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

Optimal. Leaf size=44 \[ \frac {1}{2} \sqrt {x+1} \sqrt {2 x+3}-\frac {\sinh ^{-1}\left (\sqrt {2} \sqrt {x+1}\right )}{2 \sqrt {2}} \]

[Out]

-1/4*arcsinh(2^(1/2)*(1+x)^(1/2))*2^(1/2)+1/2*(1+x)^(1/2)*(3+2*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1958, 50, 54, 215} \[ \frac {1}{2} \sqrt {x+1} \sqrt {2 x+3}-\frac {\sinh ^{-1}\left (\sqrt {2} \sqrt {x+1}\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(1 + x)/(3 + 2*x)],x]

[Out]

(Sqrt[1 + x]*Sqrt[3 + 2*x])/2 - ArcSinh[Sqrt[2]*Sqrt[1 + x]]/(2*Sqrt[2])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 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 \sqrt {\frac {1+x}{3+2 x}} \, dx &=\int \frac {\sqrt {1+x}}{\sqrt {3+2 x}} \, dx\\ &=\frac {1}{2} \sqrt {1+x} \sqrt {3+2 x}-\frac {1}{4} \int \frac {1}{\sqrt {1+x} \sqrt {3+2 x}} \, dx\\ &=\frac {1}{2} \sqrt {1+x} \sqrt {3+2 x}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+2 x^2}} \, dx,x,\sqrt {1+x}\right )\\ &=\frac {1}{2} \sqrt {1+x} \sqrt {3+2 x}-\frac {\sinh ^{-1}\left (\sqrt {2} \sqrt {1+x}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 71, normalized size = 1.61 \[ \frac {2 (x+1) \sqrt {2 x+3}-\sqrt {2} \sqrt {x+1} \sinh ^{-1}\left (\sqrt {2} \sqrt {x+1}\right )}{4 \sqrt {\frac {x+1}{2 x+3}} \sqrt {2 x+3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(1 + x)/(3 + 2*x)],x]

[Out]

(2*(1 + x)*Sqrt[3 + 2*x] - Sqrt[2]*Sqrt[1 + x]*ArcSinh[Sqrt[2]*Sqrt[1 + x]])/(4*Sqrt[(1 + x)/(3 + 2*x)]*Sqrt[3
+ 2*x])

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fricas [A]  time = 0.43, size = 55, normalized size = 1.25 \[ \frac {1}{2} \, {\left (2 \, x + 3\right )} \sqrt {\frac {x + 1}{2 \, x + 3}} + \frac {1}{8} \, \sqrt {2} \log \left (2 \, \sqrt {2} {\left (2 \, x + 3\right )} \sqrt {\frac {x + 1}{2 \, x + 3}} - 4 \, x - 5\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*x + 3)*sqrt((x + 1)/(2*x + 3)) + 1/8*sqrt(2)*log(2*sqrt(2)*(2*x + 3)*sqrt((x + 1)/(2*x + 3)) - 4*x - 5)

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giac [B]  time = 1.06, size = 61, normalized size = 1.39 \[ \frac {1}{8} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} + 5 \, x + 3}\right )} - 5 \right |}\right ) \mathrm {sgn}\left (2 \, x + 3\right ) + \frac {1}{2} \, \sqrt {2 \, x^{2} + 5 \, x + 3} \mathrm {sgn}\left (2 \, x + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*log(abs(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 + 5*x + 3)) - 5))*sgn(2*x + 3) + 1/2*sqrt(2*x^2 + 5*x +
3)*sgn(2*x + 3)

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maple [B]  time = 0.01, size = 76, normalized size = 1.73 \[ \frac {\sqrt {\frac {x +1}{2 x +3}}\, \left (2 x +3\right ) \left (-\sqrt {2}\, \ln \left (\sqrt {2}\, x +\frac {5 \sqrt {2}}{4}+\sqrt {2 x^{2}+5 x +3}\right )+4 \sqrt {2 x^{2}+5 x +3}\right )}{8 \sqrt {\left (2 x +3\right ) \left (x +1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((x+1)/(2*x+3))^(1/2)*(2*x+3)*(-ln(5/4*2^(1/2)+2^(1/2)*x+(2*x^2+5*x+3)^(1/2))*2^(1/2)+4*(2*x^2+5*x+3)^(1/2
))/((2*x+3)*(x+1))^(1/2)

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maxima [B]  time = 1.31, size = 80, normalized size = 1.82 \[ \frac {1}{8} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - 2 \, \sqrt {\frac {x + 1}{2 \, x + 3}}}{\sqrt {2} + 2 \, \sqrt {\frac {x + 1}{2 \, x + 3}}}\right ) - \frac {\sqrt {\frac {x + 1}{2 \, x + 3}}}{2 \, {\left (\frac {2 \, {\left (x + 1\right )}}{2 \, x + 3} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*log(-(sqrt(2) - 2*sqrt((x + 1)/(2*x + 3)))/(sqrt(2) + 2*sqrt((x + 1)/(2*x + 3)))) - 1/2*sqrt((x +
1)/(2*x + 3))/(2*(x + 1)/(2*x + 3) - 1)

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mupad [B]  time = 0.21, size = 57, normalized size = 1.30 \[ -\frac {\sqrt {2}\,\mathrm {atanh}\left (\sqrt {2}\,\sqrt {\frac {x+1}{2\,x+3}}\right )}{4}-\frac {\sqrt {\frac {x+1}{2\,x+3}}}{2\,\left (\frac {2\,x+2}{2\,x+3}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x + 1)/(2*x + 3))^(1/2),x)

[Out]

- (2^(1/2)*atanh(2^(1/2)*((x + 1)/(2*x + 3))^(1/2)))/4 - ((x + 1)/(2*x + 3))^(1/2)/(2*((2*x + 2)/(2*x + 3) - 1
))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt((x + 1)/(2*x + 3)), x)

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