3.970 \(\int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx\)

Optimal. Leaf size=82 \[ \frac {1}{2} \sqrt {x} \left (x+\sqrt {x}\right )^{3/2}-\frac {5}{12} \left (x+\sqrt {x}\right )^{3/2}+\frac {5}{32} \left (2 \sqrt {x}+1\right ) \sqrt {x+\sqrt {x}}-\frac {5}{32} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt {x+\sqrt {x}}}\right ) \]

[Out]

-5/32*arctanh(x^(1/2)/(x+x^(1/2))^(1/2))-5/12*(x+x^(1/2))^(3/2)+1/2*x^(1/2)*(x+x^(1/2))^(3/2)+5/32*(1+2*x^(1/2
))*(x+x^(1/2))^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2018, 670, 640, 612, 620, 206} \[ \frac {1}{2} \sqrt {x} \left (x+\sqrt {x}\right )^{3/2}-\frac {5}{12} \left (x+\sqrt {x}\right )^{3/2}+\frac {5}{32} \left (2 \sqrt {x}+1\right ) \sqrt {x+\sqrt {x}}-\frac {5}{32} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt {x+\sqrt {x}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*Sqrt[Sqrt[x] + x],x]

[Out]

(5*(1 + 2*Sqrt[x])*Sqrt[Sqrt[x] + x])/32 - (5*(Sqrt[x] + x)^(3/2))/12 + (Sqrt[x]*(Sqrt[x] + x)^(3/2))/2 - (5*A
rcTanh[Sqrt[x]/Sqrt[Sqrt[x] + x]])/32

Rule 206

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

Rule 612

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

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 640

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

Rule 670

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

Rule 2018

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)
/n] - 1)*(a*x^Simplify[j/n] + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && IntegerQ[Simplify[j/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

\begin {align*} \int \sqrt {x} \sqrt {\sqrt {x}+x} \, dx &=2 \operatorname {Subst}\left (\int x^2 \sqrt {x+x^2} \, dx,x,\sqrt {x}\right )\\ &=\frac {1}{2} \sqrt {x} \left (\sqrt {x}+x\right )^{3/2}-\frac {5}{4} \operatorname {Subst}\left (\int x \sqrt {x+x^2} \, dx,x,\sqrt {x}\right )\\ &=-\frac {5}{12} \left (\sqrt {x}+x\right )^{3/2}+\frac {1}{2} \sqrt {x} \left (\sqrt {x}+x\right )^{3/2}+\frac {5}{8} \operatorname {Subst}\left (\int \sqrt {x+x^2} \, dx,x,\sqrt {x}\right )\\ &=\frac {5}{32} \left (1+2 \sqrt {x}\right ) \sqrt {\sqrt {x}+x}-\frac {5}{12} \left (\sqrt {x}+x\right )^{3/2}+\frac {1}{2} \sqrt {x} \left (\sqrt {x}+x\right )^{3/2}-\frac {5}{64} \operatorname {Subst}\left (\int \frac {1}{\sqrt {x+x^2}} \, dx,x,\sqrt {x}\right )\\ &=\frac {5}{32} \left (1+2 \sqrt {x}\right ) \sqrt {\sqrt {x}+x}-\frac {5}{12} \left (\sqrt {x}+x\right )^{3/2}+\frac {1}{2} \sqrt {x} \left (\sqrt {x}+x\right )^{3/2}-\frac {5}{32} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {\sqrt {x}+x}}\right )\\ &=\frac {5}{32} \left (1+2 \sqrt {x}\right ) \sqrt {\sqrt {x}+x}-\frac {5}{12} \left (\sqrt {x}+x\right )^{3/2}+\frac {1}{2} \sqrt {x} \left (\sqrt {x}+x\right )^{3/2}-\frac {5}{32} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt {\sqrt {x}+x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 58, normalized size = 0.71 \[ \frac {1}{96} \sqrt {x+\sqrt {x}} \left (48 x^{3/2}+8 x-10 \sqrt {x}-\frac {15 \sinh ^{-1}\left (\sqrt [4]{x}\right )}{\sqrt {\sqrt {x}+1} \sqrt [4]{x}}+15\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sqrt[x] + x]*(15 - 10*Sqrt[x] + 8*x + 48*x^(3/2) - (15*ArcSinh[x^(1/4)])/(Sqrt[1 + Sqrt[x]]*x^(1/4))))/9
6

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fricas [A]  time = 2.10, size = 54, normalized size = 0.66 \[ \frac {1}{96} \, {\left (2 \, {\left (24 \, x - 5\right )} \sqrt {x} + 8 \, x + 15\right )} \sqrt {x + \sqrt {x}} + \frac {5}{128} \, \log \left (4 \, \sqrt {x + \sqrt {x}} {\left (2 \, \sqrt {x} + 1\right )} - 8 \, x - 8 \, \sqrt {x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(2*(24*x - 5)*sqrt(x) + 8*x + 15)*sqrt(x + sqrt(x)) + 5/128*log(4*sqrt(x + sqrt(x))*(2*sqrt(x) + 1) - 8*x
- 8*sqrt(x) - 1)

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giac [A]  time = 0.44, size = 50, normalized size = 0.61 \[ \frac {1}{96} \, {\left (2 \, {\left (4 \, \sqrt {x} {\left (6 \, \sqrt {x} + 1\right )} - 5\right )} \sqrt {x} + 15\right )} \sqrt {x + \sqrt {x}} + \frac {5}{64} \, \log \left (-2 \, \sqrt {x + \sqrt {x}} + 2 \, \sqrt {x} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(2*(4*sqrt(x)*(6*sqrt(x) + 1) - 5)*sqrt(x) + 15)*sqrt(x + sqrt(x)) + 5/64*log(-2*sqrt(x + sqrt(x)) + 2*sq
rt(x) + 1)

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maple [A]  time = 0.00, size = 54, normalized size = 0.66 \[ -\frac {5 \ln \left (\sqrt {x}+\frac {1}{2}+\sqrt {x +\sqrt {x}}\right )}{64}+\frac {\left (x +\sqrt {x}\right )^{\frac {3}{2}} \sqrt {x}}{2}-\frac {5 \left (x +\sqrt {x}\right )^{\frac {3}{2}}}{12}+\frac {5 \left (2 \sqrt {x}+1\right ) \sqrt {x +\sqrt {x}}}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^(1/2)*(x+x^(1/2))^(3/2)-5/12*(x+x^(1/2))^(3/2)+5/32*(2*x^(1/2)+1)*(x+x^(1/2))^(1/2)-5/64*ln(x^(1/2)+1/2+
(x+x^(1/2))^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x + \sqrt {x}} \sqrt {x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(x + x^(1/2))^(1/2),x)

[Out]

int(x^(1/2)*(x + x^(1/2))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {x} \sqrt {\sqrt {x} + x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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