3.7.42 \(\int \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} \sqrt {x} \, dx\)

Optimal. Leaf size=73 \[ \frac {1}{2} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{3/2}-\frac {1}{4} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \sqrt {x}-\frac {1}{4} \cosh ^{-1}\left (\sqrt {x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {280, 323, 330, 52} \begin {gather*} \frac {1}{2} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{3/2}-\frac {1}{4} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \sqrt {x}-\frac {1}{4} \cosh ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*Sqrt[x],x]

[Out]

-(Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*Sqrt[x])/4 + (Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*x^(3/2))/2 - ArcCosh
[Sqrt[x]]/4

Rule 52

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ArcCosh[(b*x)/a]/b, x] /; FreeQ[{a,
b, c, d}, x] && EqQ[a + c, 0] && EqQ[b - d, 0] && GtQ[a, 0]

Rule 280

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*
x)^(m + 1)*(a1 + b1*x^n)^p*(a2 + b2*x^n)^p)/(c*(m + 2*n*p + 1)), x] + Dist[(2*a1*a2*n*p)/(m + 2*n*p + 1), Int[
(c*x)^m*(a1 + b1*x^n)^(p - 1)*(a2 + b2*x^n)^(p - 1), x], x] /; FreeQ[{a1, b1, a2, b2, c, m}, x] && EqQ[a2*b1 +
a1*b2, 0] && IGtQ[2*n, 0] && GtQ[p, 0] && NeQ[m + 2*n*p + 1, 0] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x
]

Rule 323

Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(2
*n - 1)*(c*x)^(m - 2*n + 1)*(a1 + b1*x^n)^(p + 1)*(a2 + b2*x^n)^(p + 1))/(b1*b2*(m + 2*n*p + 1)), x] - Dist[(a
1*a2*c^(2*n)*(m - 2*n + 1))/(b1*b2*(m + 2*n*p + 1)), Int[(c*x)^(m - 2*n)*(a1 + b1*x^n)^p*(a2 + b2*x^n)^p, x],
x] /; FreeQ[{a1, b1, a2, b2, c, p}, x] && EqQ[a2*b1 + a1*b2, 0] && IGtQ[2*n, 0] && GtQ[m, 2*n - 1] && NeQ[m +
2*n*p + 1, 0] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]

Rule 330

Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k =
Denominator[m]}, Dist[k/c, Subst[Int[x^(k*(m + 1) - 1)*(a1 + (b1*x^(k*n))/c^n)^p*(a2 + (b2*x^(k*n))/c^n)^p, x]
, x, (c*x)^(1/k)], x]] /; FreeQ[{a1, b1, a2, b2, c, p}, x] && EqQ[a2*b1 + a1*b2, 0] && IGtQ[2*n, 0] && Fractio
nQ[m] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 87, normalized size = 1.19 \begin {gather*} \frac {\sqrt {\sqrt {x}+1} \sqrt {x} \left (2 x^{3/2}-2 x-\sqrt {x}+1\right )+2 \sqrt {1-\sqrt {x}} \sin ^{-1}\left (\frac {\sqrt {1-\sqrt {x}}}{\sqrt {2}}\right )}{4 \sqrt {\sqrt {x}-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 + Sqrt[x]]*Sqrt[x]*(1 - Sqrt[x] - 2*x + 2*x^(3/2)) + 2*Sqrt[1 - Sqrt[x]]*ArcSin[Sqrt[1 - Sqrt[x]]/Sqrt
[2]])/(4*Sqrt[-1 + Sqrt[x]])

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 1.48, size = 404, normalized size = 5.53 \begin {gather*} \frac {-4 \sqrt {\sqrt {x}+1} \left (-1136 x^{7/2}-1096 x^{5/2}+17296 x^{3/2}-4752 x^3+7240 x^2+55360 x+28224 \sqrt {x}-18816\right )-4 \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \left (-194 x^{7/2}-6079 x^{5/2}+6992 x^{3/2}-3120 x^3-104 x^2+38632 x+74488 \sqrt {x}+32592\right )+\sqrt {3} \left (-4 \sqrt {\sqrt {x}-1} \left (656 x^{7/2}+3832 x^{5/2}-10928 x^{3/2}+3408 x^3-1192 x^2-41472 x-52416 \sqrt {x}-18816\right )-4 \left (1800 x^{7/2}-1148 x^{5/2}-23268 x^{3/2}+112 x^4+3416 x^3-6678 x^2-41440 x-10872 \sqrt {x}+10864\right )\right )}{24960 x^{3/2}+\sqrt {3} \sqrt {\sqrt {x}+1} \left (-5248 x^{3/2}-22016 x-11264 \sqrt {x}+7168\right )+\sqrt {\sqrt {x}-1} \left (9088 x^{3/2}+\sqrt {3} \sqrt {\sqrt {x}+1} \left (-896 x^{3/2}-14400 x-28672 \sqrt {x}-12416\right )+47104 x+60416 \sqrt {x}+21504\right )+1552 x^2+49408 x+13312 \sqrt {x}-12416}+\tanh ^{-1}\left (\frac {\sqrt {\sqrt {x}-1}-1}{\sqrt {3}-\sqrt {\sqrt {x}+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*Sqrt[x],x]

[Out]

(-4*Sqrt[1 + Sqrt[x]]*(-18816 + 28224*Sqrt[x] + 55360*x + 17296*x^(3/2) + 7240*x^2 - 1096*x^(5/2) - 4752*x^3 -
1136*x^(7/2)) - 4*Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*(32592 + 74488*Sqrt[x] + 38632*x + 6992*x^(3/2) - 104*
x^2 - 6079*x^(5/2) - 3120*x^3 - 194*x^(7/2)) + Sqrt[3]*(-4*Sqrt[-1 + Sqrt[x]]*(-18816 - 52416*Sqrt[x] - 41472*
x - 10928*x^(3/2) - 1192*x^2 + 3832*x^(5/2) + 3408*x^3 + 656*x^(7/2)) - 4*(10864 - 10872*Sqrt[x] - 41440*x - 2
3268*x^(3/2) - 6678*x^2 - 1148*x^(5/2) + 3416*x^3 + 1800*x^(7/2) + 112*x^4)))/(-12416 + 13312*Sqrt[x] + 49408*
x + 24960*x^(3/2) + 1552*x^2 + Sqrt[3]*Sqrt[1 + Sqrt[x]]*(7168 - 11264*Sqrt[x] - 22016*x - 5248*x^(3/2)) + Sqr
t[-1 + Sqrt[x]]*(21504 + 60416*Sqrt[x] + 47104*x + 9088*x^(3/2) + Sqrt[3]*Sqrt[1 + Sqrt[x]]*(-12416 - 28672*Sq
rt[x] - 14400*x - 896*x^(3/2)))) + ArcTanh[(-1 + Sqrt[-1 + Sqrt[x]])/(Sqrt[3] - Sqrt[1 + Sqrt[x]])]

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 52, normalized size = 0.71 \begin {gather*} \frac {1}{4} \, {\left (2 \, x - 1\right )} \sqrt {x} \sqrt {\sqrt {x} + 1} \sqrt {\sqrt {x} - 1} + \frac {1}{8} \, \log \left (2 \, \sqrt {x} \sqrt {\sqrt {x} + 1} \sqrt {\sqrt {x} - 1} - 2 \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*x - 1)*sqrt(x)*sqrt(sqrt(x) + 1)*sqrt(sqrt(x) - 1) + 1/8*log(2*sqrt(x)*sqrt(sqrt(x) + 1)*sqrt(sqrt(x) -
1) - 2*x + 1)

________________________________________________________________________________________

giac [B]  time = 0.19, size = 92, normalized size = 1.26 \begin {gather*} \frac {1}{12} \, {\left ({\left (2 \, {\left (3 \, \sqrt {x} - 10\right )} {\left (\sqrt {x} + 1\right )} + 43\right )} {\left (\sqrt {x} + 1\right )} - 39\right )} \sqrt {\sqrt {x} + 1} \sqrt {\sqrt {x} - 1} + \frac {1}{3} \, {\left ({\left (2 \, \sqrt {x} - 5\right )} {\left (\sqrt {x} + 1\right )} + 9\right )} \sqrt {\sqrt {x} + 1} \sqrt {\sqrt {x} - 1} + \frac {1}{2} \, \log \left (\sqrt {\sqrt {x} + 1} - \sqrt {\sqrt {x} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((2*(3*sqrt(x) - 10)*(sqrt(x) + 1) + 43)*(sqrt(x) + 1) - 39)*sqrt(sqrt(x) + 1)*sqrt(sqrt(x) - 1) + 1/3*((
2*sqrt(x) - 5)*(sqrt(x) + 1) + 9)*sqrt(sqrt(x) + 1)*sqrt(sqrt(x) - 1) + 1/2*log(sqrt(sqrt(x) + 1) - sqrt(sqrt(
x) - 1))

________________________________________________________________________________________

maple [A]  time = 0.05, size = 52, normalized size = 0.71 \begin {gather*} -\frac {\sqrt {\sqrt {x}-1}\, \sqrt {\sqrt {x}+1}\, \left (-2 \sqrt {x -1}\, x^{\frac {3}{2}}+\ln \left (\sqrt {x}+\sqrt {x -1}\right )+\sqrt {x -1}\, \sqrt {x}\right )}{4 \sqrt {x -1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(x^(1/2)-1)^(1/2)*(x^(1/2)+1)^(1/2)*(-2*(x-1)^(1/2)*x^(3/2)+(x-1)^(1/2)*x^(1/2)+ln(x^(1/2)+(x-1)^(1/2)))/
(x-1)^(1/2)

________________________________________________________________________________________

maxima [A]  time = 0.58, size = 37, normalized size = 0.51 \begin {gather*} \frac {1}{2} \, {\left (x - 1\right )}^{\frac {3}{2}} \sqrt {x} + \frac {1}{4} \, \sqrt {x - 1} \sqrt {x} - \frac {1}{4} \, \log \left (2 \, \sqrt {x - 1} + 2 \, \sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x - 1)^(3/2)*sqrt(x) + 1/4*sqrt(x - 1)*sqrt(x) - 1/4*log(2*sqrt(x - 1) + 2*sqrt(x))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________