3.6.12 \(\int \frac {x^{-1+2 n}}{(a^2+2 a b x^n+b^2 x^{2 n})^{3/2}} \, dx\) [512]

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]

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

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Rubi [A]
time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, 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 = {1369, 270} \begin {gather*} \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 {gather*}

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 270

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]

Rule 1369

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]

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*}

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Mathematica [A]
time = 0.04, size = 40, normalized size = 0.83 \begin {gather*} \frac {\left (-a-2 b x^n\right ) \left (a+b x^n\right )}{2 b^2 n \left (\left (a+b x^n\right )^2\right )^{3/2}} \end {gather*}

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]

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

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Maple [A]
time = 0.03, size = 37, normalized size = 0.77

method	result	size
risch	\(-\frac {\sqrt {\left (a +b \,x^{n}\right )^{2}}\, \left (2 b \,x^{n}+a \right )}{2 \left (a +b \,x^{n}\right )^{3} b^{2} n}\)	\(37\)

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,method=_RETURNVERBOSE)

[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 = 0.41, size = 41, normalized size = 0.85 \begin {gather*} -\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 {gather*}

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 = 0.34, size = 41, normalized size = 0.85 \begin {gather*} -\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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2 n - 1}}{\left (\left (a + b x^{n}\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

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]

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

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

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^{2\,n-1}}{{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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