3.12.21 \(\int \sqrt {1+x} \sqrt {x+\sqrt {1+x}} \, dx\) [1121]

Optimal. Leaf size=83 \[ \frac {1}{96} (75+8 x) \sqrt {x+\sqrt {1+x}}+\frac {1}{48} \sqrt {1+x} (7+24 x) \sqrt {x+\sqrt {1+x}}+\frac {45}{64} \log \left (1+2 \sqrt {1+x}-2 \sqrt {x+\sqrt {1+x}}\right ) \]

[Out]

1/96*(75+8*x)*(x+(1+x)^(1/2))^(1/2)+1/48*(1+x)^(1/2)*(7+24*x)*(x+(1+x)^(1/2))^(1/2)+45/64*ln(1+2*(1+x)^(1/2)-2
*(x+(1+x)^(1/2))^(1/2))

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Rubi [A]
time = 0.07, antiderivative size = 103, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {756, 654, 626, 635, 212} \begin {gather*} \frac {1}{2} \sqrt {x+1} \left (x+\sqrt {x+1}\right )^{3/2}-\frac {5}{12} \left (x+\sqrt {x+1}\right )^{3/2}+\frac {9}{32} \left (2 \sqrt {x+1}+1\right ) \sqrt {x+\sqrt {x+1}}-\frac {45}{64} \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[1 + x]*Sqrt[x + Sqrt[1 + x]],x]

[Out]

(-5*(x + Sqrt[1 + x])^(3/2))/12 + (Sqrt[1 + x]*(x + Sqrt[1 + x])^(3/2))/2 + (9*Sqrt[x + Sqrt[1 + x]]*(1 + 2*Sq
rt[1 + x]))/32 - (45*ArcTanh[(1 + 2*Sqrt[1 + x])/(2*Sqrt[x + Sqrt[1 + x]])])/64

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 626

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 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 654

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

Rule 756

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

Rubi steps

\begin {align*} \int \sqrt {1+x} \sqrt {x+\sqrt {1+x}} \, dx &=2 \text {Subst}\left (\int x^2 \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \text {Subst}\left (\int \left (1-\frac {5 x}{2}\right ) \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {5}{12} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {9}{8} \text {Subst}\left (\int \sqrt {-1+x+x^2} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {5}{12} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {9}{32} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )-\frac {45}{64} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x+x^2}} \, dx,x,\sqrt {1+x}\right )\\ &=-\frac {5}{12} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {9}{32} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )-\frac {45}{32} \text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1+2 \sqrt {1+x}}{\sqrt {x+\sqrt {1+x}}}\right )\\ &=-\frac {5}{12} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {1}{2} \sqrt {1+x} \left (x+\sqrt {1+x}\right )^{3/2}+\frac {9}{32} \sqrt {x+\sqrt {1+x}} \left (1+2 \sqrt {1+x}\right )-\frac {45}{64} \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 = 0.09, size = 74, normalized size = 0.89 \begin {gather*} \frac {1}{96} \sqrt {x+\sqrt {1+x}} \left (67-34 \sqrt {1+x}+8 (1+x)+48 (1+x)^{3/2}\right )+\frac {45}{64} \log \left (-1-2 \sqrt {1+x}+2 \sqrt {x+\sqrt {1+x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x]*Sqrt[x + Sqrt[1 + x]],x]

[Out]

(Sqrt[x + Sqrt[1 + x]]*(67 - 34*Sqrt[1 + x] + 8*(1 + x) + 48*(1 + x)^(3/2)))/96 + (45*Log[-1 - 2*Sqrt[1 + x] +
2*Sqrt[x + Sqrt[1 + x]]])/64

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Maple [A]
time = 0.44, size = 68, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {\sqrt {1+x}\, \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{2}-\frac {5 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{12}+\frac {9 \left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{32}-\frac {45 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{64}\)	\(68\)
default	\(\frac {\sqrt {1+x}\, \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{2}-\frac {5 \left (x +\sqrt {1+x}\right )^{\frac {3}{2}}}{12}+\frac {9 \left (2 \sqrt {1+x}+1\right ) \sqrt {x +\sqrt {1+x}}}{32}-\frac {45 \ln \left (\frac {1}{2}+\sqrt {1+x}+\sqrt {x +\sqrt {1+x}}\right )}{64}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(1+x)^(1/2)*(x+(1+x)^(1/2))^(3/2)-5/12*(x+(1+x)^(1/2))^(3/2)+9/32*(2*(1+x)^(1/2)+1)*(x+(1+x)^(1/2))^(1/2)-
45/64*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*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.62, size = 64, normalized size = 0.77 \begin {gather*} \frac {1}{96} \, {\left (2 \, {\left (24 \, x + 7\right )} \sqrt {x + 1} + 8 \, x + 75\right )} \sqrt {x + \sqrt {x + 1}} + \frac {45}{128} \, \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((1+x)^(1/2)*(x+(1+x)^(1/2))^(1/2),x, algorithm="fricas")

[Out]

1/96*(2*(24*x + 7)*sqrt(x + 1) + 8*x + 75)*sqrt(x + sqrt(x + 1)) + 45/128*log(4*sqrt(x + sqrt(x + 1))*(2*sqrt(
x + 1) + 1) - 8*x - 8*sqrt(x + 1) - 5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 82, normalized size = 0.99 \begin {gather*} \frac {1}{96} \, {\left (2 \, {\left (4 \, \sqrt {x + 1} {\left (6 \, \sqrt {x + 1} + 1\right )} - 65\right )} \sqrt {x + 1} + 19\right )} \sqrt {x + \sqrt {x + 1}} + \frac {1}{2} \, \sqrt {x + \sqrt {x + 1}} {\left (2 \, \sqrt {x + 1} + 1\right )} + \frac {45}{64} \, \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((1+x)^(1/2)*(x+(1+x)^(1/2))^(1/2),x, algorithm="giac")

[Out]

1/96*(2*(4*sqrt(x + 1)*(6*sqrt(x + 1) + 1) - 65)*sqrt(x + 1) + 19)*sqrt(x + sqrt(x + 1)) + 1/2*sqrt(x + sqrt(x
+ 1))*(2*sqrt(x + 1) + 1) + 45/64*log(-2*sqrt(x + sqrt(x + 1)) + 2*sqrt(x + 1) + 1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \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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