3.2639 \(\int \frac {x^{-1-\frac {n}{2}}}{a+b x^n} \, dx\)

Optimal. Leaf size=50 \[ \frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x^{-n/2}}{\sqrt {b}}\right )}{a^{3/2} n}-\frac {2 x^{-n/2}}{a n} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {345, 193, 321, 205} \[ \frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x^{-n/2}}{\sqrt {b}}\right )}{a^{3/2} n}-\frac {2 x^{-n/2}}{a n} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 - n/2)/(a + b*x^n),x]

[Out]

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

Rule 193

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]
&& IntegerQ[p]

Rule 205

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

Rule 321

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,
c, n, m, p, x]

Rule 345

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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 32, normalized size = 0.64 \[ -\frac {2 x^{-n/2} \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\frac {b x^n}{a}\right )}{a n} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 - n/2)/(a + b*x^n),x]

[Out]

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

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fricas [A]  time = 0.78, size = 130, normalized size = 2.60 \[ \left [-\frac {2 \, x x^{-\frac {1}{2} \, n - 1} - \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} x^{-n - 2} + 2 \, a x x^{-\frac {1}{2} \, n - 1} \sqrt {-\frac {b}{a}} - b}{a x^{2} x^{-n - 2} + b}\right )}{a n}, -\frac {2 \, {\left (x x^{-\frac {1}{2} \, n - 1} + \sqrt {\frac {b}{a}} \arctan \left (\frac {\sqrt {\frac {b}{a}}}{x x^{-\frac {1}{2} \, n - 1}}\right )\right )}}{a n}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*x*x^(-1/2*n - 1) - sqrt(-b/a)*log((a*x^2*x^(-n - 2) + 2*a*x*x^(-1/2*n - 1)*sqrt(-b/a) - b)/(a*x^2*x^(-n -
2) + b)))/(a*n), -2*(x*x^(-1/2*n - 1) + sqrt(b/a)*arctan(sqrt(b/a)/(x*x^(-1/2*n - 1))))/(a*n)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 79, normalized size = 1.58 \[ -\frac {2 x^{-\frac {n}{2}}}{a n}+\frac {\sqrt {-a b}\, \ln \left (x^{\frac {n}{2}}-\frac {\sqrt {-a b}}{b}\right )}{a^{2} n}-\frac {\sqrt {-a b}\, \ln \left (x^{\frac {n}{2}}+\frac {\sqrt {-a b}}{b}\right )}{a^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(n/2 + 1)*(a + b*x^n)),x)

[Out]

int(1/(x^(n/2 + 1)*(a + b*x^n)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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