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

Optimal result

Integrand size = 18, antiderivative size = 12 \[ \int \frac {1+3 x^2}{\sqrt {-1+x+x^3}} \, dx=2 \sqrt {-1+x+x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1602} \[ \int \frac {1+3 x^2}{\sqrt {-1+x+x^3}} \, dx=2 \sqrt {x^3+x-1} \]

[In]

Int[(1 + 3*x^2)/Sqrt[-1 + x + x^3],x]

[Out]

2*Sqrt[-1 + x + x^3]

Rule 1602

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*x^(p - q +
1)*(Qq^(m + 1)/((p + m*q + 1)*Coeff[Qq, x, q])), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = 2 \sqrt {-1+x+x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.00 \[ \int \frac {1+3 x^2}{\sqrt {-1+x+x^3}} \, dx=2 \sqrt {-1+x+x^3} \]

[In]

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

[Out]

2*Sqrt[-1 + x + x^3]

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.92

method	result	size
gosper	\(2 \sqrt {x^{3}+x -1}\)	\(11\)
derivativedivides	\(2 \sqrt {x^{3}+x -1}\)	\(11\)
default	\(2 \sqrt {x^{3}+x -1}\)	\(11\)
trager	\(2 \sqrt {x^{3}+x -1}\)	\(11\)
risch	\(2 \sqrt {x^{3}+x -1}\)	\(11\)
elliptic	\(2 \sqrt {x^{3}+x -1}\)	\(11\)
pseudoelliptic	\(2 \sqrt {x^{3}+x -1}\)	\(11\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1+3 x^2}{\sqrt {-1+x+x^3}} \, dx=2 \, \sqrt {x^{3} + x - 1} \]

[In]

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

[Out]

2*sqrt(x^3 + x - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1+3 x^2}{\sqrt {-1+x+x^3}} \, dx=2 \sqrt {x^{3} + x - 1} \]

[In]

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

[Out]

2*sqrt(x**3 + x - 1)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1+3 x^2}{\sqrt {-1+x+x^3}} \, dx=2 \, \sqrt {x^{3} + x - 1} \]

[In]

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

[Out]

2*sqrt(x^3 + x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1+3 x^2}{\sqrt {-1+x+x^3}} \, dx=2 \, \sqrt {x^{3} + x - 1} \]

[In]

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

[Out]

2*sqrt(x^3 + x - 1)

Mupad [B] (verification not implemented)

Time = 4.96 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.83 \[ \int \frac {1+3 x^2}{\sqrt {-1+x+x^3}} \, dx=2\,\sqrt {x^3+x-1} \]

[In]

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

[Out]

2*(x + x^3 - 1)^(1/2)