3.6.46 \(\int (a^2+2 a b x^n+b^2 x^{2 n})^{\frac {-1-2 n}{2 n}} \, dx\) [546]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 102, normalized size of antiderivative = 1.00, 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 = {1357, 198, 197} \begin {gather*} \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)} \end {gather*}

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 197

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]

Rule 198

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 1357

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]

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] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.05, size = 59, normalized size = 0.58 \begin {gather*} \frac {x \left (\left (a+b x^n\right )^2\right )^{\left .-\frac {1}{2}\right /n} \left (1+\frac {b x^n}{a}\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} \end {gather*}

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)))

________________________________________________________________________________________

Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (a^{2}+2 a b \,x^{n}+b^{2} x^{2 n}\right )^{-\frac {1+2 n}{2 n}}\, dx\]

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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 82, normalized size = 0.80 \begin {gather*} \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 {gather*}

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))

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{\frac {n+\frac {1}{2}}{n}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________