Integrand size = 26, antiderivative size = 251 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{16}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{15 x^{15} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 x^{12} \left (a+b x^3\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^6 \left (a+b x^3\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}+\frac {b^5 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]
-1/15*a^5*((b*x^3+a)^2)^(1/2)/x^15/(b*x^3+a)-5/12*a^4*b*((b*x^3+a)^2)^(1/2 )/x^12/(b*x^3+a)-10/9*a^3*b^2*((b*x^3+a)^2)^(1/2)/x^9/(b*x^3+a)-5/3*a^2*b^ 3*((b*x^3+a)^2)^(1/2)/x^6/(b*x^3+a)-5/3*a*b^4*((b*x^3+a)^2)^(1/2)/x^3/(b*x ^3+a)+b^5*ln(x)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)
Time = 0.49 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{16}} \, dx=\frac {1}{360} \left (-\frac {\sqrt {\left (a+b x^3\right )^2} \left (12 a^4+63 a^3 b x^3+137 a^2 b^2 x^6+163 a b^3 x^9+137 b^4 x^{12}\right )}{x^{15}}+\frac {\sqrt {a^2} \left (12 a^4+75 a^3 b x^3+200 a^2 b^2 x^6+300 a b^3 x^9+300 b^4 x^{12}\right )}{x^{15}}-120 b^5 \text {arctanh}\left (\frac {b x^3}{\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}}\right )-\frac {120 \sqrt {a^2} b^5 \log \left (x^3\right )}{a}+\frac {60 \sqrt {a^2} b^5 \log \left (a \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{a}+\frac {60 \sqrt {a^2} b^5 \log \left (a \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{a}\right ) \]
(-((Sqrt[(a + b*x^3)^2]*(12*a^4 + 63*a^3*b*x^3 + 137*a^2*b^2*x^6 + 163*a*b ^3*x^9 + 137*b^4*x^12))/x^15) + (Sqrt[a^2]*(12*a^4 + 75*a^3*b*x^3 + 200*a^ 2*b^2*x^6 + 300*a*b^3*x^9 + 300*b^4*x^12))/x^15 - 120*b^5*ArcTanh[(b*x^3)/ (Sqrt[a^2] - Sqrt[(a + b*x^3)^2])] - (120*Sqrt[a^2]*b^5*Log[x^3])/a + (60* Sqrt[a^2]*b^5*Log[a*(Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2])])/a + (60*Sq rt[a^2]*b^5*Log[a*(Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2])])/a)/360
Time = 0.25 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.39, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{16}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^5 \left (b x^3+a\right )^5}{x^{16}}dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^{16}}dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^{18}}dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a^5}{x^{18}}+\frac {5 b a^4}{x^{15}}+\frac {10 b^2 a^3}{x^{12}}+\frac {10 b^3 a^2}{x^9}+\frac {5 b^4 a}{x^6}+\frac {b^5}{x^3}\right )dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {a^5}{5 x^{15}}-\frac {5 a^4 b}{4 x^{12}}-\frac {10 a^3 b^2}{3 x^9}-\frac {5 a^2 b^3}{x^6}-\frac {5 a b^4}{x^3}+b^5 \log \left (x^3\right )\right )}{3 \left (a+b x^3\right )}\) |
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/5*a^5/x^15 - (5*a^4*b)/(4*x^12) - (10 *a^3*b^2)/(3*x^9) - (5*a^2*b^3)/x^6 - (5*a*b^4)/x^3 + b^5*Log[x^3]))/(3*(a + b*x^3))
3.1.79.3.1 Defintions of rubi rules used
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 = 0.63 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.29
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (-5 \ln \left (b \,x^{3}\right ) b^{5} x^{15}+a \left (25 b^{4} x^{12}+25 a \,b^{3} x^{9}+\frac {50}{3} a^{2} b^{2} x^{6}+\frac {25}{4} a^{3} b \,x^{3}+a^{4}\right )\right )}{15 x^{15}}\) | \(72\) |
default | \(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}} \left (180 b^{5} \ln \left (x \right ) x^{15}-300 a \,b^{4} x^{12}-300 a^{2} b^{3} x^{9}-200 a^{3} b^{2} x^{6}-75 a^{4} b \,x^{3}-12 a^{5}\right )}{180 \left (b \,x^{3}+a \right )^{5} x^{15}}\) | \(82\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{15} a^{5}-\frac {5}{12} a^{4} b \,x^{3}-\frac {10}{9} a^{3} b^{2} x^{6}-\frac {5}{3} a^{2} b^{3} x^{9}-\frac {5}{3} a \,b^{4} x^{12}\right )}{\left (b \,x^{3}+a \right ) x^{15}}+\frac {b^{5} \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\) | \(98\) |
-1/15*csgn(b*x^3+a)*(-5*ln(b*x^3)*b^5*x^15+a*(25*b^4*x^12+25*a*b^3*x^9+50/ 3*a^2*b^2*x^6+25/4*a^3*b*x^3+a^4))/x^15
Time = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{16}} \, dx=\frac {180 \, b^{5} x^{15} \log \left (x\right ) - 300 \, a b^{4} x^{12} - 300 \, a^{2} b^{3} x^{9} - 200 \, a^{3} b^{2} x^{6} - 75 \, a^{4} b x^{3} - 12 \, a^{5}}{180 \, x^{15}} \]
1/180*(180*b^5*x^15*log(x) - 300*a*b^4*x^12 - 300*a^2*b^3*x^9 - 200*a^3*b^ 2*x^6 - 75*a^4*b*x^3 - 12*a^5)/x^15
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{16}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{16}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 374 vs. \(2 (175) = 350\).
Time = 0.23 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.49 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{16}} \, dx=\frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{6} x^{3}}{6 \, a^{2}} + \frac {1}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} b^{5} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {1}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b^{5} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{6} x^{3}}{12 \, a^{4}} + \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{5}}{2 \, a} + \frac {7 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{5}}{36 \, a^{3}} - \frac {2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{5}}{45 \, a^{5}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{4}}{9 \, a^{4} x^{3}} + \frac {2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{3}}{45 \, a^{5} x^{6}} - \frac {11 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{2}}{180 \, a^{4} x^{9}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b}{20 \, a^{3} x^{12}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{15 \, a^{2} x^{15}} \]
1/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^6*x^3/a^2 + 1/3*(-1)^(2*b^2*x^3 + 2* a*b)*b^5*log(2*b^2*x^3 + 2*a*b) - 1/3*(-1)^(2*a*b*x^3 + 2*a^2)*b^5*log(2*a *b*x/abs(x) + 2*a^2/(x^2*abs(x))) + 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2) *b^6*x^3/a^4 + 1/2*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^5/a + 7/36*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^5/a^3 - 2/45*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b ^5/a^5 - 1/9*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^4/(a^4*x^3) + 2/45*(b^2*x ^6 + 2*a*b*x^3 + a^2)^(7/2)*b^3/(a^5*x^6) - 11/180*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b^2/(a^4*x^9) + 1/20*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b/(a^3*x ^12) - 1/15*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)/(a^2*x^15)
Time = 0.30 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.49 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{16}} \, dx=b^{5} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {137 \, b^{5} x^{15} \mathrm {sgn}\left (b x^{3} + a\right ) + 300 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 300 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 200 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 75 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 12 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{180 \, x^{15}} \]
b^5*log(abs(x))*sgn(b*x^3 + a) - 1/180*(137*b^5*x^15*sgn(b*x^3 + a) + 300* a*b^4*x^12*sgn(b*x^3 + a) + 300*a^2*b^3*x^9*sgn(b*x^3 + a) + 200*a^3*b^2*x ^6*sgn(b*x^3 + a) + 75*a^4*b*x^3*sgn(b*x^3 + a) + 12*a^5*sgn(b*x^3 + a))/x ^15
Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{16}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x^{16}} \,d x \]