3.2558 \(\int \frac{(5-x) (3+2 x)^{7/2}}{(2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=98 \[ -\frac{(139 x+121) (2 x+3)^{5/2}}{3 \left (3 x^2+5 x+2\right )}+\frac{826}{27} (2 x+3)^{3/2}+\frac{1358}{27} \sqrt{2 x+3}-154 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{2800}{27} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

(1358*Sqrt[3 + 2*x])/27 + (826*(3 + 2*x)^(3/2))/27 - ((3 + 2*x)^(5/2)*(121 + 139*x))/(3*(2 + 5*x + 3*x^2)) - 1
54*ArcTanh[Sqrt[3 + 2*x]] + (2800*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/27

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Rubi [A]  time = 0.0811127, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {818, 824, 826, 1166, 207} \[ -\frac{(139 x+121) (2 x+3)^{5/2}}{3 \left (3 x^2+5 x+2\right )}+\frac{826}{27} (2 x+3)^{3/2}+\frac{1358}{27} \sqrt{2 x+3}-154 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{2800}{27} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((5 - x)*(3 + 2*x)^(7/2))/(2 + 5*x + 3*x^2)^2,x]

[Out]

(1358*Sqrt[3 + 2*x])/27 + (826*(3 + 2*x)^(3/2))/27 - ((3 + 2*x)^(5/2)*(121 + 139*x))/(3*(2 + 5*x + 3*x^2)) - 1
54*ArcTanh[Sqrt[3 + 2*x]] + (2800*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/27

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 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)^{7/2}}{\left (2+5 x+3 x^2\right )^2} \, dx &=-\frac{(3+2 x)^{5/2} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}+\frac{1}{3} \int \frac{(3+2 x)^{3/2} (182+413 x)}{2+5 x+3 x^2} \, dx\\ &=\frac{826}{27} (3+2 x)^{3/2}-\frac{(3+2 x)^{5/2} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}+\frac{1}{9} \int \frac{\sqrt{3+2 x} (-14+679 x)}{2+5 x+3 x^2} \, dx\\ &=\frac{1358}{27} \sqrt{3+2 x}+\frac{826}{27} (3+2 x)^{3/2}-\frac{(3+2 x)^{5/2} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}+\frac{1}{27} \int \frac{-2842-763 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=\frac{1358}{27} \sqrt{3+2 x}+\frac{826}{27} (3+2 x)^{3/2}-\frac{(3+2 x)^{5/2} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}+\frac{2}{27} \operatorname{Subst}\left (\int \frac{-3395-763 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=\frac{1358}{27} \sqrt{3+2 x}+\frac{826}{27} (3+2 x)^{3/2}-\frac{(3+2 x)^{5/2} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}+462 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )-\frac{14000}{27} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=\frac{1358}{27} \sqrt{3+2 x}+\frac{826}{27} (3+2 x)^{3/2}-\frac{(3+2 x)^{5/2} (121+139 x)}{3 \left (2+5 x+3 x^2\right )}-154 \tanh ^{-1}\left (\sqrt{3+2 x}\right )+\frac{2800}{27} \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0944341, size = 81, normalized size = 0.83 \[ \frac{1}{81} \left (2800 \sqrt{15} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )-\frac{3 \sqrt{2 x+3} \left (48 x^3-400 x^2+1843 x+2129\right )}{3 x^2+5 x+2}\right )-154 \tanh ^{-1}\left (\sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((5 - x)*(3 + 2*x)^(7/2))/(2 + 5*x + 3*x^2)^2,x]

[Out]

-154*ArcTanh[Sqrt[3 + 2*x]] + ((-3*Sqrt[3 + 2*x]*(2129 + 1843*x - 400*x^2 + 48*x^3))/(2 + 5*x + 3*x^2) + 2800*
Sqrt[15]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/81

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Maple [A]  time = 0.019, size = 104, normalized size = 1.1 \begin{align*} -{\frac{8}{27} \left ( 3+2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{184}{27}\sqrt{3+2\,x}}-{\frac{4250}{81}\sqrt{3+2\,x} \left ( 2\,x+{\frac{4}{3}} \right ) ^{-1}}+{\frac{2800\,\sqrt{15}}{81}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-6\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-1}-77\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-1}+77\,\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)^(7/2)/(3*x^2+5*x+2)^2,x)

[Out]

-8/27*(3+2*x)^(3/2)+184/27*(3+2*x)^(1/2)-4250/81*(3+2*x)^(1/2)/(2*x+4/3)+2800/81*arctanh(1/5*15^(1/2)*(3+2*x)^
(1/2))*15^(1/2)-6/(1+(3+2*x)^(1/2))-77*ln(1+(3+2*x)^(1/2))-6/(-1+(3+2*x)^(1/2))+77*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.46223, size = 157, normalized size = 1.6 \begin{align*} -\frac{8}{27} \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - \frac{1400}{81} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) + \frac{184}{27} \, \sqrt{2 \, x + 3} - \frac{2 \,{\left (2611 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 2935 \, \sqrt{2 \, x + 3}\right )}}{27 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 77 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 77 \, \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)^(7/2)/(3*x^2+5*x+2)^2,x, algorithm="maxima")

[Out]

-8/27*(2*x + 3)^(3/2) - 1400/81*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 184
/27*sqrt(2*x + 3) - 2/27*(2611*(2*x + 3)^(3/2) - 2935*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 77*log(sqrt
(2*x + 3) + 1) + 77*log(sqrt(2*x + 3) - 1)

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Fricas [A]  time = 1.60452, size = 366, normalized size = 3.73 \begin{align*} \frac{1400 \, \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 ) - 6237 \,{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) + 6237 \,{\left (3 \, x^{2} + 5 \, x + 2\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) - 3 \,{\left (48 \, x^{3} - 400 \, x^{2} + 1843 \, x + 2129\right )} \sqrt{2 \, x + 3}}{81 \,{\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)^(7/2)/(3*x^2+5*x+2)^2,x, algorithm="fricas")

[Out]

1/81*(1400*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)) - 6237*(
3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) + 1) + 6237*(3*x^2 + 5*x + 2)*log(sqrt(2*x + 3) - 1) - 3*(48*x^3 - 400*x^2
+ 1843*x + 2129)*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)**(7/2)/(3*x**2+5*x+2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.13587, size = 162, normalized size = 1.65 \begin{align*} -\frac{8}{27} \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - \frac{1400}{81} \, \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{184}{27} \, \sqrt{2 \, x + 3} - \frac{2 \,{\left (2611 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 2935 \, \sqrt{2 \, x + 3}\right )}}{27 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - 77 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 77 \, \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)^(7/2)/(3*x^2+5*x+2)^2,x, algorithm="giac")

[Out]

-8/27*(2*x + 3)^(3/2) - 1400/81*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3
))) + 184/27*sqrt(2*x + 3) - 2/27*(2611*(2*x + 3)^(3/2) - 2935*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 77
*log(sqrt(2*x + 3) + 1) + 77*log(abs(sqrt(2*x + 3) - 1))