Integrand size = 26, antiderivative size = 119 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {a^2 \left (a+b x^3\right )^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 b^3}-\frac {2 a \left (a+b x^3\right )^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{21 b^3}+\frac {\left (a+b x^3\right )^7 \sqrt {a^2+2 a b x^3+b^2 x^6}}{24 b^3} \] Output:
1/18*a^2*(b*x^3+a)^5*((b*x^3+a)^2)^(1/2)/b^3-2/21*a*(b*x^3+a)^6*((b*x^3+a) ^2)^(1/2)/b^3+1/24*(b*x^3+a)^7*((b*x^3+a)^2)^(1/2)/b^3
Time = 1.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.70 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^9 \sqrt {\left (a+b x^3\right )^2} \left (56 a^5+210 a^4 b x^3+336 a^3 b^2 x^6+280 a^2 b^3 x^9+120 a b^4 x^{12}+21 b^5 x^{15}\right )}{504 \left (a+b x^3\right )} \] Input:
Integrate[x^8*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
Output:
(x^9*Sqrt[(a + b*x^3)^2]*(56*a^5 + 210*a^4*b*x^3 + 336*a^3*b^2*x^6 + 280*a ^2*b^3*x^9 + 120*a*b^4*x^12 + 21*b^5*x^15))/(504*(a + b*x^3))
Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.74, 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.
\(\displaystyle \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int b^5 x^8 \left (b x^3+a\right )^5dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^8 \left (b x^3+a\right )^5dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int x^6 \left (b x^3+a\right )^5dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {\left (b x^3+a\right )^7}{b^2}-\frac {2 a \left (b x^3+a\right )^6}{b^2}+\frac {a^2 \left (b x^3+a\right )^5}{b^2}\right )dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (\frac {a^2 \left (a+b x^3\right )^6}{6 b^3}+\frac {\left (a+b x^3\right )^8}{8 b^3}-\frac {2 a \left (a+b x^3\right )^7}{7 b^3}\right )}{3 \left (a+b x^3\right )}\) |
Input:
Int[x^8*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*((a^2*(a + b*x^3)^6)/(6*b^3) - (2*a*(a + b*x^3)^7)/(7*b^3) + (a + b*x^3)^8/(8*b^3)))/(3*(a + b*x^3))
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 = 0.15 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35
method | result | size |
pseudoelliptic | \(\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{6} \left (21 b^{2} x^{6}-6 a \,x^{3} b +a^{2}\right )}{504 b^{3}}\) | \(42\) |
gosper | \(\frac {x^{9} \left (21 b^{5} x^{15}+120 a \,b^{4} x^{12}+280 a^{2} b^{3} x^{9}+336 a^{3} b^{2} x^{6}+210 a^{4} b \,x^{3}+56 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{504 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{9} \left (21 b^{5} x^{15}+120 a \,b^{4} x^{12}+280 a^{2} b^{3} x^{9}+336 a^{3} b^{2} x^{6}+210 a^{4} b \,x^{3}+56 a^{5}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}{504 \left (b \,x^{3}+a \right )^{5}}\) | \(80\) |
orering | \(\frac {x^{9} \left (21 b^{5} x^{15}+120 a \,b^{4} x^{12}+280 a^{2} b^{3} x^{9}+336 a^{3} b^{2} x^{6}+210 a^{4} b \,x^{3}+56 a^{5}\right ) \left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {5}{2}}}{504 \left (b \,x^{3}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} b^{3} x^{18}}{9 \left (b \,x^{3}+a \right )}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a \,b^{4} x^{21}}{21 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{5} x^{9}}{9 b \,x^{3}+9 a}+\frac {2 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} b^{2} x^{15}}{3 \left (b \,x^{3}+a \right )}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{4} b \,x^{12}}{12 \left (b \,x^{3}+a \right )}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{5} x^{24}}{24 b \,x^{3}+24 a}\) | \(178\) |
Input:
int(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/504*csgn(b*x^3+a)*(b*x^3+a)^6*(21*b^2*x^6-6*a*b*x^3+a^2)/b^3
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.48 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{24} \, b^{5} x^{24} + \frac {5}{21} \, a b^{4} x^{21} + \frac {5}{9} \, a^{2} b^{3} x^{18} + \frac {2}{3} \, a^{3} b^{2} x^{15} + \frac {5}{12} \, a^{4} b x^{12} + \frac {1}{9} \, a^{5} x^{9} \] Input:
integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")
Output:
1/24*b^5*x^24 + 5/21*a*b^4*x^21 + 5/9*a^2*b^3*x^18 + 2/3*a^3*b^2*x^15 + 5/ 12*a^4*b*x^12 + 1/9*a^5*x^9
\[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^{8} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}\, dx \] Input:
integrate(x**8*(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)
Output:
Integral(x**8*((a + b*x**3)**2)**(5/2), x)
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.96 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a^{2} x^{3}}{18 \, b^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} x^{3}}{24 \, b^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} a^{3}}{18 \, b^{3}} - \frac {3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} a}{56 \, b^{3}} \] Input:
integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")
Output:
1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a^2*x^3/b^2 + 1/24*(b^2*x^6 + 2*a*b *x^3 + a^2)^(7/2)*x^3/b^2 + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*a^3/b^3 - 3/56*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*a/b^3
Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {1}{24} \, b^{5} x^{24} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{21} \, a b^{4} x^{21} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{9} \, a^{2} b^{3} x^{18} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {2}{3} \, a^{3} b^{2} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{12} \, a^{4} b x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {1}{9} \, a^{5} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) \] Input:
integrate(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")
Output:
1/24*b^5*x^24*sgn(b*x^3 + a) + 5/21*a*b^4*x^21*sgn(b*x^3 + a) + 5/9*a^2*b^ 3*x^18*sgn(b*x^3 + a) + 2/3*a^3*b^2*x^15*sgn(b*x^3 + a) + 5/12*a^4*b*x^12* sgn(b*x^3 + a) + 1/9*a^5*x^9*sgn(b*x^3 + a)
Timed out. \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\int x^8\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2} \,d x \] Input:
int(x^8*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)
Output:
int(x^8*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2), x)
Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.50 \[ \int x^8 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2} \, dx=\frac {x^{9} \left (21 b^{5} x^{15}+120 a \,b^{4} x^{12}+280 a^{2} b^{3} x^{9}+336 a^{3} b^{2} x^{6}+210 a^{4} b \,x^{3}+56 a^{5}\right )}{504} \] Input:
int(x^8*(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)
Output:
(x**9*(56*a**5 + 210*a**4*b*x**3 + 336*a**3*b**2*x**6 + 280*a**2*b**3*x**9 + 120*a*b**4*x**12 + 21*b**5*x**15))/504