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

Optimal. Leaf size=97 \[ \sqrt {x+1} \sqrt {x+\sqrt {x+1}}-\frac {3}{2} \sqrt {x+\sqrt {x+1}}+\frac {1}{4} \log \left (2 \sqrt {x+1}-2 \sqrt {x+\sqrt {x+1}}+1\right )-4 \tan ^{-1}\left (\sqrt {x+1}-\sqrt {x+\sqrt {x+1}}+1\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 91, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {814, 843, 621, 206, 724, 204} \begin {gather*} -\frac {1}{2} \sqrt {x+\sqrt {x+1}} \left (3-2 \sqrt {x+1}\right )-2 \tan ^{-1}\left (\frac {\sqrt {x+1}+3}{2 \sqrt {x+\sqrt {x+1}}}\right )-\frac {1}{4} \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]]/(1 + Sqrt[1 + x]),x]

[Out]

-1/2*((3 - 2*Sqrt[1 + x])*Sqrt[x + Sqrt[1 + x]]) - 2*ArcTan[(3 + Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])] - Arc
Tanh[(1 + 2*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 621

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 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 814

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

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {x+\sqrt {1+x}}}{1+\sqrt {1+x}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x \sqrt {-1+x+x^2}}{1+x} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {1}{2} \left (3-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {-\frac {7}{2}+\frac {x}{2}}{(1+x) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {1}{2} \left (3-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )+2 \operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {1}{2} \left (3-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1+2 \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )-4 \operatorname {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,\frac {-3-\sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )\\ &=-\frac {1}{2} \left (3-2 \sqrt {1+x}\right ) \sqrt {x+\sqrt {1+x}}-2 \tan ^{-1}\left (\frac {3+\sqrt {1+x}}{2 \sqrt {x+\sqrt {1+x}}}\right )-\frac {1}{4} \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 = 0.07, size = 93, normalized size = 0.96 \begin {gather*} \frac {1}{4} \left (2 \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}-3\right )+8 \tan ^{-1}\left (\frac {-\sqrt {x+1}-3}{2 \sqrt {x+\sqrt {x+1}}}\right )-\tanh ^{-1}\left (\frac {2 \sqrt {x+1}+1}{2 \sqrt {x+\sqrt {x+1}}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x + Sqrt[1 + x]]*(-3 + 2*Sqrt[1 + x]) + 8*ArcTan[(-3 - Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])] - ArcTa
nh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/4

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IntegrateAlgebraic [A]  time = 0.22, size = 87, normalized size = 0.90 \begin {gather*} \frac {1}{2} \sqrt {x+\sqrt {1+x}} \left (-3+2 \sqrt {1+x}\right )-4 \tan ^{-1}\left (1+\sqrt {1+x}-\sqrt {x+\sqrt {1+x}}\right )+\frac {1}{4} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x + Sqrt[1 + x]]/(1 + Sqrt[1 + x]),x]

[Out]

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

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fricas [A]  time = 1.93, size = 82, normalized size = 0.85 \begin {gather*} \frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} - 3\right )} + 2 \, \arctan \left (\frac {2 \, \sqrt {x + \sqrt {x + 1}} {\left (\sqrt {x + 1} - 3\right )}}{x - 8}\right ) + \frac {1}{8} \, \log \left (4 \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} - 8 \, x - 8 \, \sqrt {x + 1} - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) - 3) + 2*arctan(2*sqrt(x + sqrt(x + 1))*(sqrt(x + 1) - 3)/(x - 8)) +
1/8*log(4*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) + 1) - 8*x - 8*sqrt(x + 1) - 5)

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giac [A]  time = 0.17, size = 65, normalized size = 0.67 \begin {gather*} \frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} - 3\right )} + 4 \, \arctan \left (\sqrt {x + \sqrt {x + 1}} - \sqrt {x + 1} - 1\right ) + \frac {1}{4} \, \log \left (-2 \, \sqrt {x + \sqrt {x + 1}} + 2 \, \sqrt {x + 1} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 125, normalized size = 1.29

method	result	size
derivativedivides	\(\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{2}-\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}-2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}+\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )+2 \arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )\)	\(125\)
default	\(\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{2}-\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}-2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}+\ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}\right )+2 \arctan \left (\frac {-3-\sqrt {1+x}}{2 \sqrt {\left (1+\sqrt {1+x}\right )^{2}-\sqrt {1+x}-2}}\right )\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(1+x)^(1/2)+1)*(x+(1+x)^(1/2))^(1/2)-5/4*ln(1/2+(1+x)^(1/2)+(x+(1+x)^(1/2))^(1/2))-2*((1+(1+x)^(1/2))^2
-(1+x)^(1/2)-2)^(1/2)+ln(1/2+(1+x)^(1/2)+((1+(1+x)^(1/2))^2-(1+x)^(1/2)-2)^(1/2))+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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x + \sqrt {x + 1}}}{\sqrt {x + 1} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + sqrt(x + 1))/(sqrt(x + 1) + 1), 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}}}{\sqrt {x+1}+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 + 1)^(1/2) + 1),x)

[Out]

int((x + (x + 1)^(1/2))^(1/2)/((x + 1)^(1/2) + 1), 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}}}{\sqrt {x + 1} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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