3.1402 \(\int \frac{5-x}{(3+2 x) \sqrt{2+3 x^2}} \, dx\)

Optimal. Leaf size=52 \[ -\frac{13 \tanh ^{-1}\left (\frac{4-9 x}{\sqrt{35} \sqrt{3 x^2+2}}\right )}{2 \sqrt{35}}-\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{2 \sqrt{3}} \]

[Out]

-ArcSinh[Sqrt[3/2]*x]/(2*Sqrt[3]) - (13*ArcTanh[(4 - 9*x)/(Sqrt[35]*Sqrt[2 + 3*x^2])])/(2*Sqrt[35])

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Rubi [A]  time = 0.0274531, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {844, 215, 725, 206} \[ -\frac{13 \tanh ^{-1}\left (\frac{4-9 x}{\sqrt{35} \sqrt{3 x^2+2}}\right )}{2 \sqrt{35}}-\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(5 - x)/((3 + 2*x)*Sqrt[2 + 3*x^2]),x]

[Out]

-ArcSinh[Sqrt[3/2]*x]/(2*Sqrt[3]) - (13*ArcTanh[(4 - 9*x)/(Sqrt[35]*Sqrt[2 + 3*x^2])])/(2*Sqrt[35])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 215

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

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

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])

Rubi steps

\begin{align*} \int \frac{5-x}{(3+2 x) \sqrt{2+3 x^2}} \, dx &=-\left (\frac{1}{2} \int \frac{1}{\sqrt{2+3 x^2}} \, dx\right )+\frac{13}{2} \int \frac{1}{(3+2 x) \sqrt{2+3 x^2}} \, dx\\ &=-\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{2 \sqrt{3}}-\frac{13}{2} \operatorname{Subst}\left (\int \frac{1}{35-x^2} \, dx,x,\frac{4-9 x}{\sqrt{2+3 x^2}}\right )\\ &=-\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{2 \sqrt{3}}-\frac{13 \tanh ^{-1}\left (\frac{4-9 x}{\sqrt{35} \sqrt{2+3 x^2}}\right )}{2 \sqrt{35}}\\ \end{align*}

Mathematica [A]  time = 0.0165614, size = 52, normalized size = 1. \[ -\frac{13 \tanh ^{-1}\left (\frac{4-9 x}{\sqrt{35} \sqrt{3 x^2+2}}\right )}{2 \sqrt{35}}-\frac{\sinh ^{-1}\left (\sqrt{\frac{3}{2}} x\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(5 - x)/((3 + 2*x)*Sqrt[2 + 3*x^2]),x]

[Out]

-ArcSinh[Sqrt[3/2]*x]/(2*Sqrt[3]) - (13*ArcTanh[(4 - 9*x)/(Sqrt[35]*Sqrt[2 + 3*x^2])])/(2*Sqrt[35])

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Maple [A]  time = 0.006, size = 44, normalized size = 0.9 \begin{align*} -{\frac{\sqrt{3}}{6}{\it Arcsinh} \left ({\frac{x\sqrt{6}}{2}} \right ) }-{\frac{13\,\sqrt{35}}{70}{\it Artanh} \left ({\frac{ \left ( 8-18\,x \right ) \sqrt{35}}{35}{\frac{1}{\sqrt{12\, \left ( x+3/2 \right ) ^{2}-36\,x-19}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-x)/(3+2*x)/(3*x^2+2)^(1/2),x)

[Out]

-1/6*arcsinh(1/2*x*6^(1/2))*3^(1/2)-13/70*35^(1/2)*arctanh(2/35*(4-9*x)*35^(1/2)/(12*(x+3/2)^2-36*x-19)^(1/2))

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Maxima [A]  time = 1.54959, size = 63, normalized size = 1.21 \begin{align*} -\frac{1}{6} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{2} \, \sqrt{6} x\right ) + \frac{13}{70} \, \sqrt{35} \operatorname{arsinh}\left (\frac{3 \, \sqrt{6} x}{2 \,{\left | 2 \, x + 3 \right |}} - \frac{2 \, \sqrt{6}}{3 \,{\left | 2 \, x + 3 \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3+2*x)/(3*x^2+2)^(1/2),x, algorithm="maxima")

[Out]

-1/6*sqrt(3)*arcsinh(1/2*sqrt(6)*x) + 13/70*sqrt(35)*arcsinh(3/2*sqrt(6)*x/abs(2*x + 3) - 2/3*sqrt(6)/abs(2*x
+ 3))

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Fricas [B]  time = 1.78949, size = 213, normalized size = 4.1 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (\sqrt{3} \sqrt{3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + \frac{13}{140} \, \sqrt{35} \log \left (-\frac{\sqrt{35} \sqrt{3 \, x^{2} + 2}{\left (9 \, x - 4\right )} + 93 \, x^{2} - 36 \, x + 43}{4 \, x^{2} + 12 \, x + 9}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3+2*x)/(3*x^2+2)^(1/2),x, algorithm="fricas")

[Out]

1/12*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + 13/140*sqrt(35)*log(-(sqrt(35)*sqrt(3*x^2 + 2)*(9*x
- 4) + 93*x^2 - 36*x + 43)/(4*x^2 + 12*x + 9))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x}{2 x \sqrt{3 x^{2} + 2} + 3 \sqrt{3 x^{2} + 2}}\, dx - \int - \frac{5}{2 x \sqrt{3 x^{2} + 2} + 3 \sqrt{3 x^{2} + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3+2*x)/(3*x**2+2)**(1/2),x)

[Out]

-Integral(x/(2*x*sqrt(3*x**2 + 2) + 3*sqrt(3*x**2 + 2)), x) - Integral(-5/(2*x*sqrt(3*x**2 + 2) + 3*sqrt(3*x**
2 + 2)), x)

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Giac [B]  time = 1.26026, size = 122, normalized size = 2.35 \begin{align*} \frac{1}{6} \, \sqrt{3} \log \left (-\sqrt{3} x + \sqrt{3 \, x^{2} + 2}\right ) + \frac{13}{70} \, \sqrt{35} \log \left (-\frac{{\left | -2 \, \sqrt{3} x - \sqrt{35} - 3 \, \sqrt{3} + 2 \, \sqrt{3 \, x^{2} + 2} \right |}}{2 \, \sqrt{3} x - \sqrt{35} + 3 \, \sqrt{3} - 2 \, \sqrt{3 \, x^{2} + 2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)/(3+2*x)/(3*x^2+2)^(1/2),x, algorithm="giac")

[Out]

1/6*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2)) + 13/70*sqrt(35)*log(-abs(-2*sqrt(3)*x - sqrt(35) - 3*sqrt(3) +
2*sqrt(3*x^2 + 2))/(2*sqrt(3)*x - sqrt(35) + 3*sqrt(3) - 2*sqrt(3*x^2 + 2)))