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

Optimal. Leaf size=110 \[ -\frac {\sqrt {x+1} \sqrt {x+\sqrt {x+1}}}{x}+\frac {3}{2} \tan ^{-1}\left (\frac {\sqrt {x+1}-\sqrt {x+\sqrt {x+1}}}{\sqrt {x+1}-\sqrt {x+\sqrt {x+1}}+2}\right )-\frac {1}{2} \tanh ^{-1}\left (-\sqrt {x+1}+\sqrt {x+\sqrt {x+1}}+1\right ) \]

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Rubi [A]  time = 0.37, antiderivative size = 96, normalized size of antiderivative = 0.87, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {976, 1021, 1033, 724, 206, 204} \begin {gather*} -\frac {\left (x+\sqrt {x+1}\right )^{3/2}}{x}+\sqrt {x+\sqrt {x+1}}+\frac {3}{4} \tan ^{-1}\left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )+\frac {1}{4} \tanh ^{-1}\left (\frac {1-3 \sqrt {x+1}}{2 \sqrt {x+\sqrt {x+1}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 976

Int[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((2*a*c^2*e + c*(2
*c^2*d - c*(2*a*f))*x)*(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1))/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p +
1)), x] - Dist[1/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1)), Int[(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Si
mp[2*c*((c*d - a*f)^2 - (-(a*e))*(c*e))*(p + 1) - (2*c^2*d - c*(2*a*f))*(a*f*(p + 1) - c*d*(p + 2)) - e*(-2*a*
c^2*e)*(p + q + 2) + (2*f*(2*a*c^2*e)*(p + q + 2) - (2*c^2*d - c*(2*a*f))*(-(c*e*(2*p + q + 4))))*x + c*f*(2*c
^2*d - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, q}, x] && NeQ[e^2 - 4*d*f, 0] && Lt
Q[p, -1] && NeQ[a*c*e^2 + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &&  !IGtQ[q, 0]

Rule 1021

Int[((g_.) + (h_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[(h*(a + b*x + c*x^2)^p*(d + f*x^2)^(q + 1))/(2*f*(p + q + 1)), x] - Dist[1/(2*f*(p + q + 1)), Int[(a + b*x +
c*x^2)^(p - 1)*(d + f*x^2)^q*Simp[h*p*(b*d) + a*(-2*g*f)*(p + q + 1) + (2*h*p*(c*d - a*f) + b*(-2*g*f)*(p + q
+ 1))*x + (h*p*(-(b*f)) + c*(-2*g*f)*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, g, h, q}, x] && NeQ
[b^2 - 4*a*c, 0] && GtQ[p, 0] && NeQ[p + q + 1, 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)]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 88, normalized size = 0.80 \begin {gather*} \frac {1}{4} \left (-\frac {4 \sqrt {x+1} \sqrt {x+\sqrt {x+1}}}{x}+3 \tan ^{-1}\left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )+\tanh ^{-1}\left (\frac {1-3 \sqrt {x+1}}{2 \sqrt {x+\sqrt {x+1}}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.28, size = 84, normalized size = 0.76 \begin {gather*} -\frac {\sqrt {1+x} \sqrt {x+\sqrt {1+x}}}{x}+\frac {3}{2} \tan ^{-1}\left (1+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}\right )-\frac {1}{2} \tanh ^{-1}\left (1-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 2.44, size = 87, normalized size = 0.79 \begin {gather*} -\frac {3 \, x \arctan \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} - 3\right )}}{x - 8}\right ) - 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}} \sqrt {x + 1}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.43, size = 188, normalized size = 1.71 \begin {gather*} -\frac {{\left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1}\right )}^{3} - 2 \, {\left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1}\right )}^{2} + 5 \, \sqrt {x + \sqrt {x + 1}} - 5 \, \sqrt {x + 1} + 1}{{\left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1}\right )}^{4} - 2 \, {\left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1}\right )}^{2} + 4 \, \sqrt {x + \sqrt {x + 1}} - 4 \, \sqrt {x + 1}} - \frac {3}{2} \, \arctan \left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} - 1\right ) - \frac {1}{4} \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} + 2 \right |}\right ) + \frac {1}{4} \, \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((sqrt(x + sqrt(x + 1)) - sqrt(x + 1))^3 - 2*(sqrt(x + sqrt(x + 1)) - sqrt(x + 1))^2 + 5*sqrt(x + sqrt(x + 1)
) - 5*sqrt(x + 1) + 1)/((sqrt(x + sqrt(x + 1)) - sqrt(x + 1))^4 - 2*(sqrt(x + sqrt(x + 1)) - sqrt(x + 1))^2 +
4*sqrt(x + sqrt(x + 1)) - 4*sqrt(x + 1)) - 3/2*arctan(sqrt(x + sqrt(x + 1)) - sqrt(x + 1) - 1) - 1/4*log(abs(s
qrt(x + sqrt(x + 1)) - sqrt(x + 1) + 2)) + 1/4*log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1)))

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maple [B]  time = 0.07, size = 298, normalized size = 2.71

method	result	size
derivativedivides	\(-\frac {\left (\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2\right )^{\frac {3}{2}}}{2 \left (-1+\sqrt {1+x}\right )}+\frac {\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}{4}-\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}\right )}{4}-\frac {\arctanh \left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}\right )}{4}+\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}{4}+\frac {\left (\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2\right )^{\frac {3}{2}}}{2+2 \sqrt {1+x}}+\frac {3 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )}{4}-\frac {3 \arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )}{4}-\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}{4}\)	\(298\)
default	\(-\frac {\left (\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2\right )^{\frac {3}{2}}}{2 \left (-1+\sqrt {1+x}\right )}+\frac {\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}{4}-\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}\right )}{4}-\frac {\arctanh \left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}\right )}{4}+\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}{4}+\frac {\left (\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2\right )^{\frac {3}{2}}}{2+2 \sqrt {1+x}}+\frac {3 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}{4}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )}{4}-\frac {3 \arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )}{4}-\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}{4}\)	\(298\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+(1+x)^(1/2))^(1/2)/x^2/(1+x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x + \sqrt {x + 1}}}{\sqrt {x + 1} x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x+\sqrt {x+1}}}{x^2\,\sqrt {x+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x + \sqrt {x + 1}}}{x^{2} \sqrt {x + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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