3.1529 \(\int \frac {1}{\sqrt {2+b x} \sqrt {3+b x}} \, dx\)

Optimal. Leaf size=15 \[ \frac {2 \sinh ^{-1}\left (\sqrt {b x+2}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 15, 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 = {63, 215} \[ \frac {2 \sinh ^{-1}\left (\sqrt {b x+2}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcSinh[Sqrt[2 + b*x]])/b

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ \frac {2 \sinh ^{-1}\left (\sqrt {b x+2}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcSinh[Sqrt[2 + b*x]])/b

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fricas [B]  time = 0.43, size = 28, normalized size = 1.87 \[ -\frac {\log \left (-2 \, b x + 2 \, \sqrt {b x + 3} \sqrt {b x + 2} - 5\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-2*b*x + 2*sqrt(b*x + 3)*sqrt(b*x + 2) - 5)/b

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giac [A]  time = 1.01, size = 23, normalized size = 1.53 \[ -\frac {2 \, \log \left (\sqrt {b x + 3} - \sqrt {b x + 2}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 66, normalized size = 4.40 \[ \frac {\sqrt {\left (b x +2\right ) \left (b x +3\right )}\, \ln \left (\frac {b^{2} x +\frac {5}{2} b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}+5 b x +6}\right )}{\sqrt {b x +2}\, \sqrt {b x +3}\, \sqrt {b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*x+2)*(b*x+3))^(1/2)/(b*x+2)^(1/2)/(b*x+3)^(1/2)*ln((5/2*b+b^2*x)/(b^2)^(1/2)+(b^2*x^2+5*b*x+6)^(1/2))/(b^2
)^(1/2)

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maxima [B]  time = 1.38, size = 33, normalized size = 2.20 \[ \frac {\log \left (2 \, b^{2} x + 2 \, \sqrt {b^{2} x^{2} + 5 \, b x + 6} b + 5 \, b\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(2*b^2*x + 2*sqrt(b^2*x^2 + 5*b*x + 6)*b + 5*b)/b

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mupad [B]  time = 0.29, size = 47, normalized size = 3.13 \[ -\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {3}-\sqrt {b\,x+3}\right )}{\left (\sqrt {2}-\sqrt {b\,x+2}\right )\,\sqrt {-b^2}}\right )}{\sqrt {-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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