Integrand size = 26, antiderivative size = 196 \[ \int \frac {x^{14}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {4 a^3}{3 b^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a^4}{6 b^5 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a x^3 \left (a+b x^3\right )}{b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {x^6 \left (a+b x^3\right )}{6 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {2 a^2 \left (a+b x^3\right ) \log \left (a+b x^3\right )}{b^5 \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
4/3*a^3/b^5/((b*x^3+a)^2)^(1/2)-1/6*a^4/b^5/(b*x^3+a)/((b*x^3+a)^2)^(1/2)- a*x^3*(b*x^3+a)/b^4/((b*x^3+a)^2)^(1/2)+1/6*x^6*(b*x^3+a)/b^3/((b*x^3+a)^2 )^(1/2)+2*a^2*(b*x^3+a)*ln(b*x^3+a)/b^5/((b*x^3+a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.47 \[ \int \frac {x^{14}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {7 a^4+2 a^3 b x^3-11 a^2 b^2 x^6-4 a b^3 x^9+b^4 x^{12}+12 a^2 \left (a+b x^3\right )^2 \log \left (a+b x^3\right )}{6 b^5 \left (a+b x^3\right ) \sqrt {\left (a+b x^3\right )^2}} \] Input:
Integrate[x^14/(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]
Output:
(7*a^4 + 2*a^3*b*x^3 - 11*a^2*b^2*x^6 - 4*a*b^3*x^9 + b^4*x^12 + 12*a^2*(a + b*x^3)^2*Log[a + b*x^3])/(6*b^5*(a + b*x^3)*Sqrt[(a + b*x^3)^2])
Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.54, 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.
\(\displaystyle \int \frac {x^{14}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^3\right ) \int \frac {x^{14}}{b^3 \left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^{14}}{\left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^{12}}{\left (b x^3+a\right )^3}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \left (\frac {a^4}{b^4 \left (b x^3+a\right )^3}-\frac {4 a^3}{b^4 \left (b x^3+a\right )^2}+\frac {6 a^2}{b^4 \left (b x^3+a\right )}-\frac {3 a}{b^4}+\frac {x^3}{b^3}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^3\right ) \left (-\frac {a^4}{2 b^5 \left (a+b x^3\right )^2}+\frac {4 a^3}{b^5 \left (a+b x^3\right )}+\frac {6 a^2 \log \left (a+b x^3\right )}{b^5}-\frac {3 a x^3}{b^4}+\frac {x^6}{2 b^3}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[x^14/(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2),x]
Output:
((a + b*x^3)*((-3*a*x^3)/b^4 + x^6/(2*b^3) - a^4/(2*b^5*(a + b*x^3)^2) + ( 4*a^3)/(b^5*(a + b*x^3)) + (6*a^2*Log[a + b*x^3])/b^5))/(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 = 0.15 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.39
method | result | size |
pseudoelliptic | \(\frac {2 \,\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (a^{2} \ln \left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{2}-\left (\frac {b \,x^{3}}{2}+a \right ) \left (-\frac {1}{6} b^{2} x^{6}+a \,x^{3} b +a^{2}\right ) b \,x^{3}\right )}{b^{5} \left (b \,x^{3}+a \right )^{2}}\) | \(77\) |
default | \(\frac {\left (b^{4} x^{12}-4 a \,b^{3} x^{9}+12 \ln \left (b \,x^{3}+a \right ) a^{2} b^{2} x^{6}-2 b^{2} x^{6} a^{2}+24 \ln \left (b \,x^{3}+a \right ) a^{3} b \,x^{3}+20 b \,x^{3} a^{3}+12 \ln \left (b \,x^{3}+a \right ) a^{4}+16 a^{4}\right ) \left (b \,x^{3}+a \right )}{6 b^{5} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(115\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-b \,x^{3}+3 a \right )^{2}}{6 \left (b \,x^{3}+a \right ) b^{5}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\frac {4 a^{3} x^{3}}{3}+\frac {7 a^{4}}{6 b}\right )}{\left (b \,x^{3}+a \right )^{3} b^{4}}+\frac {2 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{2} \ln \left (b \,x^{3}+a \right )}{\left (b \,x^{3}+a \right ) b^{5}}\) | \(116\) |
Input:
int(x^14/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
2*csgn(b*x^3+a)*(a^2*ln(b*x^3+a)*(b*x^3+a)^2-(1/2*b*x^3+a)*(-1/6*b^2*x^6+a *x^3*b+a^2)*b*x^3)/b^5/(b*x^3+a)^2
Time = 0.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.53 \[ \int \frac {x^{14}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {b^{4} x^{12} - 4 \, a b^{3} x^{9} - 11 \, a^{2} b^{2} x^{6} + 2 \, a^{3} b x^{3} + 7 \, a^{4} + 12 \, {\left (a^{2} b^{2} x^{6} + 2 \, a^{3} b x^{3} + a^{4}\right )} \log \left (b x^{3} + a\right )}{6 \, {\left (b^{7} x^{6} + 2 \, a b^{6} x^{3} + a^{2} b^{5}\right )}} \] Input:
integrate(x^14/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")
Output:
1/6*(b^4*x^12 - 4*a*b^3*x^9 - 11*a^2*b^2*x^6 + 2*a^3*b*x^3 + 7*a^4 + 12*(a ^2*b^2*x^6 + 2*a^3*b*x^3 + a^4)*log(b*x^3 + a))/(b^7*x^6 + 2*a*b^6*x^3 + a ^2*b^5)
\[ \int \frac {x^{14}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {x^{14}}{\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**14/(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)
Output:
Integral(x**14/((a + b*x**3)**2)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.74 \[ \int \frac {x^{14}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {x^{9}}{6 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{2}} - \frac {5 \, a x^{6}}{6 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{3}} + \frac {4 \, a^{3} x^{3}}{{\left (x^{3} + \frac {a}{b}\right )}^{2} b^{6}} + \frac {2 \, a^{2} \log \left (x^{3} + \frac {a}{b}\right )}{b^{5}} - \frac {5 \, a^{3}}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{5}} + \frac {23 \, a^{4}}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} b^{7}} \] Input:
integrate(x^14/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")
Output:
1/6*x^9/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^2) - 5/6*a*x^6/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^3) + 4*a^3*x^3/((x^3 + a/b)^2*b^6) + 2*a^2*log(x^3 + a/ b)/b^5 - 5/3*a^3/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^5) + 23/6*a^4/((x^3 + a/b)^2*b^7)
Time = 0.13 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.59 \[ \int \frac {x^{14}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {2 \, a^{2} \log \left ({\left | b x^{3} + a \right |}\right )}{b^{5} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {b^{3} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) - 6 \, a b^{2} x^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{6 \, b^{6}} - \frac {18 \, a^{2} b^{2} x^{6} + 28 \, a^{3} b x^{3} + 11 \, a^{4}}{6 \, {\left (b x^{3} + a\right )}^{2} b^{5} \mathrm {sgn}\left (b x^{3} + a\right )} \] Input:
integrate(x^14/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")
Output:
2*a^2*log(abs(b*x^3 + a))/(b^5*sgn(b*x^3 + a)) + 1/6*(b^3*x^6*sgn(b*x^3 + a) - 6*a*b^2*x^3*sgn(b*x^3 + a))/b^6 - 1/6*(18*a^2*b^2*x^6 + 28*a^3*b*x^3 + 11*a^4)/((b*x^3 + a)^2*b^5*sgn(b*x^3 + a))
Timed out. \[ \int \frac {x^{14}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {x^{14}}{{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}} \,d x \] Input:
int(x^14/(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2),x)
Output:
int(x^14/(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2), x)
Time = 0.18 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.03 \[ \int \frac {x^{14}}{\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {12 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{4}+24 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{3} b \,x^{3}+12 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2} b^{2} x^{6}+12 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{4}+24 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{3} b \,x^{3}+12 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} b^{2} x^{6}+6 a^{4}-12 a^{2} b^{2} x^{6}-4 a \,b^{3} x^{9}+b^{4} x^{12}}{6 b^{5} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )} \] Input:
int(x^14/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x)
Output:
(12*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**4 + 24*log(a**( 2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b*x**3 + 12*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**2*x**6 + 12*log(a**(1/3) + b**(1/3)*x)*a**4 + 24*log(a**(1/3) + b**(1/3)*x)*a**3*b*x**3 + 12*log(a**( 1/3) + b**(1/3)*x)*a**2*b**2*x**6 + 6*a**4 - 12*a**2*b**2*x**6 - 4*a*b**3* x**9 + b**4*x**12)/(6*b**5*(a**2 + 2*a*b*x**3 + b**2*x**6))