3.2615 \(\int \frac{(5-x) (3+2 x)^{3/2}}{(2+5 x+3 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=143 \[ -\frac{350 \sqrt{-3 x^2-5 x-2} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right ),-\frac{2}{3}\right )}{3 \sqrt{3} \sqrt{3 x^2+5 x+2}}-\frac{2 \sqrt{2 x+3} (139 x+121)}{3 \sqrt{3 x^2+5 x+2}}+\frac{274 \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{3 \sqrt{3} \sqrt{3 x^2+5 x+2}} \]

[Out]

(-2*Sqrt[3 + 2*x]*(121 + 139*x))/(3*Sqrt[2 + 5*x + 3*x^2]) + (274*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt
[3]*Sqrt[1 + x]], -2/3])/(3*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]) - (350*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt
[3]*Sqrt[1 + x]], -2/3])/(3*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

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Rubi [A]  time = 0.0779076, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {818, 843, 718, 424, 419} \[ -\frac{2 \sqrt{2 x+3} (139 x+121)}{3 \sqrt{3 x^2+5 x+2}}-\frac{350 \sqrt{-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{3 \sqrt{3} \sqrt{3 x^2+5 x+2}}+\frac{274 \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{3 \sqrt{3} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[3 + 2*x]*(121 + 139*x))/(3*Sqrt[2 + 5*x + 3*x^2]) + (274*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt
[3]*Sqrt[1 + x]], -2/3])/(3*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]) - (350*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt
[3]*Sqrt[1 + x]], -2/3])/(3*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

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 843

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

Rule 718

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(2*Rt[b^2 - 4*a*c, 2]
*(d + e*x)^m*Sqrt[-((c*(a + b*x + c*x^2))/(b^2 - 4*a*c))])/(c*Sqrt[a + b*x + c*x^2]*((2*c*(d + e*x))/(2*c*d -
b*e - e*Rt[b^2 - 4*a*c, 2]))^m), Subst[Int[(1 + (2*e*Rt[b^2 - 4*a*c, 2]*x^2)/(2*c*d - b*e - e*Rt[b^2 - 4*a*c,
2]))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b
, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m^2, 1/4]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rubi steps

\begin{align*} \int \frac{(5-x) (3+2 x)^{3/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx &=-\frac{2 \sqrt{3+2 x} (121+139 x)}{3 \sqrt{2+5 x+3 x^2}}+\frac{2}{3} \int \frac{118+137 x}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{2 \sqrt{3+2 x} (121+139 x)}{3 \sqrt{2+5 x+3 x^2}}+\frac{137}{3} \int \frac{\sqrt{3+2 x}}{\sqrt{2+5 x+3 x^2}} \, dx-\frac{175}{3} \int \frac{1}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{2 \sqrt{3+2 x} (121+139 x)}{3 \sqrt{2+5 x+3 x^2}}+\frac{\left (274 \sqrt{-2-5 x-3 x^2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{2 x^2}{3}}}{\sqrt{1-x^2}} \, dx,x,\frac{\sqrt{6+6 x}}{\sqrt{2}}\right )}{3 \sqrt{3} \sqrt{2+5 x+3 x^2}}-\frac{\left (350 \sqrt{-2-5 x-3 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1+\frac{2 x^2}{3}}} \, dx,x,\frac{\sqrt{6+6 x}}{\sqrt{2}}\right )}{3 \sqrt{3} \sqrt{2+5 x+3 x^2}}\\ &=-\frac{2 \sqrt{3+2 x} (121+139 x)}{3 \sqrt{2+5 x+3 x^2}}+\frac{274 \sqrt{-2-5 x-3 x^2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{3 \sqrt{3} \sqrt{2+5 x+3 x^2}}-\frac{350 \sqrt{-2-5 x-3 x^2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{3 \sqrt{3} \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.3392, size = 183, normalized size = 1.28 \[ -\frac{64 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} \sqrt{\frac{3 x+2}{2 x+3}} (2 x+3)^2 \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right ),\frac{3}{5}\right )+2 \left (12 x^2+607 x+541\right ) \sqrt{2 x+3}-274 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} \sqrt{\frac{3 x+2}{2 x+3}} (2 x+3)^2 E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right )|\frac{3}{5}\right )}{9 (2 x+3) \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*Sqrt[3 + 2*x]*(541 + 607*x + 12*x^2) - 274*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^2*Sqrt[(2 + 3*x)/(3 +
2*x)]*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5] + 64*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^2*Sqrt[(
2 + 3*x)/(3 + 2*x)]*EllipticF[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5])/(9*(3 + 2*x)*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.02, size = 131, normalized size = 0.9 \begin{align*} -{\frac{1}{270\,{x}^{3}+855\,{x}^{2}+855\,x+270}\sqrt{3+2\,x}\sqrt{3\,{x}^{2}+5\,x+2} \left ( 38\,\sqrt{3+2\,x}\sqrt{15}\sqrt{-2-2\,x}\sqrt{-20-30\,x}{\it EllipticF} \left ( 1/5\,\sqrt{30\,x+45},1/3\,\sqrt{15} \right ) +137\,\sqrt{3+2\,x}\sqrt{15}\sqrt{-2-2\,x}\sqrt{-20-30\,x}{\it EllipticE} \left ( 1/5\,\sqrt{30\,x+45},1/3\,\sqrt{15} \right ) +8340\,{x}^{2}+19770\,x+10890 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5-x)*(3+2*x)^(3/2)/(3*x^2+5*x+2)^(3/2),x)

[Out]

-1/45*(3+2*x)^(1/2)*(3*x^2+5*x+2)^(1/2)*(38*(3+2*x)^(1/2)*15^(1/2)*(-2-2*x)^(1/2)*(-20-30*x)^(1/2)*EllipticF(1
/5*(30*x+45)^(1/2),1/3*15^(1/2))+137*(3+2*x)^(1/2)*15^(1/2)*(-2-2*x)^(1/2)*(-20-30*x)^(1/2)*EllipticE(1/5*(30*
x+45)^(1/2),1/3*15^(1/2))+8340*x^2+19770*x+10890)/(6*x^3+19*x^2+19*x+6)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (2 \, x + 3\right )}^{\frac{3}{2}}{\left (x - 5\right )}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{3 \, x^{2} + 5 \, x + 2}{\left (2 \, x^{2} - 7 \, x - 15\right )} \sqrt{2 \, x + 3}}{9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(3*x^2 + 5*x + 2)*(2*x^2 - 7*x - 15)*sqrt(2*x + 3)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{15 \sqrt{2 x + 3}}{3 x^{2} \sqrt{3 x^{2} + 5 x + 2} + 5 x \sqrt{3 x^{2} + 5 x + 2} + 2 \sqrt{3 x^{2} + 5 x + 2}}\, dx - \int - \frac{7 x \sqrt{2 x + 3}}{3 x^{2} \sqrt{3 x^{2} + 5 x + 2} + 5 x \sqrt{3 x^{2} + 5 x + 2} + 2 \sqrt{3 x^{2} + 5 x + 2}}\, dx - \int \frac{2 x^{2} \sqrt{2 x + 3}}{3 x^{2} \sqrt{3 x^{2} + 5 x + 2} + 5 x \sqrt{3 x^{2} + 5 x + 2} + 2 \sqrt{3 x^{2} + 5 x + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-15*sqrt(2*x + 3)/(3*x**2*sqrt(3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5*x + 2) + 2*sqrt(3*x**2 + 5*x
+ 2)), x) - Integral(-7*x*sqrt(2*x + 3)/(3*x**2*sqrt(3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5*x + 2) + 2*sqrt(3
*x**2 + 5*x + 2)), x) - Integral(2*x**2*sqrt(2*x + 3)/(3*x**2*sqrt(3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5*x +
2) + 2*sqrt(3*x**2 + 5*x + 2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (2 \, x + 3\right )}^{\frac{3}{2}}{\left (x - 5\right )}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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