∫ x−1−n2 ______________

Optimal. Leaf size=68 \[ -\frac {2 x^{-3 n/2}}{3 b n}+\frac {2 c x^{-n/2}}{b^2 n}-\frac {2 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^{-n/2}}{\sqrt {c}}\right )}{b^{5/2} n} \]

-2/3/b/n/(x^(3/2*n))+2*c/b^2/n/(x^(1/2*n))-2*c^(3/2)*arctan(b^(1/2)/(x^(1/2*n))/c^(1/2))/b^(5/2)/n

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Rubi [A]
time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1598, 369, 352, 199, 327, 211} \begin {gather*} -\frac {2 c^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} x^{-n/2}}{\sqrt {c}}\right )}{b^{5/2} n}+\frac {2 c x^{-n/2}}{b^2 n}-\frac {2 x^{-3 n/2}}{3 b n} \end {gather*}

-2/(3*b*n*x^((3*n)/2)) + (2*c)/(b^2*n*x^(n/2)) - (2*c^(3/2)*ArcTan[Sqrt[b]/(Sqrt[c]*x^(n/2))])/(b^(5/2)*n)

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n

)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p

, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/(m + 1), Subst[Int[(a + b*x^Simplify[n/(m +

1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] && !IntegerQ[n]

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[x^(m + 1)/(a*(m + 1)), x] - Dist[b/a, Int[x^Simplify

[m + n]/(a + b*x^n), x], x] /; FreeQ[{a, b, m, n}, x] && FractionQ[Simplify[(m + 1)/n]] && SumSimplerQ[m, n]

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -

p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

\begin {align*} \int \frac {x^{-1-\frac {n}{2}}}{b x^n+c x^{2 n}} \, dx &=\int \frac {x^{-1-\frac {3 n}{2}}}{b+c x^n} \, dx\\ &=-\frac {2 x^{-3 n/2}}{3 b n}-\frac {c \int \frac {x^{-1-\frac {n}{2}}}{b+c x^n} \, dx}{b}\\ &=-\frac {2 x^{-3 n/2}}{3 b n}+\frac {(2 c) \text {Subst}\left (\int \frac {1}{b+\frac {c}{x^2}} \, dx,x,x^{-n/2}\right )}{b n}\\ &=-\frac {2 x^{-3 n/2}}{3 b n}+\frac {(2 c) \text {Subst}\left (\int \frac {x^2}{c+b x^2} \, dx,x,x^{-n/2}\right )}{b n}\\ &=-\frac {2 x^{-3 n/2}}{3 b n}+\frac {2 c x^{-n/2}}{b^2 n}-\frac {\left (2 c^2\right ) \text {Subst}\left (\int \frac {1}{c+b x^2} \, dx,x,x^{-n/2}\right )}{b^2 n}\\ &=-\frac {2 x^{-3 n/2}}{3 b n}+\frac {2 c x^{-n/2}}{b^2 n}-\frac {2 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x^{-n/2}}{\sqrt {c}}\right )}{b^{5/2} n}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.03, size = 34, normalized size = 0.50 \begin {gather*} -\frac {2 x^{-3 n/2} \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-\frac {c x^n}{b}\right )}{3 b n} \end {gather*}

(-2*Hypergeometric2F1[-3/2, 1, -1/2, -((c*x^n)/b)])/(3*b*n*x^((3*n)/2))

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method	result	size

risch	\(\frac {2 c \,x^{-\frac {n}{2}}}{b^{2} n}-\frac {2 x^{-\frac {3 n}{2}}}{3 b n}+\frac {\sqrt {-b c}\, c \ln \left (x^{\frac {n}{2}}+\frac {\sqrt {-b c}}{c}\right )}{b^{3} n}-\frac {\sqrt {-b c}\, c \ln \left (x^{\frac {n}{2}}-\frac {\sqrt {-b c}}{c}\right )}{b^{3} n}\)	\(97\)

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

c^2*integrate(x^(1/2*n)/(b^2*c*x*x^n + b^3*x), x) + 2/3*(3*c*x^n - b)/(b^2*n*x^(3/2*n))

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Fricas [A]
time = 0.38, size = 161, normalized size = 2.37 \begin {gather*} \left [-\frac {2 \, b x^{3} x^{-\frac {3}{2} \, n - 3} - 6 \, c x x^{-\frac {1}{2} \, n - 1} - 3 \, c \sqrt {-\frac {c}{b}} \log \left (\frac {b x^{2} x^{-n - 2} - 2 \, b x x^{-\frac {1}{2} \, n - 1} \sqrt {-\frac {c}{b}} - c}{b x^{2} x^{-n - 2} + c}\right )}{3 \, b^{2} n}, -\frac {2 \, {\left (b x^{3} x^{-\frac {3}{2} \, n - 3} - 3 \, c x x^{-\frac {1}{2} \, n - 1} - 3 \, c \sqrt {\frac {c}{b}} \arctan \left (\frac {\sqrt {\frac {c}{b}}}{x x^{-\frac {1}{2} \, n - 1}}\right )\right )}}{3 \, b^{2} n}\right ] \end {gather*}

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

[-1/3*(2*b*x^3*x^(-3/2*n - 3) - 6*c*x*x^(-1/2*n - 1) - 3*c*sqrt(-c/b)*log((b*x^2*x^(-n - 2) - 2*b*x*x^(-1/2*n

- 1)*sqrt(-c/b) - c)/(b*x^2*x^(-n - 2) + c)))/(b^2*n), -2/3*(b*x^3*x^(-3/2*n - 3) - 3*c*x*x^(-1/2*n - 1) - 3*c

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{\frac {n}{2}+1}\,\left (b\,x^n+c\,x^{2\,n}\right )} \,d x \end {gather*}

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3.6.3 \(\int \frac {x^{-1-\frac {n}{2}}}{b x^n+c x^{2 n}} \, dx\) [503]