3.3.79 \(\int \frac {1}{\sqrt {4+2 x+x^2} (-x+x^3)} \, dx\) [279]

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

[Out]

1/2*arctanh(1/2*(4+x)/(x^2+2*x+4)^(1/2))-1/6*arctanh(1/3*(x^2+2*x+4)^(1/2)*3^(1/2))*3^(1/2)-1/14*arctanh(1/7*(
2*x+5)*7^(1/2)/(x^2+2*x+4)^(1/2))*7^(1/2)

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Rubi [A]
time = 0.19, 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 = {1607, 6857, 738, 212, 1047, 702, 213} \begin {gather*} \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}} \end {gather*}

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 212

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

Rule 213

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

Rule 702

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 738

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 1047

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 1607

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 6857

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 \text {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 \text {Subst}\left (\int \frac {1}{-12+4 x^2} \, dx,x,\sqrt {4+2 x+x^2}\right )-\text {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 = 0.18, size = 85, normalized size = 0.99 \begin {gather*} -\tanh ^{-1}\left (\frac {1}{2} \left (x-\sqrt {4+2 x+x^2}\right )\right )+\frac {\tanh ^{-1}\left (\frac {1+x-\sqrt {4+2 x+x^2}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {1-x+\sqrt {4+2 x+x^2}}{\sqrt {7}}\right )}{\sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.24, 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	\(\frac {\ln \left (\frac {2 \sqrt {x^{2}+2 x +4}+4+x}{x}\right )}{2}-\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {\sqrt {x^{2}+2 x +4}+\RootOf \left (\textit {\_Z}^{2}-3\right )}{1+x}\right )}{6}-\frac {\RootOf \left (\textit {\_Z}^{2}-7\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}-7\right ) x +7 \sqrt {x^{2}+2 x +4}+5 \RootOf \left (\textit {\_Z}^{2}-7\right )}{-1+x}\right )}{14}\)	\(101\)

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.48, size = 110, normalized size = 1.28 \begin {gather*} \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 ) \end {gather*}

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

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (66) = 132\).
time = 1.66, size = 147, normalized size = 1.71 \begin {gather*} \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 ) \end {gather*}

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

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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