3.26.41 \(\int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x^2 \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx\)

Optimal. Leaf size=213 \[ -\frac {\left (\sqrt {\sqrt {x+1}+1}+1\right )^{5/2}}{2 \left (\sqrt {x+1}-1\right ) \sqrt {\sqrt {x+1}+1}}+\frac {5 \sqrt {\sqrt {\sqrt {x+1}+1}+1}}{2 \left (\sqrt {x+1}-1\right ) \sqrt {\sqrt {x+1}+1}}-\frac {1}{4} \sqrt {17+25 \sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {\sqrt {2}-1}}\right )+\tanh ^{-1}\left (\sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )-\frac {1}{4} \sqrt {25 \sqrt {2}-17} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {1+\sqrt {2}}}\right ) \]

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Rubi [B]  time = 3.40, antiderivative size = 432, normalized size of antiderivative = 2.03, number of steps used = 19, number of rules used = 9, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.209, Rules used = {1586, 2102, 1594, 28, 2073, 207, 1178, 1166, 203} \begin {gather*} \frac {8 \left (\sqrt {\sqrt {x+1}+1}+1\right )^{3/2}}{3 \left (1-\sqrt {x+1}\right ) \sqrt {\sqrt {x+1}+1}}-\frac {\left (50-17 \sqrt {\sqrt {x+1}+1}\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}{30 \left (1-\sqrt {x+1}\right )}-\frac {24 \sqrt {\sqrt {\sqrt {x+1}+1}+1}}{5 \left (1-\sqrt {x+1}\right ) \sqrt {\sqrt {x+1}+1}}-\frac {1}{30 \left (1-\sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )}+\frac {1}{30 \left (\sqrt {\sqrt {\sqrt {x+1}+1}+1}+1\right )}-\frac {1}{60} \sqrt {10961+8989 \sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {\sqrt {2}-1}}\right )+\frac {1}{15} \sqrt {\frac {1}{2} \left (97+113 \sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {\sqrt {2}-1}}\right )+\tanh ^{-1}\left (\sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )-\frac {1}{60} \sqrt {8989 \sqrt {2}-10961} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {1+\sqrt {2}}}\right )-\frac {1}{15} \sqrt {\frac {1}{2} \left (113 \sqrt {2}-97\right )} \tanh ^{-1}\left (\frac {\sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {1+\sqrt {2}}}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(Sqrt[1 + x]*Sqrt[1 + Sqrt[1 + x]])/(x^2*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]),x]

[Out]

(-24*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])/(5*(1 - Sqrt[1 + x])*Sqrt[1 + Sqrt[1 + x]]) - ((50 - 17*Sqrt[1 + Sqrt[1
+ x]])*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])/(30*(1 - Sqrt[1 + x])) + (8*(1 + Sqrt[1 + Sqrt[1 + x]])^(3/2))/(3*(1 -
Sqrt[1 + x])*Sqrt[1 + Sqrt[1 + x]]) - 1/(30*(1 - Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])) + 1/(30*(1 + Sqrt[1 + Sqrt
[1 + Sqrt[1 + x]]])) + (Sqrt[(97 + 113*Sqrt[2])/2]*ArcTan[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]/Sqrt[-1 + Sqrt[2]]])
/15 - (Sqrt[10961 + 8989*Sqrt[2]]*ArcTan[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]/Sqrt[-1 + Sqrt[2]]])/60 + ArcTanh[Sqr
t[1 + Sqrt[1 + Sqrt[1 + x]]]] - (Sqrt[(-97 + 113*Sqrt[2])/2]*ArcTanh[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]/Sqrt[1 +
Sqrt[2]]])/15 - (Sqrt[-10961 + 8989*Sqrt[2]]*ArcTanh[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]/Sqrt[1 + Sqrt[2]]])/60

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 1166

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

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e
^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

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]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2073

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(PP /. x ->
x^2)^p*Q^q, x], x] /;  !SumQ[NonfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x^2] && PolyQ[Q, x] && ILtQ[p,
0]

Rule 2102

Int[(Pm_)*(Qn_)^(p_.), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x]}, Simp[(Coeff[Pm, x, m]*x^(m - n
+ 1)*Qn^(p + 1))/((m + n*p + 1)*Coeff[Qn, x, n]), x] + Dist[1/((m + n*p + 1)*Coeff[Qn, x, n]), Int[ExpandToSum
[(m + n*p + 1)*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*x^(m - n)*((m - n + 1)*Qn + (p + 1)*x*D[Qn, x]), x]*Qn^p,
x], x] /; LtQ[1, n, m + 1] && m + n*p + 1 < 0] /; FreeQ[p, x] && PolyQ[Pm, x] && PolyQ[Qn, x] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {\sqrt {1+x} \sqrt {1+\sqrt {1+x}}}{x^2 \sqrt {1+\sqrt {1+\sqrt {1+x}}}} \, dx &=2 \operatorname {Subst}\left (\int \frac {x^2 \sqrt {1+x}}{\left (-1+x^2\right )^2 \sqrt {1+\sqrt {1+x}}} \, dx,x,\sqrt {1+x}\right )\\ &=2 \operatorname {Subst}\left (\int \frac {x^2}{(-1+x)^2 (1+x)^{3/2} \sqrt {1+\sqrt {1+x}}} \, dx,x,\sqrt {1+x}\right )\\ &=4 \operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{x^2 \sqrt {1+x} \left (-2+x^2\right )^2} \, dx,x,\sqrt {1+\sqrt {1+x}}\right )\\ &=4 \operatorname {Subst}\left (\int \frac {(-1+x)^2 (1+x)^{3/2}}{x^2 \left (-2+x^2\right )^2} \, dx,x,\sqrt {1+\sqrt {1+x}}\right )\\ &=8 \operatorname {Subst}\left (\int \frac {x^4 \left (-2+x^2\right )^2}{\left (1+x^2-3 x^4+x^6\right )^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=-\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}}{3 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {8}{3} \operatorname {Subst}\left (\int \frac {-3 x^2-13 x^4+9 x^6}{\left (1+x^2-3 x^4+x^6\right )^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=-\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}}{3 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {8}{3} \operatorname {Subst}\left (\int \frac {x^2 \left (-3-13 x^2+9 x^4\right )}{\left (1+x^2-3 x^4+x^6\right )^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=\frac {24 \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{5 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}}{3 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}+\frac {8}{15} \operatorname {Subst}\left (\int \frac {-9+24 x^2-16 x^4}{\left (1+x^2-3 x^4+x^6\right )^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=\frac {24 \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{5 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}}{3 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {1}{30} \operatorname {Subst}\left (\int \frac {\left (12-16 x^2\right )^2}{\left (1+x^2-3 x^4+x^6\right )^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=\frac {24 \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{5 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}}{3 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {1}{30} \operatorname {Subst}\left (\int \left (\frac {1}{(-1+x)^2}+\frac {1}{(1+x)^2}+\frac {30}{-1+x^2}+\frac {8 \left (25+8 x^2\right )}{\left (-1-2 x^2+x^4\right )^2}-\frac {4 \left (-7+8 x^2\right )}{-1-2 x^2+x^4}\right ) \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=-\frac {1}{30 \left (1-\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )}+\frac {1}{30 \left (1+\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )}+\frac {24 \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{5 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}}{3 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}+\frac {2}{15} \operatorname {Subst}\left (\int \frac {-7+8 x^2}{-1-2 x^2+x^4} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-\frac {4}{15} \operatorname {Subst}\left (\int \frac {25+8 x^2}{\left (-1-2 x^2+x^4\right )^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-\operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=-\frac {1}{30 \left (1-\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )}+\frac {1}{30 \left (1+\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )}-\frac {\left (50-17 \sqrt {1+\sqrt {1+x}}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{30 \left (1+2 \left (1+\sqrt {1+\sqrt {1+x}}\right )-\left (1+\sqrt {1+\sqrt {1+x}}\right )^2\right )}+\frac {24 \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{5 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}}{3 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}+\tanh ^{-1}\left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-\frac {1}{60} \operatorname {Subst}\left (\int \frac {-266+34 x^2}{-1-2 x^2+x^4} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )+\frac {1}{30} \left (16-\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\sqrt {2}+x^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )+\frac {1}{30} \left (16+\sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-\sqrt {2}+x^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=-\frac {1}{30 \left (1-\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )}+\frac {1}{30 \left (1+\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )}-\frac {\left (50-17 \sqrt {1+\sqrt {1+x}}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{30 \left (1+2 \left (1+\sqrt {1+\sqrt {1+x}}\right )-\left (1+\sqrt {1+\sqrt {1+x}}\right )^2\right )}+\frac {24 \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{5 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}}{3 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}+\frac {1}{30} \sqrt {194+226 \sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {-1+\sqrt {2}}}\right )+\tanh ^{-1}\left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-\frac {1}{30} \sqrt {-194+226 \sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right )-\frac {1}{60} \left (17-58 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-\sqrt {2}+x^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-\frac {1}{60} \left (17+58 \sqrt {2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\sqrt {2}+x^2} \, dx,x,\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )\\ &=-\frac {1}{30 \left (1-\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )}+\frac {1}{30 \left (1+\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )}-\frac {\left (50-17 \sqrt {1+\sqrt {1+x}}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{30 \left (1+2 \left (1+\sqrt {1+\sqrt {1+x}}\right )-\left (1+\sqrt {1+\sqrt {1+x}}\right )^2\right )}+\frac {24 \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{5 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}-\frac {8 \left (1+\sqrt {1+\sqrt {1+x}}\right )^{3/2}}{3 \left (2+\sqrt {1+\sqrt {1+x}}-3 \left (1+\sqrt {1+\sqrt {1+x}}\right )^2+\left (1+\sqrt {1+\sqrt {1+x}}\right )^3\right )}+\frac {1}{30} \sqrt {194+226 \sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {-1+\sqrt {2}}}\right )-\frac {1}{60} \sqrt {10961+8989 \sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {-1+\sqrt {2}}}\right )+\tanh ^{-1}\left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-\frac {1}{30} \sqrt {-194+226 \sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right )-\frac {1}{60} \sqrt {-10961+8989 \sqrt {2}} \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 = 1.04, size = 357, normalized size = 1.68 \begin {gather*} \frac {1}{4} \left (\frac {\left (\sqrt {2}-1\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {2}-\sqrt {\sqrt {x+1}+1}}+\frac {\left (1+\sqrt {2}\right ) \sqrt {\sqrt {\sqrt {x+1}+1}+1}}{\sqrt {\sqrt {x+1}+1}+\sqrt {2}}-\frac {2}{\sqrt {\sqrt {\sqrt {x+1}+1}+1}-1}-\frac {2}{\sqrt {\sqrt {\sqrt {x+1}+1}+1}+1}-5 \sqrt {2 \left (1+\sqrt {2}\right )} \tan ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )+\frac {\tan ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )}{\left (\sqrt {2}-1\right )^{3/2}}+4 \tanh ^{-1}\left (\sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )+\frac {\tanh ^{-1}\left (\sqrt {\sqrt {2}-1} \sqrt {\sqrt {\sqrt {x+1}+1}+1}\right )}{\left (1+\sqrt {2}\right )^{3/2}}-5 \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^2*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]),x]

[Out]

(((-1 + Sqrt[2])*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])/(Sqrt[2] - Sqrt[1 + Sqrt[1 + x]]) + ((1 + Sqrt[2])*Sqrt[1 +
Sqrt[1 + Sqrt[1 + x]]])/(Sqrt[2] + Sqrt[1 + Sqrt[1 + x]]) - 2/(-1 + Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]) - 2/(1 +
Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]) + ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]]/(-1 + Sqrt[2])^(3
/2) - 5*Sqrt[2*(1 + Sqrt[2])]*ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]] + 4*ArcTanh[Sqrt[1 + S
qrt[1 + Sqrt[1 + x]]]] - 5*Sqrt[2*(-1 + Sqrt[2])]*ArcTanh[Sqrt[-1 + Sqrt[2]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]]
+ ArcTanh[Sqrt[-1 + Sqrt[2]]*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]]/(1 + Sqrt[2])^(3/2))/4

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IntegrateAlgebraic [A]  time = 0.94, size = 210, normalized size = 0.99 \begin {gather*} \frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{1-\sqrt {1+x}}+\frac {\left (3-\sqrt {1+x}\right ) \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{2 \left (-1+\sqrt {1+x}\right ) \sqrt {1+\sqrt {1+x}}}-\frac {1}{4} \sqrt {17+25 \sqrt {2}} \tan ^{-1}\left (\sqrt {1+\sqrt {2}} \sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )+\tanh ^{-1}\left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}\right )-\frac {1}{4} \sqrt {-17+25 \sqrt {2}} \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^2*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]),x]

[Out]

Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]/(1 - Sqrt[1 + x]) + ((3 - Sqrt[1 + x])*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]])/(2*(-1
+ Sqrt[1 + x])*Sqrt[1 + Sqrt[1 + x]]) - (Sqrt[17 + 25*Sqrt[2]]*ArcTan[Sqrt[1 + Sqrt[2]]*Sqrt[1 + Sqrt[1 + Sqr
t[1 + x]]]])/4 + ArcTanh[Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]] - (Sqrt[-17 + 25*Sqrt[2]]*ArcTanh[Sqrt[-1 + Sqrt[2]]
*Sqrt[1 + Sqrt[1 + Sqrt[1 + x]]]])/4

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fricas [A]  time = 0.91, size = 255, normalized size = 1.20 \begin {gather*} \frac {4 \, x \sqrt {25 \, \sqrt {2} + 17} \arctan \left (\frac {1}{31} \, \sqrt {25 \, \sqrt {2} + 17} {\left (4 \, \sqrt {2} + 1\right )} \sqrt {\sqrt {2} + \sqrt {\sqrt {x + 1} + 1}} - \frac {1}{31} \, \sqrt {25 \, \sqrt {2} + 17} {\left (4 \, \sqrt {2} + 1\right )} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) - x \sqrt {25 \, \sqrt {2} - 17} \log \left (\sqrt {25 \, \sqrt {2} - 17} {\left (3 \, \sqrt {2} + 7\right )} + 31 \, \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) + x \sqrt {25 \, \sqrt {2} - 17} \log \left (-\sqrt {25 \, \sqrt {2} - 17} {\left (3 \, \sqrt {2} + 7\right )} + 31 \, \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}\right ) + 4 \, x \log \left (\sqrt {\sqrt {\sqrt {x + 1} + 1} + 1} + 1\right ) - 4 \, x \log \left (\sqrt {\sqrt {\sqrt {x + 1} + 1} + 1} - 1\right ) - 4 \, {\left (\sqrt {\sqrt {x + 1} + 1} {\left (\sqrt {x + 1} - 3\right )} + 2 \, \sqrt {x + 1} + 2\right )} \sqrt {\sqrt {\sqrt {x + 1} + 1} + 1}}{8 \, 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^2/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x, algorithm="fricas")

[Out]

1/8*(4*x*sqrt(25*sqrt(2) + 17)*arctan(1/31*sqrt(25*sqrt(2) + 17)*(4*sqrt(2) + 1)*sqrt(sqrt(2) + sqrt(sqrt(x +
1) + 1)) - 1/31*sqrt(25*sqrt(2) + 17)*(4*sqrt(2) + 1)*sqrt(sqrt(sqrt(x + 1) + 1) + 1)) - x*sqrt(25*sqrt(2) - 1
7)*log(sqrt(25*sqrt(2) - 17)*(3*sqrt(2) + 7) + 31*sqrt(sqrt(sqrt(x + 1) + 1) + 1)) + x*sqrt(25*sqrt(2) - 17)*l
og(-sqrt(25*sqrt(2) - 17)*(3*sqrt(2) + 7) + 31*sqrt(sqrt(sqrt(x + 1) + 1) + 1)) + 4*x*log(sqrt(sqrt(sqrt(x + 1
) + 1) + 1) + 1) - 4*x*log(sqrt(sqrt(sqrt(x + 1) + 1) + 1) - 1) - 4*(sqrt(sqrt(x + 1) + 1)*(sqrt(x + 1) - 3) +
2*sqrt(x + 1) + 2)*sqrt(sqrt(sqrt(x + 1) + 1) + 1))/x

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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^2/(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 [60]Warning, choosing root of [1,0,%%%{-2,[1]%%%}+%%%{-4,[0]%%%},%%%{-4,[1]%%%}+%%%{-4,[0]%%%},%%%{1,[2
]%%%}+%%%{-1,[1]%%%}+%%%{-1,[0]%%%}] at parameters values [-9]Warning, need to choose a branch for the root of
a polynomial with parameters. This might be wrong.The choice was done assuming [x]=[44]4*(-1/8*ln(sqrt(sqrt(s
qrt(x+1)+1)+1)-1)+1/8*ln(sqrt(sqrt(sqrt(x+1)+1)+1)+1)+sqrt(25*sqrt(2)-17)/32*ln(abs(sqrt(sqrt(sqrt(x+1)+1)+1)-
sqrt((16+16*sqrt(2))/2/8)))-sqrt(25*sqrt(2)-17)/32*ln(sqrt(sqrt(sqrt(x+1)+1)+1)+sqrt((16+16*sqrt(2))/2/8))-sqr
t(25*sqrt(2)+17)/16*atan(sqrt(sqrt(sqrt(x+1)+1)+1)/sqrt(-(16-16*sqrt(2))/2/8))+(-sqrt(sqrt(sqrt(x+1)+1)+1)*(sq
rt(sqrt(x+1)+1)+1)^2+5*sqrt(sqrt(sqrt(x+1)+1)+1))/8/((sqrt(sqrt(x+1)+1)+1)^3-3*(sqrt(sqrt(x+1)+1)+1)^2+sqrt(sq
rt(x+1)+1)+2))

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maple [A]  time = 0.26, size = 215, normalized size = 1.01

method	result	size
derivativedivides	\(-\frac {1}{2 \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-1\right )}-\frac {\ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-1\right )}{2}-\frac {1}{2 \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}+1\right )}+\frac {\ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}+1\right )}{2}+\frac {\frac {\left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{2}-\frac {3 \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{2}}{\left (1+\sqrt {1+\sqrt {1+x}}\right )^{2}-2 \sqrt {1+\sqrt {1+x}}-3}-\frac {\left (8+\sqrt {2}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right )}{8 \sqrt {1+\sqrt {2}}}+\frac {\left (-8+\sqrt {2}\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {\sqrt {2}-1}}\right )}{8 \sqrt {\sqrt {2}-1}}\)	\(215\)
default	\(-\frac {1}{2 \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-1\right )}-\frac {\ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}-1\right )}{2}-\frac {1}{2 \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}+1\right )}+\frac {\ln \left (\sqrt {1+\sqrt {1+\sqrt {1+x}}}+1\right )}{2}+\frac {\frac {\left (1+\sqrt {1+\sqrt {1+x}}\right )^{\frac {3}{2}}}{2}-\frac {3 \sqrt {1+\sqrt {1+\sqrt {1+x}}}}{2}}{\left (1+\sqrt {1+\sqrt {1+x}}\right )^{2}-2 \sqrt {1+\sqrt {1+x}}-3}-\frac {\left (8+\sqrt {2}\right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {1+\sqrt {2}}}\right )}{8 \sqrt {1+\sqrt {2}}}+\frac {\left (-8+\sqrt {2}\right ) \sqrt {2}\, \arctan \left (\frac {\sqrt {1+\sqrt {1+\sqrt {1+x}}}}{\sqrt {\sqrt {2}-1}}\right )}{8 \sqrt {\sqrt {2}-1}}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/((1+(1+(1+x)^(1/2))^(1/2))^(1/2)-1)-1/2*ln((1+(1+(1+x)^(1/2))^(1/2))^(1/2)-1)-1/2/((1+(1+(1+x)^(1/2))^(1/
2))^(1/2)+1)+1/2*ln((1+(1+(1+x)^(1/2))^(1/2))^(1/2)+1)+2*(1/4*(1+(1+(1+x)^(1/2))^(1/2))^(3/2)-3/4*(1+(1+(1+x)^
(1/2))^(1/2))^(1/2))/((1+(1+(1+x)^(1/2))^(1/2))^2-2*(1+(1+x)^(1/2))^(1/2)-3)-1/8*(8+2^(1/2))*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))+1/8*(-8+2^(1/2))*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^{2} \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^2/(1+(1+(1+x)^(1/2))^(1/2))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {\sqrt {x+1}+1}\,\sqrt {x+1}}{x^2\,\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^2*(((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^2*(((x + 1)^(1/2) + 1)^(1/2) + 1)^(1/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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**2/(1+(1+(1+x)**(1/2))**(1/2))**(1/2),x)

[Out]

Timed out

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