3.17.26 \(\int \frac {\sqrt {x+\sqrt {1+x}}}{x \sqrt {1+x}} \, dx\) [1626]

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

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 90, normalized size of antiderivative = 0.82, number of steps used = 9, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1004, 635, 212, 1047, 738, 210} \begin {gather*} -\text {ArcTan}\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 )+2 \tanh ^{-1}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 210

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

Rule 212

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

Rule 635

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

Rule 738

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 1004

Int[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (f_.)*(x_)^2), x_Symbol] :> Dist[c/f, Int[1/Sqrt[a + b*x +
c*x^2], x], x] - Dist[1/f, Int[(c*d - a*f - b*f*x)/(Sqrt[a + b*x + c*x^2]*(d + f*x^2)), x], x] /; FreeQ[{a, b,
c, d, f}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1047

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

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Mathematica [A]
time = 0.11, size = 84, normalized size = 0.76 \begin {gather*} -2 \text {ArcTan}\left (1+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}\right )-2 \tanh ^{-1}\left (1-\sqrt {1+x}+\sqrt {x+\sqrt {1+x}}\right )-2 \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2*ArcTan[1 + Sqrt[1 + x] - Sqrt[x + Sqrt[1 + x]]] - 2*ArcTanh[1 - Sqrt[1 + x] + Sqrt[x + Sqrt[1 + x]]] - 2*Lo
g[-1 - 2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt[1 + x]]]

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Maple [A]
time = 0.04, size = 170, normalized size = 1.55

method	result	size
derivativedivides	\(-\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )}{2}+\arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}+\frac {3 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}\right )}{2}-\arctanh \left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}\right )\)	\(170\)
default	\(-\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}+\frac {\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )}{2}+\arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}+\frac {3 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}\right )}{2}-\arctanh \left (\frac {-1+3 \sqrt {1+x}}{2 \sqrt {\left (-1+\sqrt {1+x}\right )^{2}+3 \sqrt {1+x}-2}}\right )\)	\(170\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((1+(1+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2)+1/2*ln(1/2+(1+x)^(1/2)+((1+(1+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2))+arcta
n(1/2*(-3-(1+x)^(1/2))/((1+(1+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2))+((-1+(1+x)^(1/2))^2+3*(1+x)^(1/2)-2)^(1/2)+3/2
*ln(1/2+(1+x)^(1/2)+((-1+(1+x)^(1/2))^2+3*(1+x)^(1/2)-2)^(1/2))-arctanh(1/2*(-1+3*(1+x)^(1/2))/((-1+(1+x)^(1/2
))^2+3*(1+x)^(1/2)-2)^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.65, size = 74, normalized size = 0.67 \begin {gather*} \arctan \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} - 3\right )}}{x - 8}\right ) + \log \left (\frac {8 \, x^{2} + 2 \, {\left ({\left (4 \, x - 1\right )} \sqrt {x + 1} - 2 \, x - 1\right )} \sqrt {x + \sqrt {x + 1}} - x + 2 \, \sqrt {x + 1} + 2}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1) - 3)/(x - 8)) + log((8*x^2 + 2*((4*x - 1)*sqrt(x + 1) - 2*x - 1)*s
qrt(x + sqrt(x + 1)) - x + 2*sqrt(x + 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 \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/(1+x)**(1/2),x)

[Out]

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

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Giac [A]
time = 0.55, size = 86, normalized size = 0.78 \begin {gather*} 2 \, \arctan \left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} - 1\right ) - 2 \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1}} + 2 \, \sqrt {x + 1} + 1\right ) - \log \left ({\left | \sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} + 2 \right |}\right ) + \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/(1+x)^(1/2),x, algorithm="giac")

[Out]

2*arctan(sqrt(x + sqrt(x + 1)) - sqrt(x + 1) - 1) - 2*log(-2*sqrt(x + sqrt(x + 1)) + 2*sqrt(x + 1) + 1) - log(
abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1) + 2)) + log(abs(sqrt(x + sqrt(x + 1)) - sqrt(x + 1)))

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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\,\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*(x + 1)^(1/2)),x)

[Out]

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

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