\(\int \frac {-1+x}{(1+x) \sqrt {x+x^2+x^3}} \, dx\) [239]

Optimal result

Integrand size = 21, antiderivative size = 24 \[ \int \frac {-1+x}{(1+x) \sqrt {x+x^2+x^3}} \, dx=-2 \arctan \left (\frac {\sqrt {x+x^2+x^3}}{1+x+x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2081, 6865, 1712, 209} \[ \int \frac {-1+x}{(1+x) \sqrt {x+x^2+x^3}} \, dx=-\frac {2 \sqrt {x} \sqrt {x^2+x+1} \arctan \left (\frac {\sqrt {x}}{\sqrt {x^2+x+1}}\right )}{\sqrt {x^3+x^2+x}} \]

[In]

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

[Out]

(-2*Sqrt[x]*Sqrt[1 + x + x^2]*ArcTan[Sqrt[x]/Sqrt[1 + x + x^2]])/Sqrt[x + x^2 + 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 1712

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[
A, Subst[Int[1/(d - (b*d - 2*a*e)*x^2), x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B},
x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rule 2081

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6865

Int[(u_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k, Subst[Int[x^(k*(m + 1) - 1)*(u /. x -> x^k
), x], x, x^(1/k)], x]] /; FractionQ[m]

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {-1+x}{(1+x) \sqrt {x+x^2+x^3}} \, dx=-\frac {2 \sqrt {x} \sqrt {1+x+x^2} \arctan \left (\frac {\sqrt {x}}{\sqrt {1+x+x^2}}\right )}{\sqrt {x \left (1+x+x^2\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.53 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75

method	result	size
default	\(2 \arctan \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}}{x}\right )\)	\(18\)
pseudoelliptic	\(2 \arctan \left (\frac {\sqrt {x \left (x^{2}+x +1\right )}}{x}\right )\)	\(18\)
trager	\(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-2 \sqrt {x^{3}+x^{2}+x}}{\left (1+x \right )^{2}}\right )\)	\(45\)
elliptic	\(\frac {2 \left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x}{-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {1}{2}+\frac {i \sqrt {3}}{2}}}, \frac {\sqrt {3}\, \sqrt {i \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}\right )}{3 \sqrt {x^{3}+x^{2}+x}}-\frac {4 \left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x}{-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {x +\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {1}{2}+\frac {i \sqrt {3}}{2}}}, \frac {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\frac {1}{2}-\frac {i \sqrt {3}}{2}}, \frac {\sqrt {3}\, \sqrt {i \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}\right )}{3 \sqrt {x^{3}+x^{2}+x}\, \left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}\)	\(271\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-1+x}{(1+x) \sqrt {x+x^2+x^3}} \, dx=\arctan \left (\frac {x^{2} + 1}{2 \, \sqrt {x^{3} + x^{2} + x}}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 179, normalized size of antiderivative = 7.46 \[ \int \frac {-1+x}{(1+x) \sqrt {x+x^2+x^3}} \, dx=\frac {\sqrt {\frac {x}{-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {-\frac {x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\left (\sqrt {3}+1{}\mathrm {i}\right )\,\left (\mathrm {F}\left (\mathrm {asin}\left (\sqrt {\frac {x}{-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )-2\,\Pi \left (\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x}{-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )\right )\,1{}\mathrm {i}}{\sqrt {x^3+x^2-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,x}} \]

[In]

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

[Out]

((x/((3^(1/2)*1i)/2 - 1/2))^(1/2)*(-(x - (3^(1/2)*1i)/2 + 1/2)/((3^(1/2)*1i)/2 - 1/2))^(1/2)*((x + (3^(1/2)*1i
)/2 + 1/2)/((3^(1/2)*1i)/2 + 1/2))^(1/2)*(3^(1/2) + 1i)*(ellipticF(asin((x/((3^(1/2)*1i)/2 - 1/2))^(1/2)), -((
3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 + 1/2)) - 2*ellipticPi(1/2 - (3^(1/2)*1i)/2, asin((x/((3^(1/2)*1i)/2 - 1/
2))^(1/2)), -((3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 + 1/2)))*1i)/(x^2 + x^3 - x*((3^(1/2)*1i)/2 - 1/2)*((3^(1/
2)*1i)/2 + 1/2))^(1/2)