\(\int (a^2+2 a b x^n+b^2 x^{2 n})^{\frac {-1-n}{2 n}} \, dx\) [544]

Optimal result

Integrand size = 33, antiderivative size = 43 \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac {-1-n}{2 n}} \, dx=\frac {x \left (a+b x^n\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{-\frac {1+n}{2 n}}}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {1357, 197} \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac {-1-n}{2 n}} \, dx=\frac {x \left (a+b x^n\right ) \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{-\frac {n+1}{2 n}}}{a} \]

[In]

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

[Out]

(x*(a + b*x^n))/(a*(a^2 + 2*a*b*x^n + b^2*x^(2*n))^((1 + n)/(2*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 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*} \text {integral}& = \left (\left (2 a b+2 b^2 x^n\right )^{-\frac {-1-n}{n}} \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac {-1-n}{2 n}}\right ) \int \left (2 a b+2 b^2 x^n\right )^{\frac {-1-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+n}{2 n}}}{a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.74 \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac {-1-n}{2 n}} \, dx=\frac {x \left (a+b x^n\right ) \left (\left (a+b x^n\right )^2\right )^{-\frac {1+n}{2 n}}}{a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.19

method	result	size
norman	\(\left (x +\frac {b x \,{\mathrm e}^{n \ln \left (x \right )}}{a}\right ) {\mathrm e}^{\frac {\left (1+n \right ) \ln \left (\frac {1}{\sqrt {a^{2}+2 a b \,{\mathrm e}^{n \ln \left (x \right )}+b^{2} {\mathrm e}^{2 n \ln \left (x \right )}}}\right )}{n}}\)	\(51\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.05 \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac {-1-n}{2 n}} \, dx=\frac {b x x^{n} + a x}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac {n + 1}{2 \, n}} a} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (37) = 74\).

Time = 4.98 (sec) , antiderivative size = 160, normalized size of antiderivative = 3.72 \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac {-1-n}{2 n}} \, dx=\begin {cases} x \left (a^{2} + 2 a b x^{n} + b^{2} x^{2 n}\right )^{- \frac {1}{2} - \frac {1}{2 n}} + \frac {b x x^{n} \left (a^{2} + 2 a b x^{n} + b^{2} x^{2 n}\right )^{- \frac {1}{2} - \frac {1}{2 n}}}{a} & \text {for}\: a \neq 0 \\- x \left (b^{2} x^{2 n}\right )^{-1 - \frac {1}{n}} \left (b^{2} x^{2 n}\right )^{\frac {1}{2} + \frac {1}{2 n}} + x \left (b^{2} x^{2 n}\right )^{- \frac {1}{2} - \frac {1}{2 n}} - \frac {x \left (b^{2} x^{2 n}\right )^{-1 - \frac {1}{n}} \left (b^{2} x^{2 n}\right )^{\frac {1}{2} + \frac {1}{2 n}}}{n} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*(a**2 + 2*a*b*x**n + b**2*x**(2*n))**(-1/2 - 1/(2*n)) + b*x*x**n*(a**2 + 2*a*b*x**n + b**2*x**(2*
n))**(-1/2 - 1/(2*n))/a, Ne(a, 0)), (-x*(b**2*x**(2*n))**(-1 - 1/n)*(b**2*x**(2*n))**(1/2 + 1/(2*n)) + x*(b**2
*x**(2*n))**(-1/2 - 1/(2*n)) - x*(b**2*x**(2*n))**(-1 - 1/n)*(b**2*x**(2*n))**(1/2 + 1/(2*n))/n, True))

Maxima [F]

\[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac {-1-n}{2 n}} \, dx=\int { \frac {1}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac {n + 1}{2 \, n}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac {-1-n}{2 n}} \, dx=\int { \frac {1}{{\left (b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}\right )}^{\frac {n + 1}{2 \, n}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \left (a^2+2 a b x^n+b^2 x^{2 n}\right )^{\frac {-1-n}{2 n}} \, dx=\int \frac {1}{{\left (a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n\right )}^{\frac {\frac {n}{2}+\frac {1}{2}}{n}}} \,d x \]

[In]

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

[Out]

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