Integrand size = 13, antiderivative size = 153 \[ \int \frac {x^6}{\sqrt {-1+x^3}} \, dx=\frac {16}{55} x \sqrt {-1+x^3}+\frac {2}{11} x^4 \sqrt {-1+x^3}-\frac {32 \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right ),-7+4 \sqrt {3}\right )}{55 \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}} \] Output:
16/55*x*(x^3-1)^(1/2)+2/11*x^4*(x^3-1)^(1/2)-32/165*(1/2*6^(1/2)-1/2*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 = 10.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.34 \[ \int \frac {x^6}{\sqrt {-1+x^3}} \, dx=\frac {2 x \left (-8+3 x^3+5 x^6+8 \sqrt {1-x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},x^3\right )\right )}{55 \sqrt {-1+x^3}} \] Input:
Integrate[x^6/Sqrt[-1 + x^3],x]
Output:
(2*x*(-8 + 3*x^3 + 5*x^6 + 8*Sqrt[1 - x^3]*Hypergeometric2F1[1/3, 1/2, 4/3 , x^3]))/(55*Sqrt[-1 + x^3])
Time = 0.40 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {843, 843, 760}
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.
\(\displaystyle \int \frac {x^6}{\sqrt {x^3-1}} \, dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {8}{11} \int \frac {x^3}{\sqrt {x^3-1}}dx+\frac {2}{11} \sqrt {x^3-1} x^4\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {8}{11} \left (\frac {2}{5} \int \frac {1}{\sqrt {x^3-1}}dx+\frac {2}{5} \sqrt {x^3-1} x\right )+\frac {2}{11} \sqrt {x^3-1} x^4\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {8}{11} \left (\frac {2}{5} x \sqrt {x^3-1}-\frac {4 \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}\right )+\frac {2}{11} \sqrt {x^3-1} x^4\) |
Input:
Int[x^6/Sqrt[-1 + x^3],x]
Output:
(2*x^4*Sqrt[-1 + x^3])/11 + (8*((2*x*Sqrt[-1 + x^3])/5 - (4*Sqrt[2 - 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]])/(5*3^(1/4)*Sqrt[-((1 - x )/(1 - Sqrt[3] - x)^2)]*Sqrt[-1 + x^3])))/11
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.93 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.22
method | result | size |
meijerg | \(\frac {\sqrt {-\operatorname {signum}\left (x^{3}-1\right )}\, x^{7} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {7}{3}\right ], \left [\frac {10}{3}\right ], x^{3}\right )}{7 \sqrt {\operatorname {signum}\left (x^{3}-1\right )}}\) | \(33\) |
risch | \(\frac {2 x \left (5 x^{3}+8\right ) \sqrt {x^{3}-1}}{55}+\frac {32 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{55 \sqrt {x^{3}-1}}\) | \(134\) |
default | \(\frac {2 x^{4} \sqrt {x^{3}-1}}{11}+\frac {16 x \sqrt {x^{3}-1}}{55}+\frac {32 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{55 \sqrt {x^{3}-1}}\) | \(139\) |
elliptic | \(\frac {2 x^{4} \sqrt {x^{3}-1}}{11}+\frac {16 x \sqrt {x^{3}-1}}{55}+\frac {32 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{55 \sqrt {x^{3}-1}}\) | \(139\) |
Input:
int(x^6/(x^3-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/7/signum(x^3-1)^(1/2)*(-signum(x^3-1))^(1/2)*x^7*hypergeom([1/2,7/3],[10 /3],x^3)
Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.16 \[ \int \frac {x^6}{\sqrt {-1+x^3}} \, dx=\frac {2}{55} \, {\left (5 \, x^{4} + 8 \, x\right )} \sqrt {x^{3} - 1} + \frac {32}{55} \, {\rm weierstrassPInverse}\left (0, 4, x\right ) \] Input:
integrate(x^6/(x^3-1)^(1/2),x, algorithm="fricas")
Output:
2/55*(5*x^4 + 8*x)*sqrt(x^3 - 1) + 32/55*weierstrassPInverse(0, 4, x)
Time = 0.50 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.18 \[ \int \frac {x^6}{\sqrt {-1+x^3}} \, dx=- \frac {i x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {10}{3}\right )} \] Input:
integrate(x**6/(x**3-1)**(1/2),x)
Output:
-I*x**7*gamma(7/3)*hyper((1/2, 7/3), (10/3,), x**3)/(3*gamma(10/3))
\[ \int \frac {x^6}{\sqrt {-1+x^3}} \, dx=\int { \frac {x^{6}}{\sqrt {x^{3} - 1}} \,d x } \] Input:
integrate(x^6/(x^3-1)^(1/2),x, algorithm="maxima")
Output:
integrate(x^6/sqrt(x^3 - 1), x)
\[ \int \frac {x^6}{\sqrt {-1+x^3}} \, dx=\int { \frac {x^{6}}{\sqrt {x^{3} - 1}} \,d x } \] Input:
integrate(x^6/(x^3-1)^(1/2),x, algorithm="giac")
Output:
integrate(x^6/sqrt(x^3 - 1), x)
Time = 0.05 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.17 \[ \int \frac {x^6}{\sqrt {-1+x^3}} \, dx=\frac {16\,x\,\sqrt {x^3-1}}{55}+\frac {2\,x^4\,\sqrt {x^3-1}}{11}-\frac {32\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{55\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \] Input:
int(x^6/(x^3 - 1)^(1/2),x)
Output:
(16*x*(x^3 - 1)^(1/2))/55 + (2*x^4*(x^3 - 1)^(1/2))/11 - (32*((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)*1 i)/2 + 3/2))^(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)))/(55*(((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))
\[ \int \frac {x^6}{\sqrt {-1+x^3}} \, dx=\frac {2 \sqrt {x^{3}-1}\, x^{4}}{11}+\frac {16 \sqrt {x^{3}-1}\, x}{55}+\frac {16 \left (\int \frac {\sqrt {x^{3}-1}}{x^{3}-1}d x \right )}{55} \] Input:
int(x^6/(x^3-1)^(1/2),x)
Output:
(2*(5*sqrt(x**3 - 1)*x**4 + 8*sqrt(x**3 - 1)*x + 8*int(sqrt(x**3 - 1)/(x** 3 - 1),x)))/55