3.513 \(\int \frac{x^{-1+2 n}}{(a^2+2 a b x^n+b^2 x^{2 n})^{5/2}} \, dx\)

Optimal. Leaf size=88 \[ \frac{a}{4 b^2 n \left (a+b x^n\right )^3 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}-\frac{1}{3 b^2 n \left (a+b x^n\right )^2 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

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

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Rubi [A]  time = 0.0515943, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {1355, 266, 43} \[ \frac{a}{4 b^2 n \left (a+b x^n\right )^3 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}-\frac{1}{3 b^2 n \left (a+b x^n\right )^2 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + 2*n)/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(5/2),x]

[Out]

a/(4*b^2*n*(a + b*x^n)^3*Sqrt[a^2 + 2*a*b*x^n + b^2*x^(2*n)]) - 1/(3*b^2*n*(a + b*x^n)^2*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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^{-1+2 n}}{\left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x^n\right )\right ) \int \frac{x^{-1+2 n}}{\left (a b+b^2 x^n\right )^5} \, dx}{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac{\left (b^4 \left (a b+b^2 x^n\right )\right ) \operatorname{Subst}\left (\int \frac{x}{\left (a b+b^2 x\right )^5} \, dx,x,x^n\right )}{n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac{\left (b^4 \left (a b+b^2 x^n\right )\right ) \operatorname{Subst}\left (\int \left (-\frac{a}{b^6 (a+b x)^5}+\frac{1}{b^6 (a+b x)^4}\right ) \, dx,x,x^n\right )}{n \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ &=\frac{a}{4 b^2 n \left (a+b x^n\right )^3 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}-\frac{1}{3 b^2 n \left (a+b x^n\right )^2 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}}\\ \end{align*}

Mathematica [A]  time = 0.0327514, size = 40, normalized size = 0.45 \[ -\frac{a+4 b x^n}{12 b^2 n \left (a+b x^n\right )^3 \sqrt{\left (a+b x^n\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(-1 + 2*n)/(a^2 + 2*a*b*x^n + b^2*x^(2*n))^(5/2),x]

[Out]

-(a + 4*b*x^n)/(12*b^2*n*(a + b*x^n)^3*Sqrt[(a + b*x^n)^2])

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Maple [A]  time = 0.024, size = 37, normalized size = 0.4 \begin{align*} -{\frac{4\,b{x}^{n}+a}{12\, \left ( a+b{x}^{n} \right ) ^{5}{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))^(5/2),x)

[Out]

-1/12*((a+b*x^n)^2)^(1/2)/(a+b*x^n)^5*(4*b*x^n+a)/b^2/n

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Maxima [A]  time = 1.02756, size = 93, normalized size = 1.06 \begin{align*} -\frac{4 \, b x^{n} + a}{12 \,{\left (b^{6} n x^{4 \, n} + 4 \, a b^{5} n x^{3 \, n} + 6 \, a^{2} b^{4} n x^{2 \, n} + 4 \, a^{3} b^{3} n x^{n} + a^{4} 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))^(5/2),x, algorithm="maxima")

[Out]

-1/12*(4*b*x^n + a)/(b^6*n*x^(4*n) + 4*a*b^5*n*x^(3*n) + 6*a^2*b^4*n*x^(2*n) + 4*a^3*b^3*n*x^n + a^4*b^2*n)

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Fricas [A]  time = 1.56607, size = 147, normalized size = 1.67 \begin{align*} -\frac{4 \, b x^{n} + a}{12 \,{\left (b^{6} n x^{4 \, n} + 4 \, a b^{5} n x^{3 \, n} + 6 \, a^{2} b^{4} n x^{2 \, n} + 4 \, a^{3} b^{3} n x^{n} + a^{4} 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))^(5/2),x, algorithm="fricas")

[Out]

-1/12*(4*b*x^n + a)/(b^6*n*x^(4*n) + 4*a*b^5*n*x^(3*n) + 6*a^2*b^4*n*x^(2*n) + 4*a^3*b^3*n*x^n + a^4*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))**(5/2),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{5}{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))^(5/2),x, algorithm="giac")

[Out]

integrate(x^(2*n - 1)/(b^2*x^(2*n) + 2*a*b*x^n + a^2)^(5/2), x)