3.2590 \(\int \frac{(5-x) (2+5 x+3 x^2)^{3/2}}{(3+2 x)^{7/2}} \, dx\)

Optimal. Leaf size=175 \[ -\frac{243 \sqrt{3} \sqrt{-3 x^2-5 x-2} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right ),-\frac{2}{3}\right )}{20 \sqrt{3 x^2+5 x+2}}+\frac{(337 x+408) \left (3 x^2+5 x+2\right )^{3/2}}{75 (2 x+3)^{5/2}}-\frac{(181 x+614) \sqrt{3 x^2+5 x+2}}{50 \sqrt{2 x+3}}+\frac{2779 \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{100 \sqrt{3} \sqrt{3 x^2+5 x+2}} \]

[Out]

-((614 + 181*x)*Sqrt[2 + 5*x + 3*x^2])/(50*Sqrt[3 + 2*x]) + ((408 + 337*x)*(2 + 5*x + 3*x^2)^(3/2))/(75*(3 + 2
*x)^(5/2)) + (2779*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(100*Sqrt[3]*Sqrt[2 +
5*x + 3*x^2]) - (243*Sqrt[3]*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(20*Sqrt[2 +
5*x + 3*x^2])

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Rubi [A]  time = 0.10381, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {810, 812, 843, 718, 424, 419} \[ \frac{(337 x+408) \left (3 x^2+5 x+2\right )^{3/2}}{75 (2 x+3)^{5/2}}-\frac{(181 x+614) \sqrt{3 x^2+5 x+2}}{50 \sqrt{2 x+3}}-\frac{243 \sqrt{3} \sqrt{-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{20 \sqrt{3 x^2+5 x+2}}+\frac{2779 \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{100 \sqrt{3} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

Int[((5 - x)*(2 + 5*x + 3*x^2)^(3/2))/(3 + 2*x)^(7/2),x]

[Out]

-((614 + 181*x)*Sqrt[2 + 5*x + 3*x^2])/(50*Sqrt[3 + 2*x]) + ((408 + 337*x)*(2 + 5*x + 3*x^2)^(3/2))/(75*(3 + 2
*x)^(5/2)) + (2779*Sqrt[-2 - 5*x - 3*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(100*Sqrt[3]*Sqrt[2 +
5*x + 3*x^2]) - (243*Sqrt[3]*Sqrt[-2 - 5*x - 3*x^2]*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(20*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 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*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] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (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) \left (2+5 x+3 x^2\right )^{3/2}}{(3+2 x)^{7/2}} \, dx &=\frac{(408+337 x) \left (2+5 x+3 x^2\right )^{3/2}}{75 (3+2 x)^{5/2}}-\frac{1}{50} \int \frac{(472+543 x) \sqrt{2+5 x+3 x^2}}{(3+2 x)^{3/2}} \, dx\\ &=-\frac{(614+181 x) \sqrt{2+5 x+3 x^2}}{50 \sqrt{3+2 x}}+\frac{(408+337 x) \left (2+5 x+3 x^2\right )^{3/2}}{75 (3+2 x)^{5/2}}+\frac{1}{300} \int \frac{7038+8337 x}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{(614+181 x) \sqrt{2+5 x+3 x^2}}{50 \sqrt{3+2 x}}+\frac{(408+337 x) \left (2+5 x+3 x^2\right )^{3/2}}{75 (3+2 x)^{5/2}}+\frac{2779}{200} \int \frac{\sqrt{3+2 x}}{\sqrt{2+5 x+3 x^2}} \, dx-\frac{729}{40} \int \frac{1}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{(614+181 x) \sqrt{2+5 x+3 x^2}}{50 \sqrt{3+2 x}}+\frac{(408+337 x) \left (2+5 x+3 x^2\right )^{3/2}}{75 (3+2 x)^{5/2}}+\frac{\left (2779 \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 )}{100 \sqrt{3} \sqrt{2+5 x+3 x^2}}-\frac{\left (243 \sqrt{3} \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 )}{20 \sqrt{2+5 x+3 x^2}}\\ &=-\frac{(614+181 x) \sqrt{2+5 x+3 x^2}}{50 \sqrt{3+2 x}}+\frac{(408+337 x) \left (2+5 x+3 x^2\right )^{3/2}}{75 (3+2 x)^{5/2}}+\frac{2779 \sqrt{-2-5 x-3 x^2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{100 \sqrt{3} \sqrt{2+5 x+3 x^2}}-\frac{243 \sqrt{3} \sqrt{-2-5 x-3 x^2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{20 \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.360746, size = 192, normalized size = 1.1 \[ \frac{-592 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} (2 x+3)^{7/2} \sqrt{\frac{3 x+2}{2 x+3}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right ),\frac{3}{5}\right )-900 x^5+16800 x^4+100610 x^3+190440 x^2+147790 x+2779 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} (2 x+3)^{7/2} \sqrt{\frac{3 x+2}{2 x+3}} E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right )|\frac{3}{5}\right )+40260}{300 (2 x+3)^{5/2} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((5 - x)*(2 + 5*x + 3*x^2)^(3/2))/(3 + 2*x)^(7/2),x]

[Out]

(40260 + 147790*x + 190440*x^2 + 100610*x^3 + 16800*x^4 - 900*x^5 + 2779*Sqrt[5]*Sqrt[(1 + x)/(3 + 2*x)]*(3 +
2*x)^(7/2)*Sqrt[(2 + 3*x)/(3 + 2*x)]*EllipticE[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5] - 592*Sqrt[5]*Sqrt[(1 + x
)/(3 + 2*x)]*(3 + 2*x)^(7/2)*Sqrt[(2 + 3*x)/(3 + 2*x)]*EllipticF[ArcSin[Sqrt[5/3]/Sqrt[3 + 2*x]], 3/5])/(300*(
3 + 2*x)^(5/2)*Sqrt[2 + 5*x + 3*x^2])

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Maple [B]  time = 0.017, size = 301, normalized size = 1.7 \begin{align*} -{\frac{1}{3000} \left ( 3464\,\sqrt{15}{\it EllipticF} \left ( 1/5\,\sqrt{30\,x+45},1/3\,\sqrt{15} \right ){x}^{2}\sqrt{3+2\,x}\sqrt{-2-2\,x}\sqrt{-20-30\,x}+11116\,\sqrt{15}{\it EllipticE} \left ( 1/5\,\sqrt{30\,x+45},1/3\,\sqrt{15} \right ){x}^{2}\sqrt{3+2\,x}\sqrt{-2-2\,x}\sqrt{-20-30\,x}+10392\,\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}+33348\,\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}+7794\,\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 ) +25011\,\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 ) +9000\,{x}^{5}+498960\,{x}^{4}+2106380\,{x}^{3}+3375700\,{x}^{2}+2357120\,x+597840 \right ) \left ( 3+2\,x \right ) ^{-{\frac{5}{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)^(3/2)/(3+2*x)^(7/2),x)

[Out]

-1/3000*(3464*15^(1/2)*EllipticF(1/5*(30*x+45)^(1/2),1/3*15^(1/2))*x^2*(3+2*x)^(1/2)*(-2-2*x)^(1/2)*(-20-30*x)
^(1/2)+11116*15^(1/2)*EllipticE(1/5*(30*x+45)^(1/2),1/3*15^(1/2))*x^2*(3+2*x)^(1/2)*(-2-2*x)^(1/2)*(-20-30*x)^
(1/2)+10392*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)+33348*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)+
7794*(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))+25011*
(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))+9000*x^5+49
8960*x^4+2106380*x^3+3375700*x^2+2357120*x+597840)/(3*x^2+5*x+2)^(1/2)/(3+2*x)^(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{3}{2}}{\left (x - 5\right )}}{{\left (2 \, x + 3\right )}^{\frac{7}{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)^(3/2)/(3+2*x)^(7/2),x, algorithm="maxima")

[Out]

-integrate((3*x^2 + 5*x + 2)^(3/2)*(x - 5)/(2*x + 3)^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (3 \, x^{3} - 10 \, x^{2} - 23 \, x - 10\right )} \sqrt{3 \, x^{2} + 5 \, x + 2} \sqrt{2 \, x + 3}}{16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5-x)*(3*x^2+5*x+2)^(3/2)/(3+2*x)^(7/2),x, algorithm="fricas")

[Out]

integral(-(3*x^3 - 10*x^2 - 23*x - 10)*sqrt(3*x^2 + 5*x + 2)*sqrt(2*x + 3)/(16*x^4 + 96*x^3 + 216*x^2 + 216*x
+ 81), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{10 \sqrt{3 x^{2} + 5 x + 2}}{8 x^{3} \sqrt{2 x + 3} + 36 x^{2} \sqrt{2 x + 3} + 54 x \sqrt{2 x + 3} + 27 \sqrt{2 x + 3}}\, dx - \int - \frac{23 x \sqrt{3 x^{2} + 5 x + 2}}{8 x^{3} \sqrt{2 x + 3} + 36 x^{2} \sqrt{2 x + 3} + 54 x \sqrt{2 x + 3} + 27 \sqrt{2 x + 3}}\, dx - \int - \frac{10 x^{2} \sqrt{3 x^{2} + 5 x + 2}}{8 x^{3} \sqrt{2 x + 3} + 36 x^{2} \sqrt{2 x + 3} + 54 x \sqrt{2 x + 3} + 27 \sqrt{2 x + 3}}\, dx - \int \frac{3 x^{3} \sqrt{3 x^{2} + 5 x + 2}}{8 x^{3} \sqrt{2 x + 3} + 36 x^{2} \sqrt{2 x + 3} + 54 x \sqrt{2 x + 3} + 27 \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)**(3/2)/(3+2*x)**(7/2),x)

[Out]

-Integral(-10*sqrt(3*x**2 + 5*x + 2)/(8*x**3*sqrt(2*x + 3) + 36*x**2*sqrt(2*x + 3) + 54*x*sqrt(2*x + 3) + 27*s
qrt(2*x + 3)), x) - Integral(-23*x*sqrt(3*x**2 + 5*x + 2)/(8*x**3*sqrt(2*x + 3) + 36*x**2*sqrt(2*x + 3) + 54*x
*sqrt(2*x + 3) + 27*sqrt(2*x + 3)), x) - Integral(-10*x**2*sqrt(3*x**2 + 5*x + 2)/(8*x**3*sqrt(2*x + 3) + 36*x
**2*sqrt(2*x + 3) + 54*x*sqrt(2*x + 3) + 27*sqrt(2*x + 3)), x) - Integral(3*x**3*sqrt(3*x**2 + 5*x + 2)/(8*x**
3*sqrt(2*x + 3) + 36*x**2*sqrt(2*x + 3) + 54*x*sqrt(2*x + 3) + 27*sqrt(2*x + 3)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{3}{2}}{\left (x - 5\right )}}{{\left (2 \, x + 3\right )}^{\frac{7}{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)^(3/2)/(3+2*x)^(7/2),x, algorithm="giac")

[Out]

integrate(-(3*x^2 + 5*x + 2)^(3/2)*(x - 5)/(2*x + 3)^(7/2), x)