\(\int \frac {1}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx\) [532]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \frac {1}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {x \left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1357, 251} \[ \int \frac {1}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {x \left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[In]

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

[Out]

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

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 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}& = \frac {\left (2 a b+2 b^2 x^n\right ) \int \frac {1}{2 a b+2 b^2 x^n} \, dx}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ & = \frac {x \left (a+b x^n\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\frac {x \left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a \sqrt {\left (a+b x^n\right )^2}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {1}{\sqrt {a^{2}+2 a b \,x^{n}+b^{2} x^{2 n}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {1}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int { \frac {1}{\sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {1}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int \frac {1}{\sqrt {a^{2} + 2 a b x^{n} + b^{2} x^{2 n}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int { \frac {1}{\sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int { \frac {1}{\sqrt {b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=\int \frac {1}{\sqrt {a^2+b^2\,x^{2\,n}+2\,a\,b\,x^n}} \,d x \]

[In]

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

[Out]

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