\(\int \frac {1+\sqrt {3}-x}{x \sqrt {-1+x^3}} \, dx\) [154]

Optimal result

Integrand size = 23, antiderivative size = 142 \[ \int \frac {1+\sqrt {3}-x}{x \sqrt {-1+x^3}} \, dx=\frac {2}{3} \left (1+\sqrt {3}\right ) \arctan \left (\sqrt {-1+x^3}\right )+\frac {2 \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {1+x+x^2}{\left (1-\sqrt {3}-x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-x}{1-\sqrt {3}-x}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}} \]

[Out]

2/3*arctan((x^3-1)^(1/2))*(1+3^(1/2))+2/3*(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 = 0.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1846, 272, 65, 209, 225} \[ \int \frac {1+\sqrt {3}-x}{x \sqrt {-1+x^3}} \, dx=\frac {2 \sqrt {2-\sqrt {3}} (1-x) \sqrt {\frac {x^2+x+1}{\left (-x-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-x+\sqrt {3}+1}{-x-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (-x-\sqrt {3}+1\right )^2}} \sqrt {x^3-1}}+\frac {2}{3} \left (1+\sqrt {3}\right ) \arctan \left (\sqrt {x^3-1}\right ) \]

[In]

Int[(1 + Sqrt[3] - x)/(x*Sqrt[-1 + x^3]),x]

[Out]

(2*(1 + Sqrt[3])*ArcTan[Sqrt[-1 + x^3]])/3 + (2*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]])/(3^(1/4)*Sqrt[-((1 - x)/(1 - Sqrt[
3] - x)^2)]*Sqrt[-1 + x^3])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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]]

Rule 1846

Int[(Pq_)/((x_)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[Coeff[Pq, x, 0], Int[1/(x*Sqrt[a + b*x^n]), x
], x] + Int[ExpandToSum[(Pq - Coeff[Pq, x, 0])/x, x]/Sqrt[a + b*x^n], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] &
& IGtQ[n, 0] && NeQ[Coeff[Pq, x, 0], 0]

Rubi steps \begin{align*} \text {integral}& = \left (1+\sqrt {3}\right ) \int \frac {1}{x \sqrt {-1+x^3}} \, dx-\int \frac {1}{\sqrt {-1+x^3}} \, dx \\ & = \frac {2 \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 )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}}+\frac {1}{3} \left (1+\sqrt {3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,x^3\right ) \\ & = \frac {2 \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 )}{\sqrt [4]{3} \sqrt {-\frac {1-x}{\left (1-\sqrt {3}-x\right )^2}} \sqrt {-1+x^3}}+\frac {1}{3} \left (2 \left (1+\sqrt {3}\right )\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x^3}\right ) \\ & = \frac {2}{3} \left (1+\sqrt {3}\right ) \tan ^{-1}\left (\sqrt {-1+x^3}\right )+\frac {2 \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 )}{\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 = 10.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.41 \[ \int \frac {1+\sqrt {3}-x}{x \sqrt {-1+x^3}} \, dx=\frac {2}{3} \left (1+\sqrt {3}\right ) \arctan \left (\sqrt {-1+x^3}\right )-\frac {x \sqrt {1-x^3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},x^3\right )}{\sqrt {-1+x^3}} \]

[In]

Integrate[(1 + Sqrt[3] - x)/(x*Sqrt[-1 + x^3]),x]

[Out]

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

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}+\frac {2 \arctan \left (\sqrt {x^{3}-1}\right ) \left (1+\sqrt {3}\right )}{3}\)	\(132\)
elliptic	\(-\frac {2 \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\, \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, F\left (\sqrt {\frac {x -1}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, \sqrt {\frac {\frac {3}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}-1}}+\frac {2 \arctan \left (\sqrt {x^{3}-1}\right ) \left (1+\sqrt {3}\right )}{3}\)	\(132\)
meijerg	\(\frac {\sqrt {-\operatorname {signum}\left (x^{3}-1\right )}\, \left (-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )+\left (-2 \ln \left (2\right )+3 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }\right )}{3 \sqrt {\pi }\, \sqrt {\operatorname {signum}\left (x^{3}-1\right )}}-\frac {\sqrt {-\operatorname {signum}\left (x^{3}-1\right )}\, x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};x^{3}\right )}{\sqrt {\operatorname {signum}\left (x^{3}-1\right )}}+\frac {\sqrt {3}\, \sqrt {-\operatorname {signum}\left (x^{3}-1\right )}\, \left (-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{3}+1}}{2}\right )+\left (-2 \ln \left (2\right )+3 \ln \left (x \right )+i \pi \right ) \sqrt {\pi }\right )}{3 \sqrt {\pi }\, \sqrt {\operatorname {signum}\left (x^{3}-1\right )}}\)	\(155\)

[In]

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

[Out]

-2*(-3/2-1/2*I*3^(1/2))*((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2)*((x+1/2-1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2)*(
(x+1/2+1/2*I*3^(1/2))/(3/2+1/2*I*3^(1/2)))^(1/2)/(x^3-1)^(1/2)*EllipticF(((x-1)/(-3/2-1/2*I*3^(1/2)))^(1/2),((
3/2+1/2*I*3^(1/2))/(3/2-1/2*I*3^(1/2)))^(1/2))+2/3*arctan((x^3-1)^(1/2))*(1+3^(1/2))

Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.37 \[ \int \frac {1+\sqrt {3}-x}{x \sqrt {-1+x^3}} \, dx=\frac {1}{3} \, \sqrt {2 \, \sqrt {3} + 4} \arctan \left (-\frac {{\left (x^{3} - \sqrt {3} {\left (x^{3} - 2\right )} - 2\right )} \sqrt {2 \, \sqrt {3} + 4}}{4 \, \sqrt {x^{3} - 1}}\right ) - 2 \, {\rm weierstrassPInverse}\left (0, 4, x\right ) \]

[In]

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

[Out]

1/3*sqrt(2*sqrt(3) + 4)*arctan(-1/4*(x^3 - sqrt(3)*(x^3 - 2) - 2)*sqrt(2*sqrt(3) + 4)/sqrt(x^3 - 1)) - 2*weier
strassPInverse(0, 4, x)

Sympy [A] (verification not implemented)

Time = 3.81 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.66 \[ \int \frac {1+\sqrt {3}-x}{x \sqrt {-1+x^3}} \, dx=\frac {i x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {x^{3}} \right )}}{3 \Gamma \left (\frac {4}{3}\right )} + \begin {cases} \frac {2 i \operatorname {acosh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {2 \operatorname {asin}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} & \text {otherwise} \end {cases} + \sqrt {3} \left (\begin {cases} \frac {2 i \operatorname {acosh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \frac {2 \operatorname {asin}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

I*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), x**3)/(3*gamma(4/3)) + Piecewise((2*I*acosh(x**(-3/2))/3, 1/Abs(x**3)
> 1), (-2*asin(x**(-3/2))/3, True)) + sqrt(3)*Piecewise((2*I*acosh(x**(-3/2))/3, 1/Abs(x**3) > 1), (-2*asin(x
**(-3/2))/3, True))

Maxima [F]

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

[In]

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

[Out]

-integrate((x - sqrt(3) - 1)/(sqrt(x^3 - 1)*x), x)

Giac [F]

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

[In]

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

[Out]

integrate(-(x - sqrt(3) - 1)/(sqrt(x^3 - 1)*x), x)

Mupad [B] (verification not implemented)

Time = 18.89 (sec) , antiderivative size = 334, normalized size of antiderivative = 2.35 \[ \int \frac {1+\sqrt {3}-x}{x \sqrt {-1+x^3}} \, dx=\frac {2\,\sqrt {3}\,\mathrm {atan}\left (\sqrt {x^3-1}\right )}{3}+\frac {2\,\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 )}{\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 )}}-\frac {2\,\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}}}\,\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 )}{\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((3^(1/2) - x + 1)/(x*(x^3 - 1)^(1/2)),x)

[Out]

(2*3^(1/2)*atan((x^3 - 1)^(1/2)))/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)
*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 - 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) - (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)*ellipticPi((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)))/(((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)