Integrand size = 26, antiderivative size = 169 \[ \int \frac {x^{11}}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {a^2 x^3 \left (a+b x^3\right )}{3 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a x^6 \left (a+b x^3\right )}{6 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {x^9 \left (a+b x^3\right )}{9 b \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a^3 \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
1/3*a^2*x^3*(b*x^3+a)/b^3/((b*x^3+a)^2)^(1/2)-1/6*a*x^6*(b*x^3+a)/b^2/((b* x^3+a)^2)^(1/2)+1/9*x^9*(b*x^3+a)/b/((b*x^3+a)^2)^(1/2)-1/3*a^3*(b*x^3+a)* ln(b*x^3+a)/b^4/((b*x^3+a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.40 \[ \int \frac {x^{11}}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\left (a+b x^3\right ) \left (b x^3 \left (6 a^2-3 a b x^3+2 b^2 x^6\right )-6 a^3 \log \left (a+b x^3\right )\right )}{18 b^4 \sqrt {\left (a+b x^3\right )^2}} \] Input:
Integrate[x^11/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]
Output:
((a + b*x^3)*(b*x^3*(6*a^2 - 3*a*b*x^3 + 2*b^2*x^6) - 6*a^3*Log[a + b*x^3] ))/(18*b^4*Sqrt[(a + b*x^3)^2])
Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.48, 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.
\(\displaystyle \int \frac {x^{11}}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b \left (a+b x^3\right ) \int \frac {x^{11}}{b \left (b x^3+a\right )}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^{11}}{b x^3+a}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^9}{b x^3+a}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \left (\frac {x^6}{b}-\frac {a x^3}{b^2}-\frac {a^3}{b^3 \left (b x^3+a\right )}+\frac {a^2}{b^3}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^3\right ) \left (-\frac {a^3 \log \left (a+b x^3\right )}{b^4}+\frac {a^2 x^3}{b^3}-\frac {a x^6}{2 b^2}+\frac {x^9}{3 b}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[x^11/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]
Output:
((a + b*x^3)*((a^2*x^3)/b^3 - (a*x^6)/(2*b^2) + x^9/(3*b) - (a^3*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)])
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.38
method | result | size |
default | \(-\frac {\left (b \,x^{3}+a \right ) \left (-2 x^{9} b^{3}+3 a \,x^{6} b^{2}-6 a^{2} x^{3} b +6 a^{3} \ln \left (b \,x^{3}+a \right )\right )}{18 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{4}}\) | \(64\) |
pseudoelliptic | \(\frac {-6 a^{3} \operatorname {csgn}\left (b \,x^{3}+a \right ) \ln \left (b \,x^{3}+a \right )+\left (2 x^{9} b^{3}-3 a \,x^{6} b^{2}+6 a^{2} x^{3} b +11 a^{3}\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{18 b^{4}}\) | \(69\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\frac {1}{9} b^{2} x^{9}-\frac {1}{6} a b \,x^{6}+\frac {1}{3} a^{2} x^{3}\right )}{\left (b \,x^{3}+a \right ) b^{3}}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{3} \ln \left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right ) b^{4}}\) | \(86\) |
Input:
int(x^11/((b*x^3+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/18*(b*x^3+a)*(-2*x^9*b^3+3*a*x^6*b^2-6*a^2*x^3*b+6*a^3*ln(b*x^3+a))/((b *x^3+a)^2)^(1/2)/b^4
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.27 \[ \int \frac {x^{11}}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {2 \, b^{3} x^{9} - 3 \, a b^{2} x^{6} + 6 \, a^{2} b x^{3} - 6 \, a^{3} \log \left (b x^{3} + a\right )}{18 \, b^{4}} \] Input:
integrate(x^11/((b*x^3+a)^2)^(1/2),x, algorithm="fricas")
Output:
1/18*(2*b^3*x^9 - 3*a*b^2*x^6 + 6*a^2*b*x^3 - 6*a^3*log(b*x^3 + a))/b^4
\[ \int \frac {x^{11}}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {x^{11}}{\sqrt {\left (a + b x^{3}\right )^{2}}}\, dx \] Input:
integrate(x**11/((b*x**3+a)**2)**(1/2),x)
Output:
Integral(x**11/sqrt((a + b*x**3)**2), x)
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.56 \[ \int \frac {x^{11}}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {5 \, a x^{6}}{18 \, b^{2}} + \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} x^{6}}{9 \, b^{2}} + \frac {5 \, a^{2} x^{3}}{9 \, b^{3}} - \frac {a^{3} \log \left (x^{3} + \frac {a}{b}\right )}{3 \, b^{4}} - \frac {2 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2}}{9 \, b^{4}} \] Input:
integrate(x^11/((b*x^3+a)^2)^(1/2),x, algorithm="maxima")
Output:
-5/18*a*x^6/b^2 + 1/9*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*x^6/b^2 + 5/9*a^2*x^ 3/b^3 - 1/3*a^3*log(x^3 + a/b)/b^4 - 2/9*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a ^2/b^4
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.47 \[ \int \frac {x^{11}}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {a^{3} \log \left ({\left | b x^{3} + a \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, b^{4}} + \frac {2 \, b^{2} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) - 3 \, a b x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 6 \, a^{2} x^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{18 \, b^{3}} \] Input:
integrate(x^11/((b*x^3+a)^2)^(1/2),x, algorithm="giac")
Output:
-1/3*a^3*log(abs(b*x^3 + a))*sgn(b*x^3 + a)/b^4 + 1/18*(2*b^2*x^9*sgn(b*x^ 3 + a) - 3*a*b*x^6*sgn(b*x^3 + a) + 6*a^2*x^3*sgn(b*x^3 + a))/b^3
Timed out. \[ \int \frac {x^{11}}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {x^{11}}{\sqrt {{\left (b\,x^3+a\right )}^2}} \,d x \] Input:
int(x^11/((a + b*x^3)^2)^(1/2),x)
Output:
int(x^11/((a + b*x^3)^2)^(1/2), x)
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.43 \[ \int \frac {x^{11}}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {-6 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{3}-6 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{3}+6 a^{2} b \,x^{3}-3 a \,b^{2} x^{6}+2 b^{3} x^{9}}{18 b^{4}} \] Input:
int(x^11/((b*x^3+a)^2)^(1/2),x)
Output:
( - 6*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3 - 6*log(a** (1/3) + b**(1/3)*x)*a**3 + 6*a**2*b*x**3 - 3*a*b**2*x**6 + 2*b**3*x**9)/(1 8*b**4)