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[Sqrt[1 + 2*x]/Sqrt[2]]

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Rubi [A]  time = 0.0052886, antiderivative size = 16, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0097572, 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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Maple [B]  time = 0.005, size = 57, normalized size = 3.6 \begin{align*}{\frac{\sqrt{4}}{4}\sqrt{ \left ( 1+2\,x \right ) \left ( 3+2\,x \right ) }\ln \left ({\frac{ \left ( 4+4\,x \right ) \sqrt{4}}{4}}+\sqrt{4\,{x}^{2}+8\,x+3} \right ){\frac{1}{\sqrt{1+2\,x}}}{\frac{1}{\sqrt{3+2\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.28276, size = 30, normalized size = 1.88 \begin{align*} \frac{1}{2} \, \log \left (8 \, x + 4 \, \sqrt{4 \, x^{2} + 8 \, x + 3} + 8\right ) \end{align*}

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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Fricas [A]  time = 1.47466, size = 66, normalized size = 4.12 \begin{align*} -\frac{1}{2} \, \log \left (\sqrt{2 \, x + 3} \sqrt{2 \, x + 1} - 2 \, x - 2\right ) \end{align*}

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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Sympy [A]  time = 1.16709, size = 27, normalized size = 1.69 \begin{align*} \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} \end{align*}

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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Giac [A]  time = 1.45895, size = 28, normalized size = 1.75 \begin{align*} -\log \left ({\left | -\sqrt{2 \, x + 3} + \sqrt{2 \, x + 1} \right |}\right ) \end{align*}

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(abs(-sqrt(2*x + 3) + sqrt(2*x + 1)))