Integrand size = 26, antiderivative size = 167 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{19}} \, dx=-\frac {a^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{18 x^{18} \left (a+b x^3\right )}-\frac {a^2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{5 x^{15} \left (a+b x^3\right )}-\frac {a b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{4 x^{12} \left (a+b x^3\right )}-\frac {b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )} \] Output:
-1/18*a^3*((b*x^3+a)^2)^(1/2)/x^18/(b*x^3+a)-1/5*a^2*b*((b*x^3+a)^2)^(1/2) /x^15/(b*x^3+a)-1/4*a*b^2*((b*x^3+a)^2)^(1/2)/x^12/(b*x^3+a)-1/9*b^3*((b*x ^3+a)^2)^(1/2)/x^9/(b*x^3+a)
Time = 1.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.37 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{19}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (10 a^3+36 a^2 b x^3+45 a b^2 x^6+20 b^3 x^9\right )}{180 x^{18} \left (a+b x^3\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^19,x]
Output:
-1/180*(Sqrt[(a + b*x^3)^2]*(10*a^3 + 36*a^2*b*x^3 + 45*a*b^2*x^6 + 20*b^3 *x^9))/(x^18*(a + b*x^3))
Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.47, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{19}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^3 \left (b x^3+a\right )^3}{x^{19}}dx}{b^3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^3}{x^{19}}dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^3}{x^{21}}dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a^3}{x^{21}}+\frac {3 b a^2}{x^{18}}+\frac {3 b^2 a}{x^{15}}+\frac {b^3}{x^{12}}\right )dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^3}{6 x^{18}}-\frac {3 a^2 b}{5 x^{15}}-\frac {3 a b^2}{4 x^{12}}-\frac {b^3}{3 x^9}\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}\) |
Input:
Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^19,x]
Output:
((-1/6*a^3/x^18 - (3*a^2*b)/(5*x^15) - (3*a*b^2)/(4*x^12) - b^3/(3*x^9))*S qrt[a^2 + 2*a*b*x^3 + b^2*x^6])/(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, 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 = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.26
method | result | size |
pseudoelliptic | \(-\frac {\left (2 x^{9} b^{3}+\frac {9}{2} a \,x^{6} b^{2}+\frac {18}{5} a^{2} x^{3} b +a^{3}\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{18 x^{18}}\) | \(44\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{18} a^{3}-\frac {1}{5} a^{2} x^{3} b -\frac {1}{4} a \,x^{6} b^{2}-\frac {1}{9} x^{9} b^{3}\right )}{\left (b \,x^{3}+a \right ) x^{18}}\) | \(57\) |
gosper | \(-\frac {\left (20 x^{9} b^{3}+45 a \,x^{6} b^{2}+36 a^{2} x^{3} b +10 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{180 x^{18} \left (b \,x^{3}+a \right )^{3}}\) | \(58\) |
default | \(-\frac {\left (20 x^{9} b^{3}+45 a \,x^{6} b^{2}+36 a^{2} x^{3} b +10 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{180 x^{18} \left (b \,x^{3}+a \right )^{3}}\) | \(58\) |
orering | \(-\frac {\left (20 x^{9} b^{3}+45 a \,x^{6} b^{2}+36 a^{2} x^{3} b +10 a^{3}\right ) \left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {3}{2}}}{180 x^{18} \left (b \,x^{3}+a \right )^{3}}\) | \(67\) |
Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^19,x,method=_RETURNVERBOSE)
Output:
-1/18*(2*x^9*b^3+9/2*a*x^6*b^2+18/5*a^2*x^3*b+a^3)*csgn(b*x^3+a)/x^18
Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{19}} \, dx=-\frac {20 \, b^{3} x^{9} + 45 \, a b^{2} x^{6} + 36 \, a^{2} b x^{3} + 10 \, a^{3}}{180 \, x^{18}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^19,x, algorithm="fricas")
Output:
-1/180*(20*b^3*x^9 + 45*a*b^2*x^6 + 36*a^2*b*x^3 + 10*a^3)/x^18
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{19}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{19}}\, dx \] Input:
integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**19,x)
Output:
Integral(((a + b*x**3)**2)**(3/2)/x**19, x)
Time = 0.04 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.26 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{19}} \, dx=\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{6}}{12 \, a^{6}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{5}}{12 \, a^{5} x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{4}}{12 \, a^{6} x^{6}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{3}}{12 \, a^{5} x^{9}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{2}}{12 \, a^{4} x^{12}} + \frac {7 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b}{90 \, a^{3} x^{15}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{18 \, a^{2} x^{18}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^19,x, algorithm="maxima")
Output:
1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^6/a^6 + 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^5/(a^5*x^3) - 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^4/( a^6*x^6) + 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^3/(a^5*x^9) - 1/12*(b^ 2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^2/(a^4*x^12) + 7/90*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b/(a^3*x^15) - 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)/(a^2*x^ 18)
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{19}} \, dx=-\frac {20 \, b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 45 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 36 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 10 \, a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{180 \, x^{18}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^19,x, algorithm="giac")
Output:
-1/180*(20*b^3*x^9*sgn(b*x^3 + a) + 45*a*b^2*x^6*sgn(b*x^3 + a) + 36*a^2*b *x^3*sgn(b*x^3 + a) + 10*a^3*sgn(b*x^3 + a))/x^18
Time = 21.56 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{19}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{18\,x^{18}\,\left (b\,x^3+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{9\,x^9\,\left (b\,x^3+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{4\,x^{12}\,\left (b\,x^3+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{5\,x^{15}\,\left (b\,x^3+a\right )} \] Input:
int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^19,x)
Output:
- (a^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(18*x^18*(a + b*x^3)) - (b^3*(a^ 2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(9*x^9*(a + b*x^3)) - (a*b^2*(a^2 + b^2*x^ 6 + 2*a*b*x^3)^(1/2))/(4*x^12*(a + b*x^3)) - (a^2*b*(a^2 + b^2*x^6 + 2*a*b *x^3)^(1/2))/(5*x^15*(a + b*x^3))
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{19}} \, dx=\frac {-20 b^{3} x^{9}-45 a \,b^{2} x^{6}-36 a^{2} b \,x^{3}-10 a^{3}}{180 x^{18}} \] Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^19,x)
Output:
( - 10*a**3 - 36*a**2*b*x**3 - 45*a*b**2*x**6 - 20*b**3*x**9)/(180*x**18)