3.2500 \(\int \frac{5-x}{(3+2 x) \sqrt{2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=77 \[ \frac{13 \tanh ^{-1}\left (\frac{8 x+7}{2 \sqrt{5} \sqrt{3 x^2+5 x+2}}\right )}{2 \sqrt{5}}-\frac{\tanh ^{-1}\left (\frac{6 x+5}{2 \sqrt{3} \sqrt{3 x^2+5 x+2}}\right )}{2 \sqrt{3}} \]

[Out]

-ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])]/(2*Sqrt[3]) + (13*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 +
5*x + 3*x^2])])/(2*Sqrt[5])

________________________________________________________________________________________

Rubi [A]  time = 0.0460193, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {843, 621, 206, 724} \[ \frac{13 \tanh ^{-1}\left (\frac{8 x+7}{2 \sqrt{5} \sqrt{3 x^2+5 x+2}}\right )}{2 \sqrt{5}}-\frac{\tanh ^{-1}\left (\frac{6 x+5}{2 \sqrt{3} \sqrt{3 x^2+5 x+2}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[(5 + 6*x)/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])]/(2*Sqrt[3]) + (13*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 +
5*x + 3*x^2])])/(2*Sqrt[5])

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 724

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

Rubi steps

\begin{align*} \int \frac{5-x}{(3+2 x) \sqrt{2+5 x+3 x^2}} \, dx &=-\left (\frac{1}{2} \int \frac{1}{\sqrt{2+5 x+3 x^2}} \, dx\right )+\frac{13}{2} \int \frac{1}{(3+2 x) \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\left (13 \operatorname{Subst}\left (\int \frac{1}{20-x^2} \, dx,x,\frac{-7-8 x}{\sqrt{2+5 x+3 x^2}}\right )\right )-\operatorname{Subst}\left (\int \frac{1}{12-x^2} \, dx,x,\frac{5+6 x}{\sqrt{2+5 x+3 x^2}}\right )\\ &=-\frac{\tanh ^{-1}\left (\frac{5+6 x}{2 \sqrt{3} \sqrt{2+5 x+3 x^2}}\right )}{2 \sqrt{3}}+\frac{13 \tanh ^{-1}\left (\frac{7+8 x}{2 \sqrt{5} \sqrt{2+5 x+3 x^2}}\right )}{2 \sqrt{5}}\\ \end{align*}

Mathematica [A]  time = 0.0218272, size = 72, normalized size = 0.94 \[ \frac{1}{30} \left (-39 \sqrt{5} \tanh ^{-1}\left (\frac{-8 x-7}{2 \sqrt{5} \sqrt{3 x^2+5 x+2}}\right )-5 \sqrt{3} \tanh ^{-1}\left (\frac{6 x+5}{2 \sqrt{9 x^2+15 x+6}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-39*Sqrt[5]*ArcTanh[(-7 - 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])] - 5*Sqrt[3]*ArcTanh[(5 + 6*x)/(2*Sqrt[6 + 1
5*x + 9*x^2])])/30

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 61, normalized size = 0.8 \begin{align*} -{\frac{\sqrt{3}}{6}\ln \left ({\frac{\sqrt{3}}{3} \left ({\frac{5}{2}}+3\,x \right ) }+\sqrt{3\,{x}^{2}+5\,x+2} \right ) }-{\frac{13\,\sqrt{5}}{10}{\it Artanh} \left ({\frac{2\,\sqrt{5}}{5} \left ( -{\frac{7}{2}}-4\,x \right ){\frac{1}{\sqrt{12\, \left ( x+3/2 \right ) ^{2}-16\,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+5*x+2)^(1/2),x)

[Out]

-1/6*ln(1/3*(5/2+3*x)*3^(1/2)+(3*x^2+5*x+2)^(1/2))*3^(1/2)-13/10*5^(1/2)*arctanh(2/5*(-7/2-4*x)*5^(1/2)/(12*(x
+3/2)^2-16*x-19)^(1/2))

________________________________________________________________________________________

Maxima [A]  time = 1.63645, size = 95, normalized size = 1.23 \begin{align*} -\frac{1}{6} \, \sqrt{3} \log \left (\sqrt{3} \sqrt{3 \, x^{2} + 5 \, x + 2} + 3 \, x + \frac{5}{2}\right ) - \frac{13}{10} \, \sqrt{5} \log \left (\frac{\sqrt{5} \sqrt{3 \, x^{2} + 5 \, x + 2}}{{\left | 2 \, x + 3 \right |}} + \frac{5}{2 \,{\left | 2 \, x + 3 \right |}} - 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*log(sqrt(3)*sqrt(3*x^2 + 5*x + 2) + 3*x + 5/2) - 13/10*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/
abs(2*x + 3) + 5/2/abs(2*x + 3) - 2)

________________________________________________________________________________________

Fricas [A]  time = 1.89032, size = 258, normalized size = 3.35 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (-4 \, \sqrt{3} \sqrt{3 \, x^{2} + 5 \, x + 2}{\left (6 \, x + 5\right )} + 72 \, x^{2} + 120 \, x + 49\right ) + \frac{13}{20} \, \sqrt{5} \log \left (\frac{4 \, \sqrt{5} \sqrt{3 \, x^{2} + 5 \, x + 2}{\left (8 \, x + 7\right )} + 124 \, x^{2} + 212 \, x + 89}{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+5*x+2)^(1/2),x, algorithm="fricas")

[Out]

1/12*sqrt(3)*log(-4*sqrt(3)*sqrt(3*x^2 + 5*x + 2)*(6*x + 5) + 72*x^2 + 120*x + 49) + 13/20*sqrt(5)*log((4*sqrt
(5)*sqrt(3*x^2 + 5*x + 2)*(8*x + 7) + 124*x^2 + 212*x + 89)/(4*x^2 + 12*x + 9))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.21158, size = 144, normalized size = 1.87 \begin{align*} \frac{13}{10} \, \sqrt{5} \log \left (\frac{{\left | -4 \, \sqrt{3} x - 2 \, \sqrt{5} - 6 \, \sqrt{3} + 4 \, \sqrt{3 \, x^{2} + 5 \, x + 2} \right |}}{{\left | -4 \, \sqrt{3} x + 2 \, \sqrt{5} - 6 \, \sqrt{3} + 4 \, \sqrt{3 \, x^{2} + 5 \, x + 2} \right |}}\right ) + \frac{1}{6} \, \sqrt{3} \log \left ({\left | -6 \, \sqrt{3} x - 5 \, \sqrt{3} + 6 \, \sqrt{3 \, x^{2} + 5 \, x + 2} \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

13/10*sqrt(5)*log(abs(-4*sqrt(3)*x - 2*sqrt(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))/abs(-4*sqrt(3)*x + 2*sqr
t(5) - 6*sqrt(3) + 4*sqrt(3*x^2 + 5*x + 2))) + 1/6*sqrt(3)*log(abs(-6*sqrt(3)*x - 5*sqrt(3) + 6*sqrt(3*x^2 + 5
*x + 2)))