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

Optimal. Leaf size=184 \[ \frac {8}{5} \sqrt {x+1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}-\frac {32}{15} \sqrt {\sqrt {x+1}+1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}+\frac {88}{15} \sqrt {\sqrt {\sqrt {x+1}+1}+1}-2 \sqrt {2 \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {\sqrt {2}-1}}\right )-2 \sqrt {2 \left (\sqrt {2}-1\right )} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {1+\sqrt {2}}}\right ) \]

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Rubi [A]  time = 1.75, antiderivative size = 162, normalized size of antiderivative = 0.88, number of steps used = 10, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {1586, 898, 1287, 1093, 207, 203} \begin {gather*} \frac {8}{5} \left (\sqrt {\sqrt {x+1}+1}+1\right )^{5/2}-\frac {16}{3} \left (\sqrt {\sqrt {x+1}+1}+1\right )^{3/2}+8 \sqrt {\sqrt {\sqrt {x+1}+1}+1}-2 \sqrt {2 \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {\sqrt {2}-1}}\right )-2 \sqrt {2 \left (\sqrt {2}-1\right )} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {1+\sqrt {2}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

8*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]] - (16*(1 + Sqrt[1 + Sqrt[1 + x]])^(3/2))/3 + (8*(1 + Sqrt[1 + Sqrt[1 + x]])^
(5/2))/5 - 2*Sqrt[2*(1 + Sqrt[2])]*ArcTan[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]/Sqrt[-1 + Sqrt[2]]] - 2*Sqrt[2*(-1 +
Sqrt[2])]*ArcTanh[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]/Sqrt[1 + Sqrt[2]]]

Rule 203

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

Rule 207

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

Rule 898

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1093

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

Rule 1287

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Int[Ex
pandIntegrand[((f*x)^m*(d + e*x^2)^q)/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^
2 - 4*a*c, 0] && IntegerQ[q] && IntegerQ[m]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^2 \sqrt {1+x}}{\left (-1+x^2\right ) \sqrt {1+\sqrt {1+x}}} \, dx,x,\sqrt {1+x}\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x^2}{(-1+x) \sqrt {1+x} \sqrt {1+\sqrt {1+x}}} \, dx,x,\sqrt {1+x}\right )\\ &=4 \operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{\sqrt {1+x} \left (-2+x^2\right )} \, dx,x,\sqrt {1+\sqrt {1+x}}\right )\\ &=4 \operatorname {Subst}\left (\int \frac {(-1+x)^2 (1+x)^{3/2}}{-2+x^2} \, dx,x,\sqrt {1+\sqrt {1+x}}\right )\\ &=8 \operatorname {Subst}\left (\int \frac {x^4 \left (-2+x^2\right )^2}{-1-2 x^2+x^4} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=8 \operatorname {Subst}\left (\int \left (1-2 x^2+x^4+\frac {1}{-1-2 x^2+x^4}\right ) \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {16}{3} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}+\frac {8}{5} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{5/2}+8 \operatorname {Subst}\left (\int \frac {1}{-1-2 x^2+x^4} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {16}{3} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}+\frac {8}{5} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{5/2}+\left (2 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-\sqrt {2}+x^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-\left (2 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\sqrt {2}+x^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {16}{3} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}+\frac {8}{5} \left (1+\sqrt {1+\sqrt {1+x}}\right )^{5/2}-2 \sqrt {2 \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {-1+\sqrt {2}}}\right )-2 \sqrt {2 \left (-1+\sqrt {2}\right )} \tanh ^{-1}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right )\\ \end {align*}

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Mathematica [A]  time = 3.53, size = 146, normalized size = 0.79 \begin {gather*} 2 \left (\frac {4}{15} \sqrt {\sqrt {\sqrt {x+1}+1}+1} \left (3 \sqrt {x+1}-4 \sqrt {\sqrt {x+1}+1}+11\right )-\sqrt {2 \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )-\sqrt {2 \left (\sqrt {2}-1\right )} \tanh ^{-1}\left (\sqrt {\sqrt {2}-1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*((4*(11 + 3*Sqrt[1 + x] - 4*Sqrt[1 + Sqrt[1 + x]])*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])/15 - Sqrt[2*(1 + Sqrt[2]
)]*ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]] - Sqrt[2*(-1 + Sqrt[2])]*ArcTanh[Sqrt[-1 + Sqrt[2
]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]])

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IntegrateAlgebraic [A]  time = 0.28, size = 165, normalized size = 0.90 \begin {gather*} -\frac {32}{15} \sqrt {1+\sqrt {1+x}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}+\frac {8}{15} \left (11+3 \sqrt {1+x}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}-2 \sqrt {2 \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-2 \sqrt {2 \left (-1+\sqrt {2}\right )} \tanh ^{-1}\left (\sqrt {-1+\sqrt {2}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*Sqrt[1 + Sqrt[1 + x]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])/15 + (8*(11 + 3*Sqrt[1 + x])*Sqrt[1 + Sqrt[1 + Sqr
t[1 + x]]])/15 - 2*Sqrt[2*(1 + Sqrt[2])]*ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]] - 2*Sqrt[2*
(-1 + Sqrt[2])]*ArcTanh[Sqrt[-1 + Sqrt[2]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]]

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fricas [A]  time = 0.61, size = 186, normalized size = 1.01 \begin {gather*} 4 \, \sqrt {2} \sqrt {\sqrt {2} + 1} \arctan \left (\sqrt {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}} \sqrt {\sqrt {2} + 1} - \sqrt {\sqrt {2} + 1} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) - \sqrt {2} \sqrt {\sqrt {2} - 1} \log \left (2 \, \sqrt {2} {\left (\sqrt {2} + 2\right )} \sqrt {\sqrt {2} - 1} + 4 \, \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) + \sqrt {2} \sqrt {\sqrt {2} - 1} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} + 2\right )} \sqrt {\sqrt {2} - 1} + 4 \, \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) + \frac {8}{15} \, {\left (3 \, \sqrt {x + 1} - 4 \, \sqrt {\sqrt {x + 1} + 1} + 11\right )} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*sqrt(2)*sqrt(sqrt(2) + 1)*arctan(sqrt(sqrt(2) + sqrt(sqrt(x + 1) + 1))*sqrt(sqrt(2) + 1) - sqrt(sqrt(2) + 1)
*sqrt(sqrt(sqrt(x + 1) + 1) + 1)) - sqrt(2)*sqrt(sqrt(2) - 1)*log(2*sqrt(2)*(sqrt(2) + 2)*sqrt(sqrt(2) - 1) +
4*sqrt(sqrt(sqrt(x + 1) + 1) + 1)) + sqrt(2)*sqrt(sqrt(2) - 1)*log(-2*sqrt(2)*(sqrt(2) + 2)*sqrt(sqrt(2) - 1)
+ 4*sqrt(sqrt(sqrt(x + 1) + 1) + 1)) + 8/15*(3*sqrt(x + 1) - 4*sqrt(sqrt(x + 1) + 1) + 11)*sqrt(sqrt(sqrt(x +
1) + 1) + 1)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-2,[1
]%%%}+%%%{-4,[0]%%%},%%%{-4,[1]%%%}+%%%{-4,[0]%%%},%%%{1,[2]%%%}+%%%{-1,[1]%%%}+%%%{-1,[0]%%%}] at parameters
values [-72]Warning, choosing root of [1,0,%%%{-2,[1]%%%}+%%%{-4,[0]%%%},%%%{-4,[1]%%%}+%%%{-4,[0]%%%},%%%{1,[
2]%%%}+%%%{-1,[1]%%%}+%%%{-1,[0]%%%}] at parameters values [11]Warning, need to choose a branch for the root o
f a polynomial with parameters. This might be wrong.The choice was done assuming [x]=[91]4*(2/5*sqrt(sqrt(sqrt
(x+1)+1)+1)*(sqrt(sqrt(x+1)+1)+1)^2-4/3*sqrt(sqrt(sqrt(x+1)+1)+1)*(sqrt(sqrt(x+1)+1)+1)+2*sqrt(sqrt(sqrt(x+1)+
1)+1)+sqrt(2*(sqrt(2)-1))/4*ln(abs(sqrt(sqrt(sqrt(x+1)+1)+1)-sqrt((2+2*sqrt(2))/2)))-sqrt(2*(sqrt(2)-1))/4*ln(
sqrt(sqrt(sqrt(x+1)+1)+1)+sqrt((2+2*sqrt(2))/2))-sqrt(2*(sqrt(2)+1))/2*atan(sqrt(sqrt(sqrt(x+1)+1)+1)/sqrt(-(2
-2*sqrt(2))/2)))

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maple [A]  time = 0.20, size = 115, normalized size = 0.62

method	result	size
derivativedivides	\(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}-\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}+8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {2 \sqrt {2}\, \arctanh \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}-\frac {2 \sqrt {2}\, \arctan \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {\sqrt {2}-1}}\right )}{\sqrt {\sqrt {2}-1}}\)	\(115\)
default	\(\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {5}{2}}}{5}-\frac {16 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{3}+8 \sqrt {1+\sqrt {1+\sqrt {1+x}}}-\frac {2 \sqrt {2}\, \arctanh \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}-\frac {2 \sqrt {2}\, \arctan \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {\sqrt {2}-1}}\right )}{\sqrt {\sqrt {2}-1}}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/5*(1+(1+(1+x)^(1/2))^(1/2))^(5/2)-16/3*(1+(1+(1+x)^(1/2))^(1/2))^(3/2)+8*(1+(1+(1+x)^(1/2))^(1/2))^(1/2)-2*2
^(1/2)/(1+2^(1/2))^(1/2)*arctanh((1+(1+(1+x)^(1/2))^(1/2))^(1/2)/(1+2^(1/2))^(1/2))-2*2^(1/2)/(2^(1/2)-1)^(1/2
)*arctan((1+(1+(1+x)^(1/2))^(1/2))^(1/2)/(2^(1/2)-1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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