3.7.50 \(\int \frac {\sqrt {2+3 x}}{1-x^2} \, dx\) [650]

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

[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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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 = {714, 1144, 212, 210} \begin {gather*} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {3 x+2}}{\sqrt {5}}\right )-\text {ArcTan}\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 210

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

Rule 212

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

Rule 714

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 1144

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2/2)*(b/q + 1), Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2/2)*(b/q - 1), 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 \text {Subst}\left (\int \frac {x^2}{5+4 x^2-x^4} \, dx,x,\sqrt {2+3 x}\right )\\ &=5 \text {Subst}\left (\int \frac {1}{5-x^2} \, dx,x,\sqrt {2+3 x}\right )+\text {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 = 0.04, size = 35, normalized size = 1.00 \begin {gather*} -\tan ^{-1}\left (\sqrt {2+3 x}\right )+\sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {2+3 x}}{\sqrt {5}}\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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Maple [A]
time = 0.44, size = 29, normalized size = 0.83

method	result	size
derivativedivides	\(-\arctan \left (\sqrt {2+3 x}\right )+\arctanh \left (\frac {\sqrt {2+3 x}\, \sqrt {5}}{5}\right ) \sqrt {5}\)	\(29\)
default	\(-\arctan \left (\sqrt {2+3 x}\right )+\arctanh \left (\frac {\sqrt {2+3 x}\, \sqrt {5}}{5}\right ) \sqrt {5}\)	\(29\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {3 x \RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {2+3 x}+\RootOf \left (\textit {\_Z}^{2}+1\right )}{x +1}\right )}{2}-\frac {\RootOf \left (\textit {\_Z}^{2}-5\right ) \ln \left (\frac {3 \RootOf \left (\textit {\_Z}^{2}-5\right ) x +7 \RootOf \left (\textit {\_Z}^{2}-5\right )-10 \sqrt {2+3 x}}{x -1}\right )}{2}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.50, 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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Fricas [A]
time = 2.69, 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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Sympy [A]
time = 1.94, size = 63, normalized size = 1.80 \begin {gather*} - 5 \left (\begin {cases} - \frac {\sqrt {5} \operatorname {acoth}{\left (\frac {\sqrt {5} \sqrt {3 x + 2}}{5} \right )}}{5} & \text {for}\: x > 1 \\- \frac {\sqrt {5} \operatorname {atanh}{\left (\frac {\sqrt {5} \sqrt {3 x + 2}}{5} \right )}}{5} & \text {for}\: x < 1 \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, x > 1), (-sqrt(5)*atanh(sqrt(5)*sqrt(3*x + 2)/5)/5, x
< 1)) - atan(sqrt(3*x + 2))

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Giac [A]
time = 2.17, 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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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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