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

Optimal. Leaf size=62 \[ -\frac{\tanh ^{-1}\left (\frac{2 x+5}{\sqrt{7} \sqrt{x^2+2 x+4}}\right )}{2 \sqrt{7}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^2+2 x+4}}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.0363772, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {1033, 724, 206, 688, 207} \[ -\frac{\tanh ^{-1}\left (\frac{2 x+5}{\sqrt{7} \sqrt{x^2+2 x+4}}\right )}{2 \sqrt{7}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^2+2 x+4}}{\sqrt{3}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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])

Rubi steps

\begin{align*} \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}{-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{\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*}

Mathematica [A]  time = 0.0246336, size = 61, normalized size = 0.98 \[ \frac{1}{42} \left (-3 \sqrt{7} \tanh ^{-1}\left (\frac{2 x+5}{\sqrt{7} \sqrt{x^2+2 x+4}}\right )-7 \sqrt{3} \tanh ^{-1}\left (\frac{\sqrt{(x+1)^2+3}}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-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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Maple [A]  time = 0.011, size = 49, normalized size = 0.8 \begin{align*} -{\frac{\sqrt{3}}{6}{\it Artanh} \left ({\sqrt{3}{\frac{1}{\sqrt{ \left ( 1+x \right ) ^{2}+3}}}} \right ) }-{\frac{\sqrt{7}}{14}{\it Artanh} \left ({\frac{ \left ( 10+4\,x \right ) \sqrt{7}}{14}{\frac{1}{\sqrt{ \left ( -1+x \right ) ^{2}+3+4\,x}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.46275, size = 73, normalized size = 1.18 \begin{align*} -\frac{1}{14} \, \sqrt{7} \operatorname{arsinh}\left (\frac{4 \, \sqrt{3} x}{3 \,{\left | 2 \, x - 2 \right |}} + \frac{10 \, \sqrt{3}}{3 \,{\left | 2 \, x - 2 \right |}}\right ) - \frac{1}{6} \, \sqrt{3} \operatorname{arsinh}\left (\frac{2 \, \sqrt{3}}{{\left | 2 \, x + 2 \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/14*sqrt(7)*arcsinh(4/3*sqrt(3)*x/abs(2*x - 2) + 10/3*sqrt(3)/abs(2*x - 2)) - 1/6*sqrt(3)*arcsinh(2*sqrt(3)/
abs(2*x + 2))

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Fricas [A]  time = 2.02069, size = 211, normalized size = 3.4 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2-1)/(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.153, size = 147, normalized size = 2.37 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(x^2-1)/(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))