\(\int \frac {1}{x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx\) [534]

Optimal result

Integrand size = 28, antiderivative size = 65 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=-\frac {\left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},-\frac {1-n}{n},-\frac {b x^n}{a}\right )}{a x \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {1369, 371} \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=-\frac {\left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},-\frac {1-n}{n},-\frac {b x^n}{a}\right )}{a x \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \]

[In]

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

[Out]

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1369

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a b+b^2 x^n\right ) \int \frac {1}{x^2 \left (a b+b^2 x^n\right )} \, dx}{\sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ & = -\frac {\left (a+b x^n\right ) \, _2F_1\left (1,-\frac {1}{n};-\frac {1-n}{n};-\frac {b x^n}{a}\right )}{a x \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^2 \sqrt {a^2+2 a b x^n+b^2 x^{2 n}}} \, dx=-\frac {\left (a+b x^n\right ) \operatorname {Hypergeometric2F1}\left (1,-\frac {1}{n},1-\frac {1}{n},-\frac {b x^n}{a}\right )}{a x \sqrt {\left (a+b x^n\right )^2}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

integrate(1/x^2/(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^2), x)

Giac [F]

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

[In]

integrate(1/x^2/(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^2), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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