3.2610 \(\int \frac{5-x}{(3+2 x)^{3/2} \sqrt{2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=136 \[ -\frac{\sqrt{-3 x^2-5 x-2} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right ),-\frac{2}{3}\right )}{\sqrt{3} \sqrt{3 x^2+5 x+2}}-\frac{26 \sqrt{3 x^2+5 x+2}}{5 \sqrt{2 x+3}}+\frac{13 \sqrt{3} \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{5 \sqrt{3 x^2+5 x+2}} \]

[Out]

(-26*Sqrt[2 + 5*x + 3*x^2])/(5*Sqrt[3 + 2*x]) + (13*Sqrt[3]*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sq
rt[1 + x]], -2/3])/(5*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]],
-2/3])/(Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

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Rubi [A]  time = 0.0811321, antiderivative size = 136, 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 = {834, 843, 718, 424, 419} \[ -\frac{26 \sqrt{3 x^2+5 x+2}}{5 \sqrt{2 x+3}}-\frac{\sqrt{-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{\sqrt{3} \sqrt{3 x^2+5 x+2}}+\frac{13 \sqrt{3} \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{5 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-26*Sqrt[2 + 5*x + 3*x^2])/(5*Sqrt[3 + 2*x]) + (13*Sqrt[3]*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sq
rt[1 + x]], -2/3])/(5*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]],
-2/3])/(Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

Rule 834

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

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} \sqrt{2+5 x+3 x^2}} \, dx &=-\frac{26 \sqrt{2+5 x+3 x^2}}{5 \sqrt{3+2 x}}-\frac{2}{5} \int \frac{-28-\frac{39 x}{2}}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{26 \sqrt{2+5 x+3 x^2}}{5 \sqrt{3+2 x}}-\frac{1}{2} \int \frac{1}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx+\frac{39}{10} \int \frac{\sqrt{3+2 x}}{\sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{26 \sqrt{2+5 x+3 x^2}}{5 \sqrt{3+2 x}}-\frac{\sqrt{-2-5 x-3 x^2} \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 )}{\sqrt{3} \sqrt{2+5 x+3 x^2}}+\frac{\left (13 \sqrt{3} \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 )}{5 \sqrt{2+5 x+3 x^2}}\\ &=-\frac{26 \sqrt{2+5 x+3 x^2}}{5 \sqrt{3+2 x}}+\frac{13 \sqrt{3} \sqrt{-2-5 x-3 x^2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{5 \sqrt{2+5 x+3 x^2}}-\frac{\sqrt{-2-5 x-3 x^2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{\sqrt{3} \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.17741, size = 104, normalized size = 0.76 \[ \frac{(x+1) \sqrt{\frac{3 x+2}{2 x+3}} \left (13 E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right )|\frac{3}{5}\right )-12 \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right ),\frac{3}{5}\right )\right )}{\sqrt{5} \sqrt{\frac{x+1}{2 x+3}} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + x)*Sqrt[(2 + 3*x)/(3 + 2*x)]*(13*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5] - 12*EllipticF[ArcSin[S
qrt[5/3]/Sqrt[3 + 2*x]], 3/5]))/(Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.02, size = 131, normalized size = 1. \begin{align*}{\frac{1}{900\,{x}^{3}+2850\,{x}^{2}+2850\,x+900}\sqrt{3+2\,x}\sqrt{3\,{x}^{2}+5\,x+2} \left ( 34\,\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 ) -39\,\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 ) -2340\,{x}^{2}-3900\,x-1560 \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)^(1/2),x)

[Out]

1/150*(3+2*x)^(1/2)*(3*x^2+5*x+2)^(1/2)*(34*(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))-39*(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))-2340*x^2-3900*x-1560)/(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{x - 5}{\sqrt{3 \, x^{2} + 5 \, x + 2}{\left (2 \, x + 3\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)^(1/2),x, algorithm="maxima")

[Out]

-integrate((x - 5)/(sqrt(3*x^2 + 5*x + 2)*(2*x + 3)^(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} \sqrt{2 \, x + 3}{\left (x - 5\right )}}{12 \, x^{4} + 56 \, x^{3} + 95 \, x^{2} + 69 \, x + 18}, 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)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(3*x^2 + 5*x + 2)*sqrt(2*x + 3)*(x - 5)/(12*x^4 + 56*x^3 + 95*x^2 + 69*x + 18), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x}{2 x \sqrt{2 x + 3} \sqrt{3 x^{2} + 5 x + 2} + 3 \sqrt{2 x + 3} \sqrt{3 x^{2} + 5 x + 2}}\, dx - \int - \frac{5}{2 x \sqrt{2 x + 3} \sqrt{3 x^{2} + 5 x + 2} + 3 \sqrt{2 x + 3} \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)**(1/2),x)

[Out]

-Integral(x/(2*x*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2) + 3*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2)), x) - Integral
(-5/(2*x*sqrt(2*x + 3)*sqrt(3*x**2 + 5*x + 2) + 3*sqrt(2*x + 3)*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{x - 5}{\sqrt{3 \, x^{2} + 5 \, x + 2}{\left (2 \, x + 3\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)^(1/2),x, algorithm="giac")

[Out]

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