### 3.650 $$\int \frac{\sqrt{2+3 x}}{1-x^2} \, dx$$

Optimal. Leaf size=35 $\sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{3 x+2}}{\sqrt{5}}\right )-\tan ^{-1}\left (\sqrt{3 x+2}\right )$

[Out]

-ArcTan[Sqrt[2 + 3*x]] + Sqrt[5]*ArcTanh[Sqrt[2 + 3*x]/Sqrt[5]]

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Rubi [A]  time = 0.0203036, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.21, Rules used = {700, 1130, 206, 204} $\sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{3 x+2}}{\sqrt{5}}\right )-\tan ^{-1}\left (\sqrt{3 x+2}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[Sqrt[2 + 3*x]/(1 - x^2),x]

[Out]

-ArcTan[Sqrt[2 + 3*x]] + Sqrt[5]*ArcTanh[Sqrt[2 + 3*x]/Sqrt[5]]

Rule 700

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(c*d^2 + a*e^2 - 2*c*d
*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1130

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2*(b/q + 1))/2, Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2*(b/q - 1))/2, Int[(d*x)^(m - 2)/(b
/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sqrt{2+3 x}}{1-x^2} \, dx &=6 \operatorname{Subst}\left (\int \frac{x^2}{5+4 x^2-x^4} \, dx,x,\sqrt{2+3 x}\right )\\ &=5 \operatorname{Subst}\left (\int \frac{1}{5-x^2} \, dx,x,\sqrt{2+3 x}\right )+\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,\sqrt{2+3 x}\right )\\ &=-\tan ^{-1}\left (\sqrt{2+3 x}\right )+\sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{2+3 x}}{\sqrt{5}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0151169, size = 35, normalized size = 1. $\sqrt{5} \tanh ^{-1}\left (\frac{\sqrt{3 x+2}}{\sqrt{5}}\right )-\tan ^{-1}\left (\sqrt{3 x+2}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sqrt[2 + 3*x]/(1 - x^2),x]

[Out]

-ArcTan[Sqrt[2 + 3*x]] + Sqrt[5]*ArcTanh[Sqrt[2 + 3*x]/Sqrt[5]]

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Maple [A]  time = 0.047, size = 29, normalized size = 0.8 \begin{align*} -\arctan \left ( \sqrt{2+3\,x} \right ) +{\it Artanh} \left ({\frac{\sqrt{5}}{5}\sqrt{2+3\,x}} \right ) \sqrt{5} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arctan((2+3*x)^(1/2))+arctanh(1/5*(2+3*x)^(1/2)*5^(1/2))*5^(1/2)

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Maxima [A]  time = 1.65505, size = 61, normalized size = 1.74 \begin{align*} -\frac{1}{2} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - \sqrt{3 \, x + 2}}{\sqrt{5} + \sqrt{3 \, x + 2}}\right ) - \arctan \left (\sqrt{3 \, x + 2}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.32939, size = 116, normalized size = 3.31 \begin{align*} \frac{1}{2} \, \sqrt{5} \log \left (\frac{2 \, \sqrt{5} \sqrt{3 \, x + 2} + 3 \, x + 7}{x - 1}\right ) - \arctan \left (\sqrt{3 \, x + 2}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(5)*log((2*sqrt(5)*sqrt(3*x + 2) + 3*x + 7)/(x - 1)) - arctan(sqrt(3*x + 2))

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Sympy [A]  time = 3.09664, size = 70, normalized size = 2. \begin{align*} - 5 \left (\begin{cases} - \frac{\sqrt{5} \operatorname{acoth}{\left (\frac{\sqrt{5} \sqrt{3 x + 2}}{5} \right )}}{5} & \text{for}\: 3 x + 2 > 5 \\- \frac{\sqrt{5} \operatorname{atanh}{\left (\frac{\sqrt{5} \sqrt{3 x + 2}}{5} \right )}}{5} & \text{for}\: 3 x + 2 < 5 \end{cases}\right ) - \operatorname{atan}{\left (\sqrt{3 x + 2} \right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5*Piecewise((-sqrt(5)*acoth(sqrt(5)*sqrt(3*x + 2)/5)/5, 3*x + 2 > 5), (-sqrt(5)*atanh(sqrt(5)*sqrt(3*x + 2)/5
)/5, 3*x + 2 < 5)) - atan(sqrt(3*x + 2))

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Giac [A]  time = 1.32168, size = 65, normalized size = 1.86 \begin{align*} -\frac{1}{2} \, \sqrt{5} \log \left (\frac{{\left | -2 \, \sqrt{5} + 2 \, \sqrt{3 \, x + 2} \right |}}{2 \,{\left (\sqrt{5} + \sqrt{3 \, x + 2}\right )}}\right ) - \arctan \left (\sqrt{3 \, x + 2}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(5)*log(1/2*abs(-2*sqrt(5) + 2*sqrt(3*x + 2))/(sqrt(5) + sqrt(3*x + 2))) - arctan(sqrt(3*x + 2))