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]

-arcsinh((2+x)^(1/2))+2*arctanh(2^(1/2)*(2+x)^(1/2)/(3+x)^(1/2))*2^(1/2)+(2+x)^(1/2)*(3+x)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.07, 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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fricas [B]  time = 0.41, size = 83, normalized size = 1.54 \[ {\left (x + 3\right )} \sqrt {\frac {x + 2}{x + 3}} + \sqrt {2} \log \left (\frac {2 \, \sqrt {2} {\left (x + 3\right )} \sqrt {\frac {x + 2}{x + 3}} + 3 \, x + 7}{x + 1}\right ) - \frac {1}{2} \, \log \left (\sqrt {\frac {x + 2}{x + 3}} + 1\right ) + \frac {1}{2} \, \log \left (\sqrt {\frac {x + 2}{x + 3}} - 1\right ) \]

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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giac [B]  time = 0.30, size = 107, normalized size = 1.98 \[ -\sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sqrt {\frac {x + 2}{x + 3}} \right |}}{2 \, {\left (\sqrt {2} + 2 \, \sqrt {\frac {x + 2}{x + 3}}\right )}}\right ) - \frac {\sqrt {\frac {x + 2}{x + 3}}}{\frac {x + 2}{x + 3} - 1} - \frac {1}{2} \, \log \left (\sqrt {\frac {x + 2}{x + 3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left | \sqrt {\frac {x + 2}{x + 3}} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.02, size = 81, normalized size = 1.50 \[ \frac {\left (x +2\right ) \left (2 \sqrt {2}\, \arctanh \left (\frac {\left (3 x +7\right ) \sqrt {2}}{4 \sqrt {x^{2}+5 x +6}}\right )-\ln \left (x +\frac {5}{2}+\sqrt {x^{2}+5 x +6}\right )+2 \sqrt {x^{2}+5 x +6}\right )}{2 \sqrt {\frac {x +2}{x +3}}\, \sqrt {\left (x +3\right ) \left (x +2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.22, size = 103, normalized size = 1.91 \[ -\sqrt {2} \log \left (-\frac {\sqrt {2} - 2 \, \sqrt {\frac {x + 2}{x + 3}}}{\sqrt {2} + 2 \, \sqrt {\frac {x + 2}{x + 3}}}\right ) - \frac {\sqrt {\frac {x + 2}{x + 3}}}{\frac {x + 2}{x + 3} - 1} - \frac {1}{2} \, \log \left (\sqrt {\frac {x + 2}{x + 3}} + 1\right ) + \frac {1}{2} \, \log \left (\sqrt {\frac {x + 2}{x + 3}} - 1\right ) \]

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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mupad [B]  time = 0.09, size = 62, normalized size = 1.15 \[ 2\,\sqrt {2}\,\mathrm {atanh}\left (\sqrt {2}\,\sqrt {\frac {x+2}{x+3}}\right )-\frac {\sqrt {\frac {x+2}{x+3}}}{\frac {x+2}{x+3}-1}-\mathrm {atanh}\left (\sqrt {\frac {x+2}{x+3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*2^(1/2)*atanh(2^(1/2)*((x + 2)/(x + 3))^(1/2)) - ((x + 2)/(x + 3))^(1/2)/((x + 2)/(x + 3) - 1) - atanh(((x +
2)/(x + 3))^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\sqrt {\frac {x + 2}{x + 3}} \left (x + 1\right )}\, dx \]

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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