3.750 \(\int \frac{x}{(1+x) \sqrt{\frac{2+x}{3+x}}} \, dx\)

Optimal. Leaf size=54 \[ \sqrt{x+2} \sqrt{x+3}-\sinh ^{-1}\left (\sqrt{x+2}\right )+2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{x+2}}{\sqrt{x+3}}\right ) \]

[Out]

Sqrt[2 + x]*Sqrt[3 + x] - ArcSinh[Sqrt[2 + x]] + 2*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[2 + x])/Sqrt[3 + x]]

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Rubi [A]  time = 0.0567202, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1958, 154, 157, 54, 215, 93, 207} \[ \sqrt{x+2} \sqrt{x+3}-\sinh ^{-1}\left (\sqrt{x+2}\right )+2 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{x+2}}{\sqrt{x+3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x/((1 + x)*Sqrt[(2 + x)/(3 + x)]),x]

[Out]

Sqrt[2 + x]*Sqrt[3 + x] - ArcSinh[Sqrt[2 + x]] + 2*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[2 + x])/Sqrt[3 + x]]

Rule 1958

Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[(u*(e*(a + b*x
^n))^p)/(c + d*x^n)^p, x] /; FreeQ[{a, b, c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - (a*d)/b, 0]

Rule 154

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

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

Mathematica [A]  time = 0.0753734, size = 106, normalized size = 1.96 \[ \frac{\sqrt{x+3} \left (x^2+5 x+6\right )+2 \sqrt{2} \sqrt{x+2} \sqrt{-(x+3)^2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{x+2}}{\sqrt{-x-3}}\right )-\sqrt{x+2} (x+3) \sinh ^{-1}\left (\sqrt{x+2}\right )}{\sqrt{\frac{x+2}{x+3}} (x+3)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/((1 + x)*Sqrt[(2 + x)/(3 + x)]),x]

[Out]

(Sqrt[3 + x]*(6 + 5*x + x^2) - Sqrt[2 + x]*(3 + x)*ArcSinh[Sqrt[2 + x]] + 2*Sqrt[2]*Sqrt[2 + x]*Sqrt[-(3 + x)^
2]*ArcTan[(Sqrt[2]*Sqrt[2 + x])/Sqrt[-3 - x]])/(Sqrt[(2 + x)/(3 + x)]*(3 + x)^(3/2))

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Maple [A]  time = 0.015, size = 79, normalized size = 1.5 \begin{align*} -{\frac{2+x}{2} \left ( -2\,\sqrt{2}{\it Artanh} \left ( 1/4\,{\frac{ \left ( 7+3\,x \right ) \sqrt{2}}{\sqrt{{x}^{2}+5\,x+6}}} \right ) +\ln \left ({\frac{5}{2}}+x+\sqrt{{x}^{2}+5\,x+6} \right ) -2\,\sqrt{{x}^{2}+5\,x+6} \right ){\frac{1}{\sqrt{{\frac{2+x}{3+x}}}}}{\frac{1}{\sqrt{ \left ( 3+x \right ) \left ( 2+x \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(1+x)/((2+x)/(3+x))^(1/2),x)

[Out]

-1/2*(2+x)*(-2*2^(1/2)*arctanh(1/4*(7+3*x)*2^(1/2)/(x^2+5*x+6)^(1/2))+ln(5/2+x+(x^2+5*x+6)^(1/2))-2*(x^2+5*x+6
)^(1/2))/((2+x)/(3+x))^(1/2)/((3+x)*(2+x))^(1/2)

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Maxima [B]  time = 1.46524, size = 139, normalized size = 2.57 \begin{align*} -\sqrt{2} \log \left (-\frac{\sqrt{2} - 2 \, \sqrt{\frac{x + 2}{x + 3}}}{\sqrt{2} + 2 \, \sqrt{\frac{x + 2}{x + 3}}}\right ) - \frac{\sqrt{\frac{x + 2}{x + 3}}}{\frac{x + 2}{x + 3} - 1} - \frac{1}{2} \, \log \left (\sqrt{\frac{x + 2}{x + 3}} + 1\right ) + \frac{1}{2} \, \log \left (\sqrt{\frac{x + 2}{x + 3}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(1+x)/((2+x)/(3+x))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.75325, size = 243, normalized size = 4.5 \begin{align*}{\left (x + 3\right )} \sqrt{\frac{x + 2}{x + 3}} + \sqrt{2} \log \left (\frac{2 \, \sqrt{2}{\left (x + 3\right )} \sqrt{\frac{x + 2}{x + 3}} + 3 \, x + 7}{x + 1}\right ) - \frac{1}{2} \, \log \left (\sqrt{\frac{x + 2}{x + 3}} + 1\right ) + \frac{1}{2} \, \log \left (\sqrt{\frac{x + 2}{x + 3}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(1+x)/((2+x)/(3+x))^(1/2),x, algorithm="fricas")

[Out]

(x + 3)*sqrt((x + 2)/(x + 3)) + sqrt(2)*log((2*sqrt(2)*(x + 3)*sqrt((x + 2)/(x + 3)) + 3*x + 7)/(x + 1)) - 1/2
*log(sqrt((x + 2)/(x + 3)) + 1) + 1/2*log(sqrt((x + 2)/(x + 3)) - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\sqrt{\frac{x + 2}{x + 3}} \left (x + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(1+x)/((2+x)/(3+x))**(1/2),x)

[Out]

Integral(x/(sqrt((x + 2)/(x + 3))*(x + 1)), x)

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Giac [B]  time = 1.21892, size = 144, normalized size = 2.67 \begin{align*} -\sqrt{2} \log \left (\frac{{\left | -2 \, \sqrt{2} + 4 \, \sqrt{\frac{x + 2}{x + 3}} \right |}}{2 \,{\left (\sqrt{2} + 2 \, \sqrt{\frac{x + 2}{x + 3}}\right )}}\right ) - \frac{\sqrt{\frac{x + 2}{x + 3}}}{\frac{x + 2}{x + 3} - 1} - \frac{1}{2} \, \log \left (\sqrt{\frac{x + 2}{x + 3}} + 1\right ) + \frac{1}{2} \, \log \left ({\left | \sqrt{\frac{x + 2}{x + 3}} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(1+x)/((2+x)/(3+x))^(1/2),x, algorithm="giac")

[Out]

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