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

Optimal. Leaf size=70 \[ \sqrt {x+1} \sqrt {x+\sqrt {x+1}}+\frac {1}{2} \sqrt {x+\sqrt {x+1}}+\frac {5}{4} \log \left (2 \sqrt {x+1}-2 \sqrt {x+\sqrt {x+1}}+1\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 62, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {612, 621, 206} \begin {gather*} \frac {1}{2} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+1\right )-\frac {5}{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]]/Sqrt[1 + x],x]

[Out]

(Sqrt[x + Sqrt[1 + x]]*(1 + 2*Sqrt[1 + x]))/2 - (5*ArcTanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/4

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 612

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

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {x+\sqrt {1+x}}}{\sqrt {1+x}} \, dx &=2 \operatorname {Subst}\left (\int \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {1}{2} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )-\frac {5}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {1}{2} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )-\frac {5}{2} \operatorname {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1+2 \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )\\ &=\frac {1}{2} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )-\frac {5}{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 = 0.03, size = 64, normalized size = 0.91 \begin {gather*} 2 \left (\frac {1}{4} \sqrt {x+\sqrt {x+1}} \left (2 \sqrt {x+1}+1\right )-\frac {5}{8} \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]]/Sqrt[1 + x],x]

[Out]

2*((Sqrt[x + Sqrt[1 + x]]*(1 + 2*Sqrt[1 + x]))/4 - (5*ArcTanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/
8)

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IntegrateAlgebraic [A]  time = 0.13, size = 60, normalized size = 0.86 \begin {gather*} \frac {1}{2} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )+\frac {5}{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]]/Sqrt[1 + x],x]

[Out]

(Sqrt[x + Sqrt[1 + x]]*(1 + 2*Sqrt[1 + x]))/2 + (5*Log[-1 - 2*Sqrt[1 + x] + 2*Sqrt[x + Sqrt[1 + x]]])/4

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fricas [A]  time = 0.90, size = 56, normalized size = 0.80 \begin {gather*} \frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} + \frac {5}{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+x)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(x + sqrt(x + 1))*(2*sqrt(x + 1) + 1) + 5/8*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 = 0.26, size = 44, normalized size = 0.63 \begin {gather*} \frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} + \frac {5}{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+x)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 41, normalized size = 0.59

method result size
derivativedivides \(\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{2}-\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}\) \(41\)
default \(\frac {\left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{2}-\frac {5 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{4}\) \(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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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}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x + sqrt(x + 1))/sqrt(x + 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}} \,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),x)

[Out]

int((x + (x + 1)^(1/2))^(1/2)/(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}}}{\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)/(1+x)**(1/2),x)

[Out]

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

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