Integrand size = 13, antiderivative size = 294 \[ \int \frac {1}{x^5 \sqrt {-1+x^3}} \, dx=\frac {5 \sqrt {-1+x^3}}{8 \left (1-\sqrt {3}-x\right )}+\frac {\sqrt {-1+x^3}}{4 x^4}+\frac {5 \sqrt {-1+x^3}}{8 x}-\frac {5 \sqrt [4]{3} \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} E\left (\arcsin \left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right )|-7+4 \sqrt {3}\right )}{16 \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}}+\frac {5 (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right ),-7+4 \sqrt {3}\right )}{4 \sqrt {2} \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}} \] Output:
5*(x^3-1)^(1/2)/(8-8*3^(1/2)-8*x)+1/4*(x^3-1)^(1/2)/x^4+5/8*(x^3-1)^(1/2)/ x-5/16*3^(1/4)*(1/2*6^(1/2)+1/2*2^(1/2))*(1-x)*((x^2+x+1)/(1-3^(1/2)-x)^2) ^(1/2)*EllipticE((1+3^(1/2)-x)/(1-3^(1/2)-x),2*I-I*3^(1/2))/(-(1-x)/(1-3^( 1/2)-x)^2)^(1/2)/(x^3-1)^(1/2)+5/24*2^(1/2)*(1-x)*((x^2+x+1)/(1-3^(1/2)-x) ^2)^(1/2)*EllipticF((1+3^(1/2)-x)/(1-3^(1/2)-x),2*I-I*3^(1/2))*3^(3/4)/(-( 1-x)/(1-3^(1/2)-x)^2)^(1/2)/(x^3-1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.14 \[ \int \frac {1}{x^5 \sqrt {-1+x^3}} \, dx=-\frac {\sqrt {1-x^3} \operatorname {Hypergeometric2F1}\left (-\frac {4}{3},\frac {1}{2},-\frac {1}{3},x^3\right )}{4 x^4 \sqrt {-1+x^3}} \] Input:
Integrate[1/(x^5*Sqrt[-1 + x^3]),x]
Output:
-1/4*(Sqrt[1 - x^3]*Hypergeometric2F1[-4/3, 1/2, -1/3, x^3])/(x^4*Sqrt[-1 + x^3])
Time = 0.68 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {847, 847, 833, 760, 2418}
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.
| \(\displaystyle \int \frac {1}{x^5 \sqrt {x^3-1}} \, dx\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle \frac {5}{8} \int \frac {1}{x^2 \sqrt {x^3-1}}dx+\frac {\sqrt {x^3-1}}{4 x^4}\) |
| \(\Big \downarrow \) 847 |
| \(\displaystyle \frac {5}{8} \left (\frac {\sqrt {x^3-1}}{x}-\frac {1}{2} \int \frac {x}{\sqrt {x^3-1}}dx\right )+\frac {\sqrt {x^3-1}}{4 x^4}\) |
| \(\Big \downarrow \) 833 |
| \(\displaystyle \frac {5}{8} \left (\frac {1}{2} \left (\int \frac {-x+\sqrt {3}+1}{\sqrt {x^3-1}}dx-\left (1+\sqrt {3}\right ) \int \frac {1}{\sqrt {x^3-1}}dx\right )+\frac {\sqrt {x^3-1}}{x}\right )+\frac {\sqrt {x^3-1}}{4 x^4}\) |
| \(\Big \downarrow \) 760 |
| \(\displaystyle \frac {5}{8} \left (\frac {1}{2} \left (\int \frac {-x+\sqrt {3}+1}{\sqrt {x^3-1}}dx+\frac {2 \sqrt {2-\sqrt {3}} \left (1+\sqrt {3}\right ) (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}\right )+\frac {\sqrt {x^3-1}}{x}\right )+\frac {\sqrt {x^3-1}}{4 x^4}\) |
| \(\Big \downarrow \) 2418 |
| \(\displaystyle \frac {5}{8} \left (\frac {1}{2} \left (\frac {2 \sqrt {2-\sqrt {3}} \left (1+\sqrt {3}\right ) (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}-\frac {\sqrt [4]{3} \sqrt {2+\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} E\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{\sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}+\frac {2 \sqrt {x^3-1}}{-x-\sqrt {3}+1}\right )+\frac {\sqrt {x^3-1}}{x}\right )+\frac {\sqrt {x^3-1}}{4 x^4}\) |
Input:
Int[1/(x^5*Sqrt[-1 + x^3]),x]
Output:
Sqrt[-1 + x^3]/(4*x^4) + (5*(Sqrt[-1 + x^3]/x + ((2*Sqrt[-1 + x^3])/(1 - S qrt[3] - x) - (3^(1/4)*Sqrt[2 + Sqrt[3]]*(1 - x)*Sqrt[(1 + x + x^2)/(1 - S qrt[3] - x)^2]*EllipticE[ArcSin[(1 + Sqrt[3] - x)/(1 - Sqrt[3] - x)], -7 + 4*Sqrt[3]])/(Sqrt[-((1 - x)/(1 - Sqrt[3] - x)^2)]*Sqrt[-1 + x^3]) + (2*Sq rt[2 - Sqrt[3]]*(1 + Sqrt[3])*(1 - x)*Sqrt[(1 + x + x^2)/(1 - Sqrt[3] - x) ^2]*EllipticF[ArcSin[(1 + Sqrt[3] - x)/(1 - Sqrt[3] - x)], -7 + 4*Sqrt[3]] )/(3^(1/4)*Sqrt[-((1 - x)/(1 - Sqrt[3] - x)^2)]*Sqrt[-1 + x^3]))/2))/8
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) *s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3] ], s = Denom[Rt[b/a, 3]]}, Simp[(-(1 + Sqrt[3]))*(s/r) Int[1/Sqrt[a + b*x ^3], x], x] + Simp[1/r Int[((1 + Sqrt[3])*s + r*x)/Sqrt[a + b*x^3], x], x ]] /; FreeQ[{a, b}, x] && NegQ[a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N umer[Simplify[(1 + Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 + Sqrt[3])*(d/c) ]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 - Sqrt[3])*s + r*x))), x] + S imp[3^(1/4)*Sqrt[2 + Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( (1 - Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[(-s)*((s + r*x)/((1 - S qrt[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 + Sqrt[3])*s + r*x)/((1 - Sqrt[ 3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && NegQ[a] && EqQ[b*c^3 - 2*(5 + 3*Sqrt[3])*a*d^3, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.84 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.11
| method | result | size |
| meijerg | \(-\frac {\sqrt {-\operatorname {signum}\left (x^{3}-1\right )}\, \operatorname {hypergeom}\left (\left [-\frac {4}{3}, \frac {1}{2}\right ], \left [-\frac {1}{3}\right ], x^{3}\right )}{4 \sqrt {\operatorname {signum}\left (x^{3}-1\right )}\, x^{4}}\) | \(33\) |
| default | \(\frac {\sqrt {x^{3}-1}}{4 x^{4}}+\frac {5 \sqrt {x^{3}-1}}{8 x}-\frac {5 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{8 \sqrt {x^{3}-1}}\) | \(198\) |
| risch | \(\frac {5 x^{6}-3 x^{3}-2}{8 x^{4} \sqrt {x^{3}-1}}-\frac {5 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{8 \sqrt {x^{3}-1}}\) | \(198\) |
| elliptic | \(\frac {\sqrt {x^{3}-1}}{4 x^{4}}+\frac {5 \sqrt {x^{3}-1}}{8 x}-\frac {5 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \left (\left (\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticE}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )+\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \operatorname {EllipticF}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )\right )}{8 \sqrt {x^{3}-1}}\) | \(198\) |
Input:
int(1/x^5/(x^3-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/4/signum(x^3-1)^(1/2)*(-signum(x^3-1))^(1/2)/x^4*hypergeom([-4/3,1/2],[ -1/3],x^3)
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.11 \[ \int \frac {1}{x^5 \sqrt {-1+x^3}} \, dx=\frac {5 \, x^{4} {\rm weierstrassZeta}\left (0, 4, {\rm weierstrassPInverse}\left (0, 4, x\right )\right ) + {\left (5 \, x^{3} + 2\right )} \sqrt {x^{3} - 1}}{8 \, x^{4}} \] Input:
integrate(1/x^5/(x^3-1)^(1/2),x, algorithm="fricas")
Output:
1/8*(5*x^4*weierstrassZeta(0, 4, weierstrassPInverse(0, 4, x)) + (5*x^3 + 2)*sqrt(x^3 - 1))/x^4
Time = 0.61 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.12 \[ \int \frac {1}{x^5 \sqrt {-1+x^3}} \, dx=- \frac {i \Gamma \left (- \frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {4}{3}, \frac {1}{2} \\ - \frac {1}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 x^{4} \Gamma \left (- \frac {1}{3}\right )} \] Input:
integrate(1/x**5/(x**3-1)**(1/2),x)
Output:
-I*gamma(-4/3)*hyper((-4/3, 1/2), (-1/3,), x**3)/(3*x**4*gamma(-1/3))
\[ \int \frac {1}{x^5 \sqrt {-1+x^3}} \, dx=\int { \frac {1}{\sqrt {x^{3} - 1} x^{5}} \,d x } \] Input:
integrate(1/x^5/(x^3-1)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(x^3 - 1)*x^5), x)
\[ \int \frac {1}{x^5 \sqrt {-1+x^3}} \, dx=\int { \frac {1}{\sqrt {x^{3} - 1} x^{5}} \,d x } \] Input:
integrate(1/x^5/(x^3-1)^(1/2),x, algorithm="giac")
Output:
integrate(1/(sqrt(x^3 - 1)*x^5), x)
Time = 0.07 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^5 \sqrt {-1+x^3}} \, dx=\frac {5\,\sqrt {x^3-1}}{8\,x}+\frac {\sqrt {x^3-1}}{4\,x^4}+\frac {5\,\left (\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-\left (-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\mathrm {E}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}}{8\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:
int(1/(x^5*(x^3 - 1)^(1/2)),x)
Output:
(5*(x^3 - 1)^(1/2))/(8*x) + (x^3 - 1)^(1/2)/(4*x^4) + (5*(((3^(1/2)*1i)/2 - 1/2)*ellipticF(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)) - ((3^(1/2)*1i)/2 - 3/2)*ellipticE(a sin((-(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)))*((3^(1/2)*1i)/2 + 3/2)*(-(x - (3^(1/2)*1i)/2 + 1/2)/( (3^(1/2)*1i)/2 - 3/2))^(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))/(8*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) - x*(((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) + x^3)^(1/2))
\[ \int \frac {1}{x^5 \sqrt {-1+x^3}} \, dx=\int \frac {\sqrt {x^{3}-1}}{x^{8}-x^{5}}d x \] Input:
int(1/x^5/(x^3-1)^(1/2),x)
Output:
int(sqrt(x**3 - 1)/(x**8 - x**5),x)