Integrand size = 15, antiderivative size = 71 \[ \int \frac {1}{x^{10} \sqrt {-1-x^3}} \, dx=\frac {\sqrt {-1-x^3}}{9 x^9}-\frac {5 \sqrt {-1-x^3}}{36 x^6}+\frac {5 \sqrt {-1-x^3}}{24 x^3}-\frac {5}{24} \arctan \left (\sqrt {-1-x^3}\right ) \] Output:
1/9*(-x^3-1)^(1/2)/x^9-5/36*(-x^3-1)^(1/2)/x^6+5/24*(-x^3-1)^(1/2)/x^3-5/2 4*arctan((-x^3-1)^(1/2))
Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.66 \[ \int \frac {1}{x^{10} \sqrt {-1-x^3}} \, dx=\frac {\sqrt {-1-x^3} \left (8-10 x^3+15 x^6\right )}{72 x^9}-\frac {5}{24} \arctan \left (\sqrt {-1-x^3}\right ) \] Input:
Integrate[1/(x^10*Sqrt[-1 - x^3]),x]
Output:
(Sqrt[-1 - x^3]*(8 - 10*x^3 + 15*x^6))/(72*x^9) - (5*ArcTan[Sqrt[-1 - x^3] ])/24
Time = 0.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {798, 52, 52, 52, 73, 217}
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.
\(\displaystyle \int \frac {1}{x^{10} \sqrt {-x^3-1}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^{12} \sqrt {-x^3-1}}dx^3\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (\frac {\sqrt {-x^3-1}}{3 x^9}-\frac {5}{6} \int \frac {1}{x^9 \sqrt {-x^3-1}}dx^3\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (\frac {\sqrt {-x^3-1}}{3 x^9}-\frac {5}{6} \left (\frac {\sqrt {-x^3-1}}{2 x^6}-\frac {3}{4} \int \frac {1}{x^6 \sqrt {-x^3-1}}dx^3\right )\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (\frac {\sqrt {-x^3-1}}{3 x^9}-\frac {5}{6} \left (\frac {\sqrt {-x^3-1}}{2 x^6}-\frac {3}{4} \left (\frac {\sqrt {-x^3-1}}{x^3}-\frac {1}{2} \int \frac {1}{x^3 \sqrt {-x^3-1}}dx^3\right )\right )\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {\sqrt {-x^3-1}}{3 x^9}-\frac {5}{6} \left (\frac {\sqrt {-x^3-1}}{2 x^6}-\frac {3}{4} \left (\int \frac {1}{-x^6-1}d\sqrt {-x^3-1}+\frac {\sqrt {-x^3-1}}{x^3}\right )\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (\frac {\sqrt {-x^3-1}}{3 x^9}-\frac {5}{6} \left (\frac {\sqrt {-x^3-1}}{2 x^6}-\frac {3}{4} \left (\frac {\sqrt {-x^3-1}}{x^3}-\arctan \left (\sqrt {-x^3-1}\right )\right )\right )\right )\) |
Input:
Int[1/(x^10*Sqrt[-1 - x^3]),x]
Output:
(Sqrt[-1 - x^3]/(3*x^9) - (5*(Sqrt[-1 - x^3]/(2*x^6) - (3*(Sqrt[-1 - x^3]/ x^3 - ArcTan[Sqrt[-1 - x^3]]))/4))/6)/3
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 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]]
Time = 1.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.62
method | result | size |
pseudoelliptic | \(\frac {\left (15 x^{6}-10 x^{3}+8\right ) \sqrt {-x^{3}-1}-15 \arctan \left (\sqrt {-x^{3}-1}\right ) x^{9}}{72 x^{9}}\) | \(44\) |
risch | \(-\frac {15 x^{9}+5 x^{6}-2 x^{3}+8}{72 x^{9} \sqrt {-x^{3}-1}}-\frac {5 \arctan \left (\sqrt {-x^{3}-1}\right )}{24}\) | \(45\) |
default | \(\frac {\sqrt {-x^{3}-1}}{9 x^{9}}-\frac {5 \sqrt {-x^{3}-1}}{36 x^{6}}+\frac {5 \sqrt {-x^{3}-1}}{24 x^{3}}-\frac {5 \arctan \left (\sqrt {-x^{3}-1}\right )}{24}\) | \(56\) |
elliptic | \(\frac {\sqrt {-x^{3}-1}}{9 x^{9}}-\frac {5 \sqrt {-x^{3}-1}}{36 x^{6}}+\frac {5 \sqrt {-x^{3}-1}}{24 x^{3}}-\frac {5 \arctan \left (\sqrt {-x^{3}-1}\right )}{24}\) | \(56\) |
trager | \(\frac {\left (15 x^{6}-10 x^{3}+8\right ) \sqrt {-x^{3}-1}}{72 x^{9}}-\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \sqrt {-x^{3}-1}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{3}}\right )}{48}\) | \(72\) |
meijerg | \(-\frac {i \left (-\frac {\sqrt {\pi }}{3 x^{9}}+\frac {\sqrt {\pi }}{4 x^{6}}-\frac {3 \sqrt {\pi }}{8 x^{3}}-\frac {5 \left (\frac {37}{30}-2 \ln \left (2\right )+3 \ln \left (x \right )\right ) \sqrt {\pi }}{16}+\frac {\sqrt {\pi }\, \left (148 x^{9}+144 x^{6}-96 x^{3}+128\right )}{384 x^{9}}-\frac {\sqrt {\pi }\, \left (240 x^{6}-160 x^{3}+128\right ) \sqrt {x^{3}+1}}{384 x^{9}}+\frac {5 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {x^{3}+1}}{2}\right )}{8}\right )}{3 \sqrt {\pi }}\) | \(116\) |
Input:
int(1/x^10/(-x^3-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/72*((15*x^6-10*x^3+8)*(-x^3-1)^(1/2)-15*arctan((-x^3-1)^(1/2))*x^9)/x^9
Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^{10} \sqrt {-1-x^3}} \, dx=-\frac {15 \, x^{9} \arctan \left (\sqrt {-x^{3} - 1}\right ) - {\left (15 \, x^{6} - 10 \, x^{3} + 8\right )} \sqrt {-x^{3} - 1}}{72 \, x^{9}} \] Input:
integrate(1/x^10/(-x^3-1)^(1/2),x, algorithm="fricas")
Output:
-1/72*(15*x^9*arctan(sqrt(-x^3 - 1)) - (15*x^6 - 10*x^3 + 8)*sqrt(-x^3 - 1 ))/x^9
Result contains complex when optimal does not.
Time = 7.68 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^{10} \sqrt {-1-x^3}} \, dx=- \frac {5 i \operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{24} + \frac {5 i}{24 x^{\frac {3}{2}} \sqrt {1 + \frac {1}{x^{3}}}} + \frac {5 i}{72 x^{\frac {9}{2}} \sqrt {1 + \frac {1}{x^{3}}}} - \frac {i}{36 x^{\frac {15}{2}} \sqrt {1 + \frac {1}{x^{3}}}} + \frac {i}{9 x^{\frac {21}{2}} \sqrt {1 + \frac {1}{x^{3}}}} \] Input:
integrate(1/x**10/(-x**3-1)**(1/2),x)
Output:
-5*I*asinh(x**(-3/2))/24 + 5*I/(24*x**(3/2)*sqrt(1 + x**(-3))) + 5*I/(72*x **(9/2)*sqrt(1 + x**(-3))) - I/(36*x**(15/2)*sqrt(1 + x**(-3))) + I/(9*x** (21/2)*sqrt(1 + x**(-3)))
Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^{10} \sqrt {-1-x^3}} \, dx=\frac {15 \, {\left (-x^{3} - 1\right )}^{\frac {5}{2}} + 40 \, {\left (-x^{3} - 1\right )}^{\frac {3}{2}} + 33 \, \sqrt {-x^{3} - 1}}{72 \, {\left ({\left (x^{3} + 1\right )}^{3} + 3 \, x^{3} - 3 \, {\left (x^{3} + 1\right )}^{2} + 2\right )}} - \frac {5}{24} \, \arctan \left (\sqrt {-x^{3} - 1}\right ) \] Input:
integrate(1/x^10/(-x^3-1)^(1/2),x, algorithm="maxima")
Output:
1/72*(15*(-x^3 - 1)^(5/2) + 40*(-x^3 - 1)^(3/2) + 33*sqrt(-x^3 - 1))/((x^3 + 1)^3 + 3*x^3 - 3*(x^3 + 1)^2 + 2) - 5/24*arctan(sqrt(-x^3 - 1))
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^{10} \sqrt {-1-x^3}} \, dx=\frac {15 \, {\left (x^{3} + 1\right )}^{2} \sqrt {-x^{3} - 1} + 40 \, {\left (-x^{3} - 1\right )}^{\frac {3}{2}} + 33 \, \sqrt {-x^{3} - 1}}{72 \, x^{9}} - \frac {5}{24} \, \arctan \left (\sqrt {-x^{3} - 1}\right ) \] Input:
integrate(1/x^10/(-x^3-1)^(1/2),x, algorithm="giac")
Output:
1/72*(15*(x^3 + 1)^2*sqrt(-x^3 - 1) + 40*(-x^3 - 1)^(3/2) + 33*sqrt(-x^3 - 1))/x^9 - 5/24*arctan(sqrt(-x^3 - 1))
Time = 0.03 (sec) , antiderivative size = 223, normalized size of antiderivative = 3.14 \[ \int \frac {1}{x^{10} \sqrt {-1-x^3}} \, dx=\frac {5\,\sqrt {-x^3-1}}{24\,x^3}-\frac {5\,\sqrt {-x^3-1}}{36\,x^6}+\frac {\sqrt {-x^3-1}}{9\,x^9}+\frac {5\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {x^3+1}\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{8\,\sqrt {-x^3-1}\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:
int(1/(x^10*(- x^3 - 1)^(1/2)),x)
Output:
(5*(- x^3 - 1)^(1/2))/(24*x^3) - (5*(- x^3 - 1)^(1/2))/(36*x^6) + (- x^3 - 1)^(1/2)/(9*x^9) + (5*((3^(1/2)*1i)/2 + 3/2)*(x^3 + 1)^(1/2)*((x + (3^(1/ 2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + 1)/((3^(1/2)*1i)/2 + 3 /2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*ellip ticPi((3^(1/2)*1i)/2 + 3/2, asin(((x + 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(8*(- x^3 - 1)^(1/2)*(x^3 - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2))
Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^{10} \sqrt {-1-x^3}} \, dx=\frac {i \left (30 \sqrt {x^{3}+1}\, x^{6}-20 \sqrt {x^{3}+1}\, x^{3}+16 \sqrt {x^{3}+1}+15 \,\mathrm {log}\left (\sqrt {x^{3}+1}-1\right ) x^{9}-15 \,\mathrm {log}\left (\sqrt {x^{3}+1}+1\right ) x^{9}\right )}{144 x^{9}} \] Input:
int(1/x^10/(-x^3-1)^(1/2),x)
Output:
(i*(30*sqrt(x**3 + 1)*x**6 - 20*sqrt(x**3 + 1)*x**3 + 16*sqrt(x**3 + 1) + 15*log(sqrt(x**3 + 1) - 1)*x**9 - 15*log(sqrt(x**3 + 1) + 1)*x**9))/(144*x **9)