Integrand size = 26, antiderivative size = 252 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{13}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 x^{12} \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac {5 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^6 \left (a+b x^3\right )}-\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}+\frac {b^5 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {5 a b^4 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]
-1/12*a^5*((b*x^3+a)^2)^(1/2)/x^12/(b*x^3+a)-5/9*a^4*b*((b*x^3+a)^2)^(1/2) /x^9/(b*x^3+a)-5/3*a^3*b^2*((b*x^3+a)^2)^(1/2)/x^6/(b*x^3+a)-10/3*a^2*b^3* ((b*x^3+a)^2)^(1/2)/x^3/(b*x^3+a)+1/3*b^5*x^3*((b*x^3+a)^2)^(1/2)/(b*x^3+a )+5*a*b^4*ln(x)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)
Leaf count is larger than twice the leaf count of optimal. \(696\) vs. \(2(252)=504\).
Time = 0.94 (sec) , antiderivative size = 696, normalized size of antiderivative = 2.76 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{13}} \, dx=\frac {48 a^6 \sqrt {a^2}+368 a^5 \sqrt {a^2} b x^3+1280 a^4 \sqrt {a^2} b^2 x^6+2880 a^3 \sqrt {a^2} b^3 x^9+2677 \left (a^2\right )^{3/2} b^4 x^{12}+565 a \sqrt {a^2} b^5 x^{15}-192 \sqrt {a^2} b^6 x^{18}-48 a^6 \sqrt {\left (a+b x^3\right )^2}-320 a^5 b x^3 \sqrt {\left (a+b x^3\right )^2}-960 a^4 b^2 x^6 \sqrt {\left (a+b x^3\right )^2}-1920 a^3 b^3 x^9 \sqrt {\left (a+b x^3\right )^2}-757 a^2 b^4 x^{12} \sqrt {\left (a+b x^3\right )^2}+192 a b^5 x^{15} \sqrt {\left (a+b x^3\right )^2}-960 a b^4 x^{12} \left (a^2+a b x^3-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right ) \text {arctanh}\left (\frac {b x^3}{\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}}\right )-960 b^4 x^{12} \left (\left (a^2\right )^{3/2}+a \sqrt {a^2} b x^3-a^2 \sqrt {\left (a+b x^3\right )^2}\right ) \log \left (x^3\right )+480 \left (a^2\right )^{3/2} b^4 x^{12} \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+480 a \sqrt {a^2} b^5 x^{15} \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-480 a^2 b^4 x^{12} \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+480 \left (a^2\right )^{3/2} b^4 x^{12} \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+480 a \sqrt {a^2} b^5 x^{15} \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )-480 a^2 b^4 x^{12} \sqrt {\left (a+b x^3\right )^2} \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )}{576 x^{12} \left (a^2+a b x^3-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )} \]
(48*a^6*Sqrt[a^2] + 368*a^5*Sqrt[a^2]*b*x^3 + 1280*a^4*Sqrt[a^2]*b^2*x^6 + 2880*a^3*Sqrt[a^2]*b^3*x^9 + 2677*(a^2)^(3/2)*b^4*x^12 + 565*a*Sqrt[a^2]* b^5*x^15 - 192*Sqrt[a^2]*b^6*x^18 - 48*a^6*Sqrt[(a + b*x^3)^2] - 320*a^5*b *x^3*Sqrt[(a + b*x^3)^2] - 960*a^4*b^2*x^6*Sqrt[(a + b*x^3)^2] - 1920*a^3* b^3*x^9*Sqrt[(a + b*x^3)^2] - 757*a^2*b^4*x^12*Sqrt[(a + b*x^3)^2] + 192*a *b^5*x^15*Sqrt[(a + b*x^3)^2] - 960*a*b^4*x^12*(a^2 + a*b*x^3 - Sqrt[a^2]* Sqrt[(a + b*x^3)^2])*ArcTanh[(b*x^3)/(Sqrt[a^2] - Sqrt[(a + b*x^3)^2])] - 960*b^4*x^12*((a^2)^(3/2) + a*Sqrt[a^2]*b*x^3 - a^2*Sqrt[(a + b*x^3)^2])*L og[x^3] + 480*(a^2)^(3/2)*b^4*x^12*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3 )^2]] + 480*a*Sqrt[a^2]*b^5*x^15*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^ 2]] - 480*a^2*b^4*x^12*Sqrt[(a + b*x^3)^2]*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] + 480*(a^2)^(3/2)*b^4*x^12*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]] + 480*a*Sqrt[a^2]*b^5*x^15*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b *x^3)^2]] - 480*a^2*b^4*x^12*Sqrt[(a + b*x^3)^2]*Log[Sqrt[a^2] + b*x^3 - S qrt[(a + b*x^3)^2]])/(576*x^12*(a^2 + a*b*x^3 - Sqrt[a^2]*Sqrt[(a + b*x^3) ^2]))
Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.38, 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.
| \(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{13}} \, dx\) |
| \(\Big \downarrow \) 1384 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^5 \left (b x^3+a\right )^5}{x^{13}}dx}{b^5 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^{13}}dx}{a+b x^3}\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^{15}}dx^3}{3 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a^5}{x^{15}}+\frac {5 b a^4}{x^{12}}+\frac {10 b^2 a^3}{x^9}+\frac {10 b^3 a^2}{x^6}+\frac {5 b^4 a}{x^3}+b^5\right )dx^3}{3 \left (a+b x^3\right )}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {a^5}{4 x^{12}}-\frac {5 a^4 b}{3 x^9}-\frac {5 a^3 b^2}{x^6}-\frac {10 a^2 b^3}{x^3}+5 a b^4 \log \left (x^3\right )+b^5 x^3\right )}{3 \left (a+b x^3\right )}\) |
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/4*a^5/x^12 - (5*a^4*b)/(3*x^9) - (5*a ^3*b^2)/x^6 - (10*a^2*b^3)/x^3 + b^5*x^3 + 5*a*b^4*Log[x^3]))/(3*(a + b*x^ 3))
3.1.76.3.1 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 = 0.33 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.32
| method | result | size |
| pseudoelliptic | \(-\frac {\left (-4 b^{5} x^{15}-20 \ln \left (b \,x^{3}\right ) a \,b^{4} x^{12}-4 a \,b^{4} x^{12}+40 a^{2} b^{3} x^{9}+20 a^{3} b^{2} x^{6}+\frac {20 a^{4} b \,x^{3}}{3}+a^{5}\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{12 x^{12}}\) | \(81\) |
| default | \(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}} \left (12 b^{5} x^{15}+180 b^{4} a \ln \left (x \right ) x^{12}-120 a^{2} b^{3} x^{9}-60 a^{3} b^{2} x^{6}-20 a^{4} b \,x^{3}-3 a^{5}\right )}{36 \left (b \,x^{3}+a \right )^{5} x^{12}}\) | \(82\) |
| risch | \(\frac {b^{5} x^{3} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{3 b \,x^{3}+3 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {10}{3} a^{2} b^{3} x^{9}-\frac {5}{3} a^{3} b^{2} x^{6}-\frac {5}{9} a^{4} b \,x^{3}-\frac {1}{12} a^{5}\right )}{\left (b \,x^{3}+a \right ) x^{12}}+\frac {5 a \,b^{4} \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\) | \(119\) |
-1/12*(-4*b^5*x^15-20*ln(b*x^3)*a*b^4*x^12-4*a*b^4*x^12+40*a^2*b^3*x^9+20* a^3*b^2*x^6+20/3*a^4*b*x^3+a^5)*csgn(b*x^3+a)/x^12
Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{13}} \, dx=\frac {12 \, b^{5} x^{15} + 180 \, a b^{4} x^{12} \log \left (x\right ) - 120 \, a^{2} b^{3} x^{9} - 60 \, a^{3} b^{2} x^{6} - 20 \, a^{4} b x^{3} - 3 \, a^{5}}{36 \, x^{12}} \]
1/36*(12*b^5*x^15 + 180*a*b^4*x^12*log(x) - 120*a^2*b^3*x^9 - 60*a^3*b^2*x ^6 - 20*a^4*b*x^3 - 3*a^5)/x^12
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{13}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{13}}\, dx \]
Time = 0.22 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{13}} \, dx=\frac {5 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{5} x^{3}}{6 \, a} + \frac {5}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a b^{4} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {5}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a b^{4} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {5 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{5} x^{3}}{12 \, a^{3}} + \frac {5}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{4} + \frac {35 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{4}}{36 \, a^{2}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{4}}{9 \, a^{4}} - \frac {2 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{3}}{9 \, a^{3} x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b^{2}}{9 \, a^{4} x^{6}} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b}{36 \, a^{3} x^{9}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{12 \, a^{2} x^{12}} \]
5/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^5*x^3/a + 5/3*(-1)^(2*b^2*x^3 + 2*a* b)*a*b^4*log(2*b^2*x^3 + 2*a*b) - 5/3*(-1)^(2*a*b*x^3 + 2*a^2)*a*b^4*log(2 *a*b*x/abs(x) + 2*a^2/(x^2*abs(x))) + 5/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/ 2)*b^5*x^3/a^3 + 5/2*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^4 + 35/36*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*b^4/a^2 + 1/9*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b ^4/a^4 - 2/9*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^3/(a^3*x^3) - 1/9*(b^2*x^ 6 + 2*a*b*x^3 + a^2)^(7/2)*b^2/(a^4*x^6) + 1/36*(b^2*x^6 + 2*a*b*x^3 + a^2 )^(7/2)*b/(a^3*x^9) - 1/12*(b^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)/(a^2*x^12)
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{13}} \, dx=\frac {1}{3} \, b^{5} x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 5 \, a b^{4} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {125 \, a b^{4} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + 120 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 60 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 20 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 3 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{36 \, x^{12}} \]
1/3*b^5*x^3*sgn(b*x^3 + a) + 5*a*b^4*log(abs(x))*sgn(b*x^3 + a) - 1/36*(12 5*a*b^4*x^12*sgn(b*x^3 + a) + 120*a^2*b^3*x^9*sgn(b*x^3 + a) + 60*a^3*b^2* x^6*sgn(b*x^3 + a) + 20*a^4*b*x^3*sgn(b*x^3 + a) + 3*a^5*sgn(b*x^3 + a))/x ^12
Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{13}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x^{13}} \,d x \]