Integrand size = 26, antiderivative size = 208 \[ \int \frac {1}{x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {a+b x^3}{9 a x^9 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {b \left (a+b x^3\right )}{6 a^2 x^6 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b^2 \left (a+b x^3\right )}{3 a^3 x^3 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {b^3 \left (a+b x^3\right ) \log (x)}{a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {b^3 \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^4 \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:
-1/9*(b*x^3+a)/a/x^9/((b*x^3+a)^2)^(1/2)+1/6*b*(b*x^3+a)/a^2/x^6/((b*x^3+a )^2)^(1/2)-1/3*b^2*(b*x^3+a)/a^3/x^3/((b*x^3+a)^2)^(1/2)-b^3*(b*x^3+a)*ln( x)/a^4/((b*x^3+a)^2)^(1/2)+1/3*b^3*(b*x^3+a)*ln(b*x^3+a)/a^4/((b*x^3+a)^2) ^(1/2)
Time = 0.74 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {1}{36} \left (\frac {\sqrt {\left (a+b x^3\right )^2} \left (-2 a^2+5 a b x^3-11 b^2 x^6\right )}{a^4 x^9}+\frac {2 a^2-3 a b x^3+6 b^2 x^6}{\left (a^2\right )^{3/2} x^9}+\frac {12 \sqrt {a^2} b^3 \log \left (x^3\right )}{a^5}-\frac {6 \left (-a+\sqrt {a^2}\right ) b^3 \log \left (\sqrt {a^2}-b x^3-\sqrt {\left (a+b x^3\right )^2}\right )}{a^5}-\frac {6 \left (a+\sqrt {a^2}\right ) b^3 \log \left (\sqrt {a^2}+b x^3-\sqrt {\left (a+b x^3\right )^2}\right )}{a^5}\right ) \] Input:
Integrate[1/(x^10*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]),x]
Output:
((Sqrt[(a + b*x^3)^2]*(-2*a^2 + 5*a*b*x^3 - 11*b^2*x^6))/(a^4*x^9) + (2*a^ 2 - 3*a*b*x^3 + 6*b^2*x^6)/((a^2)^(3/2)*x^9) + (12*Sqrt[a^2]*b^3*Log[x^3]) /a^5 - (6*(-a + Sqrt[a^2])*b^3*Log[Sqrt[a^2] - b*x^3 - Sqrt[(a + b*x^3)^2] ])/a^5 - (6*(a + Sqrt[a^2])*b^3*Log[Sqrt[a^2] + b*x^3 - Sqrt[(a + b*x^3)^2 ]])/a^5)/36
Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.45, 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.
\(\displaystyle \int \frac {1}{x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b \left (a+b x^3\right ) \int \frac {1}{b x^{10} \left (b x^3+a\right )}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^{10} \left (b x^3+a\right )}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {1}{x^{12} \left (b x^3+a\right )}dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\left (a+b x^3\right ) \int \left (\frac {b^4}{a^4 \left (b x^3+a\right )}-\frac {b^3}{a^4 x^3}+\frac {b^2}{a^3 x^6}-\frac {b}{a^2 x^9}+\frac {1}{a x^{12}}\right )dx^3}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^3\right ) \left (-\frac {b^3 \log \left (x^3\right )}{a^4}+\frac {b^3 \log \left (a+b x^3\right )}{a^4}-\frac {b^2}{a^3 x^3}+\frac {b}{2 a^2 x^6}-\frac {1}{3 a x^9}\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}}\) |
Input:
Int[1/(x^10*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]),x]
Output:
((a + b*x^3)*(-1/3*1/(a*x^9) + b/(2*a^2*x^6) - b^2/(a^3*x^3) - (b^3*Log[x^ 3])/a^4 + (b^3*Log[a + b*x^3])/a^4))/(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 = 71, normalized size of antiderivative = 0.34
method | result | size |
pseudoelliptic | \(\frac {\left (6 b^{3} \ln \left (b \,x^{3}+a \right ) x^{9}-6 \ln \left (b \,x^{3}\right ) b^{3} x^{9}-6 a \,x^{6} b^{2}+3 a^{2} x^{3} b -2 a^{3}\right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{18 a^{4} x^{9}}\) | \(71\) |
default | \(\frac {\left (b \,x^{3}+a \right ) \left (6 b^{3} \ln \left (b \,x^{3}+a \right ) x^{9}-18 b^{3} \ln \left (x \right ) x^{9}-6 a \,x^{6} b^{2}+3 a^{2} x^{3} b -2 a^{3}\right )}{18 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, a^{4} x^{9}}\) | \(77\) |
risch | \(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (-\frac {b^{2} x^{6}}{3 a^{3}}+\frac {b \,x^{3}}{6 a^{2}}-\frac {1}{9 a}\right )}{\left (b \,x^{3}+a \right ) x^{9}}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{3} \ln \left (x \right )}{\left (b \,x^{3}+a \right ) a^{4}}+\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{3} \ln \left (-b \,x^{3}-a \right )}{3 \left (b \,x^{3}+a \right ) a^{4}}\) | \(121\) |
Input:
int(1/x^10/((b*x^3+a)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/18*(6*b^3*ln(b*x^3+a)*x^9-6*ln(b*x^3)*b^3*x^9-6*a*x^6*b^2+3*a^2*x^3*b-2* a^3)*csgn(b*x^3+a)/a^4/x^9
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.28 \[ \int \frac {1}{x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {6 \, b^{3} x^{9} \log \left (b x^{3} + a\right ) - 18 \, b^{3} x^{9} \log \left (x\right ) - 6 \, a b^{2} x^{6} + 3 \, a^{2} b x^{3} - 2 \, a^{3}}{18 \, a^{4} x^{9}} \] Input:
integrate(1/x^10/((b*x^3+a)^2)^(1/2),x, algorithm="fricas")
Output:
1/18*(6*b^3*x^9*log(b*x^3 + a) - 18*b^3*x^9*log(x) - 6*a*b^2*x^6 + 3*a^2*b *x^3 - 2*a^3)/(a^4*x^9)
Timed out. \[ \int \frac {1}{x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\text {Timed out} \] Input:
integrate(1/x**10/((b*x**3+a)**2)**(1/2),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\left (-1\right )^{2 \, a b x^{3} + 2 \, a^{2}} b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{x^{2} {\left | x \right |}}\right )}{3 \, a^{4}} - \frac {11 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b^{2}}{18 \, a^{4} x^{3}} + \frac {5 \, \sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}} b}{18 \, a^{3} x^{6}} - \frac {\sqrt {b^{2} x^{6} + 2 \, a b x^{3} + a^{2}}}{9 \, a^{2} x^{9}} \] Input:
integrate(1/x^10/((b*x^3+a)^2)^(1/2),x, algorithm="maxima")
Output:
1/3*(-1)^(2*a*b*x^3 + 2*a^2)*b^3*log(2*a*b*x/abs(x) + 2*a^2/(x^2*abs(x)))/ a^4 - 11/18*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)*b^2/(a^4*x^3) + 5/18*sqrt(b^2* x^6 + 2*a*b*x^3 + a^2)*b/(a^3*x^6) - 1/9*sqrt(b^2*x^6 + 2*a*b*x^3 + a^2)/( a^2*x^9)
Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.38 \[ \int \frac {1}{x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {1}{18} \, {\left (\frac {6 \, b^{3} \log \left ({\left | b x^{3} + a \right |}\right )}{a^{4}} - \frac {18 \, b^{3} \log \left ({\left | x \right |}\right )}{a^{4}} + \frac {11 \, b^{3} x^{9} - 6 \, a b^{2} x^{6} + 3 \, a^{2} b x^{3} - 2 \, a^{3}}{a^{4} x^{9}}\right )} \mathrm {sgn}\left (b x^{3} + a\right ) \] Input:
integrate(1/x^10/((b*x^3+a)^2)^(1/2),x, algorithm="giac")
Output:
1/18*(6*b^3*log(abs(b*x^3 + a))/a^4 - 18*b^3*log(abs(x))/a^4 + (11*b^3*x^9 - 6*a*b^2*x^6 + 3*a^2*b*x^3 - 2*a^3)/(a^4*x^9))*sgn(b*x^3 + a)
Timed out. \[ \int \frac {1}{x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {1}{x^{10}\,\sqrt {{\left (b\,x^3+a\right )}^2}} \,d x \] Input:
int(1/(x^10*((a + b*x^3)^2)^(1/2)),x)
Output:
int(1/(x^10*((a + b*x^3)^2)^(1/2)), x)
Time = 0.17 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x^{10} \sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {6 \,\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) b^{3} x^{9}+6 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) b^{3} x^{9}-18 \,\mathrm {log}\left (x \right ) b^{3} x^{9}-2 a^{3}+3 a^{2} b \,x^{3}-6 a \,b^{2} x^{6}}{18 a^{4} x^{9}} \] Input:
int(1/x^10/((b*x^3+a)^2)^(1/2),x)
Output:
(6*log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*b**3*x**9 + 6*log(a **(1/3) + b**(1/3)*x)*b**3*x**9 - 18*log(x)*b**3*x**9 - 2*a**3 + 3*a**2*b* x**3 - 6*a*b**2*x**6)/(18*a**4*x**9)