3.2570 \(\int \frac{(5-x) \sqrt{3+2 x}}{(2+5 x+3 x^2)^3} \, dx\)

Optimal. Leaf size=102 \[ -\frac{\sqrt{2 x+3} (35 x+29)}{2 \left (3 x^2+5 x+2\right )^2}+\frac{3 \sqrt{2 x+3} (1063 x+878)}{10 \left (3 x^2+5 x+2\right )}+730 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{4713}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

[Out]

-(Sqrt[3 + 2*x]*(29 + 35*x))/(2*(2 + 5*x + 3*x^2)^2) + (3*Sqrt[3 + 2*x]*(878 + 1063*x))/(10*(2 + 5*x + 3*x^2))
+ 730*ArcTanh[Sqrt[3 + 2*x]] - (4713*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/5

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Rubi [A]  time = 0.0623808, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {820, 822, 826, 1166, 207} \[ -\frac{\sqrt{2 x+3} (35 x+29)}{2 \left (3 x^2+5 x+2\right )^2}+\frac{3 \sqrt{2 x+3} (1063 x+878)}{10 \left (3 x^2+5 x+2\right )}+730 \tanh ^{-1}\left (\sqrt{2 x+3}\right )-\frac{4713}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[3 + 2*x]*(29 + 35*x))/(2*(2 + 5*x + 3*x^2)^2) + (3*Sqrt[3 + 2*x]*(878 + 1063*x))/(10*(2 + 5*x + 3*x^2))
+ 730*ArcTanh[Sqrt[3 + 2*x]] - (4713*Sqrt[3/5]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/5

Rule 820

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

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 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) \sqrt{3+2 x}}{\left (2+5 x+3 x^2\right )^3} \, dx &=-\frac{\sqrt{3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}-\frac{1}{2} \int \frac{286+175 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )^2} \, dx\\ &=-\frac{\sqrt{3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac{3 \sqrt{3+2 x} (878+1063 x)}{10 \left (2+5 x+3 x^2\right )}+\frac{1}{10} \int \frac{6839+3189 x}{\sqrt{3+2 x} \left (2+5 x+3 x^2\right )} \, dx\\ &=-\frac{\sqrt{3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac{3 \sqrt{3+2 x} (878+1063 x)}{10 \left (2+5 x+3 x^2\right )}+\frac{1}{5} \operatorname{Subst}\left (\int \frac{4111+3189 x^2}{5-8 x^2+3 x^4} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{\sqrt{3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac{3 \sqrt{3+2 x} (878+1063 x)}{10 \left (2+5 x+3 x^2\right )}-2190 \operatorname{Subst}\left (\int \frac{1}{-3+3 x^2} \, dx,x,\sqrt{3+2 x}\right )+\frac{14139}{5} \operatorname{Subst}\left (\int \frac{1}{-5+3 x^2} \, dx,x,\sqrt{3+2 x}\right )\\ &=-\frac{\sqrt{3+2 x} (29+35 x)}{2 \left (2+5 x+3 x^2\right )^2}+\frac{3 \sqrt{3+2 x} (878+1063 x)}{10 \left (2+5 x+3 x^2\right )}+730 \tanh ^{-1}\left (\sqrt{3+2 x}\right )-\frac{4713}{5} \sqrt{\frac{3}{5}} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{3+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.162053, size = 81, normalized size = 0.79 \[ \frac{1}{50} \left (\frac{5 \sqrt{2 x+3} \left (9567 x^3+23847 x^2+19373 x+5123\right )}{\left (3 x^2+5 x+2\right )^2}-9426 \sqrt{15} \tanh ^{-1}\left (\sqrt{\frac{3}{5}} \sqrt{2 x+3}\right )\right )+730 \tanh ^{-1}\left (\sqrt{2 x+3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

730*ArcTanh[Sqrt[3 + 2*x]] + ((5*Sqrt[3 + 2*x]*(5123 + 19373*x + 23847*x^2 + 9567*x^3))/(2 + 5*x + 3*x^2)^2 -
9426*Sqrt[15]*ArcTanh[Sqrt[3/5]*Sqrt[3 + 2*x]])/50

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Maple [A]  time = 0.019, size = 124, normalized size = 1.2 \begin{align*} 162\,{\frac{1}{ \left ( 6\,x+4 \right ) ^{2}} \left ({\frac{503\, \left ( 3+2\,x \right ) ^{3/2}}{90}}-{\frac{179\,\sqrt{3+2\,x}}{18}} \right ) }-{\frac{4713\,\sqrt{15}}{25}{\it Artanh} \left ({\frac{\sqrt{15}}{5}\sqrt{3+2\,x}} \right ) }-3\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-2}+56\, \left ( 1+\sqrt{3+2\,x} \right ) ^{-1}+365\,\ln \left ( 1+\sqrt{3+2\,x} \right ) +3\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-2}+56\, \left ( -1+\sqrt{3+2\,x} \right ) ^{-1}-365\,\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)^(1/2)/(3*x^2+5*x+2)^3,x)

[Out]

162*(503/90*(3+2*x)^(3/2)-179/18*(3+2*x)^(1/2))/(6*x+4)^2-4713/25*arctanh(1/5*15^(1/2)*(3+2*x)^(1/2))*15^(1/2)
-3/(1+(3+2*x)^(1/2))^2+56/(1+(3+2*x)^(1/2))+365*ln(1+(3+2*x)^(1/2))+3/(-1+(3+2*x)^(1/2))^2+56/(-1+(3+2*x)^(1/2
))-365*ln(-1+(3+2*x)^(1/2))

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Maxima [A]  time = 1.43789, size = 181, normalized size = 1.77 \begin{align*} \frac{4713}{50} \, \sqrt{15} \log \left (-\frac{\sqrt{15} - 3 \, \sqrt{2 \, x + 3}}{\sqrt{15} + 3 \, \sqrt{2 \, x + 3}}\right ) + \frac{9567 \,{\left (2 \, x + 3\right )}^{\frac{7}{2}} - 38409 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} + 49637 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 20555 \, \sqrt{2 \, x + 3}}{5 \,{\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 )}} + 365 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 365 \, \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)^(1/2)/(3*x^2+5*x+2)^3,x, algorithm="maxima")

[Out]

4713/50*sqrt(15)*log(-(sqrt(15) - 3*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/5*(9567*(2*x + 3)^(7/2) -
38409*(2*x + 3)^(5/2) + 49637*(2*x + 3)^(3/2) - 20555*sqrt(2*x + 3))/(9*(2*x + 3)^4 - 48*(2*x + 3)^3 + 94*(2*
x + 3)^2 - 160*x - 215) + 365*log(sqrt(2*x + 3) + 1) - 365*log(sqrt(2*x + 3) - 1)

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Fricas [B]  time = 1.85116, size = 479, normalized size = 4.7 \begin{align*} \frac{4713 \, \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 ) + 18250 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt{2 \, x + 3} + 1\right ) - 18250 \,{\left (9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4\right )} \log \left (\sqrt{2 \, x + 3} - 1\right ) + 5 \,{\left (9567 \, x^{3} + 23847 \, x^{2} + 19373 \, x + 5123\right )} \sqrt{2 \, x + 3}}{50 \,{\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)^(1/2)/(3*x^2+5*x+2)^3,x, algorithm="fricas")

[Out]

1/50*(4713*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)) + 18250*(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4)*log(sqrt(2*x + 3) + 1) - 18250*(9*x^4 + 30*x^3 + 37*x
^2 + 20*x + 4)*log(sqrt(2*x + 3) - 1) + 5*(9567*x^3 + 23847*x^2 + 19373*x + 5123)*sqrt(2*x + 3))/(9*x^4 + 30*x
^3 + 37*x^2 + 20*x + 4)

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Sympy [A]  time = 159.335, size = 388, normalized size = 3.8 \begin{align*} - 2712 \left (\begin{cases} \frac{\sqrt{15} \left (- \frac{\log{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} - 1 \right )}}{4} + \frac{\log{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} + 1 \right )}}{4} - \frac{1}{4 \left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} + 1\right )} - \frac{1}{4 \left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} - 1\right )}\right )}{75} & \text{for}\: x \geq - \frac{3}{2} \wedge x < - \frac{2}{3} \end{cases}\right ) + 2040 \left (\begin{cases} \frac{\sqrt{15} \left (\frac{3 \log{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} - 1 \right )}}{16} - \frac{3 \log{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} + 1 \right )}}{16} + \frac{3}{16 \left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} + 1\right )} + \frac{1}{16 \left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} + 1\right )^{2}} + \frac{3}{16 \left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} - 1\right )} - \frac{1}{16 \left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} - 1\right )^{2}}\right )}{375} & \text{for}\: x \geq - \frac{3}{2} \wedge x < - \frac{2}{3} \end{cases}\right ) + 2526 \left (\begin{cases} - \frac{\sqrt{15} \operatorname{acoth}{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} \right )}}{15} & \text{for}\: 2 x + 3 > \frac{5}{3} \\- \frac{\sqrt{15} \operatorname{atanh}{\left (\frac{\sqrt{15} \sqrt{2 x + 3}}{5} \right )}}{15} & \text{for}\: 2 x + 3 < \frac{5}{3} \end{cases}\right ) - 365 \log{\left (\sqrt{2 x + 3} - 1 \right )} + 365 \log{\left (\sqrt{2 x + 3} + 1 \right )} + \frac{56}{\sqrt{2 x + 3} + 1} - \frac{3}{\left (\sqrt{2 x + 3} + 1\right )^{2}} + \frac{56}{\sqrt{2 x + 3} - 1} + \frac{3}{\left (\sqrt{2 x + 3} - 1\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3+2*x)**(1/2)/(3*x**2+5*x+2)**3,x)

[Out]

-2712*Piecewise((sqrt(15)*(-log(sqrt(15)*sqrt(2*x + 3)/5 - 1)/4 + log(sqrt(15)*sqrt(2*x + 3)/5 + 1)/4 - 1/(4*(
sqrt(15)*sqrt(2*x + 3)/5 + 1)) - 1/(4*(sqrt(15)*sqrt(2*x + 3)/5 - 1)))/75, (x >= -3/2) & (x < -2/3))) + 2040*P
iecewise((sqrt(15)*(3*log(sqrt(15)*sqrt(2*x + 3)/5 - 1)/16 - 3*log(sqrt(15)*sqrt(2*x + 3)/5 + 1)/16 + 3/(16*(s
qrt(15)*sqrt(2*x + 3)/5 + 1)) + 1/(16*(sqrt(15)*sqrt(2*x + 3)/5 + 1)**2) + 3/(16*(sqrt(15)*sqrt(2*x + 3)/5 - 1
)) - 1/(16*(sqrt(15)*sqrt(2*x + 3)/5 - 1)**2))/375, (x >= -3/2) & (x < -2/3))) + 2526*Piecewise((-sqrt(15)*aco
th(sqrt(15)*sqrt(2*x + 3)/5)/15, 2*x + 3 > 5/3), (-sqrt(15)*atanh(sqrt(15)*sqrt(2*x + 3)/5)/15, 2*x + 3 < 5/3)
) - 365*log(sqrt(2*x + 3) - 1) + 365*log(sqrt(2*x + 3) + 1) + 56/(sqrt(2*x + 3) + 1) - 3/(sqrt(2*x + 3) + 1)**
2 + 56/(sqrt(2*x + 3) - 1) + 3/(sqrt(2*x + 3) - 1)**2

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Giac [A]  time = 1.0936, size = 162, normalized size = 1.59 \begin{align*} \frac{4713}{50} \, \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{9567 \,{\left (2 \, x + 3\right )}^{\frac{7}{2}} - 38409 \,{\left (2 \, x + 3\right )}^{\frac{5}{2}} + 49637 \,{\left (2 \, x + 3\right )}^{\frac{3}{2}} - 20555 \, \sqrt{2 \, x + 3}}{5 \,{\left (3 \,{\left (2 \, x + 3\right )}^{2} - 16 \, x - 19\right )}^{2}} + 365 \, \log \left (\sqrt{2 \, x + 3} + 1\right ) - 365 \, \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)^(1/2)/(3*x^2+5*x+2)^3,x, algorithm="giac")

[Out]

4713/50*sqrt(15)*log(1/2*abs(-2*sqrt(15) + 6*sqrt(2*x + 3))/(sqrt(15) + 3*sqrt(2*x + 3))) + 1/5*(9567*(2*x + 3
)^(7/2) - 38409*(2*x + 3)^(5/2) + 49637*(2*x + 3)^(3/2) - 20555*sqrt(2*x + 3))/(3*(2*x + 3)^2 - 16*x - 19)^2 +
365*log(sqrt(2*x + 3) + 1) - 365*log(abs(sqrt(2*x + 3) - 1))