3.2560 \(\int \frac{(5-x) (3+2 x)^{3/2}}{(2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=72 \[ -\frac{\sqrt{2 x+3} (139 x+121)}{3 \left (3 x^2+5 x+2\right )}-106 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{248}{3} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

-(Sqrt[3 + 2*x]*(121 + 139*x))/(3*(2 + 5*x + 3*x^2)) - 106*ArcTanh[Sqrt[3 + 2*x]] + (248*Sqrt[5/3]*ArcTanh[Sqr
t[3/5]*Sqrt[3 + 2*x]])/3

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Rubi [A]  time = 0.0463592, antiderivative size = 72, 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 = {818, 826, 1166, 207} \[ -\frac{\sqrt{2 x+3} (139 x+121)}{3 \left (3 x^2+5 x+2\right )}-106 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{248}{3} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[3 + 2*x]*(121 + 139*x))/(3*(2 + 5*x + 3*x^2)) - 106*ArcTanh[Sqrt[3 + 2*x]] + (248*Sqrt[5/3]*ArcTanh[Sqr
t[3/5]*Sqrt[3 + 2*x]])/3

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 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}}{\left (2+5 x+3 x^2\right )^2} \, dx &=-\frac{\sqrt{3+2 x} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}+\frac{1}{3} \int \frac{-302-143 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{\sqrt{3+2 x} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}+\frac{2}{3} \operatorname{Subst}\left (\int \frac{-175-143 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{\sqrt{3+2 x} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}+318 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )-\frac{1240}{3} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{\sqrt{3+2 x} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}-106 \tanh ^{-1}\left (\sqrt{3+2 x}\right )+\frac{248}{3} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0598495, size = 70, normalized size = 0.97 \[ \frac{1}{9} \left (-\frac{3 \sqrt{2 x+3} (139 x+121)}{3 x^2+5 x+2}-954 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+248 \sqrt{15} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*Sqrt[3 + 2*x]*(121 + 139*x))/(2 + 5*x + 3*x^2) - 954*ArcTanh[Sqrt[3 + 2*x]] + 248*Sqrt[15]*ArcTanh[Sqrt[3
/5]*Sqrt[3 + 2*x]])/9

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Maple [A]  time = 0.016, size = 86, normalized size = 1.2 \begin{align*} -{\frac{170}{9}\sqrt{3+2\,x} \left ( 2\,x+{\frac{4}{3}} \right ) ^{-1}}+{\frac{248\,\sqrt{15}}{9}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-6\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-1}-53\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-1}+53\,\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)^2,x)

[Out]

-170/9*(3+2*x)^(1/2)/(2*x+4/3)+248/9*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-6/(1+(3+2*x)^(1/2))-53*ln(1+
(3+2*x)^(1/2))-6/(-1+(3+2*x)^(1/2))+53*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.47368, size = 132, normalized size = 1.83 \begin{align*} -\frac{124}{9} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) - \frac{2 \,{\left (139 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 175 \, \sqrt{2 \, x + 3}\right )}}{3 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 53 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 53 \, \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)^2,x, algorithm="maxima")

[Out]

-124/9*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/3*(139*(2*x + 3)^(3/2) - 1
75*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 53*log(sqrt(2*x + 3) + 1) + 53*log(sqrt(2*x + 3) - 1)

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Fricas [B]  time = 1.59148, size = 332, normalized size = 4.61 \begin{align*} \frac{124 \, \sqrt{5} \sqrt{3}{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 477 \,{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) + 477 \,{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) - 3 \,{\left (139 \, x + 121\right )} \sqrt{2 \, x + 3}}{9 \,{\left (3 \, x^{2} + 5 \, x + 2\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)^2,x, algorithm="fricas")

[Out]

1/9*(124*sqrt(5)*sqrt(3)*(3*x^2 + 5*x + 2)*log((sqrt(5)*sqrt(3)*sqrt(2*x + 3) + 3*x + 7)/(3*x + 2)) - 477*(3*x
^2 + 5*x + 2)*log(sqrt(2*x + 3) + 1) + 477*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) - 1) - 3*(139*x + 121)*sqrt(2*x
+ 3))/(3*x^2 + 5*x + 2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.09199, size = 138, normalized size = 1.92 \begin{align*} -\frac{124}{9} \, \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{2 \,{\left (139 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 175 \, \sqrt{2 \, x + 3}\right )}}{3 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 53 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 53 \, \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)^2,x, algorithm="giac")

[Out]

-124/9*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/3*(139*(2*x + 3)^
(3/2) - 175*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 53*log(sqrt(2*x + 3) + 1) + 53*log(abs(sqrt(2*x + 3)
- 1))