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

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

[Out]

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

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Rubi [A]  time = 0.0347027, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1584, 345, 193, 321, 205} \[ \frac{2 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{b} x^{-n/2}}{\sqrt{c}}\right )}{b^{3/2} n}-\frac{2 x^{-n/2}}{b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1584

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]

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]

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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.0071318, size = 32, normalized size = 0.64 \[ -\frac{2 x^{-n/2} \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{c x^n}{b}\right )}{b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.062, size = 79, normalized size = 1.6 \begin{align*} -2\,{\frac{1}{bn{x}^{n/2}}}+{\frac{1}{{b}^{2}n}\sqrt{-bc}\ln \left ({x}^{{\frac{n}{2}}}-{\frac{1}{c}\sqrt{-bc}} \right ) }-{\frac{1}{{b}^{2}n}\sqrt{-bc}\ln \left ({x}^{{\frac{n}{2}}}+{\frac{1}{c}\sqrt{-bc}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -c \int \frac{x^{\frac{1}{2} \, n}}{b c x x^{n} + b^{2} x}\,{d x} - \frac{2}{b n x^{\frac{1}{2} \, n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.66699, size = 323, normalized size = 6.46 \begin{align*} \left [\frac{x x^{\frac{1}{2} \, n - 1} \sqrt{-\frac{c}{b}} \log \left (\frac{c x^{2} x^{n - 2} - 2 \, b x x^{\frac{1}{2} \, n - 1} \sqrt{-\frac{c}{b}} - b}{c x^{2} x^{n - 2} + b}\right ) - 2}{b n x x^{\frac{1}{2} \, n - 1}}, \frac{2 \,{\left (x x^{\frac{1}{2} \, n - 1} \sqrt{\frac{c}{b}} \arctan \left (\frac{b \sqrt{\frac{c}{b}}}{c x x^{\frac{1}{2} \, n - 1}}\right ) - 1\right )}}{b n x x^{\frac{1}{2} \, n - 1}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{1}{2} \, n - 1}}{c x^{2 \, n} + b x^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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