3.2567 \(\int \frac{(5-x) (3+2 x)^{7/2}}{(2+5 x+3 x^2)^3} \, dx\)

Optimal. Leaf size=100 \[ -\frac{(139 x+121) (2 x+3)^{5/2}}{6 \left (3 x^2+5 x+2\right )^2}+\frac{7 (619 x+546) \sqrt{2 x+3}}{6 \left (3 x^2+5 x+2\right )}+1582 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-1225 \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

-((3 + 2*x)^(5/2)*(121 + 139*x))/(6*(2 + 5*x + 3*x^2)^2) + (7*Sqrt[3 + 2*x]*(546 + 619*x))/(6*(2 + 5*x + 3*x^2
)) + 1582*ArcTanh[Sqrt[3 + 2*x]] - 1225*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]

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Rubi [A]  time = 0.0632812, antiderivative size = 100, 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 = {818, 826, 1166, 207} \[ -\frac{(139 x+121) (2 x+3)^{5/2}}{6 \left (3 x^2+5 x+2\right )^2}+\frac{7 (619 x+546) \sqrt{2 x+3}}{6 \left (3 x^2+5 x+2\right )}+1582 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-1225 \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3 + 2*x)^(5/2)*(121 + 139*x))/(6*(2 + 5*x + 3*x^2)^2) + (7*Sqrt[3 + 2*x]*(546 + 619*x))/(6*(2 + 5*x + 3*x^2
)) + 1582*ArcTanh[Sqrt[3 + 2*x]] - 1225*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]

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)^{7/2}}{\left (2+5 x+3 x^2\right )^3} \, dx &=-\frac{(3+2 x)^{5/2} (121+139 x)}{6 \left (2+5 x+3 x^2\right )^2}+\frac{1}{6} \int \frac{(-658-147 x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac{(3+2 x)^{5/2} (121+139 x)}{6 \left (2+5 x+3 x^2\right )^2}+\frac{7 \sqrt{3+2 x} (546+619 x)}{6 \left (2+5 x+3 x^2\right )}+\frac{1}{18} \int \frac{26649+12411 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{(3+2 x)^{5/2} (121+139 x)}{6 \left (2+5 x+3 x^2\right )^2}+\frac{7 \sqrt{3+2 x} (546+619 x)}{6 \left (2+5 x+3 x^2\right )}+\frac{1}{9} \operatorname{Subst}\left (\int \frac{16065+12411 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{(3+2 x)^{5/2} (121+139 x)}{6 \left (2+5 x+3 x^2\right )^2}+\frac{7 \sqrt{3+2 x} (546+619 x)}{6 \left (2+5 x+3 x^2\right )}-4746 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )+6125 \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{(3+2 x)^{5/2} (121+139 x)}{6 \left (2+5 x+3 x^2\right )^2}+\frac{7 \sqrt{3+2 x} (546+619 x)}{6 \left (2+5 x+3 x^2\right )}+1582 \tanh ^{-1}\left (\sqrt{3+2 x}\right )-1225 \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0904416, size = 80, normalized size = 0.8 \[ \frac{\sqrt{2 x+3} \left (12443 x^3+30979 x^2+25073 x+6555\right )}{6 \left (3 x^2+5 x+2\right )^2}+1582 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-1225 \sqrt{\frac{5}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3 + 2*x]*(6555 + 25073*x + 30979*x^2 + 12443*x^3))/(6*(2 + 5*x + 3*x^2)^2) + 1582*ArcTanh[Sqrt[3 + 2*x]]
- 1225*Sqrt[5/3]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]

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Maple [A]  time = 0.019, size = 124, normalized size = 1.2 \begin{align*} 450\,{\frac{1}{ \left ( 6\,x+4 \right ) ^{2}} \left ({\frac{299\, \left ( 3+2\,x \right ) ^{3/2}}{54}}-{\frac{185\,\sqrt{3+2\,x}}{18}} \right ) }-{\frac{1225\,\sqrt{15}}{3}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-3\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-2}+92\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-1}+791\,\ln \left ( 1+\sqrt{3+2\,x} \right ) +3\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-2}+92\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-1}-791\,\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)^3,x)

[Out]

450*(299/54*(3+2*x)^(3/2)-185/18*(3+2*x)^(1/2))/(6*x+4)^2-1225/3*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-
3/(1+(3+2*x)^(1/2))^2+92/(1+(3+2*x)^(1/2))+791*ln(1+(3+2*x)^(1/2))+3/(-1+(3+2*x)^(1/2))^2+92/(-1+(3+2*x)^(1/2)
)-791*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.43916, size = 181, normalized size = 1.81 \begin{align*} \frac{1225}{6} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) + \frac{12443 \,{\left (2 \, x + 3\right )}^{\frac{7}{2}} - 50029 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} + 64505 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 26775 \, \sqrt{2 \, x + 3}}{3 \,{\left (9 \,{\left (2 \, x + 3\right )}^{4} - 48 \,{\left (2 \, x + 3\right )}^{3} + 94 \,{\left (2 \, x + 3\right )}^{2} - 160 \, x - 215\right )}} + 791 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 791 \, \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)^3,x, algorithm="maxima")

[Out]

1225/6*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/3*(12443*(2*x + 3)^(7/2) -
50029*(2*x + 3)^(5/2) + 64505*(2*x + 3)^(3/2) - 26775*sqrt(2*x + 3))/(9*(2*x + 3)^4 - 48*(2*x + 3)^3 + 94*(2*
x + 3)^2 - 160*x - 215) + 791*log(sqrt(2*x + 3) + 1) - 791*log(sqrt(2*x + 3) - 1)

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Fricas [B]  time = 1.59164, size = 474, normalized size = 4.74 \begin{align*} \frac{1225 \, \sqrt{5} \sqrt{3}{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (-\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} - 3 \, x - 7}{3 \, x + 2}\right ) + 4746 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) - 4746 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) +{\left (12443 \, x^{3} + 30979 \, x^{2} + 25073 \, x + 6555\right )} \sqrt{2 \, x + 3}}{6 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\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)^3,x, algorithm="fricas")

[Out]

1/6*(1225*sqrt(5)*sqrt(3)*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(-(sqrt(5)*sqrt(3)*sqrt(2*x + 3) - 3*x - 7)/
(3*x + 2)) + 4746*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(sqrt(2*x + 3) + 1) - 4746*(9*x^4 + 30*x^3 + 37*x^2
+ 20*x + 4)*log(sqrt(2*x + 3) - 1) + (12443*x^3 + 30979*x^2 + 25073*x + 6555)*sqrt(2*x + 3))/(9*x^4 + 30*x^3 +
37*x^2 + 20*x + 4)

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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.09047, size = 162, normalized size = 1.62 \begin{align*} \frac{1225}{6} \, \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{12443 \,{\left (2 \, x + 3\right )}^{\frac{7}{2}} - 50029 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} + 64505 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 26775 \, \sqrt{2 \, x + 3}}{3 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 791 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 791 \, \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)^3,x, algorithm="giac")

[Out]

1225/6*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/3*(12443*(2*x + 3
)^(7/2) - 50029*(2*x + 3)^(5/2) + 64505*(2*x + 3)^(3/2) - 26775*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19)^2 +
791*log(sqrt(2*x + 3) + 1) - 791*log(abs(sqrt(2*x + 3) - 1))