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

Optimal. Leaf size=74 \[ -\frac{4 \sqrt{x} \sqrt{x+1}}{3 (1-3 x)}+\frac{8}{9 (1-3 x)}+\frac{5}{9} \log (1-3 x)-\frac{8}{9} \sinh ^{-1}\left (\sqrt{x}\right )+\frac{10}{9} \tanh ^{-1}\left (\frac{2 \sqrt{x}}{\sqrt{x+1}}\right ) \]

[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

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Rubi [A]  time = 0.0559415, antiderivative size = 74, normalized size of antiderivative = 1., 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} \[ -\frac{4 \sqrt{x} \sqrt{x+1}}{3 (1-3 x)}+\frac{8}{9 (1-3 x)}+\frac{5}{9} \log (1-3 x)-\frac{8}{9} \sinh ^{-1}\left (\sqrt{x}\right )+\frac{10}{9} \tanh ^{-1}\left (\frac{2 \sqrt{x}}{\sqrt{x+1}}\right ) \]

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 6742

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

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 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 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 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 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])

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*}

Mathematica [A]  time = 0.132287, size = 126, normalized size = 1.7 \[ \frac{12 x^{3/2}+12 \sqrt{x}-8 \sqrt{x+1}+15 \sqrt{x+1} x \log (1-3 x)-5 \sqrt{x+1} \log (1-3 x)+10 \sqrt{-x-1} (3 x-1) \tan ^{-1}\left (\frac{2 \sqrt{x}}{\sqrt{-x-1}}\right )-8 \sqrt{x+1} (3 x-1) \sinh ^{-1}\left (\sqrt{x}\right )}{9 \sqrt{x+1} (3 x-1)} \]

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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Maple [B]  time = 0.014, size = 115, normalized size = 1.6 \begin{align*} -{\frac{8}{27\,x-9}}+{\frac{5\,\ln \left ( 3\,x-1 \right ) }{9}}-{\frac{1}{27\,x-9}\sqrt{x}\sqrt{1+x} \left ( 12\,\ln \left ( 1/2+x+\sqrt{x \left ( 1+x \right ) } \right ) x-15\,{\it Artanh} \left ( 1/4\,{\frac{1+5\,x}{\sqrt{x \left ( 1+x \right ) }}} \right ) x-4\,\ln \left ( 1/2+x+\sqrt{x \left ( 1+x \right ) } \right ) +5\,{\it Artanh} \left ( 1/4\,{\frac{1+5\,x}{\sqrt{x \left ( 1+x \right ) }}} \right ) -12\,\sqrt{x \left ( 1+x \right ) } \right ){\frac{1}{\sqrt{x \left ( 1+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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Fricas [A]  time = 2.12849, size = 309, normalized size = 4.18 \begin{align*} -\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{align*}

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

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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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

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