Integrand size = 26, antiderivative size = 255 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^5 x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 \left (a+b x^2\right )}+\frac {5 a^4 b x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {5 a^3 b^2 x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 \left (a+b x^2\right )}+\frac {5 a^2 b^3 x^{18} \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {a b^4 x^{20} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {b^5 x^{22} \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 \left (a+b x^2\right )} \]
1/12*a^5*x^12*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/14*a^4*b*x^14*((b*x^2+a)^2)^ (1/2)/(b*x^2+a)+5/8*a^3*b^2*x^16*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/9*a^2*b^3 *x^18*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/4*a*b^4*x^20*((b*x^2+a)^2)^(1/2)/(b* x^2+a)+1/22*b^5*x^22*((b*x^2+a)^2)^(1/2)/(b*x^2+a)
Time = 0.90 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.53 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^{12} \left (462 a^5+1980 a^4 b x^2+3465 a^3 b^2 x^4+3080 a^2 b^3 x^6+1386 a b^4 x^8+252 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{5544 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]
(x^12*(462*a^5 + 1980*a^4*b*x^2 + 3465*a^3*b^2*x^4 + 3080*a^2*b^3*x^6 + 13 86*a*b^4*x^8 + 252*b^5*x^10)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b *x^2)^2])))/(5544*(-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 = 0.41, 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.
\(\displaystyle \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^5 x^{11} \left (b x^2+a\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^{11} \left (b x^2+a\right )^5dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^{10} \left (b x^2+a\right )^5dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^5 x^{20}+5 a b^4 x^{18}+10 a^2 b^3 x^{16}+10 a^3 b^2 x^{14}+5 a^4 b x^{12}+a^5 x^{10}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {a^5 x^{12}}{6}+\frac {5}{7} a^4 b x^{14}+\frac {5}{4} a^3 b^2 x^{16}+\frac {10}{9} a^2 b^3 x^{18}+\frac {1}{2} a b^4 x^{20}+\frac {b^5 x^{22}}{11}\right )}{2 \left (a+b x^2\right )}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((a^5*x^12)/6 + (5*a^4*b*x^14)/7 + (5*a^3 *b^2*x^16)/4 + (10*a^2*b^3*x^18)/9 + (a*b^4*x^20)/2 + (b^5*x^22)/11))/(2*( a + b*x^2))
3.6.86.3.1 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 = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.26
method | result | size |
pseudoelliptic | \(\frac {x^{12} \operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\frac {6}{11} x^{10} b^{5}+3 a \,x^{8} b^{4}+\frac {20}{3} a^{2} x^{6} b^{3}+\frac {15}{2} a^{3} x^{4} b^{2}+\frac {30}{7} x^{2} a^{4} b +a^{5}\right )}{12}\) | \(66\) |
gosper | \(\frac {x^{12} \left (252 x^{10} b^{5}+1386 a \,x^{8} b^{4}+3080 a^{2} x^{6} b^{3}+3465 a^{3} x^{4} b^{2}+1980 x^{2} a^{4} b +462 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{5544 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{12} \left (252 x^{10} b^{5}+1386 a \,x^{8} b^{4}+3080 a^{2} x^{6} b^{3}+3465 a^{3} x^{4} b^{2}+1980 x^{2} a^{4} b +462 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{5544 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
risch | \(\frac {a^{5} x^{12} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{12 b \,x^{2}+12 a}+\frac {5 a^{4} b \,x^{14} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{14 \left (b \,x^{2}+a \right )}+\frac {5 a^{3} b^{2} x^{16} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{8 \left (b \,x^{2}+a \right )}+\frac {5 a^{2} b^{3} x^{18} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{9 \left (b \,x^{2}+a \right )}+\frac {a \,b^{4} x^{20} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 b \,x^{2}+4 a}+\frac {b^{5} x^{22} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{22 b \,x^{2}+22 a}\) | \(178\) |
1/12*x^12*csgn(b*x^2+a)*(6/11*x^10*b^5+3*a*x^8*b^4+20/3*a^2*x^6*b^3+15/2*a ^3*x^4*b^2+30/7*x^2*a^4*b+a^5)
Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{22} \, b^{5} x^{22} + \frac {1}{4} \, a b^{4} x^{20} + \frac {5}{9} \, a^{2} b^{3} x^{18} + \frac {5}{8} \, a^{3} b^{2} x^{16} + \frac {5}{14} \, a^{4} b x^{14} + \frac {1}{12} \, a^{5} x^{12} \]
1/22*b^5*x^22 + 1/4*a*b^4*x^20 + 5/9*a^2*b^3*x^18 + 5/8*a^3*b^2*x^16 + 5/1 4*a^4*b*x^14 + 1/12*a^5*x^12
\[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{11} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{22} \, b^{5} x^{22} + \frac {1}{4} \, a b^{4} x^{20} + \frac {5}{9} \, a^{2} b^{3} x^{18} + \frac {5}{8} \, a^{3} b^{2} x^{16} + \frac {5}{14} \, a^{4} b x^{14} + \frac {1}{12} \, a^{5} x^{12} \]
1/22*b^5*x^22 + 1/4*a*b^4*x^20 + 5/9*a^2*b^3*x^18 + 5/8*a^3*b^2*x^16 + 5/1 4*a^4*b*x^14 + 1/12*a^5*x^12
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.41 \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{22} \, b^{5} x^{22} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{4} \, a b^{4} x^{20} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{9} \, a^{2} b^{3} x^{18} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{8} \, a^{3} b^{2} x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{14} \, a^{4} b x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{12} \, a^{5} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) \]
1/22*b^5*x^22*sgn(b*x^2 + a) + 1/4*a*b^4*x^20*sgn(b*x^2 + a) + 5/9*a^2*b^3 *x^18*sgn(b*x^2 + a) + 5/8*a^3*b^2*x^16*sgn(b*x^2 + a) + 5/14*a^4*b*x^14*s gn(b*x^2 + a) + 1/12*a^5*x^12*sgn(b*x^2 + a)
Timed out. \[ \int x^{11} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{11}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]