\(\int \frac {-2-2 x+x^2}{(-1+3 x+x^2) \sqrt {-1+x^3}} \, dx\) [374]

Optimal result

Integrand size = 28, antiderivative size = 31 \[ \int \frac {-2-2 x+x^2}{\left (-1+3 x+x^2\right ) \sqrt {-1+x^3}} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-1+x^3}}{1+x+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2170, 209} \[ \int \frac {-2-2 x+x^2}{\left (-1+3 x+x^2\right ) \sqrt {-1+x^3}} \, dx=\sqrt {2} \arctan \left (\frac {\sqrt {2} (1-x)}{\sqrt {x^3-1}}\right ) \]

[In]

Int[(-2 - 2*x + x^2)/((-1 + 3*x + x^2)*Sqrt[-1 + x^3]),x]

[Out]

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

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 2170

Int[((f_) + (g_.)*(x_) + (h_.)*(x_)^2)/(((c_) + (d_.)*(x_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^3]), x_Symbo
l] :> Dist[-2*g*h, Subst[Int[1/(2*e*h - (b*d*f - 2*a*e*h)*x^2), x], x, (1 + 2*h*(x/g))/Sqrt[a + b*x^3]], x] /;
FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b*d*f - 2*a*e*h, 0] && EqQ[b*g^3 - 8*a*h^3, 0] && EqQ[g^2 + 2*f*h,
0] && EqQ[b*d*f + b*c*g - 4*a*e*h, 0]

Rubi steps \begin{align*} \text {integral}& = 4 \text {Subst}\left (\int \frac {1}{2+4 x^2} \, dx,x,\frac {1-x}{\sqrt {-1+x^3}}\right ) \\ & = \sqrt {2} \arctan \left (\frac {\sqrt {2} (1-x)}{\sqrt {-1+x^3}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 1.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-2-2 x+x^2}{\left (-1+3 x+x^2\right ) \sqrt {-1+x^3}} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-1+x^3}}{1+x+x^2}\right ) \]

[In]

Integrate[(-2 - 2*x + x^2)/((-1 + 3*x + x^2)*Sqrt[-1 + x^3]),x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.92 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.87

method	result	size
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )+4 \sqrt {x^{3}-1}}{x^{2}+3 x -1}\right )}{2}\)	\(58\)
default	\(\text {Expression too large to display}\)	\(1641\)
elliptic	\(\text {Expression too large to display}\)	\(1850\)

[In]

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

[Out]

-1/2*RootOf(_Z^2+2)*ln((RootOf(_Z^2+2)*x^2-RootOf(_Z^2+2)*x+3*RootOf(_Z^2+2)+4*(x^3-1)^(1/2))/(x^2+3*x-1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {-2-2 x+x^2}{\left (-1+3 x+x^2\right ) \sqrt {-1+x^3}} \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{2} - x + 3\right )}}{4 \, \sqrt {x^{3} - 1}}\right ) \]

[In]

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

[Out]

1/2*sqrt(2)*arctan(1/4*sqrt(2)*(x^2 - x + 3)/sqrt(x^3 - 1))

Sympy [F]

\[ \int \frac {-2-2 x+x^2}{\left (-1+3 x+x^2\right ) \sqrt {-1+x^3}} \, dx=\int \frac {x^{2} - 2 x - 2}{\sqrt {\left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x^{2} + 3 x - 1\right )}\, dx \]

[In]

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

[Out]

Integral((x**2 - 2*x - 2)/(sqrt((x - 1)*(x**2 + x + 1))*(x**2 + 3*x - 1)), x)

Maxima [F]

\[ \int \frac {-2-2 x+x^2}{\left (-1+3 x+x^2\right ) \sqrt {-1+x^3}} \, dx=\int { \frac {x^{2} - 2 \, x - 2}{\sqrt {x^{3} - 1} {\left (x^{2} + 3 \, x - 1\right )}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {-2-2 x+x^2}{\left (-1+3 x+x^2\right ) \sqrt {-1+x^3}} \, dx=\int { \frac {x^{2} - 2 \, x - 2}{\sqrt {x^{3} - 1} {\left (x^{2} + 3 \, x - 1\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 273, normalized size of antiderivative = 8.81 \[ \int \frac {-2-2 x+x^2}{\left (-1+3 x+x^2\right ) \sqrt {-1+x^3}} \, dx=\frac {\left (3+\sqrt {3}\,1{}\mathrm {i}\right )\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (-\mathrm {F}\left (\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )+\Pi \left (\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {\sqrt {13}}{2}+\frac {5}{2}};\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )+\Pi \left (-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {\sqrt {13}}{2}-\frac {5}{2}};\mathrm {asin}\left (\sqrt {-\frac {x-1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )}{\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x+\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \]

[In]

int(-(2*x - x^2 + 2)/((x^3 - 1)^(1/2)*(3*x + x^2 - 1)),x)

[Out]

((3^(1/2)*1i + 3)*(-(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)/(13^(1/2)/
2 + 5/2), 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)) - ell
ipticF(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)) + ellipt
icPi(-((3^(1/2)*1i)/2 + 3/2)/(13^(1/2)/2 - 5/2), 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)