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

Optimal. Leaf size=123 \[ \frac {\sqrt {x} \left (4-4 \sqrt {x+1}\right )}{3 x+\left (6 \sqrt {x+1}-6\right ) \sqrt {x}-3 \sqrt {x+1}+3}-\frac {2}{9} \log \left (-\sqrt {x}+\sqrt {x+1}-1\right )-2 \log \left (\sqrt {x}+\sqrt {x+1}-1\right )+\frac {10}{9} \log \left (x+\left (2 \sqrt {x+1}-2\right ) \sqrt {x}-\sqrt {x+1}+1\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 74, normalized size of antiderivative = 0.60, number of steps used = 8, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {6742, 97, 157, 54, 215, 93, 207} \begin {gather*} -\frac {4 \sqrt {x} \sqrt {x+1}}{3 (1-3 x)}+\frac {8}{9 (1-3 x)}+\frac {5}{9} \log (1-3 x)-\frac {8}{9} \sinh ^{-1}\left (\sqrt {x}\right )+\frac {10}{9} \tanh ^{-1}\left (\frac {2 \sqrt {x}}{\sqrt {x+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*Sqrt[x] + Sqrt[1 + x])^(-2),x]

[Out]

8/(9*(1 - 3*x)) - (4*Sqrt[x]*Sqrt[1 + x])/(3*(1 - 3*x)) - (8*ArcSinh[Sqrt[x]])/9 + (10*ArcTanh[(2*Sqrt[x])/Sqr
t[1 + x]])/9 + (5*Log[1 - 3*x])/9

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
 a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
 :> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

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 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
 x] && GtQ[a, 0] && PosQ[b]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {1}{\left (2 \sqrt {x}+\sqrt {1+x}\right )^2} \, dx &=\int \left (\frac {8}{3 (-1+3 x)^2}-\frac {4 \sqrt {x} \sqrt {1+x}}{(-1+3 x)^2}+\frac {5}{3 (-1+3 x)}\right ) \, dx\\ &=\frac {8}{9 (1-3 x)}+\frac {5}{9} \log (1-3 x)-4 \int \frac {\sqrt {x} \sqrt {1+x}}{(-1+3 x)^2} \, dx\\ &=\frac {8}{9 (1-3 x)}-\frac {4 \sqrt {x} \sqrt {1+x}}{3 (1-3 x)}+\frac {5}{9} \log (1-3 x)-\frac {4}{3} \int \frac {\frac {1}{2}+x}{\sqrt {x} \sqrt {1+x} (-1+3 x)} \, dx\\ &=\frac {8}{9 (1-3 x)}-\frac {4 \sqrt {x} \sqrt {1+x}}{3 (1-3 x)}+\frac {5}{9} \log (1-3 x)-\frac {4}{9} \int \frac {1}{\sqrt {x} \sqrt {1+x}} \, dx-\frac {10}{9} \int \frac {1}{\sqrt {x} \sqrt {1+x} (-1+3 x)} \, dx\\ &=\frac {8}{9 (1-3 x)}-\frac {4 \sqrt {x} \sqrt {1+x}}{3 (1-3 x)}+\frac {5}{9} \log (1-3 x)-\frac {8}{9} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,\sqrt {x}\right )-\frac {20}{9} \operatorname {Subst}\left (\int \frac {1}{-1+4 x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {1+x}}\right )\\ &=\frac {8}{9 (1-3 x)}-\frac {4 \sqrt {x} \sqrt {1+x}}{3 (1-3 x)}-\frac {8}{9} \sinh ^{-1}\left (\sqrt {x}\right )+\frac {10}{9} \tanh ^{-1}\left (\frac {2 \sqrt {x}}{\sqrt {1+x}}\right )+\frac {5}{9} \log (1-3 x)\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 126, normalized size = 1.02 \begin {gather*} \frac {12 x^{3/2}+12 \sqrt {x}-8 \sqrt {x+1}+15 \sqrt {x+1} x \log (1-3 x)-5 \sqrt {x+1} \log (1-3 x)+10 \sqrt {-x-1} (3 x-1) \tan ^{-1}\left (\frac {2 \sqrt {x}}{\sqrt {-x-1}}\right )-8 \sqrt {x+1} (3 x-1) \sinh ^{-1}\left (\sqrt {x}\right )}{9 \sqrt {x+1} (3 x-1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*Sqrt[x] + Sqrt[1 + x])^(-2),x]

[Out]

(12*Sqrt[x] + 12*x^(3/2) - 8*Sqrt[1 + x] - 8*Sqrt[1 + x]*(-1 + 3*x)*ArcSinh[Sqrt[x]] + 10*Sqrt[-1 - x]*(-1 + 3
*x)*ArcTan[(2*Sqrt[x])/Sqrt[-1 - x]] - 5*Sqrt[1 + x]*Log[1 - 3*x] + 15*x*Sqrt[1 + x]*Log[1 - 3*x])/(9*Sqrt[1 +
 x]*(-1 + 3*x))

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IntegrateAlgebraic [A]  time = 0.08, size = 78, normalized size = 0.63 \begin {gather*} -\frac {8}{9 (-1+3 x)}+\frac {4 \sqrt {x} \sqrt {1+x}}{3 (-1+3 x)}-\frac {2}{9} \log \left (-\sqrt {x}+\sqrt {1+x}\right )+\frac {10}{9} \log \left (1-x+\sqrt {x} \sqrt {1+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2*Sqrt[x] + Sqrt[1 + x])^(-2),x]

[Out]

-8/(9*(-1 + 3*x)) + (4*Sqrt[x]*Sqrt[1 + x])/(3*(-1 + 3*x)) - (2*Log[-Sqrt[x] + Sqrt[1 + x]])/9 + (10*Log[1 - x
 + Sqrt[x]*Sqrt[1 + x]])/9

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fricas [A]  time = 0.49, size = 105, normalized size = 0.85 \begin {gather*} -\frac {5 \, {\left (3 \, x - 1\right )} \log \left (3 \, \sqrt {x + 1} \sqrt {x} - 3 \, x - 1\right ) - 4 \, {\left (3 \, x - 1\right )} \log \left (2 \, \sqrt {x + 1} \sqrt {x} - 2 \, x - 1\right ) - 5 \, {\left (3 \, x - 1\right )} \log \left (\sqrt {x + 1} \sqrt {x} - x + 1\right ) - 5 \, {\left (3 \, x - 1\right )} \log \left (3 \, x - 1\right ) - 12 \, \sqrt {x + 1} \sqrt {x} - 12 \, x + 12}{9 \, {\left (3 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(5*(3*x - 1)*log(3*sqrt(x + 1)*sqrt(x) - 3*x - 1) - 4*(3*x - 1)*log(2*sqrt(x + 1)*sqrt(x) - 2*x - 1) - 5*
(3*x - 1)*log(sqrt(x + 1)*sqrt(x) - x + 1) - 5*(3*x - 1)*log(3*x - 1) - 12*sqrt(x + 1)*sqrt(x) - 12*x + 12)/(3
*x - 1)

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
]%%%}+%%%{-2,[0]%%%},0,1] at parameters values [-89]Warning, choosing root of [1,0,%%%{-4,[1]%%%}+%%%{-2,[0]%%
%},0,1] at parameters values [63]2*((-5*(x+1)+4)/6/(3*(x+1)-4)+5/18*ln(abs(3*(x+1)-4))+2*(1/9*ln((sqrt(x)-sqrt
(x+1))^2)+5/36*ln(abs((sqrt(x)-sqrt(x+1))^2-3))-5/36*ln(abs(3*(sqrt(x)-sqrt(x+1))^2-1))-(10*(sqrt(x)-sqrt(x+1)
)^2-6)/9/(3*(sqrt(x)-sqrt(x+1))^4-10*(sqrt(x)-sqrt(x+1))^2+3)))

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maple [A]  time = 0.03, size = 115, normalized size = 0.93

method result size
default \(-\frac {8}{9 \left (-1+3 x \right )}+\frac {5 \ln \left (-1+3 x \right )}{9}-\frac {\sqrt {x}\, \sqrt {1+x}\, \left (12 \ln \left (\frac {1}{2}+x +\sqrt {x \left (1+x \right )}\right ) x -15 \arctanh \left (\frac {5 x +1}{4 \sqrt {x \left (1+x \right )}}\right ) x -4 \ln \left (\frac {1}{2}+x +\sqrt {x \left (1+x \right )}\right )+5 \arctanh \left (\frac {5 x +1}{4 \sqrt {x \left (1+x \right )}}\right )-12 \sqrt {x \left (1+x \right )}\right )}{9 \sqrt {x \left (1+x \right )}\, \left (-1+3 x \right )}\) \(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/9/(-1+3*x)+5/9*ln(-1+3*x)-1/9*x^(1/2)*(1+x)^(1/2)*(12*ln(1/2+x+(x*(1+x))^(1/2))*x-15*arctanh(1/4*(5*x+1)/(x
*(1+x))^(1/2))*x-4*ln(1/2+x+(x*(1+x))^(1/2))+5*arctanh(1/4*(5*x+1)/(x*(1+x))^(1/2))-12*(x*(1+x))^(1/2))/(x*(1+
x))^(1/2)/(-1+3*x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (\sqrt {x + 1} + 2 \, \sqrt {x}\right )}^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.19, size = 82, normalized size = 0.67 \begin {gather*} \frac {10\,\mathrm {atanh}\left (\frac {2662400\,\sqrt {x}}{81\,\left (\frac {665600\,x}{81\,{\left (\sqrt {x+1}-1\right )}^2}+\frac {665600}{81}\right )\,\left (\sqrt {x+1}-1\right )}\right )}{9}+\frac {5\,\ln \left (x-\frac {1}{3}\right )}{9}-\frac {16\,\mathrm {atanh}\left (\frac {\sqrt {x}}{\sqrt {x+1}-1}\right )}{9}-\frac {8}{27\,\left (x-\frac {1}{3}\right )}+\frac {4\,\sqrt {x}\,\sqrt {x+1}}{3\,\left (3\,x-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(10*atanh((2662400*x^(1/2))/(81*((665600*x)/(81*((x + 1)^(1/2) - 1)^2) + 665600/81)*((x + 1)^(1/2) - 1))))/9 +
 (5*log(x - 1/3))/9 - (16*atanh(x^(1/2)/((x + 1)^(1/2) - 1)))/9 - 8/(27*(x - 1/3)) + (4*x^(1/2)*(x + 1)^(1/2))
/(3*(3*x - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (2 \sqrt {x} + \sqrt {x + 1}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((2*sqrt(x) + sqrt(x + 1))**(-2), x)

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