Integrand size = 26, antiderivative size = 160 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=-\frac {a^3 \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^4}+\frac {a^2 \left (a+b x^3\right )^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{7 b^4}-\frac {a \left (a+b x^3\right )^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{8 b^4}+\frac {\left (a+b x^3\right )^8 \sqrt {a^2+2 a b x^3+b^2 x^6}}{27 b^4} \]
-1/18*a^3*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/b^4+1/7*a^2*(b*x^3+a)^6*((b*x^3+ a)^2)^(1/2)/b^4-1/8*a*(b*x^3+a)^7*((b*x^3+a)^2)^(1/2)/b^4+1/27*(b*x^3+a)^8 *((b*x^3+a)^2)^(1/2)/b^4
Time = 0.81 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.84 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^{12} \left (126 a^5+504 a^4 b x^3+840 a^3 b^2 x^6+720 a^2 b^3 x^9+315 a b^4 x^{12}+56 b^5 x^{15}\right ) \left (\sqrt {a^2} b x^3+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{1512 \left (-a^2-a b x^3+\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )} \]
(x^12*(126*a^5 + 504*a^4*b*x^3 + 840*a^3*b^2*x^6 + 720*a^2*b^3*x^9 + 315*a *b^4*x^12 + 56*b^5*x^15)*(Sqrt[a^2]*b*x^3 + a*(Sqrt[a^2] - Sqrt[(a + b*x^3 )^2])))/(1512*(-a^2 - a*b*x^3 + Sqrt[a^2]*Sqrt[(a + b*x^3)^2]))
Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.67, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 798, 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.
\(\displaystyle \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int b^5 x^{11} \left (b x^3+a\right )^5dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^{11} \left (b x^3+a\right )^5dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^9 \left (b x^3+a\right )^5dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {\left (b x^3+a\right )^8}{b^3}-\frac {3 a \left (b x^3+a\right )^7}{b^3}+\frac {3 a^2 \left (b x^3+a\right )^6}{b^3}-\frac {a^3 \left (b x^3+a\right )^5}{b^3}\right )dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {a^3 \left (a+b x^3\right )^6}{6 b^4}+\frac {3 a^2 \left (a+b x^3\right )^7}{7 b^4}+\frac {\left (a+b x^3\right )^9}{9 b^4}-\frac {3 a \left (a+b x^3\right )^8}{8 b^4}\right )}{3 \left (a+b x^3\right )}\) |
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/6*(a^3*(a + b*x^3)^6)/b^4 + (3*a^2*(a + b*x^3)^7)/(7*b^4) - (3*a*(a + b*x^3)^8)/(8*b^4) + (a + b*x^3)^9/(9*b^4) ))/(3*(a + b*x^3))
3.1.52.3.1 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_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
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 = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.33
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{6} \left (-56 b^{3} x^{9}+21 b^{2} x^{6} a -6 a^{2} b \,x^{3}+a^{3}\right )}{1512 b^{4}}\) | \(53\) |
gosper | \(\frac {x^{12} \left (56 b^{5} x^{15}+315 a \,b^{4} x^{12}+720 a^{2} b^{3} x^{9}+840 a^{3} b^{2} x^{6}+504 a^{4} b \,x^{3}+126 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1512 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{12} \left (56 b^{5} x^{15}+315 a \,b^{4} x^{12}+720 a^{2} b^{3} x^{9}+840 a^{3} b^{2} x^{6}+504 a^{4} b \,x^{3}+126 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{1512 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
risch | \(\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{4} a \,x^{24}}{24 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{5} x^{27}}{27 b \,x^{3}+27 a}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} b^{2} x^{18}}{9 \left (b \,x^{3}+a \right )}+\frac {10 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} b^{3} x^{21}}{21 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{5} x^{12}}{12 b \,x^{3}+12 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \,a^{4} x^{15}}{3 b \,x^{3}+3 a}\) | \(178\) |
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{27} \, b^{5} x^{27} + \frac {5}{24} \, a b^{4} x^{24} + \frac {10}{21} \, a^{2} b^{3} x^{21} + \frac {5}{9} \, a^{3} b^{2} x^{18} + \frac {1}{3} \, a^{4} b x^{15} + \frac {1}{12} \, a^{5} x^{12} \]
1/27*b^5*x^27 + 5/24*a*b^4*x^24 + 10/21*a^2*b^3*x^21 + 5/9*a^3*b^2*x^18 + 1/3*a^4*b*x^15 + 1/12*a^5*x^12
\[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{11} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.91 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} x^{6}}{27 \, b^{2}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a^{3} x^{3}}{18 \, b^{3}} - \frac {11 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} a x^{3}}{216 \, b^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a^{4}}{18 \, b^{4}} + \frac {83 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} a^{2}}{1512 \, b^{4}} \]
1/27*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*x^6/b^2 - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a^3*x^3/b^3 - 11/216*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*a*x^3 /b^3 - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a^4/b^4 + 83/1512*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*a^2/b^4
Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.66 \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{27} \, b^{5} x^{27} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{24} \, a b^{4} x^{24} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{21} \, a^{2} b^{3} x^{21} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{9} \, a^{3} b^{2} x^{18} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{3} \, a^{4} b x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{12} \, a^{5} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) \]
1/27*b^5*x^27*sgn(b*x^3 + a) + 5/24*a*b^4*x^24*sgn(b*x^3 + a) + 10/21*a^2* b^3*x^21*sgn(b*x^3 + a) + 5/9*a^3*b^2*x^18*sgn(b*x^3 + a) + 1/3*a^4*b*x^15 *sgn(b*x^3 + a) + 1/12*a^5*x^12*sgn(b*x^3 + a)
Timed out. \[ \int x^{11} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{11}\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \]