3.6.70 \(\int \frac {\sqrt {2+3 x}}{1-x^2} \, dx\)

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

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Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {700, 1130, 206, 204} \begin {gather*} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {3 x+2}}{\sqrt {5}}\right )-\tan ^{-1}\left (\sqrt {3 x+2}\right ) \end {gather*}

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 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])

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 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]

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*}

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Mathematica [A]  time = 0.02, size = 35, normalized size = 1.00 \begin {gather*} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {3 x+2}}{\sqrt {5}}\right )-\tan ^{-1}\left (\sqrt {3 x+2}\right ) \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.05, size = 35, normalized size = 1.00 \begin {gather*} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {3 x+2}}{\sqrt {5}}\right )-\tan ^{-1}\left (\sqrt {3 x+2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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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fricas [A]  time = 0.40, size = 40, normalized size = 1.14 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.22, size = 48, normalized size = 1.37 \begin {gather*} -\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 {gather*}

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))

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maple [A]  time = 0.06, size = 29, normalized size = 0.83 \begin {gather*} \sqrt {5}\, \arctanh \left (\frac {\sqrt {3 x +2}\, \sqrt {5}}{5}\right )-\arctan \left (\sqrt {3 x +2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.99, size = 45, normalized size = 1.29 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 0.40, size = 28, normalized size = 0.80 \begin {gather*} \sqrt {5}\,\mathrm {atanh}\left (\frac {\sqrt {5}\,\sqrt {3\,x+2}}{5}\right )-\mathrm {atan}\left (\sqrt {3\,x+2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + 2)^(1/2)/(x^2 - 1),x)

[Out]

5^(1/2)*atanh((5^(1/2)*(3*x + 2)^(1/2))/5) - atan((3*x + 2)^(1/2))

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sympy [A]  time = 4.60, size = 70, normalized size = 2.00 \begin {gather*} - 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 {gather*}

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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