Integrand size = 26, antiderivative size = 231 \[ \int \frac {1}{x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {b^2}{a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {b^2}{6 a^3 \left (a+b x^3\right ) \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a+b x^3}{6 a^3 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {b \left (a+b x^3\right )}{a^4 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {6 b^2 \left (a+b x^3\right ) \log (x)}{a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {2 b^2 \left (a+b x^3\right ) \log \left (a+b x^3\right )}{a^5 \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
b^2/a^4/((b*x^3+a)^2)^(1/2)+1/6*b^2/a^3/(b*x^3+a)/((b*x^3+a)^2)^(1/2)-1/6* (b*x^3+a)/a^3/x^6/((b*x^3+a)^2)^(1/2)+b*(b*x^3+a)/a^4/x^3/((b*x^3+a)^2)^(1 /2)+6*b^2*(b*x^3+a)*ln(x)/a^5/((b*x^3+a)^2)^(1/2)-2*b^2*(b*x^3+a)*ln(b*x^3 +a)/a^5/((b*x^3+a)^2)^(1/2)
Time = 1.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {a \left (-a^3+4 a^2 b x^3+18 a b^2 x^6+12 b^3 x^9\right )+36 b^2 x^6 \left (a+b x^3\right )^2 \log (x)-12 b^2 x^6 \left (a+b x^3\right )^2 \log \left (a+b x^3\right )}{6 a^5 x^6 \left (a+b x^3\right ) \sqrt {\left (a+b x^3\right )^2}} \] Input:
Integrate[1/(x^7*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)),x]
Output:
(a*(-a^3 + 4*a^2*b*x^3 + 18*a*b^2*x^6 + 12*b^3*x^9) + 36*b^2*x^6*(a + b*x^ 3)^2*Log[x] - 12*b^2*x^6*(a + b*x^3)^2*Log[a + b*x^3])/(6*a^5*x^6*(a + b*x ^3)*Sqrt[(a + b*x^3)^2])
Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.51, 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.
\(\displaystyle \int \frac {1}{x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^3\right ) \int \frac {1}{b^3 x^7 \left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^7 \left (b x^3+a\right )^3}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^9 \left (b x^3+a\right )^3}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \left (-\frac {6 b^3}{a^5 \left (b x^3+a\right )}-\frac {3 b^3}{a^4 \left (b x^3+a\right )^2}-\frac {b^3}{a^3 \left (b x^3+a\right )^3}+\frac {6 b^2}{a^5 x^3}-\frac {3 b}{a^4 x^6}+\frac {1}{a^3 x^9}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {6 b^2 \log \left (x^3\right )}{a^5}-\frac {6 b^2 \log \left (a+b x^3\right )}{a^5}+\frac {3 b^2}{a^4 \left (a+b x^3\right )}+\frac {3 b}{a^4 x^3}+\frac {b^2}{2 a^3 \left (a+b x^3\right )^2}-\frac {1}{2 a^3 x^6}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[1/(x^7*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)),x]
Output:
((a + b*x^3)*(-1/2*1/(a^3*x^6) + (3*b)/(a^4*x^3) + b^2/(2*a^3*(a + b*x^3)^ 2) + (3*b^2)/(a^4*(a + b*x^3)) + (6*b^2*Log[x^3])/a^5 - (6*b^2*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 = 0.13 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.45
method | result | size |
pseudoelliptic | \(-\frac {\left (12 b^{2} x^{6} \left (b \,x^{3}+a \right )^{2} \ln \left (b \,x^{3}+a \right )-12 b^{2} x^{6} \left (b \,x^{3}+a \right )^{2} \ln \left (b \,x^{3}\right )+a \left (2 b \,x^{3}+a \right ) \left (-6 b^{2} x^{6}-6 a \,x^{3} b +a^{2}\right )\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{6 \left (b \,x^{3}+a \right )^{2} a^{5} x^{6}}\) | \(104\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\frac {2 b^{3} x^{9}}{a^{4}}+\frac {3 b^{2} x^{6}}{a^{3}}+\frac {2 b \,x^{3}}{3 a^{2}}-\frac {1}{6 a}\right )}{\left (b \,x^{3}+a \right )^{3} x^{6}}+\frac {6 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{2} \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{5}}-\frac {2 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{2} \ln \left (b \,x^{3}+a \right )}{\left (b \,x^{3}+a \right ) a^{5}}\) | \(129\) |
default | \(-\frac {\left (12 \ln \left (b \,x^{3}+a \right ) b^{4} x^{12}-36 \ln \left (x \right ) b^{4} x^{12}+24 \ln \left (b \,x^{3}+a \right ) a \,b^{3} x^{9}-72 \ln \left (x \right ) a \,b^{3} x^{9}-12 a \,b^{3} x^{9}+12 \ln \left (b \,x^{3}+a \right ) a^{2} b^{2} x^{6}-36 \ln \left (x \right ) a^{2} b^{2} x^{6}-18 b^{2} x^{6} a^{2}-4 b \,x^{3} a^{3}+a^{4}\right ) \left (b \,x^{3}+a \right )}{6 x^{6} a^{5} {\left (\left (b \,x^{3}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(146\) |
Input:
int(1/x^7/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/6*(12*b^2*x^6*(b*x^3+a)^2*ln(b*x^3+a)-12*b^2*x^6*(b*x^3+a)^2*ln(b*x^3)+ a*(2*b*x^3+a)*(-6*b^2*x^6-6*a*b*x^3+a^2))*csgn(b*x^3+a)/(b*x^3+a)^2/a^5/x^ 6
Time = 0.07 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.58 \[ \int \frac {1}{x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {12 \, a b^{3} x^{9} + 18 \, a^{2} b^{2} x^{6} + 4 \, a^{3} b x^{3} - a^{4} - 12 \, {\left (b^{4} x^{12} + 2 \, a b^{3} x^{9} + a^{2} b^{2} x^{6}\right )} \log \left (b x^{3} + a\right ) + 36 \, {\left (b^{4} x^{12} + 2 \, a b^{3} x^{9} + a^{2} b^{2} x^{6}\right )} \log \left (x\right )}{6 \, {\left (a^{5} b^{2} x^{12} + 2 \, a^{6} b x^{9} + a^{7} x^{6}\right )}} \] Input:
integrate(1/x^7/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")
Output:
1/6*(12*a*b^3*x^9 + 18*a^2*b^2*x^6 + 4*a^3*b*x^3 - a^4 - 12*(b^4*x^12 + 2* a*b^3*x^9 + a^2*b^2*x^6)*log(b*x^3 + a) + 36*(b^4*x^12 + 2*a*b^3*x^9 + a^2 *b^2*x^6)*log(x))/(a^5*b^2*x^12 + 2*a^6*b*x^9 + a^7*x^6)
\[ \int \frac {1}{x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {1}{x^{7} \left (\left (a + b x^{3}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/x**7/(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)
Output:
Integral(1/(x**7*((a + b*x**3)**2)**(3/2)), x)
Time = 0.04 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {2 \, \left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{a^{5}} + \frac {2 \, b^{2}}{\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{4}} + \frac {1}{6 \, {\left (x^{3} + \frac {a}{b}\right )}^{2} a^{3}} + \frac {5 \, b}{6 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{3} x^{3}} - \frac {1}{6 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} a^{2} x^{6}} \] Input:
integrate(1/x^7/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")
Output:
-2*(-1)^(2*a*b*x^3 + 2*a^2)*b^2*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x)))/a ^5 + 2*b^2/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^4) + 1/6/((x^3 + a/b)^2*a^3) + 5/6*b/(sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*a^3*x^3) - 1/6/(sqrt(b^2*x^6 + 2 *a*b*x^3 + a^2)*a^2*x^6)
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=-\frac {2 \, b^{2} \log \left ({\left | b x^{3} + a \right |}\right )}{a^{5} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {6 \, b^{2} \log \left ({\left | x \right |}\right )}{a^{5} \mathrm {sgn}\left (b x^{3} + a\right )} + \frac {12 \, b^{3} x^{9} + 18 \, a b^{2} x^{6} + 4 \, a^{2} b x^{3} - a^{3}}{6 \, {\left (b x^{6} + a x^{3}\right )}^{2} a^{4} \mathrm {sgn}\left (b x^{3} + a\right )} \] Input:
integrate(1/x^7/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")
Output:
-2*b^2*log(abs(b*x^3 + a))/(a^5*sgn(b*x^3 + a)) + 6*b^2*log(abs(x))/(a^5*s gn(b*x^3 + a)) + 1/6*(12*b^3*x^9 + 18*a*b^2*x^6 + 4*a^2*b*x^3 - a^3)/((b*x ^6 + a*x^3)^2*a^4*sgn(b*x^3 + a))
Timed out. \[ \int \frac {1}{x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\int \frac {1}{x^7\,{\left (a^2+2\,a\,b\,x^3+b^2\,x^6\right )}^{3/2}} \,d x \] Input:
int(1/(x^7*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)),x)
Output:
int(1/(x^7*(a^2 + b^2*x^6 + 2*a*b*x^3)^(3/2)), x)
Time = 0.19 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^7 \left (a^2+2 a b x^3+b^2 x^6\right )^{3/2}} \, dx=\frac {-12 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a^{2} b^{2} x^{6}-24 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a \,b^{3} x^{9}-12 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b^{4} x^{12}-12 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} b^{2} x^{6}-24 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a \,b^{3} x^{9}-12 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{4} x^{12}+36 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{6}+72 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{9}+36 \,\mathrm {log}\left (x \right ) b^{4} x^{12}-a^{4}+4 a^{3} b \,x^{3}+12 a^{2} b^{2} x^{6}-6 b^{4} x^{12}}{6 a^{5} x^{6} \left (b^{2} x^{6}+2 a b \,x^{3}+a^{2}\right )} \] Input:
int(1/x^7/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x)
Output:
( - 12*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a**2*b**2*x**6 - 24*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**2*b**2*x**6 - 24*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**2*b**2*x** 6 + 72*log(x)*a*b**3*x**9 + 36*log(x)*b**4*x**12 - a**4 + 4*a**3*b*x**3 + 12*a**2*b**2*x**6 - 6*b**4*x**12)/(6*a**5*x**6*(a**2 + 2*a*b*x**3 + b**2*x **6))