3.1064 \(\int \frac{(2-5 x) x^{7/2}}{(2+5 x+3 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=197 \[ -\frac{20 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}+\frac{2 (95 x+74) x^{5/2}}{3 \sqrt{3 x^2+5 x+2}}-\frac{64}{3} \sqrt{3 x^2+5 x+2} x^{3/2}+20 \sqrt{3 x^2+5 x+2} \sqrt{x}-\frac{24 (3 x+2) \sqrt{x}}{\sqrt{3 x^2+5 x+2}}+\frac{24 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}} \]

[Out]

(-24*Sqrt[x]*(2 + 3*x))/Sqrt[2 + 5*x + 3*x^2] + (2*x^(5/2)*(74 + 95*x))/(3*Sqrt[2 + 5*x + 3*x^2]) + 20*Sqrt[x]
*Sqrt[2 + 5*x + 3*x^2] - (64*x^(3/2)*Sqrt[2 + 5*x + 3*x^2])/3 + (24*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*El
lipticE[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2] - (20*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[
ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2]

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Rubi [A]  time = 0.140085, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {818, 832, 839, 1189, 1100, 1136} \[ \frac{2 (95 x+74) x^{5/2}}{3 \sqrt{3 x^2+5 x+2}}-\frac{64}{3} \sqrt{3 x^2+5 x+2} x^{3/2}+20 \sqrt{3 x^2+5 x+2} \sqrt{x}-\frac{24 (3 x+2) \sqrt{x}}{\sqrt{3 x^2+5 x+2}}-\frac{20 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}+\frac{24 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-24*Sqrt[x]*(2 + 3*x))/Sqrt[2 + 5*x + 3*x^2] + (2*x^(5/2)*(74 + 95*x))/(3*Sqrt[2 + 5*x + 3*x^2]) + 20*Sqrt[x]
*Sqrt[2 + 5*x + 3*x^2] - (64*x^(3/2)*Sqrt[2 + 5*x + 3*x^2])/3 + (24*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*El
lipticE[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2] - (20*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[
ArcTan[Sqrt[x]], -1/2])/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 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 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^{7/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx &=\frac{2 x^{5/2} (74+95 x)}{3 \sqrt{2+5 x+3 x^2}}+\frac{2}{3} \int \frac{(-185-240 x) x^{3/2}}{\sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{2 x^{5/2} (74+95 x)}{3 \sqrt{2+5 x+3 x^2}}-\frac{64}{3} x^{3/2} \sqrt{2+5 x+3 x^2}+\frac{4}{45} \int \frac{\sqrt{x} \left (720+\frac{2025 x}{2}\right )}{\sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{2 x^{5/2} (74+95 x)}{3 \sqrt{2+5 x+3 x^2}}+20 \sqrt{x} \sqrt{2+5 x+3 x^2}-\frac{64}{3} x^{3/2} \sqrt{2+5 x+3 x^2}+\frac{8}{405} \int \frac{-\frac{2025}{2}-\frac{3645 x}{2}}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=\frac{2 x^{5/2} (74+95 x)}{3 \sqrt{2+5 x+3 x^2}}+20 \sqrt{x} \sqrt{2+5 x+3 x^2}-\frac{64}{3} x^{3/2} \sqrt{2+5 x+3 x^2}+\frac{16}{405} \operatorname{Subst}\left (\int \frac{-\frac{2025}{2}-\frac{3645 x^2}{2}}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 x^{5/2} (74+95 x)}{3 \sqrt{2+5 x+3 x^2}}+20 \sqrt{x} \sqrt{2+5 x+3 x^2}-\frac{64}{3} x^{3/2} \sqrt{2+5 x+3 x^2}-40 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )-72 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{24 \sqrt{x} (2+3 x)}{\sqrt{2+5 x+3 x^2}}+\frac{2 x^{5/2} (74+95 x)}{3 \sqrt{2+5 x+3 x^2}}+20 \sqrt{x} \sqrt{2+5 x+3 x^2}-\frac{64}{3} x^{3/2} \sqrt{2+5 x+3 x^2}+\frac{24 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{2+5 x+3 x^2}}-\frac{20 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{2+5 x+3 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.180552, size = 156, normalized size = 0.79 \[ \frac{12 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )-2 \left (x^4-4 x^3+22 x^2+120 x+72\right )-72 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )}{3 \sqrt{x} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(72 + 120*x + 22*x^2 - 4*x^3 + x^4) - (72*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*Ar
cSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (12*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqr
t[2/3]/Sqrt[x]], 3/2])/(3*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.042, size = 117, normalized size = 0.6 \begin{align*}{\frac{2}{3} \left ( 8\,\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 ) -6\,\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}^{4}+4\,{x}^{3}+86\,{x}^{2}+60\,x \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{3\,{x}^{2}+5\,x+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(8*(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))-6*(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^4+4*x^3+86*x^2+60*x)/x^(1/2)/(3*x^2+5
*x+2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (5 \, x^{4} - 2 \, x^{3}\right )} \sqrt{3 \, x^{2} + 5 \, x + 2} \sqrt{x}}{9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(5*x^4 - 2*x^3)*sqrt(3*x^2 + 5*x + 2)*sqrt(x)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4), 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**(7/2)/(3*x**2+5*x+2)**(3/2),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{7}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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