3.2600 \(\int \frac{(5-x) (2+5 x+3 x^2)^{5/2}}{(3+2 x)^{7/2}} \, dx\)

Optimal. Leaf size=207 \[ -\frac{12857 \sqrt{-3 x^2-5 x-2} \text{EllipticF}\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right ),-\frac{2}{3}\right )}{56 \sqrt{3} \sqrt{3 x^2+5 x+2}}-\frac{(5 x+53) \left (3 x^2+5 x+2\right )^{5/2}}{35 (2 x+3)^{5/2}}+\frac{(879 x+2291) \left (3 x^2+5 x+2\right )^{3/2}}{210 (2 x+3)^{3/2}}-\frac{(3117 x+10763) \sqrt{3 x^2+5 x+2}}{140 \sqrt{2 x+3}}+\frac{2333 \sqrt{3} \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{40 \sqrt{3 x^2+5 x+2}} \]

[Out]

-((10763 + 3117*x)*Sqrt[2 + 5*x + 3*x^2])/(140*Sqrt[3 + 2*x]) + ((2291 + 879*x)*(2 + 5*x + 3*x^2)^(3/2))/(210*
(3 + 2*x)^(3/2)) - ((53 + 5*x)*(2 + 5*x + 3*x^2)^(5/2))/(35*(3 + 2*x)^(5/2)) + (2333*Sqrt[3]*Sqrt[-2 - 5*x - 3
*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(40*Sqrt[2 + 5*x + 3*x^2]) - (12857*Sqrt[-2 - 5*x - 3*x^2]
*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(56*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

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Rubi [A]  time = 0.132171, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {812, 843, 718, 424, 419} \[ -\frac{(5 x+53) \left (3 x^2+5 x+2\right )^{5/2}}{35 (2 x+3)^{5/2}}+\frac{(879 x+2291) \left (3 x^2+5 x+2\right )^{3/2}}{210 (2 x+3)^{3/2}}-\frac{(3117 x+10763) \sqrt{3 x^2+5 x+2}}{140 \sqrt{2 x+3}}-\frac{12857 \sqrt{-3 x^2-5 x-2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{56 \sqrt{3} \sqrt{3 x^2+5 x+2}}+\frac{2333 \sqrt{3} \sqrt{-3 x^2-5 x-2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{x+1}\right )|-\frac{2}{3}\right )}{40 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((10763 + 3117*x)*Sqrt[2 + 5*x + 3*x^2])/(140*Sqrt[3 + 2*x]) + ((2291 + 879*x)*(2 + 5*x + 3*x^2)^(3/2))/(210*
(3 + 2*x)^(3/2)) - ((53 + 5*x)*(2 + 5*x + 3*x^2)^(5/2))/(35*(3 + 2*x)^(5/2)) + (2333*Sqrt[3]*Sqrt[-2 - 5*x - 3
*x^2]*EllipticE[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(40*Sqrt[2 + 5*x + 3*x^2]) - (12857*Sqrt[-2 - 5*x - 3*x^2]
*EllipticF[ArcSin[Sqrt[3]*Sqrt[1 + x]], -2/3])/(56*Sqrt[3]*Sqrt[2 + 5*x + 3*x^2])

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 )^{5/2}}{(3+2 x)^{7/2}} \, dx &=-\frac{(53+5 x) \left (2+5 x+3 x^2\right )^{5/2}}{35 (3+2 x)^{5/2}}-\frac{1}{14} \int \frac{(-245-293 x) \left (2+5 x+3 x^2\right )^{3/2}}{(3+2 x)^{5/2}} \, dx\\ &=\frac{(2291+879 x) \left (2+5 x+3 x^2\right )^{3/2}}{210 (3+2 x)^{3/2}}-\frac{(53+5 x) \left (2+5 x+3 x^2\right )^{5/2}}{35 (3+2 x)^{5/2}}+\frac{1}{140} \int \frac{(-7939-9351 x) \sqrt{2+5 x+3 x^2}}{(3+2 x)^{3/2}} \, dx\\ &=-\frac{(10763+3117 x) \sqrt{2+5 x+3 x^2}}{140 \sqrt{3+2 x}}+\frac{(2291+879 x) \left (2+5 x+3 x^2\right )^{3/2}}{210 (3+2 x)^{3/2}}-\frac{(53+5 x) \left (2+5 x+3 x^2\right )^{5/2}}{35 (3+2 x)^{5/2}}-\frac{1}{840} \int \frac{-124041-146979 x}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{(10763+3117 x) \sqrt{2+5 x+3 x^2}}{140 \sqrt{3+2 x}}+\frac{(2291+879 x) \left (2+5 x+3 x^2\right )^{3/2}}{210 (3+2 x)^{3/2}}-\frac{(53+5 x) \left (2+5 x+3 x^2\right )^{5/2}}{35 (3+2 x)^{5/2}}+\frac{6999}{80} \int \frac{\sqrt{3+2 x}}{\sqrt{2+5 x+3 x^2}} \, dx-\frac{12857}{112} \int \frac{1}{\sqrt{3+2 x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{(10763+3117 x) \sqrt{2+5 x+3 x^2}}{140 \sqrt{3+2 x}}+\frac{(2291+879 x) \left (2+5 x+3 x^2\right )^{3/2}}{210 (3+2 x)^{3/2}}-\frac{(53+5 x) \left (2+5 x+3 x^2\right )^{5/2}}{35 (3+2 x)^{5/2}}-\frac{\left (12857 \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 )}{56 \sqrt{3} \sqrt{2+5 x+3 x^2}}+\frac{\left (2333 \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 )}{40 \sqrt{2+5 x+3 x^2}}\\ &=-\frac{(10763+3117 x) \sqrt{2+5 x+3 x^2}}{140 \sqrt{3+2 x}}+\frac{(2291+879 x) \left (2+5 x+3 x^2\right )^{3/2}}{210 (3+2 x)^{3/2}}-\frac{(53+5 x) \left (2+5 x+3 x^2\right )^{5/2}}{35 (3+2 x)^{5/2}}+\frac{2333 \sqrt{3} \sqrt{-2-5 x-3 x^2} E\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{40 \sqrt{2+5 x+3 x^2}}-\frac{12857 \sqrt{-2-5 x-3 x^2} F\left (\sin ^{-1}\left (\sqrt{3} \sqrt{1+x}\right )|-\frac{2}{3}\right )}{56 \sqrt{3} \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.449724, size = 202, normalized size = 0.98 \[ \frac{-10422 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} (2 x+3)^{7/2} \sqrt{\frac{3 x+2}{2 x+3}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right ),\frac{3}{5}\right )-3240 x^7+12744 x^6+41220 x^5+335988 x^4+1717690 x^3+3262382 x^2+2556580 x+48993 \sqrt{5} \sqrt{\frac{x+1}{2 x+3}} (2 x+3)^{7/2} \sqrt{\frac{3 x+2}{2 x+3}} E\left (\sin ^{-1}\left (\frac{\sqrt{\frac{5}{3}}}{\sqrt{2 x+3}}\right )|\frac{3}{5}\right )+701136}{840 (2 x+3)^{5/2} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(701136 + 2556580*x + 3262382*x^2 + 1717690*x^3 + 335988*x^4 + 41220*x^5 + 12744*x^6 - 3240*x^7 + 48993*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] - 10422*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])/(840*(3 + 2*x)^(5/2)*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.019, size = 311, normalized size = 1.5 \begin{align*} -{\frac{1}{8400} \left ( 61168\,\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}+195972\,\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}+32400\,{x}^{7}+183504\,\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}+587916\,\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}-127440\,{x}^{6}+137628\,\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 ) +440937\,\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 ) -412200\,{x}^{5}+8398440\,{x}^{4}[phone]\,{x}^{3}[phone]\,{x}^{2}[phone]\,x+10626120 \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)^(5/2)/(3+2*x)^(7/2),x)

[Out]

-1/8400*(61168*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)+195972*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)+32400*x^7+183504*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)+587916*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)*(-2
0-30*x)^(1/2)-127440*x^6+137628*(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))+440937*(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))-412200*x^5+8398440*x^4+37695260*x^3+60462880*x^2+42044540*x+10626120)/(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{5}{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)^(5/2)/(3+2*x)^(7/2),x, algorithm="maxima")

[Out]

-integrate((3*x^2 + 5*x + 2)^(5/2)*(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 (9 \, x^{5} - 15 \, x^{4} - 113 \, x^{3} - 165 \, x^{2} - 96 \, x - 20\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)^(5/2)/(3+2*x)^(7/2),x, algorithm="fricas")

[Out]

integral(-(9*x^5 - 15*x^4 - 113*x^3 - 165*x^2 - 96*x - 20)*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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{5}{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)^(5/2)/(3+2*x)^(7/2),x, algorithm="giac")

[Out]

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