3.23.90 \(\int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx\)

Optimal. Leaf size=174 \[ \frac {32 \left (13 c d-12 d^3\right ) \sqrt {\sqrt {\sqrt {a x+b}+c}+d}}{105 a}-\frac {32 \left (5 c-6 d^2\right ) \sqrt {\sqrt {a x+b}+c} \sqrt {\sqrt {\sqrt {a x+b}+c}+d}}{105 a}+\sqrt {a x+b} \left (\frac {8 \sqrt {\sqrt {a x+b}+c} \sqrt {\sqrt {\sqrt {a x+b}+c}+d}}{7 a}-\frac {48 d \sqrt {\sqrt {\sqrt {a x+b}+c}+d}}{35 a}\right ) \]

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Rubi [A]  time = 0.16, antiderivative size = 127, normalized size of antiderivative = 0.73, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {371, 1398, 772} \begin {gather*} -\frac {8 \left (c-3 d^2\right ) \left (\sqrt {\sqrt {a x+b}+c}+d\right )^{3/2}}{3 a}+\frac {8 d \left (c-d^2\right ) \sqrt {\sqrt {\sqrt {a x+b}+c}+d}}{a}+\frac {8 \left (\sqrt {\sqrt {a x+b}+c}+d\right )^{7/2}}{7 a}-\frac {24 d \left (\sqrt {\sqrt {a x+b}+c}+d\right )^{5/2}}{5 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[d + Sqrt[c + Sqrt[b + a*x]]],x]

[Out]

(8*d*(c - d^2)*Sqrt[d + Sqrt[c + Sqrt[b + a*x]]])/a - (8*(c - 3*d^2)*(d + Sqrt[c + Sqrt[b + a*x]])^(3/2))/(3*a
) - (24*d*(d + Sqrt[c + Sqrt[b + a*x]])^(5/2))/(5*a) + (8*(d + Sqrt[c + Sqrt[b + a*x]])^(7/2))/(7*a)

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(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]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {d+\sqrt {c+\sqrt {b+a x}}}} \, dx &=\frac {2 \operatorname {Subst}\left (\int \frac {x}{\sqrt {d+\sqrt {c+x}}} \, dx,x,\sqrt {b+a x}\right )}{a}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {-c+x}{\sqrt {d+\sqrt {x}}} \, dx,x,c+\sqrt {b+a x}\right )}{a}\\ &=\frac {4 \operatorname {Subst}\left (\int \frac {x \left (-c+x^2\right )}{\sqrt {d+x}} \, dx,x,\sqrt {c+\sqrt {b+a x}}\right )}{a}\\ &=\frac {4 \operatorname {Subst}\left (\int \left (-\frac {d \left (-c+d^2\right )}{\sqrt {d+x}}+\left (-c+3 d^2\right ) \sqrt {d+x}-3 d (d+x)^{3/2}+(d+x)^{5/2}\right ) \, dx,x,\sqrt {c+\sqrt {b+a x}}\right )}{a}\\ &=\frac {8 d \left (c-d^2\right ) \sqrt {d+\sqrt {c+\sqrt {b+a x}}}}{a}-\frac {8 \left (c-3 d^2\right ) \left (d+\sqrt {c+\sqrt {b+a x}}\right )^{3/2}}{3 a}-\frac {24 d \left (d+\sqrt {c+\sqrt {b+a x}}\right )^{5/2}}{5 a}+\frac {8 \left (d+\sqrt {c+\sqrt {b+a x}}\right )^{7/2}}{7 a}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 114, normalized size = 0.66 \begin {gather*} \frac {8 \sqrt {\sqrt {\sqrt {a x+b}+c}+d} \left (24 d^2 \sqrt {\sqrt {a x+b}+c}-20 c \sqrt {\sqrt {a x+b}+c}+15 \sqrt {a x+b} \sqrt {\sqrt {a x+b}+c}-18 d \sqrt {a x+b}+52 c d-48 d^3\right )}{105 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[d + Sqrt[c + Sqrt[b + a*x]]],x]

[Out]

(8*Sqrt[d + Sqrt[c + Sqrt[b + a*x]]]*(52*c*d - 48*d^3 - 18*d*Sqrt[b + a*x] - 20*c*Sqrt[c + Sqrt[b + a*x]] + 24
*d^2*Sqrt[c + Sqrt[b + a*x]] + 15*Sqrt[b + a*x]*Sqrt[c + Sqrt[b + a*x]]))/(105*a)

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IntegrateAlgebraic [A]  time = 0.12, size = 114, normalized size = 0.66 \begin {gather*} -\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 \left (26 c d-24 d^3-9 d \sqrt {b+a x}\right ) \sqrt {d+\sqrt {c+\sqrt {b+a x}}}}{105 a} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/Sqrt[d + Sqrt[c + Sqrt[b + a*x]]],x]

[Out]

(-8*(20*c - 24*d^2 - 15*Sqrt[b + a*x])*Sqrt[c + Sqrt[b + a*x]]*Sqrt[d + Sqrt[c + Sqrt[b + a*x]]])/(105*a) + (1
6*(26*c*d - 24*d^3 - 9*d*Sqrt[b + a*x])*Sqrt[d + Sqrt[c + Sqrt[b + a*x]]])/(105*a)

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fricas [A]  time = 0.66, size = 71, normalized size = 0.41 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2),x, algorithm="fricas")

[Out]

-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)*sqr
t(d + sqrt(c + sqrt(a*x + b)))/a

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giac [A]  time = 4.50, size = 190, normalized size = 1.09 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2),x, algorithm="giac")

[Out]

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))) + 105*c*sqrt(d + sqrt(c + sqrt(a*x + b)))*d*sgn(sqrt(c + sqrt(a*x + b))))
/a

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maple [A]  time = 0.12, size = 93, normalized size = 0.53

method	result	size
derivativedivides	\(\frac {\frac {8 \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {7}{2}}}{7}-\frac {24 d \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {5}{2}}}{5}+\frac {8 \left (3 d^{2}-c \right ) \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {3}{2}}}{3}-8 \left (d^{2}-c \right ) d \sqrt {d +\sqrt {c +\sqrt {a x +b}}}}{a}\)	\(93\)
default	\(\frac {\frac {8 \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {7}{2}}}{7}-\frac {24 d \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {5}{2}}}{5}+\frac {8 \left (3 d^{2}-c \right ) \left (d +\sqrt {c +\sqrt {a x +b}}\right )^{\frac {3}{2}}}{3}-8 \left (d^{2}-c \right ) d \sqrt {d +\sqrt {c +\sqrt {a x +b}}}}{a}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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^2-c)*d*(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2))

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maxima [A]  time = 0.31, size = 92, normalized size = 0.53 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+(c+(a*x+b)^(1/2))^(1/2))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d + (c + (b + a*x)^(1/2))^(1/2))^(1/2),x)

[Out]

int(1/(d + (c + (b + a*x)^(1/2))^(1/2))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+(c+(a*x+b)**(1/2))**(1/2))**(1/2),x)

[Out]

Integral(1/sqrt(d + sqrt(c + sqrt(a*x + b))), x)

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