Integrand size = 26, antiderivative size = 201 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^4 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 b^5}-\frac {2 a^3 \left (a+b x^2\right )^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 b^5}+\frac {3 a^2 \left (a+b x^2\right )^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 b^5}-\frac {2 a \left (a+b x^2\right )^8 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 b^5}+\frac {\left (a+b x^2\right )^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{20 b^5} \]
1/12*a^4*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/b^5-2/7*a^3*(b*x^2+a)^6*((b*x^2+a )^2)^(1/2)/b^5+3/8*a^2*(b*x^2+a)^7*((b*x^2+a)^2)^(1/2)/b^5-2/9*a*(b*x^2+a) ^8*((b*x^2+a)^2)^(1/2)/b^5+1/20*(b*x^2+a)^9*((b*x^2+a)^2)^(1/2)/b^5
Time = 0.83 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.67 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^{10} \left (252 a^5+1050 a^4 b x^2+1800 a^3 b^2 x^4+1575 a^2 b^3 x^6+700 a b^4 x^8+126 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{2520 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]
(x^10*(252*a^5 + 1050*a^4*b*x^2 + 1800*a^3*b^2*x^4 + 1575*a^2*b^3*x^6 + 70 0*a*b^4*x^8 + 126*b^5*x^10)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b* x^2)^2])))/(2520*(-a^2 - a*b*x^2 + Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))
Time = 0.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.52, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 243, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^5 x^9 \left (b x^2+a\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^9 \left (b x^2+a\right )^5dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^8 \left (b x^2+a\right )^5dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^5 x^{18}+5 a b^4 x^{16}+10 a^2 b^3 x^{14}+10 a^3 b^2 x^{12}+5 a^4 b x^{10}+a^5 x^8\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {a^5 x^{10}}{5}+\frac {5}{6} a^4 b x^{12}+\frac {10}{7} a^3 b^2 x^{14}+\frac {5}{4} a^2 b^3 x^{16}+\frac {5}{9} a b^4 x^{18}+\frac {b^5 x^{20}}{10}\right )}{2 \left (a+b x^2\right )}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((a^5*x^10)/5 + (5*a^4*b*x^12)/6 + (10*a^ 3*b^2*x^14)/7 + (5*a^2*b^3*x^16)/4 + (5*a*b^4*x^18)/9 + (b^5*x^20)/10))/(2 *(a + b*x^2))
3.6.87.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(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] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.34
method | result | size |
pseudoelliptic | \(\frac {x^{10} \left (126 x^{10} b^{5}+700 a \,x^{8} b^{4}+1575 a^{2} x^{6} b^{3}+1800 a^{3} x^{4} b^{2}+1050 x^{2} a^{4} b +252 a^{5}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2520}\) | \(68\) |
gosper | \(\frac {x^{10} \left (126 x^{10} b^{5}+700 a \,x^{8} b^{4}+1575 a^{2} x^{6} b^{3}+1800 a^{3} x^{4} b^{2}+1050 x^{2} a^{4} b +252 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{2520 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{10} \left (126 x^{10} b^{5}+700 a \,x^{8} b^{4}+1575 a^{2} x^{6} b^{3}+1800 a^{3} x^{4} b^{2}+1050 x^{2} a^{4} b +252 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{2520 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{5} x^{10}}{10 b \,x^{2}+10 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,a^{4} x^{12}}{12 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{14}}{7 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{16}}{8 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{4} a \,x^{18}}{18 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{5} x^{20}}{20 b \,x^{2}+20 a}\) | \(178\) |
1/2520*x^10*(126*b^5*x^10+700*a*b^4*x^8+1575*a^2*b^3*x^6+1800*a^3*b^2*x^4+ 1050*a^4*b*x^2+252*a^5)*csgn(b*x^2+a)
Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.28 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{20} \, b^{5} x^{20} + \frac {5}{18} \, a b^{4} x^{18} + \frac {5}{8} \, a^{2} b^{3} x^{16} + \frac {5}{7} \, a^{3} b^{2} x^{14} + \frac {5}{12} \, a^{4} b x^{12} + \frac {1}{10} \, a^{5} x^{10} \]
1/20*b^5*x^20 + 5/18*a*b^4*x^18 + 5/8*a^2*b^3*x^16 + 5/7*a^3*b^2*x^14 + 5/ 12*a^4*b*x^12 + 1/10*a^5*x^10
\[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{9} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.28 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{20} \, b^{5} x^{20} + \frac {5}{18} \, a b^{4} x^{18} + \frac {5}{8} \, a^{2} b^{3} x^{16} + \frac {5}{7} \, a^{3} b^{2} x^{14} + \frac {5}{12} \, a^{4} b x^{12} + \frac {1}{10} \, a^{5} x^{10} \]
1/20*b^5*x^20 + 5/18*a*b^4*x^18 + 5/8*a^2*b^3*x^16 + 5/7*a^3*b^2*x^14 + 5/ 12*a^4*b*x^12 + 1/10*a^5*x^10
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.52 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{20} \, b^{5} x^{20} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{18} \, a b^{4} x^{18} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{8} \, a^{2} b^{3} x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{7} \, a^{3} b^{2} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{12} \, a^{4} b x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{10} \, a^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) \]
1/20*b^5*x^20*sgn(b*x^2 + a) + 5/18*a*b^4*x^18*sgn(b*x^2 + a) + 5/8*a^2*b^ 3*x^16*sgn(b*x^2 + a) + 5/7*a^3*b^2*x^14*sgn(b*x^2 + a) + 5/12*a^4*b*x^12* sgn(b*x^2 + a) + 1/10*a^5*x^10*sgn(b*x^2 + a)
Timed out. \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^9\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]