3.2607 \(\int \frac{(5-x) (3+2 x)^{3/2}}{\sqrt{2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=165 \[ -\frac{163 \sqrt{-3 x^2-5 x-2} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right ),-\frac{2}{3}\right )}{27 \sqrt{3} \sqrt{3 x^2+5 x+2}}-\frac{2}{15} \sqrt{3 x^2+5 x+2} (2 x+3)^{3/2}+\frac{326}{135} \sqrt{3 x^2+5 x+2} \sqrt{2 x+3}+\frac{2743 \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{135 \sqrt{3} \sqrt{3 x^2+5 x+2}} \]

[Out]

(326*Sqrt[3 + 2*x]*Sqrt[2 + 5*x + 3*x^2])/135 - (2*(3 + 2*x)^(3/2)*Sqrt[2 + 5*x + 3*x^2])/15 + (2743*Sqrt[-2 -
5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(135*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]) - (163*Sqrt[-2
- 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(27*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

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Rubi [A]  time = 0.100454, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {832, 843, 718, 424, 419} \[ -\frac{2}{15} \sqrt{3 x^2+5 x+2} (2 x+3)^{3/2}+\frac{326}{135} \sqrt{3 x^2+5 x+2} \sqrt{2 x+3}-\frac{163 \sqrt{-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{27 \sqrt{3} \sqrt{3 x^2+5 x+2}}+\frac{2743 \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{135 \sqrt{3} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(326*Sqrt[3 + 2*x]*Sqrt[2 + 5*x + 3*x^2])/135 - (2*(3 + 2*x)^(3/2)*Sqrt[2 + 5*x + 3*x^2])/15 + (2743*Sqrt[-2 -
5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(135*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2]) - (163*Sqrt[-2
- 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(27*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*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] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 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}}{\sqrt{2+5 x+3 x^2}} \, dx &=-\frac{2}{15} (3+2 x)^{3/2} \sqrt{2+5 x+3 x^2}+\frac{2}{15} \int \frac{\sqrt{3+2 x} \left (126+\frac{163 x}{2}\right )}{\sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{326}{135} \sqrt{3+2 x} \sqrt{2+5 x+3 x^2}-\frac{2}{15} (3+2 x)^{3/2} \sqrt{2+5 x+3 x^2}+\frac{4}{135} \int \frac{\frac{3707}{4}+\frac{2743 x}{4}}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{326}{135} \sqrt{3+2 x} \sqrt{2+5 x+3 x^2}-\frac{2}{15} (3+2 x)^{3/2} \sqrt{2+5 x+3 x^2}-\frac{163}{54} \int \frac{1}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx+\frac{2743}{270} \int \frac{\sqrt{3+2 x}}{\sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{326}{135} \sqrt{3+2 x} \sqrt{2+5 x+3 x^2}-\frac{2}{15} (3+2 x)^{3/2} \sqrt{2+5 x+3 x^2}-\frac{\left (163 \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 )}{27 \sqrt{3} \sqrt{2+5 x+3 x^2}}+\frac{\left (2743 \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 )}{135 \sqrt{3} \sqrt{2+5 x+3 x^2}}\\ &=\frac{326}{135} \sqrt{3+2 x} \sqrt{2+5 x+3 x^2}-\frac{2}{15} (3+2 x)^{3/2} \sqrt{2+5 x+3 x^2}+\frac{2743 \sqrt{-2-5 x-3 x^2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{135 \sqrt{3} \sqrt{2+5 x+3 x^2}}-\frac{163 \sqrt{-2-5 x-3 x^2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{27 \sqrt{3} \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.298358, size = 193, normalized size = 1.17 \[ -\frac{2254 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} \sqrt{\frac{3 x+2}{2 x+3}} (2 x+3)^2 \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right ),\frac{3}{5}\right )+2 \left (324 x^4-1422 x^3-14955 x^2-21143 x-7934\right ) \sqrt{2 x+3}-2743 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} \sqrt{\frac{3 x+2}{2 x+3}} (2 x+3)^2 E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right )|\frac{3}{5}\right )}{405 (2 x+3) \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*Sqrt[3 + 2*x]*(-7934 - 21143*x - 14955*x^2 - 1422*x^3 + 324*x^4) - 2743*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] + 2254*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])/(405*(3 +
2*x)*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.014, size = 141, normalized size = 0.9 \begin{align*}{\frac{1}{24300\,{x}^{3}+76950\,{x}^{2}+76950\,x+24300}\sqrt{3+2\,x}\sqrt{3\,{x}^{2}+5\,x+2} \left ( 1928\,\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 ) -2743\,\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 ) -6480\,{x}^{4}+28440\,{x}^{3}+134520\,{x}^{2}+148560\,x+48960 \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]

1/4050*(3+2*x)^(1/2)*(3*x^2+5*x+2)^(1/2)*(1928*(3+2*x)^(1/2)*15^(1/2)*(-2-2*x)^(1/2)*(-20-30*x)^(1/2)*Elliptic
F(1/5*(30*x+45)^(1/2),1/3*15^(1/2))-2743*(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))-6480*x^4+28440*x^3+134520*x^2+148560*x+48960)/(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 )}}{\sqrt{3 \, x^{2} + 5 \, x + 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((2*x + 3)^(3/2)*(x - 5)/sqrt(3*x^2 + 5*x + 2), x)

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

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