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

Optimal. Leaf size=83 \[ -\frac{\sqrt{x+\sqrt{x+1}}}{x}-\frac{1}{4} \tan ^{-1}\left (\frac{\sqrt{x+1}+3}{2 \sqrt{x+\sqrt{x+1}}}\right )+\frac{3}{4} \tanh ^{-1}\left (\frac{1-3 \sqrt{x+1}}{2 \sqrt{x+\sqrt{x+1}}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.100013, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {1014, 1033, 724, 206, 204} \[ -\frac{\sqrt{x+\sqrt{x+1}}}{x}-\frac{1}{4} \tan ^{-1}\left (\frac{\sqrt{x+1}+3}{2 \sqrt{x+\sqrt{x+1}}}\right )+\frac{3}{4} \tanh ^{-1}\left (\frac{1-3 \sqrt{x+1}}{2 \sqrt{x+\sqrt{x+1}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x + Sqrt[1 + x]]/x^2,x]

[Out]

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

Rule 1014

Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[((a*h - g*c*x)*(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q)/(2*a*c*(p + 1)), x] + Dist[2/(4*a*c*(p + 1)), Int[(a
+ c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q - 1)*Simp[g*c*d*(2*p + 3) - a*(h*e*q) + (g*c*e*(2*p + q + 3) - a*(2*h*f*
q))*x + g*c*f*(2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f, 0] && LtQ
[p, -1] && GtQ[q, 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 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{\sqrt{x+\sqrt{1+x}}}{x^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{x \sqrt{-1+x+x^2}}{\left (-1+x^2\right )^2} \, dx,x,\sqrt{1+x}\right )\\ &=-\frac{\sqrt{x+\sqrt{1+x}}}{x}+\operatorname{Subst}\left (\int \frac{\frac{1}{2}+x}{\left (-1+x^2\right ) \sqrt{-1+x+x^2}} \, dx,x,\sqrt{1+x}\right )\\ &=-\frac{\sqrt{x+\sqrt{1+x}}}{x}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{-1+x+x^2}} \, dx,x,\sqrt{1+x}\right )+\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{-1+x+x^2}} \, dx,x,\sqrt{1+x}\right )\\ &=-\frac{\sqrt{x+\sqrt{1+x}}}{x}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-4-x^2} \, dx,x,\frac{-3-\sqrt{1+x}}{\sqrt{x+\sqrt{1+x}}}\right )-\frac{3}{2} \operatorname{Subst}\left (\int \frac{1}{4-x^2} \, dx,x,\frac{-1+3 \sqrt{1+x}}{\sqrt{x+\sqrt{1+x}}}\right )\\ &=-\frac{\sqrt{x+\sqrt{1+x}}}{x}-\frac{1}{4} \tan ^{-1}\left (\frac{3+\sqrt{1+x}}{2 \sqrt{x+\sqrt{1+x}}}\right )+\frac{3}{4} \tanh ^{-1}\left (\frac{1-3 \sqrt{1+x}}{2 \sqrt{x+\sqrt{1+x}}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0337009, size = 85, normalized size = 1.02 \[ -\frac{\sqrt{x+\sqrt{x+1}}}{x}+\frac{1}{4} \tan ^{-1}\left (\frac{-\sqrt{x+1}-3}{2 \sqrt{x+\sqrt{x+1}}}\right )-\frac{3}{4} \tanh ^{-1}\left (\frac{3 \sqrt{x+1}-1}{2 \sqrt{x+\sqrt{x+1}}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x + Sqrt[1 + x]]/x^2,x]

[Out]

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

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Maple [B]  time = 0.011, size = 298, normalized size = 3.6 \begin{align*} -{\frac{1}{2} \left ( \left ( -1+\sqrt{1+x} \right ) ^{2}-2+3\,\sqrt{1+x} \right ) ^{{\frac{3}{2}}} \left ( -1+\sqrt{1+x} \right ) ^{-1}}+{\frac{3}{4}\sqrt{ \left ( -1+\sqrt{1+x} \right ) ^{2}-2+3\,\sqrt{1+x}}}+{\frac{1}{2}\ln \left ({\frac{1}{2}}+\sqrt{1+x}+\sqrt{ \left ( -1+\sqrt{1+x} \right ) ^{2}-2+3\,\sqrt{1+x}} \right ) }-{\frac{3}{4}{\it Artanh} \left ({\frac{1}{2} \left ( -1+3\,\sqrt{1+x} \right ){\frac{1}{\sqrt{ \left ( -1+\sqrt{1+x} \right ) ^{2}-2+3\,\sqrt{1+x}}}}} \right ) }+{\frac{1}{4} \left ( 2\,\sqrt{1+x}+1 \right ) \sqrt{ \left ( -1+\sqrt{1+x} \right ) ^{2}-2+3\,\sqrt{1+x}}}-{\frac{1}{2} \left ( \left ( 1+\sqrt{1+x} \right ) ^{2}-2-\sqrt{1+x} \right ) ^{{\frac{3}{2}}} \left ( 1+\sqrt{1+x} \right ) ^{-1}}-{\frac{1}{4}\sqrt{ \left ( 1+\sqrt{1+x} \right ) ^{2}-2-\sqrt{1+x}}}-{\frac{1}{2}\ln \left ({\frac{1}{2}}+\sqrt{1+x}+\sqrt{ \left ( 1+\sqrt{1+x} \right ) ^{2}-2-\sqrt{1+x}} \right ) }+{\frac{1}{4}\arctan \left ({\frac{1}{2} \left ( -3-\sqrt{1+x} \right ){\frac{1}{\sqrt{ \left ( 1+\sqrt{1+x} \right ) ^{2}-2-\sqrt{1+x}}}}} \right ) }+{\frac{1}{4} \left ( 2\,\sqrt{1+x}+1 \right ) \sqrt{ \left ( 1+\sqrt{1+x} \right ) ^{2}-2-\sqrt{1+x}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(-1+(1+x)^(1/2))*((-1+(1+x)^(1/2))^2-2+3*(1+x)^(1/2))^(3/2)+3/4*((-1+(1+x)^(1/2))^2-2+3*(1+x)^(1/2))^(1/2
)+1/2*ln(1/2+(1+x)^(1/2)+((-1+(1+x)^(1/2))^2-2+3*(1+x)^(1/2))^(1/2))-3/4*arctanh(1/2*(-1+3*(1+x)^(1/2))/((-1+(
1+x)^(1/2))^2-2+3*(1+x)^(1/2))^(1/2))+1/4*(2*(1+x)^(1/2)+1)*((-1+(1+x)^(1/2))^2-2+3*(1+x)^(1/2))^(1/2)-1/2/(1+
(1+x)^(1/2))*((1+(1+x)^(1/2))^2-2-(1+x)^(1/2))^(3/2)-1/4*((1+(1+x)^(1/2))^2-2-(1+x)^(1/2))^(1/2)-1/2*ln(1/2+(1
+x)^(1/2)+((1+(1+x)^(1/2))^2-2-(1+x)^(1/2))^(1/2))+1/4*arctan(1/2*(-3-(1+x)^(1/2))/((1+(1+x)^(1/2))^2-2-(1+x)^
(1/2))^(1/2))+1/4*(2*(1+x)^(1/2)+1)*((1+(1+x)^(1/2))^2-2-(1+x)^(1/2))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 29.4364, size = 240, normalized size = 2.89 \begin{align*} \frac{x \arctan \left (\frac{2 \, \sqrt{x + \sqrt{x + 1}}{\left (\sqrt{x + 1} - 3\right )}}{x - 8}\right ) + 3 \, x \log \left (\frac{2 \, \sqrt{x + \sqrt{x + 1}}{\left (\sqrt{x + 1} + 1\right )} - 3 \, x - 2 \, \sqrt{x + 1} - 2}{x}\right ) - 4 \, \sqrt{x + \sqrt{x + 1}}}{4 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(x*arctan(2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1) - 3)/(x - 8)) + 3*x*log((2*sqrt(x + sqrt(x + 1))*(sqrt(x +
1) + 1) - 3*x - 2*sqrt(x + 1) - 2)/x) - 4*sqrt(x + sqrt(x + 1)))/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out