3.2564 \(\int \frac{5-x}{(3+2 x)^{5/2} (2+5 x+3 x^2)^2} \, dx\)

Optimal. Leaf size=98 \[ -\frac{3 (47 x+37)}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}-\frac{686}{25 \sqrt{2 x+3}}-\frac{262}{15 (2 x+3)^{3/2}}-10 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{936}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

-262/(15*(3 + 2*x)^(3/2)) - 686/(25*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(5*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)) - 1
0*ArcTanh[Sqrt[3 + 2*x]] + (936*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/25

________________________________________________________________________________________

Rubi [A]  time = 0.0802408, 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 = {822, 828, 826, 1166, 207} \[ -\frac{3 (47 x+37)}{5 (2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}-\frac{686}{25 \sqrt{2 x+3}}-\frac{262}{15 (2 x+3)^{3/2}}-10 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+\frac{936}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-262/(15*(3 + 2*x)^(3/2)) - 686/(25*Sqrt[3 + 2*x]) - (3*(37 + 47*x))/(5*(3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)) - 1
0*ArcTanh[Sqrt[3 + 2*x]] + (936*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/25

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)^{5/2} \left (2+5 x+3 x^2\right )^2} \, dx &=-\frac{3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-\frac{1}{5} \int \frac{730+705 x}{(3+2 x)^{5/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{262}{15 (3+2 x)^{3/2}}-\frac{3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-\frac{1}{25} \int \frac{2090+1965 x}{(3+2 x)^{3/2} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{262}{15 (3+2 x)^{3/2}}-\frac{686}{25 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-\frac{1}{125} \int \frac{5770+5145 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{262}{15 (3+2 x)^{3/2}}-\frac{686}{25 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-\frac{2}{125} \operatorname{Subst}\left (\int \frac{-3895+5145 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{262}{15 (3+2 x)^{3/2}}-\frac{686}{25 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}+30 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )-\frac{2808}{25} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{262}{15 (3+2 x)^{3/2}}-\frac{686}{25 \sqrt{3+2 x}}-\frac{3 (37+47 x)}{5 (3+2 x)^{3/2} \left (2+5 x+3 x^2\right )}-10 \tanh ^{-1}\left (\sqrt{3+2 x}\right )+\frac{936}{25} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.123823, size = 94, normalized size = 0.96 \[ \frac{1}{75} \left (-\frac{45 (47 x+37)}{(2 x+3)^{3/2} \left (3 x^2+5 x+2\right )}-\frac{2058}{\sqrt{2 x+3}}-\frac{1310}{(2 x+3)^{3/2}}-750 \tanh ^{-1}\left (\sqrt{2 x+3}\right )+2808 \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1310/(3 + 2*x)^(3/2) - 2058/Sqrt[3 + 2*x] - (45*(37 + 47*x))/((3 + 2*x)^(3/2)*(2 + 5*x + 3*x^2)) - 750*ArcTa
nh[Sqrt[3 + 2*x]] + 2808*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/75

________________________________________________________________________________________

Maple [A]  time = 0.019, size = 104, normalized size = 1.1 \begin{align*} -{\frac{104}{75} \left ( 3+2\,x \right ) ^{-{\frac{3}{2}}}}-{\frac{1624}{125}{\frac{1}{\sqrt{3+2\,x}}}}-{\frac{306}{125}\sqrt{3+2\,x} \left ( 2\,x+{\frac{4}{3}} \right ) ^{-1}}+{\frac{936\,\sqrt{15}}{125}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-6\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-1}-5\,\ln \left ( 1+\sqrt{3+2\,x} \right ) -6\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-1}+5\,\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)^(5/2)/(3*x^2+5*x+2)^2,x)

[Out]

-104/75/(3+2*x)^(3/2)-1624/125/(3+2*x)^(1/2)-306/125*(3+2*x)^(1/2)/(2*x+4/3)+936/125*arctanh(1/5*15^(1/2)*(3+2
*x)^(1/2))*15^(1/2)-6/(1+(3+2*x)^(1/2))-5*ln(1+(3+2*x)^(1/2))-6/(-1+(3+2*x)^(1/2))+5*ln(-1+(3+2*x)^(1/2))

________________________________________________________________________________________

Maxima [A]  time = 1.45849, size = 157, normalized size = 1.6 \begin{align*} -\frac{468}{125} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) - \frac{2 \,{\left (3087 \,{\left (2 \, x + 3\right )}^{3} - 6267 \,{\left (2 \, x + 3\right )}^{2} + 4040 \, x + 6320\right )}}{75 \,{\left (3 \,{\left (2 \, x + 3\right )}^{\frac{7}{2}} - 8 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} + 5 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}}\right )}} - 5 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 5 \, \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)^(5/2)/(3*x^2+5*x+2)^2,x, algorithm="maxima")

[Out]

-468/125*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 2/75*(3087*(2*x + 3)^3 - 6
267*(2*x + 3)^2 + 4040*x + 6320)/(3*(2*x + 3)^(7/2) - 8*(2*x + 3)^(5/2) + 5*(2*x + 3)^(3/2)) - 5*log(sqrt(2*x
+ 3) + 1) + 5*log(sqrt(2*x + 3) - 1)

________________________________________________________________________________________

Fricas [B]  time = 1.62132, size = 490, normalized size = 5. \begin{align*} \frac{1404 \, \sqrt{5} \sqrt{3}{\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (\frac{\sqrt{5} \sqrt{3} \sqrt{2 \, x + 3} + 3 \, x + 7}{3 \, x + 2}\right ) - 1875 \,{\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) + 1875 \,{\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) - 5 \,{\left (12348 \, x^{3} + 43032 \, x^{2} + 47767 \, x + 16633\right )} \sqrt{2 \, x + 3}}{375 \,{\left (12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/375*(1404*sqrt(5)*sqrt(3)*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*log((sqrt(5)*sqrt(3)*sqrt(2*x + 3) + 3*x +
7)/(3*x + 2)) - 1875*(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18)*log(sqrt(2*x + 3) + 1) + 1875*(12*x^4 + 56*x^3 + 9
5*x^2 + 69*x + 18)*log(sqrt(2*x + 3) - 1) - 5*(12348*x^3 + 43032*x^2 + 47767*x + 16633)*sqrt(2*x + 3))/(12*x^4
+ 56*x^3 + 95*x^2 + 69*x + 18)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.08611, size = 157, normalized size = 1.6 \begin{align*} -\frac{468}{125} \, \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 (903 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 1403 \, \sqrt{2 \, x + 3}\right )}}{125 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}} - \frac{16 \,{\left (609 \, x + 946\right )}}{375 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}}} - 5 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) + 5 \, \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)^(5/2)/(3*x^2+5*x+2)^2,x, algorithm="giac")

[Out]

-468/125*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) - 6/125*(903*(2*x +
3)^(3/2) - 1403*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19) - 16/375*(609*x + 946)/(2*x + 3)^(3/2) - 5*log(sqr
t(2*x + 3) + 1) + 5*log(abs(sqrt(2*x + 3) - 1))