3.2581 \(\int \frac{(5-x) \sqrt{2+5 x+3 x^2}}{(3+2 x)^{5/2}} \, dx\)

Optimal. Leaf size=143 \[ \frac{17 \sqrt{-3 x^2-5 x-2} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right ),-\frac{2}{3}\right )}{2 \sqrt{3} \sqrt{3 x^2+5 x+2}}+\frac{\sqrt{3 x^2+5 x+2} (119 x+146)}{15 (2 x+3)^{3/2}}-\frac{67 \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{10 \sqrt{3} \sqrt{3 x^2+5 x+2}} \]

[Out]

((146 + 119*x)*Sqrt[2 + 5*x + 3*x^2])/(15*(3 + 2*x)^(3/2)) - (67*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[
3]*Sqrt[1 + x]], -2/3])/(10*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]) + (17*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[
3]*Sqrt[1 + x]], -2/3])/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

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Rubi [A]  time = 0.0854779, 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 = {810, 843, 718, 424, 419} \[ \frac{\sqrt{3 x^2+5 x+2} (119 x+146)}{15 (2 x+3)^{3/2}}+\frac{17 \sqrt{-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{2 \sqrt{3} \sqrt{3 x^2+5 x+2}}-\frac{67 \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{10 \sqrt{3} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((146 + 119*x)*Sqrt[2 + 5*x + 3*x^2])/(15*(3 + 2*x)^(3/2)) - (67*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[
3]*Sqrt[1 + x]], -2/3])/(10*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]) + (17*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[
3]*Sqrt[1 + x]], -2/3])/(2*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

Rule 810

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*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*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] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !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) \sqrt{2+5 x+3 x^2}}{(3+2 x)^{5/2}} \, dx &=\frac{(146+119 x) \sqrt{2+5 x+3 x^2}}{15 (3+2 x)^{3/2}}-\frac{1}{30} \int \frac{174+201 x}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{(146+119 x) \sqrt{2+5 x+3 x^2}}{15 (3+2 x)^{3/2}}-\frac{67}{20} \int \frac{\sqrt{3+2 x}}{\sqrt{2+5 x+3 x^2}} \, dx+\frac{17}{4} \int \frac{1}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{(146+119 x) \sqrt{2+5 x+3 x^2}}{15 (3+2 x)^{3/2}}-\frac{\left (67 \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 )}{10 \sqrt{3} \sqrt{2+5 x+3 x^2}}+\frac{\left (17 \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 )}{2 \sqrt{3} \sqrt{2+5 x+3 x^2}}\\ &=\frac{(146+119 x) \sqrt{2+5 x+3 x^2}}{15 (3+2 x)^{3/2}}-\frac{67 \sqrt{-2-5 x-3 x^2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{10 \sqrt{3} \sqrt{2+5 x+3 x^2}}+\frac{17 \sqrt{-2-5 x-3 x^2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{2 \sqrt{3} \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.311247, size = 182, normalized size = 1.27 \[ -\frac{-16 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} (2 x+3)^{5/2} \sqrt{\frac{3 x+2}{2 x+3}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right ),\frac{3}{5}\right )+90 x^3+480 x^2+610 x+67 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} (2 x+3)^{5/2} \sqrt{\frac{3 x+2}{2 x+3}} E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right )|\frac{3}{5}\right )+220}{30 (2 x+3)^{3/2} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(220 + 610*x + 480*x^2 + 90*x^3 + 67*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(5/2)*Sqrt[(2 + 3*x)/(3 + 2*x)
]*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5] - 16*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 + 2*x)^(5/2)*Sqrt[(2
+ 3*x)/(3 + 2*x)]*EllipticF[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5])/(30*(3 + 2*x)^(3/2)*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.03, size = 203, normalized size = 1.4 \begin{align*}{\frac{1}{300} \left ( 36\,\sqrt{15}{\it EllipticF} \left ( 1/5\,\sqrt{30\,x+45},1/3\,\sqrt{15} \right ) x\sqrt{3+2\,x}\sqrt{-2-2\,x}\sqrt{-20-30\,x}+134\,\sqrt{15}{\it EllipticE} \left ( 1/5\,\sqrt{30\,x+45},1/3\,\sqrt{15} \right ) x\sqrt{3+2\,x}\sqrt{-2-2\,x}\sqrt{-20-30\,x}+54\,\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 ) +201\,\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 ) +7140\,{x}^{3}+20660\,{x}^{2}+19360\,x+5840 \right ) \left ( 3+2\,x \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{3\,{x}^{2}+5\,x+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/300*(36*15^(1/2)*EllipticF(1/5*(30*x+45)^(1/2),1/3*15^(1/2))*x*(3+2*x)^(1/2)*(-2-2*x)^(1/2)*(-20-30*x)^(1/2)
+134*15^(1/2)*EllipticE(1/5*(30*x+45)^(1/2),1/3*15^(1/2))*x*(3+2*x)^(1/2)*(-2-2*x)^(1/2)*(-20-30*x)^(1/2)+54*(
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))+201*(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))+7140*x^3+20660*x^2+
19360*x+5840)/(3*x^2+5*x+2)^(1/2)/(3+2*x)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3*x^2+5*x+2)^(1/2)/(3+2*x)^(5/2),x, algorithm="maxima")

[Out]

-integrate(sqrt(3*x^2 + 5*x + 2)*(x - 5)/(2*x + 3)^(5/2), 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 )}}{8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3*x^2+5*x+2)^(1/2)/(3+2*x)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(3*x^2 + 5*x + 2)*sqrt(2*x + 3)*(x - 5)/(8*x^3 + 36*x^2 + 54*x + 27), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{5 \sqrt{3 x^{2} + 5 x + 2}}{4 x^{2} \sqrt{2 x + 3} + 12 x \sqrt{2 x + 3} + 9 \sqrt{2 x + 3}}\, dx - \int \frac{x \sqrt{3 x^{2} + 5 x + 2}}{4 x^{2} \sqrt{2 x + 3} + 12 x \sqrt{2 x + 3} + 9 \sqrt{2 x + 3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-5*sqrt(3*x**2 + 5*x + 2)/(4*x**2*sqrt(2*x + 3) + 12*x*sqrt(2*x + 3) + 9*sqrt(2*x + 3)), x) - Integr
al(x*sqrt(3*x**2 + 5*x + 2)/(4*x**2*sqrt(2*x + 3) + 12*x*sqrt(2*x + 3) + 9*sqrt(2*x + 3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3*x^2+5*x+2)^(1/2)/(3+2*x)^(5/2),x, algorithm="giac")

[Out]

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