Integrand size = 26, antiderivative size = 223 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {1}{3 a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{12 a \left (a+b x^3\right )^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{9 a^2 \left (a+b x^3\right )^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {1}{6 a^3 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\left (a+b x^3\right ) \log (x)}{a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
1/3/a^4/((b*x^3+a)^2)^(1/2)+1/12/a/(b*x^3+a)^3/((b*x^3+a)^2)^(1/2)+1/9/a^2 /(b*x^3+a)^2/((b*x^3+a)^2)^(1/2)+1/6/a^3/(b*x^3+a)/((b*x^3+a)^2)^(1/2)+(b* x^3+a)*ln(x)/a^5/((b*x^3+a)^2)^(1/2)-1/3*(b*x^3+a)*ln(b*x^3+a)/a^5/((b*x^3 +a)^2)^(1/2)
Time = 1.06 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {a \left (25 a^3+52 a^2 b x^3+42 a b^2 x^6+12 b^3 x^9\right )+36 \left (a+b x^3\right )^4 \log (x)-12 \left (a+b x^3\right )^4 \log \left (a+b x^3\right )}{36 a^5 \left (a+b x^3\right )^3 \sqrt {\left (a+b x^3\right )^2}} \] Input:
Integrate[1/(x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)),x]
Output:
(a*(25*a^3 + 52*a^2*b*x^3 + 42*a*b^2*x^6 + 12*b^3*x^9) + 36*(a + b*x^3)^4* Log[x] - 12*(a + b*x^3)^4*Log[a + b*x^3])/(36*a^5*(a + b*x^3)^3*Sqrt[(a + b*x^3)^2])
Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.52, 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.
| \(\displaystyle \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1384 |
| \(\displaystyle \frac {b^5 \left (a+b x^3\right ) \int \frac {1}{b^5 x \left (b x^3+a\right )^5}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x \left (b x^3+a\right )^5}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^3 \left (b x^3+a\right )^5}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {\left (a+b x^3\right ) \int \left (-\frac {b}{a^5 \left (b x^3+a\right )}-\frac {b}{a^4 \left (b x^3+a\right )^2}-\frac {b}{a^3 \left (b x^3+a\right )^3}-\frac {b}{a^2 \left (b x^3+a\right )^4}-\frac {b}{a \left (b x^3+a\right )^5}+\frac {1}{a^5 x^3}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (a+b x^3\right ) \left (-\frac {\log \left (a+b x^3\right )}{a^5}+\frac {\log \left (x^3\right )}{a^5}+\frac {1}{a^4 \left (a+b x^3\right )}+\frac {1}{2 a^3 \left (a+b x^3\right )^2}+\frac {1}{3 a^2 \left (a+b x^3\right )^3}+\frac {1}{4 a \left (a+b x^3\right )^4}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[1/(x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(5/2)),x]
Output:
((a + b*x^3)*(1/(4*a*(a + b*x^3)^4) + 1/(3*a^2*(a + b*x^3)^3) + 1/(2*a^3*( a + b*x^3)^2) + 1/(a^4*(a + b*x^3)) + Log[x^3]/a^5 - Log[a + b*x^3]/a^5))/ (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 = 0.12 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.41
| method | result | size |
| pseudoelliptic | \(\frac {\operatorname {csgn}\left (b \,x^{3}+a \right ) \left (-\ln \left (b \,x^{3}+a \right ) \left (b \,x^{3}+a \right )^{4}+\ln \left (b \,x^{3}\right ) \left (b \,x^{3}+a \right )^{4}+a \,b^{3} x^{9}+\frac {7 b^{2} x^{6} a^{2}}{2}+\frac {13 b \,x^{3} a^{3}}{3}+\frac {25 a^{4}}{12}\right )}{3 \left (b \,x^{3}+a \right )^{4} a^{5}}\) | \(92\) |
| risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\frac {b^{3} x^{9}}{3 a^{4}}+\frac {7 b^{2} x^{6}}{6 a^{3}}+\frac {13 b \,x^{3}}{9 a^{2}}+\frac {25}{36 a}\right )}{\left (b \,x^{3}+a \right )^{5}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{5}}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right ) a^{5}}\) | \(119\) |
| default | \(-\frac {\left (12 \ln \left (b \,x^{3}+a \right ) b^{4} x^{12}-36 \ln \left (x \right ) b^{4} x^{12}+48 \ln \left (b \,x^{3}+a \right ) a \,b^{3} x^{9}-144 \ln \left (x \right ) a \,b^{3} x^{9}-12 a \,b^{3} x^{9}+72 \ln \left (b \,x^{3}+a \right ) a^{2} b^{2} x^{6}-216 \ln \left (x \right ) a^{2} b^{2} x^{6}-42 b^{2} x^{6} a^{2}+48 \ln \left (b \,x^{3}+a \right ) a^{3} b \,x^{3}-144 \ln \left (x \right ) a^{3} b \,x^{3}-52 b \,x^{3} a^{3}+12 \ln \left (b \,x^{3}+a \right ) a^{4}-36 \ln \left (x \right ) a^{4}-25 a^{4}\right ) \left (b \,x^{3}+a \right )}{36 a^{5} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(193\) |
Input:
int(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3*csgn(b*x^3+a)*(-ln(b*x^3+a)*(b*x^3+a)^4+ln(b*x^3)*(b*x^3+a)^4+a*b^3*x^ 9+7/2*b^2*x^6*a^2+13/3*b*x^3*a^3+25/12*a^4)/(b*x^3+a)^4/a^5
Time = 0.07 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {12 \, a b^{3} x^{9} + 42 \, a^{2} b^{2} x^{6} + 52 \, a^{3} b x^{3} + 25 \, a^{4} - 12 \, {\left (b^{4} x^{12} + 4 \, a b^{3} x^{9} + 6 \, a^{2} b^{2} x^{6} + 4 \, a^{3} b x^{3} + a^{4}\right )} \log \left (b x^{3} + a\right ) + 36 \, {\left (b^{4} x^{12} + 4 \, a b^{3} x^{9} + 6 \, a^{2} b^{2} x^{6} + 4 \, a^{3} b x^{3} + a^{4}\right )} \log \left (x\right )}{36 \, {\left (a^{5} b^{4} x^{12} + 4 \, a^{6} b^{3} x^{9} + 6 \, a^{7} b^{2} x^{6} + 4 \, a^{8} b x^{3} + a^{9}\right )}} \] Input:
integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="fricas")
Output:
1/36*(12*a*b^3*x^9 + 42*a^2*b^2*x^6 + 52*a^3*b*x^3 + 25*a^4 - 12*(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)*log(b*x^3 + a) + 36*(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)*log(x))/(a^5*b^ 4*x^12 + 4*a^6*b^3*x^9 + 6*a^7*b^2*x^6 + 4*a^8*b*x^3 + a^9)
\[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {1}{x \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/x/(b**2*x**6+2*a*b*x**3+a**2)**(5/2),x)
Output:
Integral(1/(x*((a + b*x**3)**2)**(5/2)), x)
Time = 0.05 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{3 \, a^{5}} + \frac {1}{9 \, {\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )}^{\frac {3}{2}} a^{2}} + \frac {1}{3 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{4}} + \frac {1}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} a^{3} b^{2}} + \frac {1}{12 \, {\left (x^{3} + \frac {a}{b}\right )}^{4} a b^{4}} \] Input:
integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="maxima")
Output:
-1/3*(-1)^(2*a*b*x^3 + 2*a^2)*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x)))/a^5 + 1/9/((b^2*x^6 + 2*a*b*x^3 + a^2)^(3/2)*a^2) + 1/3/(sqrt(b^2*x^6 + 2*a*b *x^3 + a^2)*a^4) + 1/6/((x^3 + a/b)^2*a^3*b^2) + 1/12/((x^3 + a/b)^4*a*b^4 )
Time = 0.14 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=-\frac {\log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{5} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {\log \left ({\left | x \right |}\right )}{a^{5} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {25 \, b^{4} x^{12} + 112 \, a b^{3} x^{9} + 192 \, a^{2} b^{2} x^{6} + 152 \, a^{3} b x^{3} + 50 \, a^{4}}{36 \, {\left (b x^{3} + a\right )}^{4} a^{5} \mathrm {sgn}\left (b x^{3} + a\right )} \] Input:
integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x, algorithm="giac")
Output:
-1/3*log(abs(b*x^3 + a))/(a^5*sgn(b*x^3 + a)) + log(abs(x))/(a^5*sgn(b*x^3 + a)) + 1/36*(25*b^4*x^12 + 112*a*b^3*x^9 + 192*a^2*b^2*x^6 + 152*a^3*b*x ^3 + 50*a^4)/((b*x^3 + a)^4*a^5*sgn(b*x^3 + a))
Timed out. \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\int \frac {1}{x\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{5/2}} \,d x \] Input:
int(1/(x*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)),x)
Output:
int(1/(x*(a^2 + b^2*x^6 + 2*a*b*x^3)^(5/2)), x)
Time = 0.18 (sec) , antiderivative size = 372, normalized size of antiderivative = 1.67 \[ \int \frac {1}{x \left (a^2+2 a b x^3+b^2 x^6\right )^{5/2}} \, dx=\frac {-12 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{4}-48 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{3} b \,x^{3}-72 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2} b^{2} x^{6}-48 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a \,b^{3} x^{9}-12 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b^{4} x^{12}-12 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{4}-48 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{3} b \,x^{3}-72 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} b^{2} x^{6}-48 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a \,b^{3} x^{9}-12 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{4} x^{12}+36 \,\mathrm {log}\left (x \right ) a^{4}+144 \,\mathrm {log}\left (x \right ) a^{3} b \,x^{3}+216 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{6}+144 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{9}+36 \,\mathrm {log}\left (x \right ) b^{4} x^{12}+22 a^{4}+40 a^{3} b \,x^{3}+24 a^{2} b^{2} x^{6}-3 b^{4} x^{12}}{36 a^{5} \left (b^{4} x^{12}+4 a \,b^{3} x^{9}+6 a^{2} b^{2} x^{6}+4 a^{3} b \,x^{3}+a^{4}\right )} \] Input:
int(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(5/2),x)
Output:
( - 12*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**4 - 48*log(a **(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**3*b*x**3 - 72*log(a**(2/ 3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**2*x**6 - 48*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a*b**3*x**9 - 12*log(a**(2/3) - b* *(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**4*x**12 - 12*log(a**(1/3) + b**(1/3) *x)*a**4 - 48*log(a**(1/3) + b**(1/3)*x)*a**3*b*x**3 - 72*log(a**(1/3) + b **(1/3)*x)*a**2*b**2*x**6 - 48*log(a**(1/3) + b**(1/3)*x)*a*b**3*x**9 - 12 *log(a**(1/3) + b**(1/3)*x)*b**4*x**12 + 36*log(x)*a**4 + 144*log(x)*a**3* b*x**3 + 216*log(x)*a**2*b**2*x**6 + 144*log(x)*a*b**3*x**9 + 36*log(x)*b* *4*x**12 + 22*a**4 + 40*a**3*b*x**3 + 24*a**2*b**2*x**6 - 3*b**4*x**12)/(3 6*a**5*(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))