3.728 \(\int \frac{1+2 \sqrt{1+x}}{x \sqrt{1+x} \sqrt{x+\sqrt{1+x}}} \, dx\)

Optimal. Leaf size=61 \[ 3 \tanh ^{-1}\left (\frac{1-3 \sqrt{x+1}}{2 \sqrt{x+\sqrt{x+1}}}\right )-\tan ^{-1}\left (\frac{\sqrt{x+1}+3}{2 \sqrt{x+\sqrt{x+1}}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.512715, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {1033, 724, 206, 204} \[ 3 \tanh ^{-1}\left (\frac{1-3 \sqrt{x+1}}{2 \sqrt{x+\sqrt{x+1}}}\right )-\tan ^{-1}\left (\frac{\sqrt{x+1}+3}{2 \sqrt{x+\sqrt{x+1}}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 204

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

Rubi steps

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

Mathematica [A]  time = 0.051333, size = 61, normalized size = 1. \[ \tan ^{-1}\left (\frac{-\sqrt{x+1}-3}{2 \sqrt{x+\sqrt{x+1}}}\right )-3 \tanh ^{-1}\left (\frac{3 \sqrt{x+1}-1}{2 \sqrt{x+\sqrt{x+1}}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.015, size = 68, normalized size = 1.1 \begin{align*} -3\,{\it Artanh} \left ( 1/2\,{\frac{3\,\sqrt{1+x}-1}{\sqrt{ \left ( \sqrt{1+x}-1 \right ) ^{2}+3\,\sqrt{1+x}-2}}} \right ) +\arctan \left ({\frac{1}{2} \left ( -3-\sqrt{1+x} \right ){\frac{1}{\sqrt{ \left ( 1+\sqrt{1+x} \right ) ^{2}-\sqrt{1+x}-2}}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*arctanh(1/2*(3*(1+x)^(1/2)-1)/(((1+x)^(1/2)-1)^2+3*(1+x)^(1/2)-2)^(1/2))+arctan(1/2*(-3-(1+x)^(1/2))/((1+(1
+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{2 \, \sqrt{x + 1} + 1}{\sqrt{x + \sqrt{x + 1}} \sqrt{x + 1} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((2*sqrt(x + 1) + 1)/(sqrt(x + sqrt(x + 1))*sqrt(x + 1)*x), x)

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Fricas [A]  time = 15.3547, size = 189, normalized size = 3.1 \begin{align*} \arctan \left (\frac{2 \, \sqrt{x + \sqrt{x + 1}}{\left (\sqrt{x + 1} - 3\right )}}{x - 8}\right ) + 3 \, \log \left (\frac{2 \, \sqrt{x + \sqrt{x + 1}}{\left (\sqrt{x + 1} + 1\right )} - 3 \, x - 2 \, \sqrt{x + 1} - 2}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1) - 3)/(x - 8)) + 3*log((2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1) + 1) -
3*x - 2*sqrt(x + 1) - 2)/x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{2 \sqrt{x + 1} + 1}{x \sqrt{x + 1} \sqrt{x + \sqrt{x + 1}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out