3.2552 \(\int \frac{(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx\)

Optimal. Leaf size=68 \[ -\frac{2}{9} (2 x+3)^{3/2}+\frac{62}{9} \sqrt{2 x+3}+12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{170}{9} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

(62*Sqrt[3 + 2*x])/9 - (2*(3 + 2*x)^(3/2))/9 + 12*ArcTanh[Sqrt[3 + 2*x]] - (170*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sq
rt[3 + 2*x]])/9

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Rubi [A]  time = 0.0597366, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {824, 826, 1166, 207} \[ -\frac{2}{9} (2 x+3)^{3/2}+\frac{62}{9} \sqrt{2 x+3}+12 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{170}{9} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(62*Sqrt[3 + 2*x])/9 - (2*(3 + 2*x)^(3/2))/9 + 12*ArcTanh[Sqrt[3 + 2*x]] - (170*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sq
rt[3 + 2*x]])/9

Rule 824

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

Rule 826

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

Rule 1166

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(5-x) (3+2 x)^{3/2}}{2+5 x+3 x^2} \, dx &=-\frac{2}{9} (3+2 x)^{3/2}+\frac{1}{3} \int \frac{\sqrt{3+2 x} (49+31 x)}{2+5 x+3 x^2} \, dx\\ &=\frac{62}{9} \sqrt{3+2 x}-\frac{2}{9} (3+2 x)^{3/2}+\frac{1}{9} \int \frac{317+263 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=\frac{62}{9} \sqrt{3+2 x}-\frac{2}{9} (3+2 x)^{3/2}+\frac{2}{9} \operatorname{Subst}\left (\int \frac{-155+263 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=\frac{62}{9} \sqrt{3+2 x}-\frac{2}{9} (3+2 x)^{3/2}-36 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )+\frac{850}{9} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=\frac{62}{9} \sqrt{3+2 x}-\frac{2}{9} (3+2 x)^{3/2}+12 \tanh ^{-1}\left (\sqrt{3+2 x}\right )-\frac{170}{9} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0296556, size = 56, normalized size = 0.82 \[ -\frac{2}{27} \left (6 \sqrt{2 x+3} (x-14)-162 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+85 \sqrt{15} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(6*(-14 + x)*Sqrt[3 + 2*x] - 162*ArcTanh[Sqrt[3 + 2*x]] + 85*Sqrt[15]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]))/2
7

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Maple [A]  time = 0.011, size = 62, normalized size = 0.9 \begin{align*} -{\frac{2}{9} \left ( 3+2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{62}{9}\sqrt{3+2\,x}}-{\frac{170\,\sqrt{15}}{27}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }+6\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\,\ln \left ( -1+\sqrt{3+2\,x} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*(3+2*x)^(3/2)+62/9*(3+2*x)^(1/2)-170/27*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)+6*ln(1+(3+2*x)^(1/2)
)-6*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.48733, size = 107, normalized size = 1.57 \begin{align*} -\frac{2}{9} \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} + \frac{85}{27} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) + \frac{62}{9} \, \sqrt{2 \, x + 3} + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt{2 \, x + 3} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*(2*x + 3)^(3/2) + 85/27*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 62/9*s
qrt(2*x + 3) + 6*log(sqrt(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)

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Fricas [A]  time = 1.55672, size = 221, normalized size = 3.25 \begin{align*} \frac{85}{27} \, \sqrt{5} \sqrt{3} \log \left (-\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) - \frac{4}{9} \, \sqrt{2 \, x + 3}{\left (x - 14\right )} + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \log \left (\sqrt{2 \, x + 3} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

85/27*sqrt(5)*sqrt(3)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)/(3*x + 2)) - 4/9*sqrt(2*x + 3)*(x - 14) +
6*log(sqrt(2*x + 3) + 1) - 6*log(sqrt(2*x + 3) - 1)

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Sympy [A]  time = 47.9847, size = 114, normalized size = 1.68 \begin{align*} - \frac{2 \left (2 x + 3\right )^{\frac{3}{2}}}{9} + \frac{62 \sqrt{2 x + 3}}{9} + \frac{850 \left (\begin{cases} - \frac{\sqrt{15} \operatorname{acoth}{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} \right )}}{15} & \text{for}\: 2 x + 3 > \frac{5}{3} \\- \frac{\sqrt{15} \operatorname{atanh}{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} \right )}}{15} & \text{for}\: 2 x + 3 < \frac{5}{3} \end{cases}\right )}{9} - 6 \log{\left (\sqrt{2 x + 3} - 1 \right )} + 6 \log{\left (\sqrt{2 x + 3} + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*x + 3)**(3/2)/9 + 62*sqrt(2*x + 3)/9 + 850*Piecewise((-sqrt(15)*acoth(sqrt(15)*sqrt(2*x + 3)/5)/15, 2*x
+ 3 > 5/3), (-sqrt(15)*atanh(sqrt(15)*sqrt(2*x + 3)/5)/15, 2*x + 3 < 5/3))/9 - 6*log(sqrt(2*x + 3) - 1) + 6*lo
g(sqrt(2*x + 3) + 1)

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Giac [A]  time = 1.10036, size = 112, normalized size = 1.65 \begin{align*} -\frac{2}{9} \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} + \frac{85}{27} \, \sqrt{15} \log \left (\frac{{\left | -2 \, \sqrt{15} + 6 \, \sqrt{2 \, x + 3} \right |}}{2 \,{\left (\sqrt{15} + 3 \, \sqrt{2 \, x + 3}\right )}}\right ) + \frac{62}{9} \, \sqrt{2 \, x + 3} + 6 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 6 \, \log \left ({\left | \sqrt{2 \, x + 3} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*(2*x + 3)^(3/2) + 85/27*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3)))
+ 62/9*sqrt(2*x + 3) + 6*log(sqrt(2*x + 3) + 1) - 6*log(abs(sqrt(2*x + 3) - 1))