3.1040 \(\int \frac{(2-5 x) \sqrt{2+5 x+3 x^2}}{x^{5/2}} \, dx\)

Optimal. Leaf size=157 \[ -\frac{21 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}-\frac{4 \sqrt{3 x^2+5 x+2} (1-5 x)}{3 x^{3/2}}-\frac{50 \sqrt{x} (3 x+2)}{3 \sqrt{3 x^2+5 x+2}}+\frac{50 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{3 \sqrt{3 x^2+5 x+2}} \]

[Out]

(-50*Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (4*(1 - 5*x)*Sqrt[2 + 5*x + 3*x^2])/(3*x^(3/2)) + (50*Sqrt
[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2]) - (21*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 = 0.09988, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {810, 839, 1189, 1100, 1136} \[ -\frac{4 \sqrt{3 x^2+5 x+2} (1-5 x)}{3 x^{3/2}}-\frac{50 \sqrt{x} (3 x+2)}{3 \sqrt{3 x^2+5 x+2}}-\frac{21 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}+\frac{50 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{3 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-50*Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (4*(1 - 5*x)*Sqrt[2 + 5*x + 3*x^2])/(3*x^(3/2)) + (50*Sqrt
[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2]) - (21*Sqrt[2]*
(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/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 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) \sqrt{2+5 x+3 x^2}}{x^{5/2}} \, dx &=-\frac{4 (1-5 x) \sqrt{2+5 x+3 x^2}}{3 x^{3/2}}-\frac{1}{3} \int \frac{63+75 x}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{4 (1-5 x) \sqrt{2+5 x+3 x^2}}{3 x^{3/2}}-\frac{2}{3} \operatorname{Subst}\left (\int \frac{63+75 x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{4 (1-5 x) \sqrt{2+5 x+3 x^2}}{3 x^{3/2}}-42 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )-50 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{50 \sqrt{x} (2+3 x)}{3 \sqrt{2+5 x+3 x^2}}-\frac{4 (1-5 x) \sqrt{2+5 x+3 x^2}}{3 x^{3/2}}+\frac{50 \sqrt{2} (1+x) \sqrt{\frac{2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{3 \sqrt{2+5 x+3 x^2}}-\frac{21 \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 = 0.210754, size = 153, normalized size = 0.97 \[ \frac{-13 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{5/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )-2 \left (45 x^3+81 x^2+40 x+4\right )-50 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{5/2} E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )}{3 x^{3/2} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(4 + 40*x + 81*x^2 + 45*x^3) - (50*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(5/2)*EllipticE[I*ArcSinh[S
qrt[2/3]/Sqrt[x]], 3/2] - (13*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(5/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/
Sqrt[x]], 3/2])/(3*x^(3/2)*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.029, size = 115, normalized size = 0.7 \begin{align*}{\frac{1}{9} \left ( 12\,\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-25\,\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+180\,{x}^{3}+264\,{x}^{2}+60\,x-24 \right ){x}^{-{\frac{3}{2}}}{\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)*(3*x^2+5*x+2)^(1/2)/x^(5/2),x)

[Out]

1/9*(12*(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-25*(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+180*x^3+264*x^2+60*x-24)/(3*x^2+5
*x+2)^(1/2)/x^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x^2+5*x+2)^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x^2+5*x+2)^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{2 \sqrt{3 x^{2} + 5 x + 2}}{x^{\frac{5}{2}}}\, dx - \int \frac{5 \sqrt{3 x^{2} + 5 x + 2}}{x^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x**2+5*x+2)**(1/2)/x**(5/2),x)

[Out]

-Integral(-2*sqrt(3*x**2 + 5*x + 2)/x**(5/2), x) - Integral(5*sqrt(3*x**2 + 5*x + 2)/x**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x^2+5*x+2)^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

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