3.485 \(\int (a+x^2)^2 (x+\sqrt{a+x^2})^n \, dx\)

Optimal. Leaf size=164 \[ -\frac{a^5 \left (\sqrt{a+x^2}+x\right )^{n-5}}{32 (5-n)}-\frac{5 a^4 \left (\sqrt{a+x^2}+x\right )^{n-3}}{32 (3-n)}-\frac{5 a^3 \left (\sqrt{a+x^2}+x\right )^{n-1}}{16 (1-n)}+\frac{5 a^2 \left (\sqrt{a+x^2}+x\right )^{n+1}}{16 (n+1)}+\frac{5 a \left (\sqrt{a+x^2}+x\right )^{n+3}}{32 (n+3)}+\frac{\left (\sqrt{a+x^2}+x\right )^{n+5}}{32 (n+5)} \]

[Out]

-(a^5*(x + Sqrt[a + x^2])^(-5 + n))/(32*(5 - n)) - (5*a^4*(x + Sqrt[a + x^2])^(-3 + n))/(32*(3 - n)) - (5*a^3*
(x + Sqrt[a + x^2])^(-1 + n))/(16*(1 - n)) + (5*a^2*(x + Sqrt[a + x^2])^(1 + n))/(16*(1 + n)) + (5*a*(x + Sqrt
[a + x^2])^(3 + n))/(32*(3 + n)) + (x + Sqrt[a + x^2])^(5 + n)/(32*(5 + n))

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Rubi [A]  time = 0.110497, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2122, 270} \[ -\frac{a^5 \left (\sqrt{a+x^2}+x\right )^{n-5}}{32 (5-n)}-\frac{5 a^4 \left (\sqrt{a+x^2}+x\right )^{n-3}}{32 (3-n)}-\frac{5 a^3 \left (\sqrt{a+x^2}+x\right )^{n-1}}{16 (1-n)}+\frac{5 a^2 \left (\sqrt{a+x^2}+x\right )^{n+1}}{16 (n+1)}+\frac{5 a \left (\sqrt{a+x^2}+x\right )^{n+3}}{32 (n+3)}+\frac{\left (\sqrt{a+x^2}+x\right )^{n+5}}{32 (n+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^5*(x + Sqrt[a + x^2])^(-5 + n))/(32*(5 - n)) - (5*a^4*(x + Sqrt[a + x^2])^(-3 + n))/(32*(3 - n)) - (5*a^3*
(x + Sqrt[a + x^2])^(-1 + n))/(16*(1 - n)) + (5*a^2*(x + Sqrt[a + x^2])^(1 + n))/(16*(1 + n)) + (5*a*(x + Sqrt
[a + x^2])^(3 + n))/(32*(3 + n)) + (x + Sqrt[a + x^2])^(5 + n)/(32*(5 + 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 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]

Rubi steps

\begin{align*} \int \left (a+x^2\right )^2 \left (x+\sqrt{a+x^2}\right )^n \, dx &=\frac{1}{32} \operatorname{Subst}\left (\int x^{-6+n} \left (a+x^2\right )^5 \, dx,x,x+\sqrt{a+x^2}\right )\\ &=\frac{1}{32} \operatorname{Subst}\left (\int \left (a^5 x^{-6+n}+5 a^4 x^{-4+n}+10 a^3 x^{-2+n}+10 a^2 x^n+5 a x^{2+n}+x^{4+n}\right ) \, dx,x,x+\sqrt{a+x^2}\right )\\ &=-\frac{a^5 \left (x+\sqrt{a+x^2}\right )^{-5+n}}{32 (5-n)}-\frac{5 a^4 \left (x+\sqrt{a+x^2}\right )^{-3+n}}{32 (3-n)}-\frac{5 a^3 \left (x+\sqrt{a+x^2}\right )^{-1+n}}{16 (1-n)}+\frac{5 a^2 \left (x+\sqrt{a+x^2}\right )^{1+n}}{16 (1+n)}+\frac{5 a \left (x+\sqrt{a+x^2}\right )^{3+n}}{32 (3+n)}+\frac{\left (x+\sqrt{a+x^2}\right )^{5+n}}{32 (5+n)}\\ \end{align*}

Mathematica [A]  time = 0.342343, size = 138, normalized size = 0.84 \[ \frac{1}{32} \left (\sqrt{a+x^2}+x\right )^{n-5} \left (\frac{10 a^2 \left (\sqrt{a+x^2}+x\right )^6}{n+1}+\frac{10 a^3 \left (\sqrt{a+x^2}+x\right )^4}{n-1}+\frac{5 a^4 \left (\sqrt{a+x^2}+x\right )^2}{n-3}+\frac{a^5}{n-5}+\frac{\left (\sqrt{a+x^2}+x\right )^{10}}{n+5}+\frac{5 a \left (\sqrt{a+x^2}+x\right )^8}{n+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x + Sqrt[a + x^2])^(-5 + n)*(a^5/(-5 + n) + (5*a^4*(x + Sqrt[a + x^2])^2)/(-3 + n) + (10*a^3*(x + Sqrt[a + x
^2])^4)/(-1 + n) + (10*a^2*(x + Sqrt[a + x^2])^6)/(1 + n) + (5*a*(x + Sqrt[a + x^2])^8)/(3 + n) + (x + Sqrt[a
+ x^2])^10/(5 + n)))/32

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Maple [C]  time = 0.044, size = 216, normalized size = 1.3 \begin{align*}{\frac{{2}^{n}{x}^{5+n}}{5+n}{\mbox{$_3$F$_2$}(-{\frac{n}{2}},{\frac{1}{2}}-{\frac{n}{2}},-{\frac{5}{2}}-{\frac{n}{2}};\,1-n,-{\frac{3}{2}}-{\frac{n}{2}};\,-{\frac{a}{{x}^{2}}})}}+{\frac{{2}^{1+n}a{x}^{3+n}}{3+n}{\mbox{$_3$F$_2$}(-{\frac{n}{2}},{\frac{1}{2}}-{\frac{n}{2}},-{\frac{3}{2}}-{\frac{n}{2}};\,1-n,-{\frac{1}{2}}-{\frac{n}{2}};\,-{\frac{a}{{x}^{2}}})}}+{\frac{n}{4\,\sqrt{\pi }}{a}^{{\frac{5}{2}}+{\frac{n}{2}}} \left ( 8\,{\frac{\sqrt{\pi }{x}^{1+n}{a}^{-1/2-n/2}}{ \left ( 1+n \right ) n \left ( -2+2\,n \right ) } \left ({\frac{an}{{x}^{2}}}+n-1 \right ) \left ( \sqrt{1+{\frac{a}{{x}^{2}}}}+1 \right ) ^{-1+n}}+4\,{\frac{\sqrt{\pi }{x}^{1+n}{a}^{-1/2-n/2}}{ \left ( 1+n \right ) n}\sqrt{1+{\frac{a}{{x}^{2}}}} \left ( \sqrt{1+{\frac{a}{{x}^{2}}}}+1 \right ) ^{-1+n}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2^n/(5+n)*x^(5+n)*hypergeom([-1/2*n,1/2-1/2*n,-5/2-1/2*n],[1-n,-3/2-1/2*n],-a/x^2)+2^(1+n)*a/(3+n)*x^(3+n)*hyp
ergeom([-1/2*n,1/2-1/2*n,-3/2-1/2*n],[1-n,-1/2-1/2*n],-a/x^2)+1/4*a^(5/2+1/2*n)/Pi^(1/2)*n*(8*Pi^(1/2)/(1+n)/n
*x^(1+n)*a^(-1/2-1/2*n)*(a/x^2*n+n-1)/(-2+2*n)*((1+a/x^2)^(1/2)+1)^(-1+n)+4*Pi^(1/2)/(1+n)/n*x^(1+n)*a^(-1/2-1
/2*n)*(1+a/x^2)^(1/2)*((1+a/x^2)^(1/2)+1)^(-1+n))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.3776, size = 363, normalized size = 2.21 \begin{align*} -\frac{{\left (5 \,{\left (n^{4} - 10 \, n^{2} + 9\right )} x^{5} + 10 \,{\left (a n^{4} - 16 \, a n^{2} + 15 \, a\right )} x^{3} + 5 \,{\left (a^{2} n^{4} - 22 \, a^{2} n^{2} + 45 \, a^{2}\right )} x -{\left (a^{2} n^{5} - 30 \, a^{2} n^{3} +{\left (n^{5} - 10 \, n^{3} + 9 \, n\right )} x^{4} + 149 \, a^{2} n + 2 \,{\left (a n^{5} - 20 \, a n^{3} + 19 \, a n\right )} x^{2}\right )} \sqrt{x^{2} + a}\right )}{\left (x + \sqrt{x^{2} + a}\right )}^{n}}{n^{6} - 35 \, n^{4} + 259 \, n^{2} - 225} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*(n^4 - 10*n^2 + 9)*x^5 + 10*(a*n^4 - 16*a*n^2 + 15*a)*x^3 + 5*(a^2*n^4 - 22*a^2*n^2 + 45*a^2)*x - (a^2*n^5
- 30*a^2*n^3 + (n^5 - 10*n^3 + 9*n)*x^4 + 149*a^2*n + 2*(a*n^5 - 20*a*n^3 + 19*a*n)*x^2)*sqrt(x^2 + a))*(x +
sqrt(x^2 + a))^n/(n^6 - 35*n^4 + 259*n^2 - 225)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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