Integrand size = 26, antiderivative size = 154 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {a^2}{b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^3}{6 b^4 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {x^3 \left (a+b x^3\right )}{3 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a \left (a+b x^3\right ) \log \left (a+b x^3\right )}{b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
-a^2/b^4/((b*x^3+a)^2)^(1/2)+1/6*a^3/b^4/(b*x^3+a)/((b*x^3+a)^2)^(1/2)+1/3 *x^3*(b*x^3+a)/b^3/((b*x^3+a)^2)^(1/2)-a*(b*x^3+a)*ln(b*x^3+a)/b^4/((b*x^3 +a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.53 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {-5 a^3-4 a^2 b x^3+4 a b^2 x^6+2 b^3 x^9-6 a \left (a+b x^3\right )^2 \log \left (a+b x^3\right )}{6 b^4 \left (a+b x^3\right ) \sqrt {\left (a+b x^3\right )^2}} \] Input:
Integrate[x^11/(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]
Output:
(-5*a^3 - 4*a^2*b*x^3 + 4*a*b^2*x^6 + 2*b^3*x^9 - 6*a*(a + b*x^3)^2*Log[a + b*x^3])/(6*b^4*(a + b*x^3)*Sqrt[(a + b*x^3)^2])
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.59, 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.
\(\displaystyle \int \frac {x^{11}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^3\right ) \int \frac {x^{11}}{b^3 \left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^{11}}{\left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^9}{\left (b x^3+a\right )^3}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \left (-\frac {a^3}{b^3 \left (b x^3+a\right )^3}+\frac {3 a^2}{b^3 \left (b x^3+a\right )^2}-\frac {3 a}{b^3 \left (b x^3+a\right )}+\frac {1}{b^3}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {a^3}{2 b^4 \left (a+b x^3\right )^2}-\frac {3 a^2}{b^4 \left (a+b x^3\right )}-\frac {3 a \log \left (a+b x^3\right )}{b^4}+\frac {x^3}{b^3}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[x^11/(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]
Output:
((a + b*x^3)*(x^3/b^3 + a^3/(2*b^4*(a + b*x^3)^2) - (3*a^2)/(b^4*(a + b*x^ 3)) - (3*a*Log[a + b*x^3])/b^4))/(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}, 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 3.
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.42
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (\ln \left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{2} a -\frac {x^{9} b^{3}}{3}-a \,x^{6} b^{2}+\frac {a^{3}}{2}\right )}{b^{4} \left (b \,x^{3}+a \right )^{2}}\) | \(65\) |
default | \(-\frac {\left (-2 x^{9} b^{3}+6 \ln \left (b \,x^{3}+a \right ) a \,b^{2} x^{6}-4 a \,x^{6} b^{2}+12 \ln \left (b \,x^{3}+a \right ) a^{2} b \,x^{3}+4 a^{2} x^{3} b +6 a^{3} \ln \left (b \,x^{3}+a \right )+5 a^{3}\right ) \left (b \,x^{3}+a \right )}{6 b^{4} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(103\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, x^{3}}{3 \left (b \,x^{3}+a \right ) b^{3}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-a^{2} x^{3}-\frac {5 a^{3}}{6 b}\right )}{\left (b \,x^{3}+a \right )^{3} b^{3}}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a \ln \left (b \,x^{3}+a \right )}{\left (b \,x^{3}+a \right ) b^{4}}\) | \(105\) |
Input:
int(x^11/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-csgn(b*x^3+a)*(ln(b*x^3+a)*(b*x^3+a)^2*a-1/3*x^9*b^3-a*x^6*b^2+1/2*a^3)/b ^4/(b*x^3+a)^2
Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.59 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {2 \, b^{3} x^{9} + 4 \, a b^{2} x^{6} - 4 \, a^{2} b x^{3} - 5 \, a^{3} - 6 \, {\left (a b^{2} x^{6} + 2 \, a^{2} b x^{3} + a^{3}\right )} \log \left (b x^{3} + a\right )}{6 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} \] Input:
integrate(x^11/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")
Output:
1/6*(2*b^3*x^9 + 4*a*b^2*x^6 - 4*a^2*b*x^3 - 5*a^3 - 6*(a*b^2*x^6 + 2*a^2* b*x^3 + a^3)*log(b*x^3 + a))/(b^6*x^6 + 2*a*b^5*x^3 + a^2*b^4)
\[ \int \frac {x^{11}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {x^{11}}{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**11/(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)
Output:
Integral(x**11/((a + b*x**3)**2)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.74 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {x^{6}}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{2}} - \frac {2 \, a^{2} x^{3}}{{\left (x^{3} + \frac {a}{b}\right )}^{2} b^{5}} - \frac {a \log \left (x^{3} + \frac {a}{b}\right )}{b^{4}} + \frac {2 \, a^{2}}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{4}} - \frac {11 \, a^{3}}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} b^{6}} \] Input:
integrate(x^11/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")
Output:
1/3*x^6/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^2) - 2*a^2*x^3/((x^3 + a/b)^2*b ^5) - a*log(x^3 + a/b)/b^4 + 2/3*a^2/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^4) - 11/6*a^3/((x^3 + a/b)^2*b^6)
Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.60 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {x^{3}}{3 \, b^{3} \mathrm {sgn}\left (b x^{3} + a\right )} - \frac {a \log \left ({\left | b x^{3} + a \right |}\right )}{b^{4} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {9 \, a b^{2} x^{6} + 12 \, a^{2} b x^{3} + 4 \, a^{3}}{6 \, {\left (b x^{3} + a\right )}^{2} b^{4} \mathrm {sgn}\left (b x^{3} + a\right )} \] Input:
integrate(x^11/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")
Output:
1/3*x^3/(b^3*sgn(b*x^3 + a)) - a*log(abs(b*x^3 + a))/(b^4*sgn(b*x^3 + a)) + 1/6*(9*a*b^2*x^6 + 12*a^2*b*x^3 + 4*a^3)/((b*x^3 + a)^2*b^4*sgn(b*x^3 + a))
Timed out. \[ \int \frac {x^{11}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {x^{11}}{{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}} \,d x \] Input:
int(x^11/(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2),x)
Output:
int(x^11/(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {-6 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{3}-12 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2} b \,x^{3}-6 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a \,b^{2} x^{6}-6 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{3}-12 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} b \,x^{3}-6 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a \,b^{2} x^{6}-3 a^{3}+6 a \,b^{2} x^{6}+2 b^{3} x^{9}}{6 b^{4} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )} \] Input:
int(x^11/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x)
Output:
( - 6*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3 - 12*log(a* *(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b*x**3 - 6*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b**2*x**6 - 6*log(a**(1/3) + b** (1/3)*x)*a**3 - 12*log(a**(1/3) + b**(1/3)*x)*a**2*b*x**3 - 6*log(a**(1/3) + b**(1/3)*x)*a*b**2*x**6 - 3*a**3 + 6*a*b**2*x**6 + 2*b**3*x**9)/(6*b**4 *(a**2 + 2*a*b*x**3 + b**2*x**6))