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

Optimal. Leaf size=55 \[ \frac{x \left (a+b x^n\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

[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)])

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Rubi [A]  time = 0.0173405, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1343, 245} \[ \frac{x \left (a+b x^n\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a \sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \]

Antiderivative was successfully verified.

[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 1343

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]

Rule 245

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])

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a^2+2 a b x^n+b^2 x^{2 n}}} \, dx &=\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]  time = 0.0103412, size = 44, normalized size = 0.8 \[ \frac{x \left (a+b x^n\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a \sqrt{\left (a+b x^n\right )^2}} \]

Antiderivative was successfully verified.

[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])

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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{\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/(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)

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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}}}\,{d x} \end{align*}

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),x, algorithm="maxima")

[Out]

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

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

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),x, algorithm="fricas")

[Out]

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

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

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),x)

[Out]

Integral(1/sqrt(a**2 + 2*a*b*x**n + b**2*x**(2*n)), 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}}}\,{d x} \end{align*}

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),x, algorithm="giac")

[Out]

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