3.1468 \(\int \frac {3+x}{\sqrt {4+3 x} (1+x^2)} \, dx\)

Optimal. Leaf size=45 \[ \sqrt {2} \tan ^{-1}\left (\sqrt {6 x+8}+3\right )-\sqrt {2} \tan ^{-1}\left (3-\sqrt {2} \sqrt {3 x+4}\right ) \]

[Out]

arctan(-3+2^(1/2)*(4+3*x)^(1/2))*2^(1/2)+arctan(3+(8+6*x)^(1/2))*2^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 45, 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} \[ \sqrt {2} \tan ^{-1}\left (\sqrt {6 x+8}+3\right )-\sqrt {2} \tan ^{-1}\left (3-\sqrt {2} \sqrt {3 x+4}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[2]*ArcTan[3 - Sqrt[2]*Sqrt[4 + 3*x]]) + Sqrt[2]*ArcTan[3 + Sqrt[8 + 6*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 {3+x}{\sqrt {4+3 x} \left (1+x^2\right )} \, dx &=2 \operatorname {Subst}\left (\int \frac {5+x^2}{25-8 x^2+x^4} \, dx,x,\sqrt {4+3 x}\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{5-3 \sqrt {2} x+x^2} \, dx,x,\sqrt {4+3 x}\right )+\operatorname {Subst}\left (\int \frac {1}{5+3 \sqrt {2} x+x^2} \, dx,x,\sqrt {4+3 x}\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{-2-x^2} \, dx,x,-3 \sqrt {2}+2 \sqrt {4+3 x}\right )\right )-2 \operatorname {Subst}\left (\int \frac {1}{-2-x^2} \, dx,x,3 \sqrt {2}+2 \sqrt {4+3 x}\right )\\ &=-\sqrt {2} \tan ^{-1}\left (3-\sqrt {2} \sqrt {4+3 x}\right )+\sqrt {2} \tan ^{-1}\left (3+\sqrt {2} \sqrt {4+3 x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 63, normalized size = 1.40 \[ \frac {1}{5} \left ((1-3 i) \sqrt {4-3 i} \tanh ^{-1}\left (\frac {\sqrt {3 x+4}}{\sqrt {4-3 i}}\right )+(1+3 i) \sqrt {4+3 i} \tanh ^{-1}\left (\frac {\sqrt {3 x+4}}{\sqrt {4+3 i}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.06, size = 22, normalized size = 0.49 \[ \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (3 \, x - 1\right )}}{2 \, \sqrt {3 \, x + 4}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 63, normalized size = 1.40 \[ \sqrt {2} \arctan \left (\frac {1}{250} \cdot 25^{\frac {3}{4}} \sqrt {10} {\left (3 \cdot 25^{\frac {1}{4}} \sqrt {10} + 10 \, \sqrt {3 \, x + 4}\right )}\right ) + \sqrt {2} \arctan \left (-\frac {1}{250} \cdot 25^{\frac {3}{4}} \sqrt {10} {\left (3 \cdot 25^{\frac {1}{4}} \sqrt {10} - 10 \, \sqrt {3 \, x + 4}\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*arctan(1/250*25^(3/4)*sqrt(10)*(3*25^(1/4)*sqrt(10) + 10*sqrt(3*x + 4))) + sqrt(2)*arctan(-1/250*25^(3
/4)*sqrt(10)*(3*25^(1/4)*sqrt(10) - 10*sqrt(3*x + 4)))

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maple [A]  time = 0.16, size = 52, normalized size = 1.16 \[ \sqrt {2}\, \arctan \left (\frac {\left (2 \sqrt {3 x +4}-3 \sqrt {2}\right ) \sqrt {2}}{2}\right )+\sqrt {2}\, \arctan \left (\frac {\left (2 \sqrt {3 x +4}+3 \sqrt {2}\right ) \sqrt {2}}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x + 3}{{\left (x^{2} + 1\right )} \sqrt {3 \, x + 4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.76, size = 38, normalized size = 0.84 \[ \sqrt {2}\,\left (\mathrm {atan}\left (\frac {\sqrt {2}\,{\left (3\,x+4\right )}^{3/2}}{10}-\frac {3\,\sqrt {6\,x+8}}{10}\right )+\mathrm {atan}\left (\frac {\sqrt {6\,x+8}}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2^(1/2)*(atan((2^(1/2)*(3*x + 4)^(3/2))/10 - (3*(6*x + 8)^(1/2))/10) + atan((6*x + 8)^(1/2)/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x + 3}{\sqrt {3 x + 4} \left (x^{2} + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x + 3)/(sqrt(3*x + 4)*(x**2 + 1)), x)

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