3.496 \(\int (a+x^2)^{3/2} (x+\sqrt {a+x^2})^n \, dx\)

Optimal. Leaf size=131 \[ -\frac {a^4 \left (\sqrt {a+x^2}+x\right )^{n-4}}{16 (4-n)}-\frac {a^3 \left (\sqrt {a+x^2}+x\right )^{n-2}}{4 (2-n)}+\frac {3 a^2 \left (\sqrt {a+x^2}+x\right )^n}{8 n}+\frac {a \left (\sqrt {a+x^2}+x\right )^{n+2}}{4 (n+2)}+\frac {\left (\sqrt {a+x^2}+x\right )^{n+4}}{16 (n+4)} \]

[Out]

-1/16*a^4*(x+(x^2+a)^(1/2))^(-4+n)/(4-n)-1/4*a^3*(x+(x^2+a)^(1/2))^(-2+n)/(2-n)+3/8*a^2*(x+(x^2+a)^(1/2))^n/n+
1/4*a*(x+(x^2+a)^(1/2))^(2+n)/(2+n)+1/16*(x+(x^2+a)^(1/2))^(4+n)/(4+n)

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Rubi [A]  time = 0.09, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2122, 270} \[ -\frac {a^4 \left (\sqrt {a+x^2}+x\right )^{n-4}}{16 (4-n)}-\frac {a^3 \left (\sqrt {a+x^2}+x\right )^{n-2}}{4 (2-n)}+\frac {3 a^2 \left (\sqrt {a+x^2}+x\right )^n}{8 n}+\frac {a \left (\sqrt {a+x^2}+x\right )^{n+2}}{4 (n+2)}+\frac {\left (\sqrt {a+x^2}+x\right )^{n+4}}{16 (n+4)} \]

Antiderivative was successfully verified.

[In]

Int[(a + x^2)^(3/2)*(x + Sqrt[a + x^2])^n,x]

[Out]

-(a^4*(x + Sqrt[a + x^2])^(-4 + n))/(16*(4 - n)) - (a^3*(x + Sqrt[a + x^2])^(-2 + n))/(4*(2 - n)) + (3*a^2*(x
+ Sqrt[a + x^2])^n)/(8*n) + (a*(x + Sqrt[a + x^2])^(2 + n))/(4*(2 + n)) + (x + Sqrt[a + x^2])^(4 + n)/(16*(4 +
n))

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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])

Rubi steps

\begin {align*} \int \left (a+x^2\right )^{3/2} \left (x+\sqrt {a+x^2}\right )^n \, dx &=\frac {1}{16} \operatorname {Subst}\left (\int x^{-5+n} \left (a+x^2\right )^4 \, dx,x,x+\sqrt {a+x^2}\right )\\ &=\frac {1}{16} \operatorname {Subst}\left (\int \left (a^4 x^{-5+n}+4 a^3 x^{-3+n}+6 a^2 x^{-1+n}+4 a x^{1+n}+x^{3+n}\right ) \, dx,x,x+\sqrt {a+x^2}\right )\\ &=-\frac {a^4 \left (x+\sqrt {a+x^2}\right )^{-4+n}}{16 (4-n)}-\frac {a^3 \left (x+\sqrt {a+x^2}\right )^{-2+n}}{4 (2-n)}+\frac {3 a^2 \left (x+\sqrt {a+x^2}\right )^n}{8 n}+\frac {a \left (x+\sqrt {a+x^2}\right )^{2+n}}{4 (2+n)}+\frac {\left (x+\sqrt {a+x^2}\right )^{4+n}}{16 (4+n)}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 111, normalized size = 0.85 \[ \frac {1}{16} \left (\sqrt {a+x^2}+x\right )^n \left (\frac {a^4}{(n-4) \left (\sqrt {a+x^2}+x\right )^4}+\frac {4 a^3}{(n-2) \left (\sqrt {a+x^2}+x\right )^2}+\frac {6 a^2}{n}+\frac {4 a \left (\sqrt {a+x^2}+x\right )^2}{n+2}+\frac {\left (\sqrt {a+x^2}+x\right )^4}{n+4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + x^2)^(3/2)*(x + Sqrt[a + x^2])^n,x]

[Out]

((x + Sqrt[a + x^2])^n*((6*a^2)/n + a^4/((-4 + n)*(x + Sqrt[a + x^2])^4) + (4*a^3)/((-2 + n)*(x + Sqrt[a + x^2
])^2) + (4*a*(x + Sqrt[a + x^2])^2)/(2 + n) + (x + Sqrt[a + x^2])^4/(4 + n)))/16

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fricas [A]  time = 0.46, size = 110, normalized size = 0.84 \[ \frac {{\left (a^{2} n^{4} + {\left (n^{4} - 4 \, n^{2}\right )} x^{4} - 16 \, a^{2} n^{2} + 2 \, {\left (a n^{4} - 10 \, a n^{2}\right )} x^{2} + 24 \, a^{2} - 4 \, {\left ({\left (n^{3} - 4 \, n\right )} x^{3} + {\left (a n^{3} - 10 \, a n\right )} x\right )} \sqrt {x^{2} + a}\right )} {\left (x + \sqrt {x^{2} + a}\right )}^{n}}{n^{5} - 20 \, n^{3} + 64 \, n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+a)^(3/2)*(x+(x^2+a)^(1/2))^n,x, algorithm="fricas")

[Out]

(a^2*n^4 + (n^4 - 4*n^2)*x^4 - 16*a^2*n^2 + 2*(a*n^4 - 10*a*n^2)*x^2 + 24*a^2 - 4*((n^3 - 4*n)*x^3 + (a*n^3 -
10*a*n)*x)*sqrt(x^2 + a))*(x + sqrt(x^2 + a))^n/(n^5 - 20*n^3 + 64*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (x^{2} + a\right )}^{\frac {3}{2}} {\left (x + \sqrt {x^{2} + a}\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+a)^(3/2)*(x+(x^2+a)^(1/2))^n,x, algorithm="giac")

[Out]

integrate((x^2 + a)^(3/2)*(x + sqrt(x^2 + a))^n, x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (x^{2}+a \right )^{\frac {3}{2}} \left (x +\sqrt {x^{2}+a}\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+a)^(3/2)*(x+(x^2+a)^(1/2))^n,x)

[Out]

int((x^2+a)^(3/2)*(x+(x^2+a)^(1/2))^n,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (x^{2} + a\right )}^{\frac {3}{2}} {\left (x + \sqrt {x^{2} + a}\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+a)^(3/2)*(x+(x^2+a)^(1/2))^n,x, algorithm="maxima")

[Out]

integrate((x^2 + a)^(3/2)*(x + sqrt(x^2 + a))^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (x^2+a\right )}^{3/2}\,{\left (x+\sqrt {x^2+a}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + x^2)^(3/2)*(x + (a + x^2)^(1/2))^n,x)

[Out]

int((a + x^2)^(3/2)*(x + (a + x^2)^(1/2))^n, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + x^{2}\right )^{\frac {3}{2}} \left (x + \sqrt {a + x^{2}}\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+a)**(3/2)*(x+(x**2+a)**(1/2))**n,x)

[Out]

Integral((a + x**2)**(3/2)*(x + sqrt(a + x**2))**n, x)

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