3.1076 \(\int \frac{(2-5 x) x^{5/2}}{(2+5 x+3 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=187 \[ -\frac{1430 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{9 \sqrt{3 x^2+5 x+2}}+\frac{2 (95 x+74) x^{3/2}}{9 \left (3 x^2+5 x+2\right )^{3/2}}-\frac{3464 (3 x+2) \sqrt{x}}{27 \sqrt{3 x^2+5 x+2}}+\frac{4 (866 x+715) \sqrt{x}}{9 \sqrt{3 x^2+5 x+2}}+\frac{3464 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{27 \sqrt{3 x^2+5 x+2}} \]

[Out]

(2*x^(3/2)*(74 + 95*x))/(9*(2 + 5*x + 3*x^2)^(3/2)) - (3464*Sqrt[x]*(2 + 3*x))/(27*Sqrt[2 + 5*x + 3*x^2]) + (4
*Sqrt[x]*(715 + 866*x))/(9*Sqrt[2 + 5*x + 3*x^2]) + (3464*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[Ar
cTan[Sqrt[x]], -1/2])/(27*Sqrt[2 + 5*x + 3*x^2]) - (1430*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[Arc
Tan[Sqrt[x]], -1/2])/(9*Sqrt[2 + 5*x + 3*x^2])

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Rubi [A]  time = 0.121059, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {818, 820, 839, 1189, 1100, 1136} \[ \frac{2 (95 x+74) x^{3/2}}{9 \left (3 x^2+5 x+2\right )^{3/2}}-\frac{3464 (3 x+2) \sqrt{x}}{27 \sqrt{3 x^2+5 x+2}}+\frac{4 (866 x+715) \sqrt{x}}{9 \sqrt{3 x^2+5 x+2}}-\frac{1430 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{9 \sqrt{3 x^2+5 x+2}}+\frac{3464 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{27 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(74 + 95*x))/(9*(2 + 5*x + 3*x^2)^(3/2)) - (3464*Sqrt[x]*(2 + 3*x))/(27*Sqrt[2 + 5*x + 3*x^2]) + (4
*Sqrt[x]*(715 + 866*x))/(9*Sqrt[2 + 5*x + 3*x^2]) + (3464*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[Ar
cTan[Sqrt[x]], -1/2])/(27*Sqrt[2 + 5*x + 3*x^2]) - (1430*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[Arc
Tan[Sqrt[x]], -1/2])/(9*Sqrt[2 + 5*x + 3*x^2])

Rule 818

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 + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 820

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

Rule 839

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f +
g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1189

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[d, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[e, Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b +
q)/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1100

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b -
q)*x^2)*Sqrt[(2*a + (b + q)*x^2)/(2*a + (b - q)*x^2)]*EllipticF[ArcTan[Rt[(b - q)/(2*a), 2]*x], (-2*q)/(b - q)
])/(2*a*Rt[(b - q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^
2 - 4*a*c, 0]

Rule 1136

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b -
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b - q)/(2*a), 2]*(2*a + (b - q)*x^2)*Sqrt[(2*a + (
b + q)*x^2)/(2*a + (b - q)*x^2)]*EllipticE[ArcTan[Rt[(b - q)/(2*a), 2]*x], (-2*q)/(b - q)])/(2*c*Sqrt[a + b*x^
2 + c*x^4]), x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{(2-5 x) x^{5/2}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx &=\frac{2 x^{3/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac{2}{9} \int \frac{\sqrt{x} (-111+40 x)}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx\\ &=\frac{2 x^{3/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac{4 \sqrt{x} (715+866 x)}{9 \sqrt{2+5 x+3 x^2}}-\frac{4}{9} \int \frac{\frac{715}{2}+433 x}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{2 x^{3/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac{4 \sqrt{x} (715+866 x)}{9 \sqrt{2+5 x+3 x^2}}-\frac{8}{9} \operatorname{Subst}\left (\int \frac{\frac{715}{2}+433 x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 x^{3/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac{4 \sqrt{x} (715+866 x)}{9 \sqrt{2+5 x+3 x^2}}-\frac{2860}{9} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )-\frac{3464}{9} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 x^{3/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}-\frac{3464 \sqrt{x} (2+3 x)}{27 \sqrt{2+5 x+3 x^2}}+\frac{4 \sqrt{x} (715+866 x)}{9 \sqrt{2+5 x+3 x^2}}+\frac{3464 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{27 \sqrt{2+5 x+3 x^2}}-\frac{1430 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{9 \sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.261043, size = 167, normalized size = 0.89 \[ \frac{-826 i \sqrt{\frac{2}{x}+2} \sqrt{\frac{2}{x}+3} \left (3 x^2+5 x+2\right ) x^{3/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )-2 \left (12825 x^3+32020 x^2+26060 x+6928\right )-3464 i \sqrt{\frac{2}{x}+2} \sqrt{\frac{2}{x}+3} \left (3 x^2+5 x+2\right ) x^{3/2} E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )}{27 \sqrt{x} \left (3 x^2+5 x+2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(6928 + 26060*x + 32020*x^2 + 12825*x^3) - (3464*I)*Sqrt[2 + 2/x]*Sqrt[3 + 2/x]*x^(3/2)*(2 + 5*x + 3*x^2)*
EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (826*I)*Sqrt[2 + 2/x]*Sqrt[3 + 2/x]*x^(3/2)*(2 + 5*x + 3*x^2)*E
llipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(27*Sqrt[x]*(2 + 5*x + 3*x^2)^(3/2))

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Maple [A]  time = 0.022, size = 297, normalized size = 1.6 \begin{align*}{\frac{2}{81\, \left ( 1+x \right ) ^{2} \left ( 2+3\,x \right ) ^{2}} \left ( 1359\,\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{6}\sqrt{-x}{\it EllipticF} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ){x}^{2}-2598\,\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{6}\sqrt{-x}{\it EllipticE} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ){x}^{2}+2265\,\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{6}\sqrt{-x}{\it EllipticF} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ) x-4330\,\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{6}\sqrt{-x}{\it EllipticE} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ) x+906\,\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{6}\sqrt{-x}{\it EllipticF} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ) -1732\,\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{6}\sqrt{-x}{\it EllipticE} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ) +46764\,{x}^{4}+117405\,{x}^{3}+96192\,{x}^{2}+25740\,x \right ) \sqrt{3\,{x}^{2}+5\,x+2}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/81*(1359*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))*x^2-2598*(6*x
+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*x^2+2265*(6*x+4)^(1/2)*(3+3*
x)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))*x-4330*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*
(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*x+906*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*Ellipti
cF(1/2*(6*x+4)^(1/2),I*2^(1/2))-1732*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2
),I*2^(1/2))+46764*x^4+117405*x^3+96192*x^2+25740*x)*(3*x^2+5*x+2)^(1/2)/x^(1/2)/(1+x)^2/(2+3*x)^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (5 \, x^{3} - 2 \, x^{2}\right )} \sqrt{3 \, x^{2} + 5 \, x + 2} \sqrt{x}}{27 \, x^{6} + 135 \, x^{5} + 279 \, x^{4} + 305 \, x^{3} + 186 \, x^{2} + 60 \, x + 8}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(5*x^3 - 2*x^2)*sqrt(3*x^2 + 5*x + 2)*sqrt(x)/(27*x^6 + 135*x^5 + 279*x^4 + 305*x^3 + 186*x^2 + 60*x
+ 8), 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((2-5*x)*x**(5/2)/(3*x**2+5*x+2)**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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