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

Optimal. Leaf size=102 \[ \frac{n x \left (a+b x^n\right )^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac{1}{2} \left (-\frac{1}{n}-2\right )}}{a^2 (n+1)}+\frac{x \left (a+b x^n\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac{1}{2} \left (-\frac{1}{n}-2\right )}}{a (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.0442048, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {1343, 192, 191} \[ \frac{n x \left (a+b x^n\right )^2 \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac{1}{2} \left (-\frac{1}{n}-2\right )}}{a^2 (n+1)}+\frac{x \left (a+b x^n\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac{1}{2} \left (-\frac{1}{n}-2\right )}}{a (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1343

Int[((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^(2*n))^p/(b + 2*c*x
^n)^(2*p), Int[(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0]

Rule 192

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

Rule 191

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

Rubi steps

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

Mathematica [C]  time = 0.0441615, size = 59, normalized size = 0.58 \[ \frac{x \left (\left (a+b x^n\right )^2\right )^{\left .-\frac{1}{2}\right /n} \left (\frac{b x^n}{a}+1\right )^{\frac{1}{n}} \, _2F_1\left (2+\frac{1}{n},\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.092, size = 0, normalized size = 0. \begin{align*} \int \left ( \left ({a}^{2}+2\,ab{x}^{n}+{b}^{2}{x}^{2\,n} \right ) ^{{\frac{1+2\,n}{2\,n}}} \right ) ^{-1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.67306, size = 171, normalized size = 1.68 \begin{align*} \frac{b^{2} n x x^{2 \, n} +{\left (2 \, a b n + a b\right )} x x^{n} +{\left (a^{2} n + a^{2}\right )} x}{{\left (a^{2} n + a^{2}\right )}{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac{2 \, n + 1}{2 \, n}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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