Integrand size = 26, antiderivative size = 252 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^4} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 x^3 \left (a+b x^3\right )}+\frac {10 a^3 b^2 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {5 a^2 b^3 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}{3 \left (a+b x^3\right )}+\frac {5 a b^4 x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}{9 \left (a+b x^3\right )}+\frac {b^5 x^{12} \sqrt {a^2+2 a b x^3+b^2 x^6}}{12 \left (a+b x^3\right )}+\frac {5 a^4 b \sqrt {a^2+2 a b x^3+b^2 x^6} \log (x)}{a+b x^3} \]
-1/3*a^5*((b*x^3+a)^2)^(1/2)/x^3/(b*x^3+a)+10/3*a^3*b^2*x^3*((b*x^3+a)^2)^ (1/2)/(b*x^3+a)+5/3*a^2*b^3*x^6*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+5/9*a*b^4*x^ 9*((b*x^3+a)^2)^(1/2)/(b*x^3+a)+1/12*b^5*x^12*((b*x^3+a)^2)^(1/2)/(b*x^3+a )+5*a^4*b*ln(x)*((b*x^3+a)^2)^(1/2)/(b*x^3+a)
Time = 1.02 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.34 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^4} \, dx=\frac {\sqrt {\left (a+b x^3\right )^2} \left (-12 a^5+120 a^3 b^2 x^6+60 a^2 b^3 x^9+20 a b^4 x^{12}+3 b^5 x^{15}+180 a^4 b x^3 \log (x)\right )}{36 x^3 \left (a+b x^3\right )} \]
(Sqrt[(a + b*x^3)^2]*(-12*a^5 + 120*a^3*b^2*x^6 + 60*a^2*b^3*x^9 + 20*a*b^ 4*x^12 + 3*b^5*x^15 + 180*a^4*b*x^3*Log[x]))/(36*x^3*(a + b*x^3))
Time = 0.24 (sec) , antiderivative size = 97, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^4} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {b^5 \left (b x^3+a\right )^5}{x^4}dx}{b^5 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^4}dx}{a+b x^3}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \frac {\left (b x^3+a\right )^5}{x^6}dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \int \left (b^5 x^9+5 a b^4 x^6+10 a^2 b^3 x^3+10 a^3 b^2+\frac {5 a^4 b}{x^3}+\frac {a^5}{x^6}\right )dx^3}{3 \left (a+b x^3\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^3+b^2 x^6} \left (-\frac {a^5}{x^3}+5 a^4 b \log \left (x^3\right )+10 a^3 b^2 x^3+5 a^2 b^3 x^6+\frac {5}{3} a b^4 x^9+\frac {b^5 x^{12}}{4}\right )}{3 \left (a+b x^3\right )}\) |
(Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]*(-(a^5/x^3) + 10*a^3*b^2*x^3 + 5*a^2*b^3* x^6 + (5*a*b^4*x^9)/3 + (b^5*x^12)/4 + 5*a^4*b*Log[x^3]))/(3*(a + b*x^3))
3.1.67.3.1 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 = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.32
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (-\frac {b^{5} x^{15}}{4}-\frac {5 a \,b^{4} x^{12}}{3}-5 a^{2} b^{3} x^{9}-10 a^{3} b^{2} x^{6}-5 \ln \left (b \,x^{3}\right ) a^{4} b \,x^{3}-\frac {77 a^{4} b \,x^{3}}{12}+a^{5}\right )}{3 x^{3}}\) | \(81\) |
default | \(\frac {{\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}} \left (3 b^{5} x^{15}+20 a \,b^{4} x^{12}+60 a^{2} b^{3} x^{9}+120 a^{3} b^{2} x^{6}+180 b \,a^{4} \ln \left (x \right ) x^{3}-12 a^{5}\right )}{36 x^{3} \left (b \,x^{3}+a \right )^{5}}\) | \(82\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{2} \left (\frac {1}{12} b^{3} x^{12}+\frac {5}{9} a \,b^{2} x^{9}+\frac {5}{3} a^{2} b \,x^{6}+\frac {10}{3} a^{3} x^{3}\right )}{b \,x^{3}+a}-\frac {a^{5} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{3 x^{3} \left (b \,x^{3}+a \right )}+\frac {5 a^{4} b \ln \left (x \right ) \sqrt {\left (b \,x^{3}+a \right )^{2}}}{b \,x^{3}+a}\) | \(117\) |
-1/3*csgn(b*x^3+a)*(-1/4*b^5*x^15-5/3*a*b^4*x^12-5*a^2*b^3*x^9-10*a^3*b^2* x^6-5*ln(b*x^3)*a^4*b*x^3-77/12*a^4*b*x^3+a^5)/x^3
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^4} \, dx=\frac {3 \, b^{5} x^{15} + 20 \, a b^{4} x^{12} + 60 \, a^{2} b^{3} x^{9} + 120 \, a^{3} b^{2} x^{6} + 180 \, a^{4} b x^{3} \log \left (x\right ) - 12 \, a^{5}}{36 \, x^{3}} \]
1/36*(3*b^5*x^15 + 20*a*b^4*x^12 + 60*a^2*b^3*x^9 + 120*a^3*b^2*x^6 + 180* a^4*b*x^3*log(x) - 12*a^5)/x^3
\[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^4} \, dx=\int \frac {\left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}{x^{4}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^4} \, dx=\frac {5}{6} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2} b^{2} x^{3} + \frac {5}{3} \, \left (-1\right )^{2 \, b^{2} x^{3} + 2 \, a b} a^{4} b \log \left (2 \, b^{2} x^{3} + 2 \, a b\right ) - \frac {5}{3} \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} a^{4} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right ) + \frac {5}{12} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} b^{2} x^{3} + \frac {5}{2} \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{3} b + \frac {35}{36} \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a b - \frac {{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {5}{2}}}{3 \, x^{3}} \]
5/6*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^2*b^2*x^3 + 5/3*(-1)^(2*b^2*x^3 + 2* a*b)*a^4*b*log(2*b^2*x^3 + 2*a*b) - 5/3*(-1)^(2*a*b*x^3 + 2*a^2)*a^4*b*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^2*x^3 + 5/2*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^3*b + 35/36*(b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a*b - 1/3*(b^2*x^6 + 2*a*b*x^3 + a^2)^(5/2)/x^3
Time = 0.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.49 \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^4} \, dx=\frac {1}{12} \, b^{5} x^{12} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{9} \, a b^{4} x^{9} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {5}{3} \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{3} + a\right ) + \frac {10}{3} \, a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + 5 \, a^{4} b \log \left ({\left | x \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right ) - \frac {5 \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{3} + a\right ) + a^{5} \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, x^{3}} \]
1/12*b^5*x^12*sgn(b*x^3 + a) + 5/9*a*b^4*x^9*sgn(b*x^3 + a) + 5/3*a^2*b^3* x^6*sgn(b*x^3 + a) + 10/3*a^3*b^2*x^3*sgn(b*x^3 + a) + 5*a^4*b*log(abs(x)) *sgn(b*x^3 + a) - 1/3*(5*a^4*b*x^3*sgn(b*x^3 + a) + a^5*sgn(b*x^3 + a))/x^ 3
Timed out. \[ \int \frac {\left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}}{x^4} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}}{x^4} \,d x \]