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

Optimal. Leaf size=65 \[ -\frac{\left (a+b x^n\right ) \, _2F_1\left (1,-\frac{1}{n};-\frac{1-n}{n};-\frac{b x^n}{a}\right )}{a x \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

[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)]))

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Rubi [A]  time = 0.0258666, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {1355, 364} \[ -\frac{\left (a+b x^n\right ) \, _2F_1\left (1,-\frac{1}{n};-\frac{1-n}{n};-\frac{b x^n}{a}\right )}{a x \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

Antiderivative was successfully verified.

[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 1355

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]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
 (ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^2 \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \, dx &=\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]  time = 0.0140003, size = 51, normalized size = 0.78 \[ -\frac{\left (a+b x^n\right ) \, _2F_1\left (1,-\frac{1}{n};1-\frac{1}{n};-\frac{b x^n}{a}\right )}{a x \sqrt{\left (a+b x^n\right )^2}} \]

Antiderivative was successfully verified.

[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]))

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Maple [F]  time = 0.017, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}}{\frac{1}{\sqrt{{a}^{2}+2\,ab{x}^{n}+{b}^{2}{x}^{2\,n}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\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\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)