Integrand size = 26, antiderivative size = 269 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {4 b}{3 a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b}{12 a^2 \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {2 b}{9 a^3 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b}{2 a^4 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a+b x^3}{3 a^5 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {5 b \left (a+b x^3\right ) \log (x)}{a^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {5 b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^6 \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
-4/3*b/a^5/((b*x^3+a)^2)^(1/2)-1/12*b/a^2/(b*x^3+a)^3/((b*x^3+a)^2)^(1/2)- 2/9*b/a^3/(b*x^3+a)^2/((b*x^3+a)^2)^(1/2)-1/2*b/a^4/(b*x^3+a)/((b*x^3+a)^2 )^(1/2)-1/3*(b*x^3+a)/a^5/x^3/((b*x^3+a)^2)^(1/2)-5*b*(b*x^3+a)*ln(x)/a^6/ ((b*x^3+a)^2)^(1/2)+5/3*b*(b*x^3+a)*ln(b*x^3+a)/a^6/((b*x^3+a)^2)^(1/2)
Time = 1.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {-a \left (12 a^4+125 a^3 b x^3+260 a^2 b^2 x^6+210 a b^3 x^9+60 b^4 x^{12}\right )-180 b x^3 \left (a+b x^3\right )^4 \log (x)+60 b x^3 \left (a+b x^3\right )^4 \log \left (a+b x^3\right )}{36 a^6 x^3 \left (a+b x^3\right )^3 \sqrt {\left (a+b x^3\right )^2}} \] Input:
Integrate[1/(x^4*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)),x]
Output:
(-(a*(12*a^4 + 125*a^3*b*x^3 + 260*a^2*b^2*x^6 + 210*a*b^3*x^9 + 60*b^4*x^ 12)) - 180*b*x^3*(a + b*x^3)^4*Log[x] + 60*b*x^3*(a + b*x^3)^4*Log[a + b*x ^3])/(36*a^6*x^3*(a + b*x^3)^3*Sqrt[(a + b*x^3)^2])
Time = 0.30 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.49, 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, 54, 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.
| \(\displaystyle \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1384 |
| \(\displaystyle \frac {b^5 \left (a+b x^3\right ) \int \frac {1}{b^5 x^4 \left (b x^3+a\right )^5}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^4 \left (b x^3+a\right )^5}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^6 \left (b x^3+a\right )^5}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {\left (a+b x^3\right ) \int \left (\frac {5 b^2}{a^6 \left (b x^3+a\right )}+\frac {4 b^2}{a^5 \left (b x^3+a\right )^2}+\frac {3 b^2}{a^4 \left (b x^3+a\right )^3}+\frac {2 b^2}{a^3 \left (b x^3+a\right )^4}+\frac {b^2}{a^2 \left (b x^3+a\right )^5}-\frac {5 b}{a^6 x^3}+\frac {1}{a^5 x^6}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (a+b x^3\right ) \left (-\frac {5 b \log \left (x^3\right )}{a^6}+\frac {5 b \log \left (a+b x^3\right )}{a^6}-\frac {4 b}{a^5 \left (a+b x^3\right )}-\frac {1}{a^5 x^3}-\frac {3 b}{2 a^4 \left (a+b x^3\right )^2}-\frac {2 b}{3 a^3 \left (a+b x^3\right )^3}-\frac {b}{4 a^2 \left (a+b x^3\right )^4}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[1/(x^4*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)),x]
Output:
((a + b*x^3)*(-(1/(a^5*x^3)) - b/(4*a^2*(a + b*x^3)^4) - (2*b)/(3*a^3*(a + b*x^3)^3) - (3*b)/(2*a^4*(a + b*x^3)^2) - (4*b)/(a^5*(a + b*x^3)) - (5*b* Log[x^3])/a^6 + (5*b*Log[a + b*x^3])/a^6))/(3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x ^6])
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[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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 = 0.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.42
| method | result | size |
| pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (-5 b \,x^{3} \left (b \,x^{3}+a \right )^{4} \ln \left (b \,x^{3}+a \right )+5 b \,x^{3} \left (b \,x^{3}+a \right )^{4} \ln \left (b \,x^{3}\right )+a \left (5 b^{4} x^{12}+\frac {35}{2} a \,b^{3} x^{9}+\frac {65}{3} b^{2} x^{6} a^{2}+\frac {125}{12} b \,x^{3} a^{3}+a^{4}\right )\right )}{3 \left (b \,x^{3}+a \right )^{4} a^{6} x^{3}}\) | \(114\) |
| risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {1}{3 a}-\frac {125 b \,x^{3}}{36 a^{2}}-\frac {65 b^{2} x^{6}}{9 a^{3}}-\frac {35 b^{3} x^{9}}{6 a^{4}}-\frac {5 b^{4} x^{12}}{3 a^{5}}\right )}{\left (b \,x^{3}+a \right )^{5} x^{3}}-\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{6}}+\frac {5 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b \ln \left (-b \,x^{3}-a \right )}{3 \left (b \,x^{3}+a \right ) a^{6}}\) | \(139\) |
| default | \(\frac {\left (60 \ln \left (b \,x^{3}+a \right ) b^{5} x^{15}-180 b^{5} \ln \left (x \right ) x^{15}+240 \ln \left (b \,x^{3}+a \right ) a \,b^{4} x^{12}-720 a \,b^{4} \ln \left (x \right ) x^{12}-60 a \,b^{4} x^{12}+360 \ln \left (b \,x^{3}+a \right ) a^{2} b^{3} x^{9}-1080 a^{2} b^{3} \ln \left (x \right ) x^{9}-210 a^{2} b^{3} x^{9}+240 \ln \left (b \,x^{3}+a \right ) a^{3} b^{2} x^{6}-720 a^{3} b^{2} \ln \left (x \right ) x^{6}-260 a^{3} b^{2} x^{6}+60 \ln \left (b \,x^{3}+a \right ) a^{4} b \,x^{3}-180 a^{4} b \ln \left (x \right ) x^{3}-125 a^{4} b \,x^{3}-12 a^{5}\right ) \left (b \,x^{3}+a \right )}{36 x^{3} a^{6} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(219\) |
Input:
int(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/3*csgn(b*x^3+a)*(-5*b*x^3*(b*x^3+a)^4*ln(b*x^3+a)+5*b*x^3*(b*x^3+a)^4*l n(b*x^3)+a*(5*b^4*x^12+35/2*a*b^3*x^9+65/3*b^2*x^6*a^2+125/12*b*x^3*a^3+a^ 4))/(b*x^3+a)^4/a^6/x^3
Time = 0.07 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {60 \, a b^{4} x^{12} + 210 \, a^{2} b^{3} x^{9} + 260 \, a^{3} b^{2} x^{6} + 125 \, a^{4} b x^{3} + 12 \, a^{5} - 60 \, {\left (b^{5} x^{15} + 4 \, a b^{4} x^{12} + 6 \, a^{2} b^{3} x^{9} + 4 \, a^{3} b^{2} x^{6} + a^{4} b x^{3}\right )} \log \left (b x^{3} + a\right ) + 180 \, {\left (b^{5} x^{15} + 4 \, a b^{4} x^{12} + 6 \, a^{2} b^{3} x^{9} + 4 \, a^{3} b^{2} x^{6} + a^{4} b x^{3}\right )} \log \left (x\right )}{36 \, {\left (a^{6} b^{4} x^{15} + 4 \, a^{7} b^{3} x^{12} + 6 \, a^{8} b^{2} x^{9} + 4 \, a^{9} b x^{6} + a^{10} x^{3}\right )}} \] Input:
integrate(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")
Output:
-1/36*(60*a*b^4*x^12 + 210*a^2*b^3*x^9 + 260*a^3*b^2*x^6 + 125*a^4*b*x^3 + 12*a^5 - 60*(b^5*x^15 + 4*a*b^4*x^12 + 6*a^2*b^3*x^9 + 4*a^3*b^2*x^6 + a^ 4*b*x^3)*log(b*x^3 + a) + 180*(b^5*x^15 + 4*a*b^4*x^12 + 6*a^2*b^3*x^9 + 4 *a^3*b^2*x^6 + a^4*b*x^3)*log(x))/(a^6*b^4*x^15 + 4*a^7*b^3*x^12 + 6*a^8*b ^2*x^9 + 4*a^9*b*x^6 + a^10*x^3)
\[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {1}{x^{4} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/x**4/(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)
Output:
Integral(1/(x**4*((a + b*x**3)**2)**(5/2)), x)
Time = 0.05 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {5 \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{3 \, a^{6}} - \frac {5 \, b}{9 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{3}} - \frac {5 \, b}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{5}} - \frac {1}{3 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{2} x^{3}} - \frac {5}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} a^{4} b} - \frac {1}{12 \, {\left (x^{3} + \frac {a}{b}\right )}^{4} a^{2} b^{3}} \] Input:
integrate(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")
Output:
5/3*(-1)^(2*a*b*x^3 + 2*a^2)*b*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x)))/a^ 6 - 5/9*b/((b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a^3) - 5/3*b/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^5) - 1/3/((b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a^2*x^3) - 5/6/((x^3 + a/b)^2*a^4*b) - 1/12/((x^3 + a/b)^4*a^2*b^3)
Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {5 \, b \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{6} \mathrm {sgn}\left (b x^{3} + a\right )} - \frac {5 \, b \log \left ({\left | x \right |}\right )}{a^{6} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {5 \, b x^{3} - a}{3 \, a^{6} x^{3} \mathrm {sgn}\left (b x^{3} + a\right )} - \frac {125 \, b^{5} x^{12} + 548 \, a b^{4} x^{9} + 912 \, a^{2} b^{3} x^{6} + 688 \, a^{3} b^{2} x^{3} + 202 \, a^{4} b}{36 \, {\left (b x^{3} + a\right )}^{4} a^{6} \mathrm {sgn}\left (b x^{3} + a\right )} \] Input:
integrate(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")
Output:
5/3*b*log(abs(b*x^3 + a))/(a^6*sgn(b*x^3 + a)) - 5*b*log(abs(x))/(a^6*sgn( b*x^3 + a)) + 1/3*(5*b*x^3 - a)/(a^6*x^3*sgn(b*x^3 + a)) - 1/36*(125*b^5*x ^12 + 548*a*b^4*x^9 + 912*a^2*b^3*x^6 + 688*a^3*b^2*x^3 + 202*a^4*b)/((b*x ^3 + a)^4*a^6*sgn(b*x^3 + a))
Timed out. \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {1}{x^4\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}} \,d x \] Input:
int(1/(x^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)),x)
Output:
int(1/(x^4*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)), x)
Time = 0.17 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {60 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{4} b \,x^{3}+240 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{3} b^{2} x^{6}+360 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2} b^{3} x^{9}+240 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a \,b^{4} x^{12}+60 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b^{5} x^{15}+60 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{4} b \,x^{3}+240 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{3} b^{2} x^{6}+360 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} b^{3} x^{9}+240 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a \,b^{4} x^{12}+60 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{5} x^{15}-180 \,\mathrm {log}\left (x \right ) a^{4} b \,x^{3}-720 \,\mathrm {log}\left (x \right ) a^{3} b^{2} x^{6}-1080 \,\mathrm {log}\left (x \right ) a^{2} b^{3} x^{9}-720 \,\mathrm {log}\left (x \right ) a \,b^{4} x^{12}-180 \,\mathrm {log}\left (x \right ) b^{5} x^{15}-12 a^{5}-110 a^{4} b \,x^{3}-200 a^{3} b^{2} x^{6}-120 a^{2} b^{3} x^{9}+15 b^{5} x^{15}}{36 a^{6} x^{3} \left (b^{4} x^{12}+4 a \,b^{3} x^{9}+6 a^{2} b^{2} x^{6}+4 a^{3} b \,x^{3}+a^{4}\right )} \] Input:
int(1/x^4/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)
Output:
(60*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**4*b*x**3 + 240* log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b**2*x**6 + 360*l og(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**3*x**9 + 240*lo g(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b**4*x**12 + 60*log(a* *(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**5*x**15 + 60*log(a**(1/3) + b**(1/3)*x)*a**4*b*x**3 + 240*log(a**(1/3) + b**(1/3)*x)*a**3*b**2*x**6 + 360*log(a**(1/3) + b**(1/3)*x)*a**2*b**3*x**9 + 240*log(a**(1/3) + b**( 1/3)*x)*a*b**4*x**12 + 60*log(a**(1/3) + b**(1/3)*x)*b**5*x**15 - 180*log( x)*a**4*b*x**3 - 720*log(x)*a**3*b**2*x**6 - 1080*log(x)*a**2*b**3*x**9 - 720*log(x)*a*b**4*x**12 - 180*log(x)*b**5*x**15 - 12*a**5 - 110*a**4*b*x** 3 - 200*a**3*b**2*x**6 - 120*a**2*b**3*x**9 + 15*b**5*x**15)/(36*a**6*x**3 *(a**4 + 4*a**3*b*x**3 + 6*a**2*b**2*x**6 + 4*a*b**3*x**9 + b**4*x**12))