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

Optimal. Leaf size=48 \[ \frac{x^{2 n}}{2 a n \left (a+b x^n\right ) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

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

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Rubi [A]  time = 0.0268692, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {1355, 264} \[ \frac{x^{2 n}}{2 a n \left (a+b x^n\right ) \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0165093, size = 35, normalized size = 0.73 \[ \frac{x^{2 n} \left (a+b x^n\right )}{2 a n \left (\left (a+b x^n\right )^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 37, normalized size = 0.8 \begin{align*} -{\frac{2\,b{x}^{n}+a}{2\, \left ( a+b{x}^{n} \right ) ^{3}{b}^{2}n}\sqrt{ \left ( a+b{x}^{n} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.990497, size = 55, normalized size = 1.15 \begin{align*} -\frac{2 \, b x^{n} + a}{2 \,{\left (b^{4} n x^{2 \, n} + 2 \, a b^{3} n x^{n} + a^{2} b^{2} n\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="maxima")

[Out]

-1/2*(2*b*x^n + a)/(b^4*n*x^(2*n) + 2*a*b^3*n*x^n + a^2*b^2*n)

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Fricas [A]  time = 1.55798, size = 86, normalized size = 1.79 \begin{align*} -\frac{2 \, b x^{n} + a}{2 \,{\left (b^{4} n x^{2 \, n} + 2 \, a b^{3} n x^{n} + a^{2} b^{2} n\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="fricas")

[Out]

-1/2*(2*b*x^n + a)/(b^4*n*x^(2*n) + 2*a*b^3*n*x^n + a^2*b^2*n)

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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+2*n)/(a**2+2*a*b*x**n+b**2*x**(2*n))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2 \, n - 1}}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(-1+2*n)/(a^2+2*a*b*x^n+b^2*x^(2*n))^(3/2),x, algorithm="giac")

[Out]

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