3.1060 \(\int \frac{2-5 x}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=129 \[ \frac{2 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}-\frac{10 \sqrt{x} (3 x+2)}{3 \sqrt{3 x^2+5 x+2}}+\frac{10 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{3 \sqrt{3 x^2+5 x+2}} \]

[Out]

(-10*Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) + (10*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcT
an[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2]) + (2*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sq
rt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2]

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Rubi [A]  time = 0.0797963, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {839, 1189, 1100, 1136} \[ -\frac{10 \sqrt{x} (3 x+2)}{3 \sqrt{3 x^2+5 x+2}}+\frac{2 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}+\frac{10 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{3 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) + (10*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcT
an[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2]) + (2*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sq
rt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2]

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{x} \sqrt{2+5 x+3 x^2}} \, dx &=2 \operatorname{Subst}\left (\int \frac{2-5 x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=4 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )-10 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{10 \sqrt{x} (2+3 x)}{3 \sqrt{2+5 x+3 x^2}}+\frac{10 \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{2 \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 = 0.168598, size = 150, normalized size = 1.16 \[ -\frac{2 x^{3/2} \left (-\frac{8 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )}{\sqrt{x}}+5 \left (\frac{2}{x^2}+\frac{5}{x}+3\right )+\frac{5 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )}{\sqrt{x}}\right )}{3 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x^(3/2)*(5*(3 + 2/x^2 + 5/x) + ((5*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*EllipticE[I*ArcSinh[Sqrt[2/3]
/Sqrt[x]], 3/2])/Sqrt[x] - ((8*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]
], 3/2])/Sqrt[x]))/(3*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.018, size = 77, normalized size = 0.6 \begin{align*}{\frac{\sqrt{6}}{9}\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{-x} \left ( 21\,{\it EllipticF} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ) -5\,{\it EllipticE} \left ( 1/2\,\sqrt{6\,x+4},i\sqrt{2} \right ) \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^(1/2)/(3*x^2+5*x+2)^(1/2),x)

[Out]

1/9/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*(21*EllipticF(1/2*(6*x+4)^(1/2)
,I*2^(1/2))-5*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((5*x - 2)/(sqrt(3*x^2 + 5*x + 2)*sqrt(x)), 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 )} \sqrt{x}}{3 \, x^{3} + 5 \, x^{2} + 2 \, x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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