3.13.96 \(\int \frac {2+x}{\sqrt {3+4 x} (1+x^2)} \, dx\)

Optimal. Leaf size=29 \[ \tan ^{-1}\left (\sqrt {4 x+3}+2\right )-\tan ^{-1}\left (2-\sqrt {4 x+3}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {827, 1161, 618, 204} \begin {gather*} \tan ^{-1}\left (\sqrt {4 x+3}+2\right )-\tan ^{-1}\left (2-\sqrt {4 x+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + x)/(Sqrt[3 + 4*x]*(1 + x^2)),x]

[Out]

-ArcTan[2 - Sqrt[3 + 4*x]] + ArcTan[2 + Sqrt[3 + 4*x]]

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 827

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

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rubi steps

\begin {align*} \int \frac {2+x}{\sqrt {3+4 x} \left (1+x^2\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {5+x^2}{25-6 x^2+x^4} \, dx,x,\sqrt {3+4 x}\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{5-4 x+x^2} \, dx,x,\sqrt {3+4 x}\right )+\operatorname {Subst}\left (\int \frac {1}{5+4 x+x^2} \, dx,x,\sqrt {3+4 x}\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,-4+2 \sqrt {3+4 x}\right )\right )-2 \operatorname {Subst}\left (\int \frac {1}{-4-x^2} \, dx,x,4+2 \sqrt {3+4 x}\right )\\ &=-\tan ^{-1}\left (2-\sqrt {3+4 x}\right )+\tan ^{-1}\left (2+\sqrt {3+4 x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 41, normalized size = 1.41 \begin {gather*} \tan ^{-1}\left (\left (\frac {1}{5}+\frac {2 i}{5}\right ) \sqrt {4 x+3}\right )-i \tanh ^{-1}\left (\left (\frac {2}{5}+\frac {i}{5}\right ) \sqrt {4 x+3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + x)/(Sqrt[3 + 4*x]*(1 + x^2)),x]

[Out]

ArcTan[(1/5 + (2*I)/5)*Sqrt[3 + 4*x]] - I*ArcTanh[(2/5 + I/5)*Sqrt[3 + 4*x]]

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IntegrateAlgebraic [A]  time = 0.03, size = 24, normalized size = 0.83 \begin {gather*} \tan ^{-1}\left (\frac {\frac {1}{2} (4 x+3)-\frac {5}{2}}{\sqrt {4 x+3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2 + x)/(Sqrt[3 + 4*x]*(1 + x^2)),x]

[Out]

ArcTan[(-5/2 + (3 + 4*x)/2)/Sqrt[3 + 4*x]]

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fricas [A]  time = 0.43, size = 14, normalized size = 0.48 \begin {gather*} \arctan \left (\frac {2 \, x - 1}{\sqrt {4 \, x + 3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(x^2+1)/(3+4*x)^(1/2),x, algorithm="fricas")

[Out]

arctan((2*x - 1)/sqrt(4*x + 3))

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giac [A]  time = 0.19, size = 21, normalized size = 0.72 \begin {gather*} \arctan \left (\sqrt {4 \, x + 3} + 2\right ) + \arctan \left (\sqrt {4 \, x + 3} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(x^2+1)/(3+4*x)^(1/2),x, algorithm="giac")

[Out]

arctan(sqrt(4*x + 3) + 2) + arctan(sqrt(4*x + 3) - 2)

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maple [A]  time = 0.05, size = 22, normalized size = 0.76 \begin {gather*} \arctan \left (-2+\sqrt {4 x +3}\right )+\arctan \left (2+\sqrt {4 x +3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+2)/(x^2+1)/(4*x+3)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.20, size = 21, normalized size = 0.72 \begin {gather*} \arctan \left (\sqrt {4 \, x + 3} + 2\right ) + \arctan \left (\sqrt {4 \, x + 3} - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(x^2+1)/(3+4*x)^(1/2),x, algorithm="maxima")

[Out]

arctan(sqrt(4*x + 3) + 2) + arctan(sqrt(4*x + 3) - 2)

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mupad [B]  time = 0.14, size = 26, normalized size = 0.90 \begin {gather*} \mathrm {atan}\left (\frac {\sqrt {4\,x+3}}{2}\right )+\mathrm {atan}\left (\frac {\left (4\,x+2\right )\,\sqrt {4\,x+3}}{10}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 2)/((4*x + 3)^(1/2)*(x^2 + 1)),x)

[Out]

atan((4*x + 3)^(1/2)/2) + atan(((4*x + 2)*(4*x + 3)^(1/2))/10)

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sympy [A]  time = 99.08, size = 26, normalized size = 0.90 \begin {gather*} \operatorname {atan}{\left (2 - \frac {5}{\sqrt {4 x + 3}} \right )} - \operatorname {atan}{\left (2 + \frac {5}{\sqrt {4 x + 3}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+x)/(x**2+1)/(3+4*x)**(1/2),x)

[Out]

atan(2 - 5/sqrt(4*x + 3)) - atan(2 + 5/sqrt(4*x + 3))

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