

[Out]

-1/(2*a*n*x^(2*n)) + b/(a^2*n*x^n) + (b^2*Log[x])/a^3 - (b^2*Log[a + b*x^n])/(a^3*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^{-3-2 (-1+n)}}{a+b x^n} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^3 (a+b x)} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{a x^3}-\frac {b}{a^2 x^2}+\frac {b^2}{a^3 x}-\frac {b^3}{a^3 (a+b x)}\right ) \, dx,x,x^n\right )}{n}\\ &=-\frac {x^{-2 n}}{2 a n}+\frac {b x^{-n}}{a^2 n}+\frac {b^2 \log (x)}{a^3}-\frac {b^2 \log \left (a+b x^n\right )}{a^3 n}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.86 \[ \frac {-2 b^2 \log \left (a+b x^n\right )+a x^{-2 n} \left (2 b x^n-a\right )+2 b^2 n \log (x)}{2 a^3 n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-3 - 2*(-1 + n))/(a + b*x^n),x]

[Out]

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

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fricas [A]  time = 0.80, size = 59, normalized size = 1.04 \[ \frac {2 \, b^{2} n x^{2 \, n} \log \relax (x) - 2 \, b^{2} x^{2 \, n} \log \left (b x^{n} + a\right ) + 2 \, a b x^{n} - a^{2}}{2 \, a^{3} n x^{2 \, n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b^2*n*x^(2*n)*log(x) - 2*b^2*x^(2*n)*log(b*x^n + a) + 2*a*b*x^n - a^2)/(a^3*n*x^(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 = 0.00, size = 69, normalized size = 1.21 \[ \left (\frac {b^{2} {\mathrm e}^{2 n \ln \relax (x )} \ln \relax (x )}{a^{3}}+\frac {b \,{\mathrm e}^{n \ln \relax (x )}}{a^{2} n}-\frac {1}{2 a n}\right ) {\mathrm e}^{-2 n \ln \relax (x )}-\frac {b^{2} \ln \left (b \,{\mathrm e}^{n \ln \relax (x )}+a \right )}{a^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.66, size = 58, normalized size = 1.02 \[ \frac {b^{2} \log \relax (x)}{a^{3}} - \frac {b^{2} \log \left (\frac {b x^{n} + a}{b}\right )}{a^{3} n} + \frac {2 \, b x^{n} - a}{2 \, a^{2} n x^{2 \, n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{x^{2\,n+1}\,\left (a+b\,x^n\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(2*n + 1)*(a + b*x^n)),x)

[Out]

int(1/(x^(2*n + 1)*(a + b*x^n)), x)

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sympy [A]  time = 145.96, size = 85, normalized size = 1.49 \[ \begin {cases} \tilde {\infty } \log {\relax (x )} & \text {for}\: a = 0 \wedge b = 0 \wedge n = 0 \\- \frac {x^{- 3 n}}{3 b n} & \text {for}\: a = 0 \\\frac {\log {\relax (x )}}{a + b} & \text {for}\: n = 0 \\- \frac {x^{- 2 n}}{2 a n} & \text {for}\: b = 0 \\- \frac {x^{- 2 n}}{2 a n} + \frac {b x^{- n}}{a^{2} n} + \frac {b^{2} \log {\relax (x )}}{a^{3}} - \frac {b^{2} \log {\left (\frac {a}{b} + x^{n} \right )}}{a^{3} n} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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