3.2565 \(\int \frac{5-x}{(3+2 x)^{7/2} (2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=111 \[ -\frac{3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}-\frac{24626}{625 \sqrt{2 x+3}}-\frac{7042}{375 (2 x+3)^{3/2}}-\frac{2114}{125 (2 x+3)^{5/2}}+14 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{15876}{625} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

-2114/(125*(3 + 2*x)^(5/2)) - 7042/(375*(3 + 2*x)^(3/2)) - 24626/(625*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(5*(3 +
2*x)^(5/2)*(2 + 5*x + 3*x^2)) + 14*ArcTanh[Sqrt[3 + 2*x]] + (15876*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]
)/625

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Rubi [A]  time = 0.0979208, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {822, 828, 826, 1166, 207} \[ -\frac{3 (47 x+37)}{5 (2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}-\frac{24626}{625 \sqrt{2 x+3}}-\frac{7042}{375 (2 x+3)^{3/2}}-\frac{2114}{125 (2 x+3)^{5/2}}+14 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{15876}{625} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-2114/(125*(3 + 2*x)^(5/2)) - 7042/(375*(3 + 2*x)^(3/2)) - 24626/(625*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(5*(3 +
2*x)^(5/2)*(2 + 5*x + 3*x^2)) + 14*ArcTanh[Sqrt[3 + 2*x]] + (15876*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]]
)/625

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] ||
IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 828

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

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 (37+47 x)}{5 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}-\frac{1}{5} \int \frac{952+987 x}{(3+2 x)^{7/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{2114}{125 (3+2 x)^{5/2}}-\frac{3 (37+47 x)}{5 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}-\frac{1}{25} \int \frac{2996+3171 x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{2114}{125 (3+2 x)^{5/2}}-\frac{7042}{375 (3+2 x)^{3/2}}-\frac{3 (37+47 x)}{5 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}-\frac{1}{125} \int \frac{9688+10563 x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{2114}{125 (3+2 x)^{5/2}}-\frac{7042}{375 (3+2 x)^{3/2}}-\frac{24626}{625 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}-\frac{1}{625} \int \frac{32564+36939 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{2114}{125 (3+2 x)^{5/2}}-\frac{7042}{375 (3+2 x)^{3/2}}-\frac{24626}{625 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}-\frac{2}{625} \operatorname{Subst}\left (\int \frac{-45689+36939 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{2114}{125 (3+2 x)^{5/2}}-\frac{7042}{375 (3+2 x)^{3/2}}-\frac{24626}{625 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}-42 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )-\frac{47628}{625} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{2114}{125 (3+2 x)^{5/2}}-\frac{7042}{375 (3+2 x)^{3/2}}-\frac{24626}{625 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 (3+2 x)^{5/2} \left (2+5 x+3 x^2\right )}+14 \tanh ^{-1}\left (\sqrt{3+2 x}\right )+\frac{15876}{625} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.211072, size = 86, normalized size = 0.77 \[ \frac{47628 \sqrt{15} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )-\frac{5 \left (886536 x^4+4348428 x^3+7782530 x^2+5977997 x+1646109\right )}{(2 x+3)^{5/2} \left (3 x^2+5 x+2\right )}}{9375}+14 \tanh ^{-1}\left (\sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

14*ArcTanh[Sqrt[3 + 2*x]] + ((-5*(1646109 + 5977997*x + 7782530*x^2 + 4348428*x^3 + 886536*x^4))/((3 + 2*x)^(5
/2)*(2 + 5*x + 3*x^2)) + 47628*Sqrt[15]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/9375

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Maple [A]  time = 0.02, size = 113, normalized size = 1. \begin{align*} -{\frac{104}{125} \left ( 3+2\,x \right ) ^{-{\frac{5}{2}}}}-{\frac{1624}{375} \left ( 3+2\,x \right ) ^{-{\frac{3}{2}}}}-{\frac{16208}{625}{\frac{1}{\sqrt{3+2\,x}}}}-{\frac{918}{625}\sqrt{3+2\,x} \left ( 2\,x+{\frac{4}{3}} \right ) ^{-1}}+{\frac{15876\,\sqrt{15}}{3125}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-6\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-1}+7\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-1}-7\,\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]

-104/125/(3+2*x)^(5/2)-1624/375/(3+2*x)^(3/2)-16208/625/(3+2*x)^(1/2)-918/625*(3+2*x)^(1/2)/(2*x+4/3)+15876/31
25*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)-6/(1+(3+2*x)^(1/2))+7*ln(1+(3+2*x)^(1/2))-6/(-1+(3+2*x)^(1/2))
-7*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.45481, size = 169, normalized size = 1.52 \begin{align*} -\frac{7938}{3125} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) - \frac{2 \,{\left (110817 \,{\left (2 \, x + 3\right )}^{4} - 242697 \,{\left (2 \, x + 3\right )}^{3} + 91420 \,{\left (2 \, x + 3\right )}^{2} + 28120 \, x + 46080\right )}}{1875 \,{\left (3 \,{\left (2 \, x + 3\right )}^{\frac{9}{2}} - 8 \,{\left (2 \, x + 3\right )}^{\frac{7}{2}} + 5 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}}\right )}} + 7 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 7 \, \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]

-7938/3125*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/1875*(110817*(2*x + 3)
^4 - 242697*(2*x + 3)^3 + 91420*(2*x + 3)^2 + 28120*x + 46080)/(3*(2*x + 3)^(9/2) - 8*(2*x + 3)^(7/2) + 5*(2*x
+ 3)^(5/2)) + 7*log(sqrt(2*x + 3) + 1) - 7*log(sqrt(2*x + 3) - 1)

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Fricas [B]  time = 1.75487, size = 594, normalized size = 5.35 \begin{align*} \frac{23814 \, \sqrt{5} \sqrt{3}{\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )} \log \left (\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) + 65625 \,{\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) - 65625 \,{\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) - 5 \,{\left (886536 \, x^{4} + 4348428 \, x^{3} + 7782530 \, x^{2} + 5977997 \, x + 1646109\right )} \sqrt{2 \, x + 3}}{9375 \,{\left (24 \, x^{5} + 148 \, x^{4} + 358 \, x^{3} + 423 \, x^{2} + 243 \, x + 54\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]

1/9375*(23814*sqrt(5)*sqrt(3)*(24*x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243*x + 54)*log((sqrt(5)*sqrt(3)*sqrt(2*
x + 3) + 3*x + 7)/(3*x + 2)) + 65625*(24*x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243*x + 54)*log(sqrt(2*x + 3) + 1
) - 65625*(24*x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243*x + 54)*log(sqrt(2*x + 3) - 1) - 5*(886536*x^4 + 4348428
*x^3 + 7782530*x^2 + 5977997*x + 1646109)*sqrt(2*x + 3))/(24*x^5 + 148*x^4 + 358*x^3 + 423*x^2 + 243*x + 54)

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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 = 1.09842, size = 169, normalized size = 1.52 \begin{align*} -\frac{7938}{3125} \, \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{6 \,{\left (4209 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 6709 \, \sqrt{2 \, x + 3}\right )}}{625 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - \frac{16 \,{\left (3039 \,{\left (2 \, x + 3\right )}^{2} + 1015 \, x + 1620\right )}}{1875 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}}} + 7 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 7 \, \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]

-7938/3125*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 6/625*(4209*(2*
x + 3)^(3/2) - 6709*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 16/1875*(3039*(2*x + 3)^2 + 1015*x + 1620)/(2
*x + 3)^(5/2) + 7*log(sqrt(2*x + 3) + 1) - 7*log(abs(sqrt(2*x + 3) - 1))