Integrand size = 26, antiderivative size = 255 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^5 x^{11} \sqrt {a^2+2 a b x^3+b^2 x^6}}{11 \left (a+b x^3\right )}+\frac {5 a^4 b x^{14} \sqrt {a^2+2 a b x^3+b^2 x^6}}{14 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^{17} \sqrt {a^2+2 a b x^3+b^2 x^6}}{17 \left (a+b x^3\right )}+\frac {a^2 b^3 x^{20} \sqrt {a^2+2 a b x^3+b^2 x^6}}{2 \left (a+b x^3\right )}+\frac {5 a b^4 x^{23} \sqrt {a^2+2 a b x^3+b^2 x^6}}{23 \left (a+b x^3\right )}+\frac {b^5 x^{26} \sqrt {a^2+2 a b x^3+b^2 x^6}}{26 \left (a+b x^3\right )} \]
1/11*a^5*x^11*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+5/14*a^4*b*x^14*((b*x^3+a)^2)^ (1/2)/(b*x^3+a)+10/17*a^3*b^2*x^17*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/2*a^2*b ^3*x^20*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+5/23*a*b^4*x^23*((b*x^3+a)^2)^(1/2)/ (b*x^3+a)+1/26*b^5*x^26*((b*x^3+a)^2)^(1/2)/(b*x^3+a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^{11} \sqrt {\left (a+b x^3\right )^2} \left (71162 a^5+279565 a^4 b x^3+460460 a^3 b^2 x^6+391391 a^2 b^3 x^9+170170 a b^4 x^{12}+30107 b^5 x^{15}\right )}{782782 \left (a+b x^3\right )} \]
(x^11*Sqrt[(a + b*x^3)^2]*(71162*a^5 + 279565*a^4*b*x^3 + 460460*a^3*b^2*x ^6 + 391391*a^2*b^3*x^9 + 170170*a*b^4*x^12 + 30107*b^5*x^15))/(782782*(a + b*x^3))
Time = 0.24 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.40, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1384, 27, 802, 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.
\(\displaystyle \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int b^5 x^{10} \left (b x^3+a\right )^5dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^{10} \left (b x^3+a\right )^5dx}{a+b x^3}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (b^5 x^{25}+5 a b^4 x^{22}+10 a^2 b^3 x^{19}+10 a^3 b^2 x^{16}+5 a^4 b x^{13}+a^5 x^{10}\right )dx}{a+b x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (\frac {a^5 x^{11}}{11}+\frac {5}{14} a^4 b x^{14}+\frac {10}{17} a^3 b^2 x^{17}+\frac {1}{2} a^2 b^3 x^{20}+\frac {5}{23} a b^4 x^{23}+\frac {b^5 x^{26}}{26}\right )}{a+b x^3}\) |
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*((a^5*x^11)/11 + (5*a^4*b*x^14)/14 + (10* a^3*b^2*x^17)/17 + (a^2*b^3*x^20)/2 + (5*a*b^4*x^23)/23 + (b^5*x^26)/26))/ (a + b*x^3)
3.1.53.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
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)])
Time = 9.72 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.31
method | result | size |
gosper | \(\frac {x^{11} \left (30107 b^{5} x^{15}+170170 a \,b^{4} x^{12}+391391 a^{2} b^{3} x^{9}+460460 a^{3} b^{2} x^{6}+279565 a^{4} b \,x^{3}+71162 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{782782 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{11} \left (30107 b^{5} x^{15}+170170 a \,b^{4} x^{12}+391391 a^{2} b^{3} x^{9}+460460 a^{3} b^{2} x^{6}+279565 a^{4} b \,x^{3}+71162 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{782782 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
risch | \(\frac {a^{5} x^{11} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{11 b \,x^{3}+11 a}+\frac {5 a^{4} b \,x^{14} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{14 \left (b \,x^{3}+a \right )}+\frac {10 a^{3} b^{2} x^{17} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{17 \left (b \,x^{3}+a \right )}+\frac {a^{2} b^{3} x^{20} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{2 b \,x^{3}+2 a}+\frac {5 a \,b^{4} x^{23} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{23 \left (b \,x^{3}+a \right )}+\frac {b^{5} x^{26} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{26 b \,x^{3}+26 a}\) | \(178\) |
1/782782*x^11*(30107*b^5*x^15+170170*a*b^4*x^12+391391*a^2*b^3*x^9+460460* a^3*b^2*x^6+279565*a^4*b*x^3+71162*a^5)*((b*x^3+a)^2)^(5/2)/(b*x^3+a)^5
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{26} \, b^{5} x^{26} + \frac {5}{23} \, a b^{4} x^{23} + \frac {1}{2} \, a^{2} b^{3} x^{20} + \frac {10}{17} \, a^{3} b^{2} x^{17} + \frac {5}{14} \, a^{4} b x^{14} + \frac {1}{11} \, a^{5} x^{11} \]
1/26*b^5*x^26 + 5/23*a*b^4*x^23 + 1/2*a^2*b^3*x^20 + 10/17*a^3*b^2*x^17 + 5/14*a^4*b*x^14 + 1/11*a^5*x^11
\[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{10} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{26} \, b^{5} x^{26} + \frac {5}{23} \, a b^{4} x^{23} + \frac {1}{2} \, a^{2} b^{3} x^{20} + \frac {10}{17} \, a^{3} b^{2} x^{17} + \frac {5}{14} \, a^{4} b x^{14} + \frac {1}{11} \, a^{5} x^{11} \]
1/26*b^5*x^26 + 5/23*a*b^4*x^23 + 1/2*a^2*b^3*x^20 + 10/17*a^3*b^2*x^17 + 5/14*a^4*b*x^14 + 1/11*a^5*x^11
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.41 \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{26} \, b^{5} x^{26} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{23} \, a b^{4} x^{23} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{2} \, a^{2} b^{3} x^{20} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{17} \, a^{3} b^{2} x^{17} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{14} \, a^{4} b x^{14} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{11} \, a^{5} x^{11} \mathrm {sgn}\left (b x^{3} + a\right ) \]
1/26*b^5*x^26*sgn(b*x^3 + a) + 5/23*a*b^4*x^23*sgn(b*x^3 + a) + 1/2*a^2*b^ 3*x^20*sgn(b*x^3 + a) + 10/17*a^3*b^2*x^17*sgn(b*x^3 + a) + 5/14*a^4*b*x^1 4*sgn(b*x^3 + a) + 1/11*a^5*x^11*sgn(b*x^3 + a)
Timed out. \[ \int x^{10} \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{10}\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \]