Integrand size = 26, antiderivative size = 84 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {\left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 a x^{15}}+\frac {b \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{60 a^2 x^{12}} \] Output:
-1/15*(b*x^3+a)^3*((b*x^3+a)^2)^(1/2)/a/x^15+1/60*b*(b*x^3+a)^3*((b*x^3+a) ^2)^(1/2)/a^2/x^12
Time = 1.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {\sqrt {\left (a+b x^3\right )^2} \left (4 a^3+15 a^2 b x^3+20 a b^2 x^6+10 b^3 x^9\right )}{60 x^{15} \left (a+b x^3\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^16,x]
Output:
-1/60*(Sqrt[(a + b*x^3)^2]*(4*a^3 + 15*a^2*b*x^3 + 20*a*b^2*x^6 + 10*b^3*x ^9))/(x^15*(a + b*x^3))
Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.89, 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, 55, 48}
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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^3 \left (b x^3+a\right )^3}{x^{16}}dx}{b^3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^3}{x^{16}}dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^3}{x^{18}}dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {b \int \frac {\left (b x^3+a\right )^3}{x^{15}}dx^3}{5 a}-\frac {\left (a+b x^3\right )^4}{5 a x^{15}}\right )}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (\frac {b \left (a+b x^3\right )^4}{20 a^2 x^{12}}-\frac {\left (a+b x^3\right )^4}{5 a x^{15}}\right )}{3 \left (a+b x^3\right )}\) |
Input:
Int[(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)/x^16,x]
Output:
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/5*(a + b*x^3)^4/(a*x^15) + (b*(a + b* x^3)^4)/(20*a^2*x^12)))/(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] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
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 = 0.52
method | result | size |
pseudoelliptic | \(-\frac {\left (\frac {5}{2} x^{9} b^{3}+5 a \,x^{6} b^{2}+\frac {15}{4} a^{2} x^{3} b +a^{3}\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{15 x^{15}}\) | \(44\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{15} a^{3}-\frac {1}{4} a^{2} x^{3} b -\frac {1}{3} a \,x^{6} b^{2}-\frac {1}{6} x^{9} b^{3}\right )}{\left (b \,x^{3}+a \right ) x^{15}}\) | \(57\) |
gosper | \(-\frac {\left (10 x^{9} b^{3}+20 a \,x^{6} b^{2}+15 a^{2} x^{3} b +4 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{60 x^{15} \left (b \,x^{3}+a \right )^{3}}\) | \(58\) |
default | \(-\frac {\left (10 x^{9} b^{3}+20 a \,x^{6} b^{2}+15 a^{2} x^{3} b +4 a^{3}\right ) {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}{60 x^{15} \left (b \,x^{3}+a \right )^{3}}\) | \(58\) |
orering | \(-\frac {\left (10 x^{9} b^{3}+20 a \,x^{6} b^{2}+15 a^{2} x^{3} b +4 a^{3}\right ) \left (b^{2} x^{6}+2 a \,x^{3} b +a^{2}\right )^{\frac {3}{2}}}{60 x^{15} \left (b \,x^{3}+a \right )^{3}}\) | \(67\) |
Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^16,x,method=_RETURNVERBOSE)
Output:
-1/15*(5/2*x^9*b^3+5*a*x^6*b^2+15/4*a^2*x^3*b+a^3)*csgn(b*x^3+a)/x^15
Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.44 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {10 \, b^{3} x^{9} + 20 \, a b^{2} x^{6} + 15 \, a^{2} b x^{3} + 4 \, a^{3}}{60 \, x^{15}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^16,x, algorithm="fricas")
Output:
-1/60*(10*b^3*x^9 + 20*a*b^2*x^6 + 15*a^2*b*x^3 + 4*a^3)/x^15
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}{x^{16}}\, dx \] Input:
integrate((b**2*x**6+2*a*b*x**3+a**2)**(3/2)/x**16,x)
Output:
Integral(((a + b*x**3)**2)**(3/2)/x**16, x)
Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (58) = 116\).
Time = 0.04 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.13 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{5}}{12 \, a^{5}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{4}}{12 \, a^{4} x^{3}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{3}}{12 \, a^{5} x^{6}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{2}}{12 \, a^{4} x^{9}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b}{12 \, a^{3} x^{12}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{15 \, a^{2} x^{15}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^16,x, algorithm="maxima")
Output:
-1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^5/a^5 - 1/12*(b^2*x^6 + 2*a*b*x^ 3 + a^2)^(3/2)*b^4/(a^4*x^3) + 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^3/ (a^5*x^6) - 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^2/(a^4*x^9) + 1/12*(b ^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b/(a^3*x^12) - 1/15*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)/(a^2*x^15)
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {10 \, b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 20 \, a b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 15 \, a^{2} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 4 \, a^{3} \mathrm {sgn}\left (b x^{3} + a\right )}{60 \, x^{15}} \] Input:
integrate((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^16,x, algorithm="giac")
Output:
-1/60*(10*b^3*x^9*sgn(b*x^3 + a) + 20*a*b^2*x^6*sgn(b*x^3 + a) + 15*a^2*b* x^3*sgn(b*x^3 + a) + 4*a^3*sgn(b*x^3 + a))/x^15
Time = 20.76 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.80 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=-\frac {a^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{15\,x^{15}\,\left (b\,x^3+a\right )}-\frac {b^3\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{6\,x^6\,\left (b\,x^3+a\right )}-\frac {a\,b^2\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{3\,x^9\,\left (b\,x^3+a\right )}-\frac {a^2\,b\,\sqrt {a^2+2\,a\,b\,x^3+b^2\,x^6}}{4\,x^{12}\,\left (b\,x^3+a\right )} \] Input:
int((a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)/x^16,x)
Output:
- (a^3*(a^2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(15*x^15*(a + b*x^3)) - (b^3*(a^ 2 + b^2*x^6 + 2*a*b*x^3)^(1/2))/(6*x^6*(a + b*x^3)) - (a*b^2*(a^2 + b^2*x^ 6 + 2*a*b*x^3)^(1/2))/(3*x^9*(a + b*x^3)) - (a^2*b*(a^2 + b^2*x^6 + 2*a*b* x^3)^(1/2))/(4*x^12*(a + b*x^3))
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.44 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}}{x^{16}} \, dx=\frac {-10 b^{3} x^{9}-20 a \,b^{2} x^{6}-15 a^{2} b \,x^{3}-4 a^{3}}{60 x^{15}} \] Input:
int((b^2*x^6+2*a*b*x^3+a^2)^(3/2)/x^16,x)
Output:
( - 4*a**3 - 15*a**2*b*x**3 - 20*a*b**2*x**6 - 10*b**3*x**9)/(60*x**15)