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

Optimal. Leaf size=16 \[ \sinh ^{-1}\left (\frac {\sqrt {2 x+1}}{\sqrt {2}}\right ) \]

[Out]

arcsinh(1/2*(1+2*x)^(1/2)*2^(1/2))

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {54, 215} \[ \sinh ^{-1}\left (\frac {\sqrt {2 x+1}}{\sqrt {2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcSinh[Sqrt[1 + 2*x]/Sqrt[2]]

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 31, normalized size = 1.94 \[ \frac {\sqrt {2 x+1} \sin ^{-1}\left (\sqrt {-x-\frac {1}{2}}\right )}{\sqrt {-2 x-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.09, size = 23, normalized size = 1.44 \[ -\frac {1}{2} \, \log \left (\sqrt {2 \, x + 3} \sqrt {2 \, x + 1} - 2 \, x - 2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.17, size = 20, normalized size = 1.25 \[ -\log \left (\sqrt {2 \, x + 3} - \sqrt {2 \, x + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(sqrt(2*x + 3) - sqrt(2*x + 1))

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maple [B]  time = 0.01, size = 57, normalized size = 3.56 \[ \frac {\sqrt {\left (2 x +1\right ) \left (2 x +3\right )}\, \sqrt {4}\, \ln \left (\frac {\left (4 x +4\right ) \sqrt {4}}{4}+\sqrt {4 x^{2}+8 x +3}\right )}{4 \sqrt {2 x +1}\, \sqrt {2 x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.41, size = 22, normalized size = 1.38 \[ \frac {1}{2} \, \log \left (8 \, x + 4 \, \sqrt {4 \, x^{2} + 8 \, x + 3} + 8\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.37, size = 28, normalized size = 1.75 \[ -2\,\mathrm {atanh}\left (\frac {\sqrt {3}-\sqrt {2\,x+3}}{\sqrt {2\,x+1}-1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.02, size = 27, normalized size = 1.69 \[ \begin {cases} \operatorname {acosh}{\left (\sqrt {x + \frac {3}{2}} \right )} & \text {for}\: \left |{x + \frac {3}{2}}\right | > 1 \\- i \operatorname {asin}{\left (\sqrt {x + \frac {3}{2}} \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((acosh(sqrt(x + 3/2)), Abs(x + 3/2) > 1), (-I*asin(sqrt(x + 3/2)), True))

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