3.10.53 \(\int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx\)

Optimal. Leaf size=72 \[ -\sqrt {2} \tanh ^{-1}\left (\frac {(x+1) \left (\sqrt {2} x-\sqrt {2}\right )}{x^3+x^2-\sqrt {x^6+2 x^5+x^4-4 x^3-5 x^2+2 x+3}-x-1}\right ) \]

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Rubi [A]  time = 0.19, antiderivative size = 67, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 7, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {6688, 6719, 1033, 724, 206, 688, 207} \begin {gather*} -\frac {\left (1-x^2\right ) \sqrt {x^2+2 x+3} \tanh ^{-1}\left (\frac {\sqrt {x^2+2 x+3}}{\sqrt {2}}\right )}{\sqrt {2} \sqrt {\left (1-x^2\right )^2 \left (x^2+2 x+3\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x)/Sqrt[3 + 2*x - 5*x^2 - 4*x^3 + x^4 + 2*x^5 + x^6],x]

[Out]

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

Rule 206

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

Rule 207

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

Rule 688

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

Rule 724

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

Rule 1033

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[-(a*c), 2]}, Dist[h/2 + (c*g)/(2*q), Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/2 - (c*g)
/(2*q), Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f
, 0] && PosQ[-(a*c)]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

\begin {align*} \int \frac {1-x}{\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}} \, dx &=\int \frac {1-x}{\sqrt {\left (-1+x^2\right )^2 \left (3+2 x+x^2\right )}} \, dx\\ &=\frac {\left (\left (-1+x^2\right ) \sqrt {3+2 x+x^2}\right ) \int \frac {1-x}{\left (-1+x^2\right ) \sqrt {3+2 x+x^2}} \, dx}{\sqrt {\left (-1+x^2\right )^2 \left (3+2 x+x^2\right )}}\\ &=-\frac {\left (\left (-1+x^2\right ) \sqrt {3+2 x+x^2}\right ) \int \frac {1}{(1+x) \sqrt {3+2 x+x^2}} \, dx}{\sqrt {\left (-1+x^2\right )^2 \left (3+2 x+x^2\right )}}\\ &=-\frac {\left (4 \left (-1+x^2\right ) \sqrt {3+2 x+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-8+4 x^2} \, dx,x,\sqrt {3+2 x+x^2}\right )}{\sqrt {\left (-1+x^2\right )^2 \left (3+2 x+x^2\right )}}\\ &=-\frac {\left (1-x^2\right ) \sqrt {3+2 x+x^2} \tanh ^{-1}\left (\frac {\sqrt {3+2 x+x^2}}{\sqrt {2}}\right )}{\sqrt {2} \sqrt {\left (1-x^2\right )^2 \left (3+2 x+x^2\right )}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 72, normalized size = 1.00 \begin {gather*} -\frac {\left (x^2-1\right ) \sqrt {x^2+2 x+3} \left (\log (x+1)-\log \left (\sqrt {2} \sqrt {x^2+2 x+3}+2\right )\right )}{\sqrt {2} \sqrt {\left (x^2-1\right )^2 \left (x^2+2 x+3\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)/Sqrt[3 + 2*x - 5*x^2 - 4*x^3 + x^4 + 2*x^5 + x^6],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.40, size = 72, normalized size = 1.00 \begin {gather*} -\sqrt {2} \tanh ^{-1}\left (\frac {(1+x) \left (-\sqrt {2}+\sqrt {2} x\right )}{-1-x+x^2+x^3-\sqrt {3+2 x-5 x^2-4 x^3+x^4+2 x^5+x^6}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 - x)/Sqrt[3 + 2*x - 5*x^2 - 4*x^3 + x^4 + 2*x^5 + x^6],x]

[Out]

-(Sqrt[2]*ArcTanh[((1 + x)*(-Sqrt[2] + Sqrt[2]*x))/(-1 - x + x^2 + x^3 - Sqrt[3 + 2*x - 5*x^2 - 4*x^3 + x^4 +
2*x^5 + x^6])])

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fricas [A]  time = 0.60, size = 58, normalized size = 0.81 \begin {gather*} \frac {1}{2} \, \sqrt {2} \log \left (\frac {\sqrt {2} {\left (x^{2} - 1\right )} + \sqrt {x^{6} + 2 \, x^{5} + x^{4} - 4 \, x^{3} - 5 \, x^{2} + 2 \, x + 3}}{x^{3} + x^{2} - x - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(2)*log((sqrt(2)*(x^2 - 1) + sqrt(x^6 + 2*x^5 + x^4 - 4*x^3 - 5*x^2 + 2*x + 3))/(x^3 + x^2 - x - 1))

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giac [A]  time = 0.53, size = 90, normalized size = 1.25 \begin {gather*} -\frac {\sqrt {2} \log \left (-\frac {{\left | -2 \, \sqrt {3} - 2 \, \sqrt {2} + 2 \, \sqrt {\frac {2}{x} + \frac {3}{x^{2}} + 1} - \frac {2 \, \sqrt {3}}{x} \right |}}{2 \, {\left (\sqrt {3} - \sqrt {2} - \sqrt {\frac {2}{x} + \frac {3}{x^{2}} + 1} + \frac {\sqrt {3}}{x}\right )}}\right )}{2 \, \mathrm {sgn}\left (-\frac {1}{x^{3}} + \frac {1}{x^{5}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2),x, algorithm="giac")

[Out]

-1/2*sqrt(2)*log(-1/2*abs(-2*sqrt(3) - 2*sqrt(2) + 2*sqrt(2/x + 3/x^2 + 1) - 2*sqrt(3)/x)/(sqrt(3) - sqrt(2) -
sqrt(2/x + 3/x^2 + 1) + sqrt(3)/x))/sgn(-1/x^3 + 1/x^5)

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maple [A]  time = 0.16, size = 64, normalized size = 0.89

method	result	size
default	\(\frac {\left (x^{2}-1\right ) \sqrt {x^{2}+2 x +3}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {2}}{\sqrt {x^{2}+2 x +3}}\right )}{2 \sqrt {x^{6}+2 x^{5}+x^{4}-4 x^{3}-5 x^{2}+2 x +3}}\)	\(64\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{2}-2\right ) x^{2}+\sqrt {x^{6}+2 x^{5}+x^{4}-4 x^{3}-5 x^{2}+2 x +3}+\RootOf \left (\textit {\_Z}^{2}-2\right )}{\left (-1+x \right ) \left (1+x \right )^{2}}\right )}{2}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-x)/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2)*(x^2-1)*(x^2+2*x+3)^(1/2)*2^(1/2)*arctanh(2^(1/2)/(x^2+2*x+3)^(1/2
))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {x - 1}{\sqrt {x^{6} + 2 \, x^{5} + x^{4} - 4 \, x^{3} - 5 \, x^{2} + 2 \, x + 3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)/(x^6+2*x^5+x^4-4*x^3-5*x^2+2*x+3)^(1/2),x, algorithm="maxima")

[Out]

-integrate((x - 1)/sqrt(x^6 + 2*x^5 + x^4 - 4*x^3 - 5*x^2 + 2*x + 3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {x-1}{\sqrt {x^6+2\,x^5+x^4-4\,x^3-5\,x^2+2\,x+3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x - 1)/(2*x - 5*x^2 - 4*x^3 + x^4 + 2*x^5 + x^6 + 3)^(1/2),x)

[Out]

int(-(x - 1)/(2*x - 5*x^2 - 4*x^3 + x^4 + 2*x^5 + x^6 + 3)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {x}{\sqrt {x^{6} + 2 x^{5} + x^{4} - 4 x^{3} - 5 x^{2} + 2 x + 3}}\, dx - \int \left (- \frac {1}{\sqrt {x^{6} + 2 x^{5} + x^{4} - 4 x^{3} - 5 x^{2} + 2 x + 3}}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-x)/(x**6+2*x**5+x**4-4*x**3-5*x**2+2*x+3)**(1/2),x)

[Out]

-Integral(x/sqrt(x**6 + 2*x**5 + x**4 - 4*x**3 - 5*x**2 + 2*x + 3), x) - Integral(-1/sqrt(x**6 + 2*x**5 + x**4
- 4*x**3 - 5*x**2 + 2*x + 3), x)

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