Integrand size = 26, antiderivative size = 252 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{10}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 x^9 \left (a+b x^3\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 x^6 \left (a+b x^3\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}+\frac {5 a b^4 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {b^5 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{6 \left (a+b x^3\right )}+\frac {10 a^2 b^3 \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]
-1/9*a^5*((b*x^3+a)^2)^(1/2)/x^9/(b*x^3+a)-5/6*a^4*b*((b*x^3+a)^2)^(1/2)/x ^6/(b*x^3+a)-10/3*a^3*b^2*((b*x^3+a)^2)^(1/2)/x^3/(b*x^3+a)+5/3*a*b^4*x^3* ((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/6*b^5*x^6*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+10 *a^2*b^3*ln(x)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)
Time = 0.75 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.11 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{10}} \, dx=\frac {1}{3} \left (\frac {\left (4 a^5+30 a^4 b x^3+120 a^3 b^2 x^6+53 a^2 b^3 x^9-60 a b^4 x^{12}-6 b^5 x^{15}\right ) \left (\sqrt {a^2} b x^3+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}\right )\right )}{12 x^9 \left (a^2+a b x^3-\sqrt {a^2} \sqrt {\left (a+b x^3\right )^2}\right )}-10 a^2 b^3 \text {arctanh}\left (\frac {b x^3}{\sqrt {a^2}-\sqrt {\left (a+b x^3\right )^2}}\right )-10 a \sqrt {a^2} b^3 \log \left (x^3\right )+5 a \sqrt {a^2} b^3 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )+5 a \sqrt {a^2} b^3 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )\right ) \]
(((4*a^5 + 30*a^4*b*x^3 + 120*a^3*b^2*x^6 + 53*a^2*b^3*x^9 - 60*a*b^4*x^12 - 6*b^5*x^15)*(Sqrt[a^2]*b*x^3 + a*(Sqrt[a^2] - Sqrt[(a + b*x^3)^2])))/(1 2*x^9*(a^2 + a*b*x^3 - Sqrt[a^2]*Sqrt[(a + b*x^3)^2])) - 10*a^2*b^3*ArcTan h[(b*x^3)/(Sqrt[a^2] - Sqrt[(a + b*x^3)^2])] - 10*a*Sqrt[a^2]*b^3*Log[x^3] + 5*a*Sqrt[a^2]*b^3*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2]] + 5*a*Sq rt[a^2]*b^3*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2]])/3
Time = 0.24 (sec) , antiderivative size = 99, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{10}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^5 \left (b x^3+a\right )^5}{x^{10}}dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^{10}}dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^{12}}dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (\frac {a^5}{x^{12}}+\frac {5 b a^4}{x^9}+\frac {10 b^2 a^3}{x^6}+\frac {10 b^3 a^2}{x^3}+5 b^4 a+b^5 x^3\right )dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {a^5}{3 x^9}-\frac {5 a^4 b}{2 x^6}-\frac {10 a^3 b^2}{x^3}+10 a^2 b^3 \log \left (x^3\right )+5 a b^4 x^3+\frac {b^5 x^6}{2}\right )}{3 \left (a+b x^3\right )}\) |
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-1/3*a^5/x^9 - (5*a^4*b)/(2*x^6) - (10*a ^3*b^2)/x^3 + 5*a*b^4*x^3 + (b^5*x^6)/2 + 10*a^2*b^3*Log[x^3]))/(3*(a + b* x^3))
3.1.73.3.1 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)])
Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.33
method | result | size |
default | \(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}} \left (3 b^{5} x^{15}+30 a \,b^{4} x^{12}+180 a^{2} b^{3} \ln \left (x \right ) x^{9}-60 a^{3} b^{2} x^{6}-15 a^{4} b \,x^{3}-2 a^{5}\right )}{18 \left (b \,x^{3}+a \right )^{5} x^{9}}\) | \(82\) |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (-\frac {3 b^{5} x^{15}}{2}-15 a \,b^{4} x^{12}-30 \ln \left (b \,x^{3}\right ) a^{2} b^{3} x^{9}-\frac {27 a^{2} b^{3} x^{9}}{2}+30 a^{3} b^{2} x^{6}+\frac {15 a^{4} b \,x^{3}}{2}+a^{5}\right )}{9 x^{9}}\) | \(83\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{3} \left (b \,x^{3}+5 a \right )^{2}}{6 b \,x^{3}+6 a}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {10}{3} a^{3} b^{2} x^{6}-\frac {5}{6} a^{4} b \,x^{3}-\frac {1}{9} a^{5}\right )}{\left (b \,x^{3}+a \right ) x^{9}}+\frac {10 a^{2} b^{3} \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\) | \(118\) |
1/18*((b*x^3+a)^2)^(5/2)*(3*b^5*x^15+30*a*b^4*x^12+180*a^2*b^3*ln(x)*x^9-6 0*a^3*b^2*x^6-15*a^4*b*x^3-2*a^5)/(b*x^3+a)^5/x^9
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{10}} \, dx=\frac {3 \, b^{5} x^{15} + 30 \, a b^{4} x^{12} + 180 \, a^{2} b^{3} x^{9} \log \left (x\right ) - 60 \, a^{3} b^{2} x^{6} - 15 \, a^{4} b x^{3} - 2 \, a^{5}}{18 \, x^{9}} \]
1/18*(3*b^5*x^15 + 30*a*b^4*x^12 + 180*a^2*b^3*x^9*log(x) - 60*a^3*b^2*x^6 - 15*a^4*b*x^3 - 2*a^5)/x^9
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{10}} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{10}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{10}} \, dx=\frac {5}{3} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{4} x^{3} + \frac {10}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a^{2} b^{3} \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {10}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a^{2} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {5 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{4} x^{3}}{6 \, a^{2}} + 5 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a b^{3} + \frac {35 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{3}}{18 \, a} + \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{3}}{18 \, a^{3}} - \frac {11 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}} b^{2}}{18 \, a^{2} x^{3}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}} b}{18 \, a^{3} x^{6}} - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {7}{2}}}{9 \, a^{2} x^{9}} \]
5/3*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^4*x^3 + 10/3*(-1)^(2*b^2*x^3 + 2*a*b )*a^2*b^3*log(2*b^2*x^3 + 2*a*b) - 10/3*(-1)^(2*a*b*x^3 + 2*a^2)*a^2*b^3*l og(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x))) + 5/6*(b^2*x^6 + 2*a*b*x^3 + a^2)^ (3/2)*b^4*x^3/a^2 + 5*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a*b^3 + 35/18*(b^2*x ^6 + 2*a*b*x^3 + a^2)^(3/2)*b^3/a + 1/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2) *b^3/a^3 - 11/18*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)*b^2/(a^2*x^3) - 1/18*(b ^2*x^6 + 2*a*b*x^3 + a^2)^(7/2)*b/(a^3*x^6) - 1/9*(b^2*x^6 + 2*a*b*x^3 + a ^2)^(7/2)/(a^2*x^9)
Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.50 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{10}} \, dx=\frac {1}{6} \, b^{5} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{3} \, a b^{4} x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 10 \, a^{2} b^{3} \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {110 \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + 60 \, a^{3} b^{2} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + 15 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 2 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{18 \, x^{9}} \]
1/6*b^5*x^6*sgn(b*x^3 + a) + 5/3*a*b^4*x^3*sgn(b*x^3 + a) + 10*a^2*b^3*log (abs(x))*sgn(b*x^3 + a) - 1/18*(110*a^2*b^3*x^9*sgn(b*x^3 + a) + 60*a^3*b^ 2*x^6*sgn(b*x^3 + a) + 15*a^4*b*x^3*sgn(b*x^3 + a) + 2*a^5*sgn(b*x^3 + a)) /x^9
Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^{10}} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x^{10}} \,d x \]