\(\int \frac {x^3}{\sqrt {-1+x^3}} \, dx\) [483]

Optimal result

Integrand size = 13, antiderivative size = 137 \[ \int \frac {x^3}{\sqrt {-1+x^3}} \, dx=\frac {2}{5} x \sqrt {-1+x^3}-\frac {4 \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right ),-7+4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}} \]

[Out]

2/5*x*(x^3-1)^(1/2)-4/15*(1-x)*EllipticF((1-x+3^(1/2))/(1-x-3^(1/2)),2*I-I*3^(1/2))*(1/2*6^(1/2)-1/2*2^(1/2))*
((x^2+x+1)/(1-x-3^(1/2))^2)^(1/2)*3^(3/4)/(x^3-1)^(1/2)/((-1+x)/(1-x-3^(1/2))^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {327, 225} \[ \int \frac {x^3}{\sqrt {-1+x^3}} \, dx=\frac {2}{5} x \sqrt {x^3-1}-\frac {4 \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}} \]

[In]

Int[x^3/Sqrt[-1 + x^3],x]

[Out]

(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])

Rule 225

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]*Sq
rt[(-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]

Rule 327

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{5} x \sqrt {-1+x^3}+\frac {2}{5} \int \frac {1}{\sqrt {-1+x^3}} \, dx \\ & = \frac {2}{5} x \sqrt {-1+x^3}-\frac {4 \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right )|-7+4 \sqrt {3}\right )}{5 \sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.32 \[ \int \frac {x^3}{\sqrt {-1+x^3}} \, dx=\frac {2 x \left (-1+x^3+\sqrt {1-x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},x^3\right )\right )}{5 \sqrt {-1+x^3}} \]

[In]

Integrate[x^3/Sqrt[-1 + x^3],x]

[Out]

(2*x*(-1 + x^3 + Sqrt[1 - x^3]*Hypergeometric2F1[1/3, 1/2, 4/3, x^3]))/(5*Sqrt[-1 + x^3])

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 4.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.24

method	result	size
meijerg	\(\frac {\sqrt {-\operatorname {signum}\left (x^{3}-1\right )}\, x^{4} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};x^{3}\right )}{4 \sqrt {\operatorname {signum}\left (x^{3}-1\right )}}\)	\(33\)
default	\(\frac {2 x \sqrt {x^{3}-1}}{5}+\frac {4 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{5 \sqrt {x^{3}-1}}\)	\(127\)
risch	\(\frac {2 x \sqrt {x^{3}-1}}{5}+\frac {4 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{5 \sqrt {x^{3}-1}}\)	\(127\)
elliptic	\(\frac {2 x \sqrt {x^{3}-1}}{5}+\frac {4 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {-1+x}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{5 \sqrt {x^{3}-1}}\)	\(127\)

[In]

int(x^3/(x^3-1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/4/signum(x^3-1)^(1/2)*(-signum(x^3-1))^(1/2)*x^4*hypergeom([1/2,4/3],[7/3],x^3)

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.12 \[ \int \frac {x^3}{\sqrt {-1+x^3}} \, dx=\frac {2}{5} \, \sqrt {x^{3} - 1} x + \frac {4}{5} \, {\rm weierstrassPInverse}\left (0, 4, x\right ) \]

[In]

integrate(x^3/(x^3-1)^(1/2),x, algorithm="fricas")

[Out]

2/5*sqrt(x^3 - 1)*x + 4/5*weierstrassPInverse(0, 4, x)

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.20 \[ \int \frac {x^3}{\sqrt {-1+x^3}} \, dx=- \frac {i x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {7}{3}\right )} \]

[In]

integrate(x**3/(x**3-1)**(1/2),x)

[Out]

-I*x**4*gamma(4/3)*hyper((1/2, 4/3), (7/3,), x**3)/(3*gamma(7/3))

Maxima [F]

\[ \int \frac {x^3}{\sqrt {-1+x^3}} \, dx=\int { \frac {x^{3}}{\sqrt {x^{3} - 1}} \,d x } \]

[In]

integrate(x^3/(x^3-1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^3/sqrt(x^3 - 1), x)

Giac [F]

\[ \int \frac {x^3}{\sqrt {-1+x^3}} \, dx=\int { \frac {x^{3}}{\sqrt {x^{3} - 1}} \,d x } \]

[In]

integrate(x^3/(x^3-1)^(1/2),x, algorithm="giac")

[Out]

integrate(x^3/sqrt(x^3 - 1), x)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.22 \[ \int \frac {x^3}{\sqrt {-1+x^3}} \, dx=\frac {2\,x\,\sqrt {x^3-1}}{5}-\frac {4\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{5\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

[In]

int(x^3/(x^3 - 1)^(1/2),x)

[Out]

(2*x*(x^3 - 1)^(1/2))/5 - (4*((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)*ellipticF(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)))/(5*(((3^(1/2)*1
i)/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))