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

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

[Out]

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

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Rubi [A]  time = 0.0465202, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {732, 843, 619, 215, 724, 206} \[ \frac{\sqrt{x^2+2 x+4}}{1-x}-\frac{2 \tanh ^{-1}\left (\frac{2 x+5}{\sqrt{7} \sqrt{x^2+2 x+4}}\right )}{\sqrt{7}}+\sinh ^{-1}\left (\frac{x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[4 + 2*x + x^2]/(-1 + x)^2,x]

[Out]

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

Rule 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ
[2*c*d - b*e, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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])

Rubi steps

\begin{align*} \int \frac{\sqrt{4+2 x+x^2}}{(-1+x)^2} \, dx &=\frac{\sqrt{4+2 x+x^2}}{1-x}+\frac{1}{2} \int \frac{2+2 x}{(-1+x) \sqrt{4+2 x+x^2}} \, dx\\ &=\frac{\sqrt{4+2 x+x^2}}{1-x}+2 \int \frac{1}{(-1+x) \sqrt{4+2 x+x^2}} \, dx+\int \frac{1}{\sqrt{4+2 x+x^2}} \, dx\\ &=\frac{\sqrt{4+2 x+x^2}}{1-x}-4 \operatorname{Subst}\left (\int \frac{1}{28-x^2} \, dx,x,\frac{10+4 x}{\sqrt{4+2 x+x^2}}\right )+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{12}}} \, dx,x,2+2 x\right )}{2 \sqrt{3}}\\ &=\frac{\sqrt{4+2 x+x^2}}{1-x}+\sinh ^{-1}\left (\frac{1+x}{\sqrt{3}}\right )-\frac{2 \tanh ^{-1}\left (\frac{5+2 x}{\sqrt{7} \sqrt{4+2 x+x^2}}\right )}{\sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0382395, size = 61, normalized size = 0.98 \[ -\frac{\sqrt{x^2+2 x+4}}{x-1}-\frac{2 \tanh ^{-1}\left (\frac{2 x+5}{\sqrt{7} \sqrt{x^2+2 x+4}}\right )}{\sqrt{7}}+\sinh ^{-1}\left (\frac{x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 91, normalized size = 1.5 \begin{align*} -{\frac{1}{-7+7\,x} \left ( \left ( -1+x \right ) ^{2}+3+4\,x \right ) ^{{\frac{3}{2}}}}+{\frac{2}{7}\sqrt{ \left ( -1+x \right ) ^{2}+3+4\,x}}+{\it Arcsinh} \left ({\frac{ \left ( 1+x \right ) \sqrt{3}}{3}} \right ) -{\frac{2\,\sqrt{7}}{7}{\it Artanh} \left ({\frac{ \left ( 10+4\,x \right ) \sqrt{7}}{14}{\frac{1}{\sqrt{ \left ( -1+x \right ) ^{2}+3+4\,x}}}} \right ) }+{\frac{2\,x+2}{14}\sqrt{ \left ( -1+x \right ) ^{2}+3+4\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.45849, size = 82, normalized size = 1.32 \begin{align*} -\frac{2}{7} \, \sqrt{7} \operatorname{arsinh}\left (\frac{2 \, \sqrt{3} x}{3 \,{\left | x - 1 \right |}} + \frac{5 \, \sqrt{3}}{3 \,{\left | x - 1 \right |}}\right ) - \frac{\sqrt{x^{2} + 2 \, x + 4}}{x - 1} + \operatorname{arsinh}\left (\frac{1}{3} \, \sqrt{3} x + \frac{1}{3} \, \sqrt{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*sqrt(7)*arcsinh(2/3*sqrt(3)*x/abs(x - 1) + 5/3*sqrt(3)/abs(x - 1)) - sqrt(x^2 + 2*x + 4)/(x - 1) + arcsin
h(1/3*sqrt(3)*x + 1/3*sqrt(3))

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Fricas [A]  time = 1.9008, size = 263, normalized size = 4.24 \begin{align*} \frac{2 \, \sqrt{7}{\left (x - 1\right )} \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 ) - 7 \,{\left (x - 1\right )} \log \left (-x + \sqrt{x^{2} + 2 \, x + 4} - 1\right ) - 7 \, x - 7 \, \sqrt{x^{2} + 2 \, x + 4} + 7}{7 \,{\left (x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(2*sqrt(7)*(x - 1)*log((sqrt(7)*(2*x + 5) + sqrt(x^2 + 2*x + 4)*(2*sqrt(7) - 7) - 4*x - 10)/(x - 1)) - 7*(
x - 1)*log(-x + sqrt(x^2 + 2*x + 4) - 1) - 7*x - 7*sqrt(x^2 + 2*x + 4) + 7)/(x - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.15964, size = 204, normalized size = 3.29 \begin{align*} -\frac{2}{7} \, \sqrt{7} \log \left (2 \, \sqrt{7} + 7 \, \sqrt{\frac{4}{x - 1} + \frac{7}{{\left (x - 1\right )}^{2}} + 1} + \frac{7 \, \sqrt{7}}{x - 1}\right ) \mathrm{sgn}\left (\frac{1}{x - 1}\right ) + \log \left (\sqrt{\frac{4}{x - 1} + \frac{7}{{\left (x - 1\right )}^{2}} + 1} + \frac{\sqrt{7}}{x - 1} + 1\right ) \mathrm{sgn}\left (\frac{1}{x - 1}\right ) - \log \left ({\left | \sqrt{\frac{4}{x - 1} + \frac{7}{{\left (x - 1\right )}^{2}} + 1} + \frac{\sqrt{7}}{x - 1} - 1 \right |}\right ) \mathrm{sgn}\left (\frac{1}{x - 1}\right ) - \sqrt{\frac{4}{x - 1} + \frac{7}{{\left (x - 1\right )}^{2}} + 1} \mathrm{sgn}\left (\frac{1}{x - 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*sqrt(7)*log(2*sqrt(7) + 7*sqrt(4/(x - 1) + 7/(x - 1)^2 + 1) + 7*sqrt(7)/(x - 1))*sgn(1/(x - 1)) + log(sqr
t(4/(x - 1) + 7/(x - 1)^2 + 1) + sqrt(7)/(x - 1) + 1)*sgn(1/(x - 1)) - log(abs(sqrt(4/(x - 1) + 7/(x - 1)^2 +
1) + sqrt(7)/(x - 1) - 1))*sgn(1/(x - 1)) - sqrt(4/(x - 1) + 7/(x - 1)^2 + 1)*sgn(1/(x - 1))