3.2501 \(\int \frac{5-x}{(3+2 x)^2 \sqrt{2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=64 \[ \frac{47 \tanh ^{-1}\left (\frac{8 x+7}{2 \sqrt{5} \sqrt{3 x^2+5 x+2}}\right )}{10 \sqrt{5}}-\frac{13 \sqrt{3 x^2+5 x+2}}{5 (2 x+3)} \]

[Out]

(-13*Sqrt[2 + 5*x + 3*x^2])/(5*(3 + 2*x)) + (47*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(10*Sqrt
[5])

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Rubi [A]  time = 0.0414352, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {806, 724, 206} \[ \frac{47 \tanh ^{-1}\left (\frac{8 x+7}{2 \sqrt{5} \sqrt{3 x^2+5 x+2}}\right )}{10 \sqrt{5}}-\frac{13 \sqrt{3 x^2+5 x+2}}{5 (2 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-13*Sqrt[2 + 5*x + 3*x^2])/(5*(3 + 2*x)) + (47*ArcTanh[(7 + 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(10*Sqrt
[5])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(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] && EqQ[Sim
plify[m + 2*p + 3], 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]

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)^2 \sqrt{2+5 x+3 x^2}} \, dx &=-\frac{13 \sqrt{2+5 x+3 x^2}}{5 (3+2 x)}+\frac{47}{10} \int \frac{1}{(3+2 x) \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{13 \sqrt{2+5 x+3 x^2}}{5 (3+2 x)}-\frac{47}{5} \operatorname{Subst}\left (\int \frac{1}{20-x^2} \, dx,x,\frac{-7-8 x}{\sqrt{2+5 x+3 x^2}}\right )\\ &=-\frac{13 \sqrt{2+5 x+3 x^2}}{5 (3+2 x)}+\frac{47 \tanh ^{-1}\left (\frac{7+8 x}{2 \sqrt{5} \sqrt{2+5 x+3 x^2}}\right )}{10 \sqrt{5}}\\ \end{align*}

Mathematica [A]  time = 0.0233595, size = 64, normalized size = 1. \[ -\frac{13 \sqrt{3 x^2+5 x+2}}{5 (2 x+3)}-\frac{47 \tanh ^{-1}\left (\frac{-8 x-7}{2 \sqrt{5} \sqrt{3 x^2+5 x+2}}\right )}{10 \sqrt{5}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-13*Sqrt[2 + 5*x + 3*x^2])/(5*(3 + 2*x)) - (47*ArcTanh[(-7 - 8*x)/(2*Sqrt[5]*Sqrt[2 + 5*x + 3*x^2])])/(10*Sqr
t[5])

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Maple [A]  time = 0.01, size = 53, normalized size = 0.8 \begin{align*} -{\frac{47\,\sqrt{5}}{50}{\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 ) }-{\frac{13}{10}\sqrt{3\, \left ( x+3/2 \right ) ^{2}-4\,x-{\frac{19}{4}}} \left ( x+{\frac{3}{2}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-47/50*5^(1/2)*arctanh(2/5*(-7/2-4*x)*5^(1/2)/(12*(x+3/2)^2-16*x-19)^(1/2))-13/10/(x+3/2)*(3*(x+3/2)^2-4*x-19/
4)^(1/2)

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Maxima [A]  time = 1.67818, size = 86, normalized size = 1.34 \begin{align*} -\frac{47}{50} \, \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 ) - \frac{13 \, \sqrt{3 \, x^{2} + 5 \, x + 2}}{5 \,{\left (2 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-47/50*sqrt(5)*log(sqrt(5)*sqrt(3*x^2 + 5*x + 2)/abs(2*x + 3) + 5/2/abs(2*x + 3) - 2) - 13/5*sqrt(3*x^2 + 5*x
+ 2)/(2*x + 3)

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Fricas [A]  time = 1.75936, size = 215, normalized size = 3.36 \begin{align*} \frac{47 \, \sqrt{5}{\left (2 \, x + 3\right )} \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 ) - 260 \, \sqrt{3 \, x^{2} + 5 \, x + 2}}{100 \,{\left (2 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/100*(47*sqrt(5)*(2*x + 3)*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)) - 260*sqrt(3*x^2 + 5*x + 2))/(2*x + 3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x}{4 x^{2} \sqrt{3 x^{2} + 5 x + 2} + 12 x \sqrt{3 x^{2} + 5 x + 2} + 9 \sqrt{3 x^{2} + 5 x + 2}}\, dx - \int - \frac{5}{4 x^{2} \sqrt{3 x^{2} + 5 x + 2} + 12 x \sqrt{3 x^{2} + 5 x + 2} + 9 \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)**2/(3*x**2+5*x+2)**(1/2),x)

[Out]

-Integral(x/(4*x**2*sqrt(3*x**2 + 5*x + 2) + 12*x*sqrt(3*x**2 + 5*x + 2) + 9*sqrt(3*x**2 + 5*x + 2)), x) - Int
egral(-5/(4*x**2*sqrt(3*x**2 + 5*x + 2) + 12*x*sqrt(3*x**2 + 5*x + 2) + 9*sqrt(3*x**2 + 5*x + 2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x - 5}{\sqrt{3 \, x^{2} + 5 \, x + 2}{\left (2 \, x + 3\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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