3.500 \(\int \frac{(x+\sqrt{a+x^2})^n}{(a+x^2)^{5/2}} \, dx\)

Optimal. Leaf size=59 \[ \frac{16 \left (\sqrt{a+x^2}+x\right )^{n+4} \, _2F_1\left (4,\frac{n+4}{2};\frac{n+6}{2};-\frac{\left (x+\sqrt{x^2+a}\right )^2}{a}\right )}{a^4 (n+4)} \]

[Out]

(16*(x + Sqrt[a + x^2])^(4 + n)*Hypergeometric2F1[4, (4 + n)/2, (6 + n)/2, -((x + Sqrt[a + x^2])^2/a)])/(a^4*(
4 + n))

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Rubi [A]  time = 0.0715277, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2122, 364} \[ \frac{16 \left (\sqrt{a+x^2}+x\right )^{n+4} \, _2F_1\left (4,\frac{n+4}{2};\frac{n+6}{2};-\frac{\left (x+\sqrt{x^2+a}\right )^2}{a}\right )}{a^4 (n+4)} \]

Antiderivative was successfully verified.

[In]

Int[(x + Sqrt[a + x^2])^n/(a + x^2)^(5/2),x]

[Out]

(16*(x + Sqrt[a + x^2])^(4 + n)*Hypergeometric2F1[4, (4 + n)/2, (6 + n)/2, -((x + Sqrt[a + x^2])^2/a)])/(a^4*(
4 + n))

Rule 2122

Int[((g_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dis
t[(1*(i/c)^m)/(2^(2*m + 1)*e*f^(2*m)), Subst[Int[(x^n*(d^2 + a*f^2 - 2*d*x + x^2)^(2*m + 1))/(-d + x)^(2*(m +
1)), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, d, e, f, g, i, n}, x] && EqQ[e^2 - c*f^2, 0] && E
qQ[c*g - a*i, 0] && IntegerQ[2*m] && (IntegerQ[m] || GtQ[i/c, 0])

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{\left (x+\sqrt{a+x^2}\right )^n}{\left (a+x^2\right )^{5/2}} \, dx &=16 \operatorname{Subst}\left (\int \frac{x^{3+n}}{\left (a+x^2\right )^4} \, dx,x,x+\sqrt{a+x^2}\right )\\ &=\frac{16 \left (x+\sqrt{a+x^2}\right )^{4+n} \, _2F_1\left (4,\frac{4+n}{2};\frac{6+n}{2};-\frac{\left (x+\sqrt{a+x^2}\right )^2}{a}\right )}{a^4 (4+n)}\\ \end{align*}

Mathematica [A]  time = 0.0325427, size = 61, normalized size = 1.03 \[ \frac{16 \left (\sqrt{a+x^2}+x\right )^{n+4} \, _2F_1\left (4,\frac{n+4}{2};\frac{n+4}{2}+1;-\frac{\left (x+\sqrt{x^2+a}\right )^2}{a}\right )}{a^4 (n+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x + Sqrt[a + x^2])^n/(a + x^2)^(5/2),x]

[Out]

(16*(x + Sqrt[a + x^2])^(4 + n)*Hypergeometric2F1[4, (4 + n)/2, 1 + (4 + n)/2, -((x + Sqrt[a + x^2])^2/a)])/(a
^4*(4 + n))

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Maple [F]  time = 0.016, size = 0, normalized size = 0. \begin{align*} \int{ \left ( x+\sqrt{{x}^{2}+a} \right ) ^{n} \left ({x}^{2}+a \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+(x^2+a)^(1/2))^n/(x^2+a)^(5/2),x)

[Out]

int((x+(x^2+a)^(1/2))^n/(x^2+a)^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x + \sqrt{x^{2} + a}\right )}^{n}}{{\left (x^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+(x^2+a)^(1/2))^n/(x^2+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((x + sqrt(x^2 + a))^n/(x^2 + a)^(5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{x^{2} + a}{\left (x + \sqrt{x^{2} + a}\right )}^{n}}{x^{6} + 3 \, a x^{4} + 3 \, a^{2} x^{2} + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+(x^2+a)^(1/2))^n/(x^2+a)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(x^2 + a)*(x + sqrt(x^2 + a))^n/(x^6 + 3*a*x^4 + 3*a^2*x^2 + a^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x + \sqrt{a + x^{2}}\right )^{n}}{\left (a + x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+(x**2+a)**(1/2))**n/(x**2+a)**(5/2),x)

[Out]

Integral((x + sqrt(a + x**2))**n/(a + x**2)**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (x + \sqrt{x^{2} + a}\right )}^{n}}{{\left (x^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x+(x^2+a)^(1/2))^n/(x^2+a)^(5/2),x, algorithm="giac")

[Out]

integrate((x + sqrt(x^2 + a))^n/(x^2 + a)^(5/2), x)