Integrand size = 26, antiderivative size = 119 \[ \int \frac {x^8}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {a^2}{12 b^3 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {2 a}{9 b^3 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {1}{6 b^3 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
-1/12*a^2/b^3/(b*x^3+a)^3/((b*x^3+a)^2)^(1/2)+2/9*a/b^3/(b*x^3+a)^2/((b*x^ 3+a)^2)^(1/2)-1/6/b^3/(b*x^3+a)/((b*x^3+a)^2)^(1/2)
Time = 0.97 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.73 \[ \int \frac {x^8}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {x^9 \left (-4 a^6-a^5 b x^3+4 a^4 \sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}-3 a^3 \sqrt {a^2} b x^3 \sqrt {\left (a+b x^3\right )^2}+3 \sqrt {a^2} b^2 x^6 \sqrt {\left (a+b x^3\right )^2} \left (a^2+b^2 x^6\right )+3 a b^3 x^9 \left (b^2 x^6-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )\right )}{36 a^6 \left (a+b x^3\right )^3 \left (\sqrt {a^2} b x^3+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right )} \] Input:
Integrate[x^8/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
Output:
(x^9*(-4*a^6 - a^5*b*x^3 + 4*a^4*Sqrt[a^2]*Sqrt[(a + b*x^3)^2] - 3*a^3*Sqr t[a^2]*b*x^3*Sqrt[(a + b*x^3)^2] + 3*Sqrt[a^2]*b^2*x^6*Sqrt[(a + b*x^3)^2] *(a^2 + b^2*x^6) + 3*a*b^3*x^9*(b^2*x^6 - Sqrt[a^2]*Sqrt[(a + b*x^3)^2]))) /(36*a^6*(a + b*x^3)^3*(Sqrt[a^2]*b*x^3 + a*(Sqrt[a^2] - Sqrt[(a + b*x^3)^ 2])))
Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.72, 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, 53, 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.
\(\displaystyle \int \frac {x^8}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^5 \left (a+b x^3\right ) \int \frac {x^8}{b^5 \left (b x^3+a\right )^5}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^8}{\left (b x^3+a\right )^5}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^6}{\left (b x^3+a\right )^5}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \left (\frac {a^2}{b^2 \left (b x^3+a\right )^5}-\frac {2 a}{b^2 \left (b x^3+a\right )^4}+\frac {1}{b^2 \left (b x^3+a\right )^3}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^3\right ) \left (-\frac {a^2}{4 b^3 \left (a+b x^3\right )^4}+\frac {2 a}{3 b^3 \left (a+b x^3\right )^3}-\frac {1}{2 b^3 \left (a+b x^3\right )^2}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[x^8/(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2),x]
Output:
((a + b*x^3)*(-1/4*a^2/(b^3*(a + b*x^3)^4) + (2*a)/(3*b^3*(a + b*x^3)^3) - 1/(2*b^3*(a + b*x^3)^2)))/(3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6])
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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35
method | result | size |
pseudoelliptic | \(-\frac {\left (6 b^{2} x^{6}+4 a \,x^{3} b +a^{2}\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{36 \left (b \,x^{3}+a \right )^{4} b^{3}}\) | \(42\) |
gosper | \(-\frac {\left (b \,x^{3}+a \right ) \left (6 b^{2} x^{6}+4 a \,x^{3} b +a^{2}\right )}{36 b^{3} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(43\) |
default | \(-\frac {\left (b \,x^{3}+a \right ) \left (6 b^{2} x^{6}+4 a \,x^{3} b +a^{2}\right )}{36 b^{3} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(43\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {x^{6}}{6 b}-\frac {a \,x^{3}}{9 b^{2}}-\frac {a^{2}}{36 b^{3}}\right )}{\left (b \,x^{3}+a \right )^{5}}\) | \(48\) |
orering | \(-\frac {\left (b \,x^{3}+a \right ) \left (6 b^{2} x^{6}+4 a \,x^{3} b +a^{2}\right )}{36 b^{3} \left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {5}{2}}}\) | \(52\) |
Input:
int(x^8/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/36/(b*x^3+a)^4*(6*b^2*x^6+4*a*b*x^3+a^2)/b^3*csgn(b*x^3+a)
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.58 \[ \int \frac {x^8}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {6 \, b^{2} x^{6} + 4 \, a b x^{3} + a^{2}}{36 \, {\left (b^{7} x^{12} + 4 \, a b^{6} x^{9} + 6 \, a^{2} b^{5} x^{6} + 4 \, a^{3} b^{4} x^{3} + a^{4} b^{3}\right )}} \] Input:
integrate(x^8/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")
Output:
-1/36*(6*b^2*x^6 + 4*a*b*x^3 + a^2)/(b^7*x^12 + 4*a*b^6*x^9 + 6*a^2*b^5*x^ 6 + 4*a^3*b^4*x^3 + a^4*b^3)
\[ \int \frac {x^8}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {x^{8}}{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] 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 = 53, normalized size of antiderivative = 0.45 \[ \int \frac {x^8}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {1}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} b^{5}} + \frac {2 \, a}{9 \, {\left (x^{3} + \frac {a}{b}\right )}^{3} b^{6}} - \frac {a^{2}}{12 \, {\left (x^{3} + \frac {a}{b}\right )}^{4} b^{7}} \] Input:
integrate(x^8/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")
Output:
-1/6/((x^3 + a/b)^2*b^5) + 2/9*a/((x^3 + a/b)^3*b^6) - 1/12*a^2/((x^3 + a/ b)^4*b^7)
Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.36 \[ \int \frac {x^8}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {6 \, b^{2} x^{6} + 4 \, a b x^{3} + a^{2}}{36 \, {\left (b x^{3} + a\right )}^{4} b^{3} \mathrm {sgn}\left (b x^{3} + a\right )} \] Input:
integrate(x^8/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")
Output:
-1/36*(6*b^2*x^6 + 4*a*b*x^3 + a^2)/((b*x^3 + a)^4*b^3*sgn(b*x^3 + a))
Time = 18.72 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.45 \[ \int \frac {x^8}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}\,\left (a^2+4\,a\,b\,x^3+6\,b^2\,x^6\right )}{36\,b^3\,{\left (b\,x^3+a\right )}^5} \] Input:
int(x^8/(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2),x)
Output:
-((a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2)*(a^2 + 6*b^2*x^6 + 4*a*b*x^3))/(36*b^3 *(a + b*x^3)^5)
Time = 0.18 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.57 \[ \int \frac {x^8}{\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {-6 b^{2} x^{6}-4 a b \,x^{3}-a^{2}}{36 b^{3} \left (b^{4} x^{12}+4 a \,b^{3} x^{9}+6 a^{2} b^{2} x^{6}+4 a^{3} b \,x^{3}+a^{4}\right )} \] Input:
int(x^8/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)
Output:
( - a**2 - 4*a*b*x**3 - 6*b**2*x**6)/(36*b**3*(a**4 + 4*a**3*b*x**3 + 6*a* *2*b**2*x**6 + 4*a*b**3*x**9 + b**4*x**12))