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

Optimal. Leaf size=86 \[ \frac {1}{2} \tanh ^{-1}\left (\frac {x+4}{2 \sqrt {x^2+2 x+4}}\right )-\frac {\tanh ^{-1}\left (\frac {2 x+5}{\sqrt {7} \sqrt {x^2+2 x+4}}\right )}{2 \sqrt {7}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x^2+2 x+4}}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

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Rubi [A]  time = 0.28, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1593, 6725, 724, 206, 1033, 688, 207} \[ \frac {1}{2} \tanh ^{-1}\left (\frac {x+4}{2 \sqrt {x^2+2 x+4}}\right )-\frac {\tanh ^{-1}\left (\frac {2 x+5}{\sqrt {7} \sqrt {x^2+2 x+4}}\right )}{2 \sqrt {7}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x^2+2 x+4}}{\sqrt {3}}\right )}{2 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(4 + x)/(2*Sqrt[4 + 2*x + x^2])]/2 - ArcTanh[(5 + 2*x)/(Sqrt[7]*Sqrt[4 + 2*x + x^2])]/(2*Sqrt[7]) - Ar
cTanh[Sqrt[4 + 2*x + x^2]/Sqrt[3]]/(2*Sqrt[3])

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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {4+2 x+x^2} \left (-x+x^3\right )} \, dx &=\int \frac {1}{x \left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx\\ &=\int \left (-\frac {1}{x \sqrt {4+2 x+x^2}}+\frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}}\right ) \, dx\\ &=-\int \frac {1}{x \sqrt {4+2 x+x^2}} \, dx+\int \frac {x}{\left (-1+x^2\right ) \sqrt {4+2 x+x^2}} \, dx\\ &=\frac {1}{2} \int \frac {1}{(-1+x) \sqrt {4+2 x+x^2}} \, dx+\frac {1}{2} \int \frac {1}{(1+x) \sqrt {4+2 x+x^2}} \, dx+2 \operatorname {Subst}\left (\int \frac {1}{16-x^2} \, dx,x,\frac {8+2 x}{\sqrt {4+2 x+x^2}}\right )\\ &=\frac {1}{2} \tanh ^{-1}\left (\frac {4+x}{2 \sqrt {4+2 x+x^2}}\right )+2 \operatorname {Subst}\left (\int \frac {1}{-12+4 x^2} \, dx,x,\sqrt {4+2 x+x^2}\right )-\operatorname {Subst}\left (\int \frac {1}{28-x^2} \, dx,x,\frac {10+4 x}{\sqrt {4+2 x+x^2}}\right )\\ &=\frac {1}{2} \tanh ^{-1}\left (\frac {4+x}{2 \sqrt {4+2 x+x^2}}\right )-\frac {\tanh ^{-1}\left (\frac {10+4 x}{2 \sqrt {7} \sqrt {4+2 x+x^2}}\right )}{2 \sqrt {7}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {4+2 x+x^2}}{\sqrt {3}}\right )}{2 \sqrt {3}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 83, normalized size = 0.97 \[ \frac {1}{42} \left (21 \tanh ^{-1}\left (\frac {x+4}{2 \sqrt {x^2+2 x+4}}\right )-3 \sqrt {7} \tanh ^{-1}\left (\frac {2 x+5}{\sqrt {7} \sqrt {x^2+2 x+4}}\right )-7 \sqrt {3} \tanh ^{-1}\left (\frac {\sqrt {(x+1)^2+3}}{\sqrt {3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*ArcTanh[(4 + x)/(2*Sqrt[4 + 2*x + x^2])] - 3*Sqrt[7]*ArcTanh[(5 + 2*x)/(Sqrt[7]*Sqrt[4 + 2*x + x^2])] - 7*
Sqrt[3]*ArcTanh[Sqrt[3 + (1 + x)^2]/Sqrt[3]])/42

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IntegrateAlgebraic [A]  time = 0.23, size = 105, normalized size = 1.22 \[ -\tanh ^{-1}\left (\frac {x}{2}-\frac {1}{2} \sqrt {x^2+2 x+4}\right )+\frac {\tanh ^{-1}\left (-\frac {\sqrt {x^2+2 x+4}}{\sqrt {3}}+\frac {x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {x^2+2 x+4}}{\sqrt {7}}-\frac {x}{\sqrt {7}}+\frac {1}{\sqrt {7}}\right )}{\sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[x/2 - Sqrt[4 + 2*x + x^2]/2] + ArcTanh[1/Sqrt[3] + x/Sqrt[3] - Sqrt[4 + 2*x + x^2]/Sqrt[3]]/Sqrt[3] -
ArcTanh[1/Sqrt[7] - x/Sqrt[7] + Sqrt[4 + 2*x + x^2]/Sqrt[7]]/Sqrt[7]

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fricas [A]  time = 0.63, size = 110, normalized size = 1.28 \[ \frac {1}{14} \, \sqrt {7} \log \left (\frac {\sqrt {7} {\left (2 \, x + 5\right )} + \sqrt {x^{2} + 2 \, x + 4} {\left (2 \, \sqrt {7} - 7\right )} - 4 \, x - 10}{x - 1}\right ) + \frac {1}{6} \, \sqrt {3} \log \left (-\frac {\sqrt {3} - \sqrt {x^{2} + 2 \, x + 4}}{x + 1}\right ) + \frac {1}{2} \, \log \left (-x + \sqrt {x^{2} + 2 \, x + 4} + 2\right ) - \frac {1}{2} \, \log \left (-x + \sqrt {x^{2} + 2 \, x + 4} - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*sqrt(7)*log((sqrt(7)*(2*x + 5) + sqrt(x^2 + 2*x + 4)*(2*sqrt(7) - 7) - 4*x - 10)/(x - 1)) + 1/6*sqrt(3)*l
og(-(sqrt(3) - sqrt(x^2 + 2*x + 4))/(x + 1)) + 1/2*log(-x + sqrt(x^2 + 2*x + 4) + 2) - 1/2*log(-x + sqrt(x^2 +
2*x + 4) - 2)

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giac [B]  time = 0.73, size = 147, normalized size = 1.71 \[ \frac {1}{14} \, \sqrt {7} \log \left (\frac {{\left | -2 \, x - 2 \, \sqrt {7} + 2 \, \sqrt {x^{2} + 2 \, x + 4} + 2 \right |}}{{\left | -2 \, x + 2 \, \sqrt {7} + 2 \, \sqrt {x^{2} + 2 \, x + 4} + 2 \right |}}\right ) + \frac {1}{6} \, \sqrt {3} \log \left (-\frac {{\left | -2 \, x - 2 \, \sqrt {3} + 2 \, \sqrt {x^{2} + 2 \, x + 4} - 2 \right |}}{2 \, {\left (x - \sqrt {3} - \sqrt {x^{2} + 2 \, x + 4} + 1\right )}}\right ) + \frac {1}{2} \, \log \left ({\left | -x + \sqrt {x^{2} + 2 \, x + 4} + 2 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | -x + \sqrt {x^{2} + 2 \, x + 4} - 2 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14*sqrt(7)*log(abs(-2*x - 2*sqrt(7) + 2*sqrt(x^2 + 2*x + 4) + 2)/abs(-2*x + 2*sqrt(7) + 2*sqrt(x^2 + 2*x + 4
) + 2)) + 1/6*sqrt(3)*log(-1/2*abs(-2*x - 2*sqrt(3) + 2*sqrt(x^2 + 2*x + 4) - 2)/(x - sqrt(3) - sqrt(x^2 + 2*x
+ 4) + 1)) + 1/2*log(abs(-x + sqrt(x^2 + 2*x + 4) + 2)) - 1/2*log(abs(-x + sqrt(x^2 + 2*x + 4) - 2))

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maple [A]  time = 0.42, size = 69, normalized size = 0.80

method	result	size
default	\(\frac {\arctanh \left (\frac {8+2 x}{4 \sqrt {x^{2}+2 x +4}}\right )}{2}-\frac {\sqrt {7}\, \arctanh \left (\frac {\left (10+4 x \right ) \sqrt {7}}{14 \sqrt {\left (-1+x \right )^{2}+3+4 x}}\right )}{14}-\frac {\sqrt {3}\, \arctanh \left (\frac {\sqrt {3}}{\sqrt {\left (1+x \right )^{2}+3}}\right )}{6}\)	\(69\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}-7\right ) \ln \left (\frac {-2 \RootOf \left (\textit {\_Z}^{2}-7\right ) x +7 \sqrt {x^{2}+2 x +4}-5 \RootOf \left (\textit {\_Z}^{2}-7\right )}{-1+x}\right )}{14}-\frac {\ln \left (\frac {-4-x +2 \sqrt {x^{2}+2 x +4}}{x}\right )}{2}+\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-3\right )+\sqrt {x^{2}+2 x +4}}{1+x}\right )}{6}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctanh(1/4*(8+2*x)/(x^2+2*x+4)^(1/2))-1/14*7^(1/2)*arctanh(1/14*(10+4*x)*7^(1/2)/((-1+x)^2+3+4*x)^(1/2))-
1/6*3^(1/2)*arctanh(3^(1/2)/((1+x)^2+3)^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} - x\right )} \sqrt {x^{2} + 2 \, x + 4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {1}{\left (x-x^3\right )\,\sqrt {x^2+2\,x+4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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