Integrand size = 21, antiderivative size = 174 \[ \int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx=\frac {32 \left (13 c d-12 d^3\right ) \sqrt {d+\sqrt {c+\sqrt {b+a x}}}}{105 a}-\frac {32 \left (5 c-6 d^2\right ) \sqrt {c+\sqrt {b+a x}} \sqrt {d+\sqrt {c+\sqrt {b+a x}}}}{105 a}+\sqrt {b+a x} \left (-\frac {48 d \sqrt {d+\sqrt {c+\sqrt {b+a x}}}}{35 a}+\frac {8 \sqrt {c+\sqrt {b+a x}} \sqrt {d+\sqrt {c+\sqrt {b+a x}}}}{7 a}\right ) \] Output:
32/105*(-12*d^3+13*c*d)*(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2)/a-32/105*(-6*d^2 +5*c)*(c+(a*x+b)^(1/2))^(1/2)*(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2)/a+(a*x+b)^ (1/2)*(-48/35*d*(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2)/a+8/7*(c+(a*x+b)^(1/2))^ (1/2)*(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2)/a)
Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.65 \[ \int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx=-\frac {8 \left (20 c-24 d^2-15 \sqrt {b+a x}\right ) \sqrt {c+\sqrt {b+a x}} \sqrt {d+\sqrt {c+\sqrt {b+a x}}}}{105 a}-\frac {16 d \left (-26 c+24 d^2+9 \sqrt {b+a x}\right ) \sqrt {d+\sqrt {c+\sqrt {b+a x}}}}{105 a} \] Input:
Integrate[1/Sqrt[d + Sqrt[c + Sqrt[b + a*x]]],x]
Output:
(-8*(20*c - 24*d^2 - 15*Sqrt[b + a*x])*Sqrt[c + Sqrt[b + a*x]]*Sqrt[d + Sq rt[c + Sqrt[b + a*x]]])/(105*a) - (16*d*(-26*c + 24*d^2 + 9*Sqrt[b + a*x]) *Sqrt[d + Sqrt[c + Sqrt[b + a*x]]])/(105*a)
Time = 0.34 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.55, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {7267, 896, 25, 1732, 522, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt {\sqrt {\sqrt {a x+b}+c}+d}} \, dx\) |
\(\Big \downarrow \) 7267 |
\(\displaystyle \frac {2 \int \frac {\sqrt {b+a x}}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}}d\sqrt {b+a x}}{a}\) |
\(\Big \downarrow \) 896 |
\(\displaystyle \frac {2 \int \frac {\sqrt {b+a x}}{\sqrt {d+\sqrt [4]{b+a x}}}d\left (c+\sqrt {b+a x}\right )}{a}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {2 \int -\frac {\sqrt {b+a x}}{\sqrt {d+\sqrt [4]{b+a x}}}d\left (c+\sqrt {b+a x}\right )}{a}\) |
\(\Big \downarrow \) 1732 |
\(\displaystyle -\frac {4 \int \frac {(-b+c-a x) \sqrt [4]{b+a x}}{\sqrt {d+\sqrt [4]{b+a x}}}d\sqrt [4]{b+a x}}{a}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle -\frac {4 \int \left (-\left (d+\sqrt [4]{b+a x}\right )^{5/2}+3 d \left (d+\sqrt [4]{b+a x}\right )^{3/2}+\left (c-3 d^2\right ) \sqrt {d+\sqrt [4]{b+a x}}+\frac {d^3-c d}{\sqrt {d+\sqrt [4]{b+a x}}}\right )d\sqrt [4]{b+a x}}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \left (\frac {2}{3} \left (c-3 d^2\right ) \left (\sqrt [4]{a x+b}+d\right )^{3/2}-2 d \left (c-d^2\right ) \sqrt {\sqrt [4]{a x+b}+d}-\frac {2}{7} \left (\sqrt [4]{a x+b}+d\right )^{7/2}+\frac {6}{5} d \left (\sqrt [4]{a x+b}+d\right )^{5/2}\right )}{a}\) |
Input:
Int[1/Sqrt[d + Sqrt[c + Sqrt[b + a*x]]],x]
Output:
(-4*(-2*d*(c - d^2)*Sqrt[d + (b + a*x)^(1/4)] + (2*(c - 3*d^2)*(d + (b + a *x)^(1/4))^(3/2))/3 + (6*d*(d + (b + a*x)^(1/4))^(5/2))/5 - (2*(d + (b + a *x)^(1/4))^(7/2))/7))/a
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1) Subst[Int[Si mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symb ol] :> With[{g = Denominator[n]}, Simp[g Subst[Int[x^(g - 1)*(d + e*x^(g* n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p, q} , x] && EqQ[n2, 2*n] && FractionQ[n]
Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si mp[lst[[2]]*lst[[4]] Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x ] /; !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.52
method | result | size |
derivativedivides | \(\frac {\frac {8 \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {7}{2}}}{7}-\frac {24 d \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {5}{2}}}{5}-\frac {8 \left (-3 d^{2}+c \right ) \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {3}{2}}}{3}-8 \sqrt {d +\sqrt {c +\sqrt {a x +b}}}\, \left (d^{2}-c \right ) d}{a}\) | \(91\) |
default | \(\frac {\frac {8 \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {7}{2}}}{7}-\frac {24 d \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {5}{2}}}{5}-\frac {8 \left (-3 d^{2}+c \right ) \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {3}{2}}}{3}-8 \sqrt {d +\sqrt {c +\sqrt {a x +b}}}\, \left (d^{2}-c \right ) d}{a}\) | \(91\) |
Input:
int(1/(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/a*(4/7*(d+(c+(a*x+b)^(1/2))^(1/2))^(7/2)-12/5*d*(d+(c+(a*x+b)^(1/2))^(1/ 2))^(5/2)-4/3*(-3*d^2+c)*(d+(c+(a*x+b)^(1/2))^(1/2))^(3/2)-4*(d+(c+(a*x+b) ^(1/2))^(1/2))^(1/2)*(d^2-c)*d)
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.41 \[ \int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx=-\frac {8 \, {\left (48 \, d^{3} - 52 \, c d - {\left (24 \, d^{2} - 20 \, c + 15 \, \sqrt {a x + b}\right )} \sqrt {c + \sqrt {a x + b}} + 18 \, \sqrt {a x + b} d\right )} \sqrt {d + \sqrt {c + \sqrt {a x + b}}}}{105 \, a} \] Input:
integrate(1/(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2),x, algorithm="fricas")
Output:
-8/105*(48*d^3 - 52*c*d - (24*d^2 - 20*c + 15*sqrt(a*x + b))*sqrt(c + sqrt (a*x + b)) + 18*sqrt(a*x + b)*d)*sqrt(d + sqrt(c + sqrt(a*x + b)))/a
Time = 0.55 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx=\begin {cases} \frac {2 \cdot \left (4 c d \sqrt {d + \sqrt {c + \sqrt {a x + b}}} - 4 d^{3} \sqrt {d + \sqrt {c + \sqrt {a x + b}}} - \frac {12 d \left (d + \sqrt {c + \sqrt {a x + b}}\right )^{\frac {5}{2}}}{5} + \frac {4 \left (- c + 3 d^{2}\right ) \left (d + \sqrt {c + \sqrt {a x + b}}\right )^{\frac {3}{2}}}{3} + \frac {4 \left (d + \sqrt {c + \sqrt {a x + b}}\right )^{\frac {7}{2}}}{7}\right )}{a} & \text {for}\: a \neq 0 \\\frac {x}{\sqrt {d + \sqrt {\sqrt {b} + c}}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(d+(c+(a*x+b)**(1/2))**(1/2))**(1/2),x)
Output:
Piecewise((2*(4*c*d*sqrt(d + sqrt(c + sqrt(a*x + b))) - 4*d**3*sqrt(d + sq rt(c + sqrt(a*x + b))) - 12*d*(d + sqrt(c + sqrt(a*x + b)))**(5/2)/5 + 4*( -c + 3*d**2)*(d + sqrt(c + sqrt(a*x + b)))**(3/2)/3 + 4*(d + sqrt(c + sqrt (a*x + b)))**(7/2)/7)/a, Ne(a, 0)), (x/sqrt(d + sqrt(sqrt(b) + c)), True))
Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx=\frac {8 \, {\left (15 \, {\left (d + \sqrt {c + \sqrt {a x + b}}\right )}^{\frac {7}{2}} - 63 \, {\left (d + \sqrt {c + \sqrt {a x + b}}\right )}^{\frac {5}{2}} d + 35 \, {\left (3 \, d^{2} - c\right )} {\left (d + \sqrt {c + \sqrt {a x + b}}\right )}^{\frac {3}{2}} - 105 \, {\left (d^{3} - c d\right )} \sqrt {d + \sqrt {c + \sqrt {a x + b}}}\right )}}{105 \, a} \] Input:
integrate(1/(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2),x, algorithm="maxima")
Output:
8/105*(15*(d + sqrt(c + sqrt(a*x + b)))^(7/2) - 63*(d + sqrt(c + sqrt(a*x + b)))^(5/2)*d + 35*(3*d^2 - c)*(d + sqrt(c + sqrt(a*x + b)))^(3/2) - 105* (d^3 - c*d)*sqrt(d + sqrt(c + sqrt(a*x + b))))/a
Time = 5.08 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx=\frac {8 \, {\left (15 \, {\left (d + \sqrt {c + \sqrt {a x + b}}\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\sqrt {c + \sqrt {a x + b}}\right ) - 63 \, {\left (d + \sqrt {c + \sqrt {a x + b}}\right )}^{\frac {5}{2}} d \mathrm {sgn}\left (\sqrt {c + \sqrt {a x + b}}\right ) + 105 \, {\left (d + \sqrt {c + \sqrt {a x + b}}\right )}^{\frac {3}{2}} d^{2} \mathrm {sgn}\left (\sqrt {c + \sqrt {a x + b}}\right ) - 105 \, \sqrt {d + \sqrt {c + \sqrt {a x + b}}} d^{3} \mathrm {sgn}\left (\sqrt {c + \sqrt {a x + b}}\right ) - 35 \, c {\left (d + \sqrt {c + \sqrt {a x + b}}\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\sqrt {c + \sqrt {a x + b}}\right ) + 105 \, c \sqrt {d + \sqrt {c + \sqrt {a x + b}}} d \mathrm {sgn}\left (\sqrt {c + \sqrt {a x + b}}\right )\right )}}{105 \, a} \] Input:
integrate(1/(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2),x, algorithm="giac")
Output:
8/105*(15*(d + sqrt(c + sqrt(a*x + b)))^(7/2)*sgn(sqrt(c + sqrt(a*x + b))) - 63*(d + sqrt(c + sqrt(a*x + b)))^(5/2)*d*sgn(sqrt(c + sqrt(a*x + b))) + 105*(d + sqrt(c + sqrt(a*x + b)))^(3/2)*d^2*sgn(sqrt(c + sqrt(a*x + b))) - 105*sqrt(d + sqrt(c + sqrt(a*x + b)))*d^3*sgn(sqrt(c + sqrt(a*x + b))) - 35*c*(d + sqrt(c + sqrt(a*x + b)))^(3/2)*sgn(sqrt(c + sqrt(a*x + b))) + 1 05*c*sqrt(d + sqrt(c + sqrt(a*x + b)))*d*sgn(sqrt(c + sqrt(a*x + b))))/a
Timed out. \[ \int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx=\int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a\,x}}}} \,d x \] Input:
int(1/(d + (c + (b + a*x)^(1/2))^(1/2))^(1/2),x)
Output:
int(1/(d + (c + (b + a*x)^(1/2))^(1/2))^(1/2), x)
Time = 0.20 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx=\frac {8 \sqrt {\sqrt {\sqrt {a x +b}+c}+d}\, \left (15 \sqrt {a x +b}\, \sqrt {\sqrt {a x +b}+c}-20 \sqrt {\sqrt {a x +b}+c}\, c +24 \sqrt {\sqrt {a x +b}+c}\, d^{2}-18 \sqrt {a x +b}\, d +52 c d -48 d^{3}\right )}{105 a} \] Input:
int(1/(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2),x)
Output:
(8*sqrt(sqrt(sqrt(a*x + b) + c) + d)*(15*sqrt(a*x + b)*sqrt(sqrt(a*x + b) + c) - 20*sqrt(sqrt(a*x + b) + c)*c + 24*sqrt(sqrt(a*x + b) + c)*d**2 - 18 *sqrt(a*x + b)*d + 52*c*d - 48*d**3))/(105*a)