3.1061 \(\int \frac{2-5 x}{x^{3/2} \sqrt{2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=146 \[ -\frac{5 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}+\frac{2 \sqrt{x} (3 x+2)}{\sqrt{3 x^2+5 x+2}}-\frac{2 \sqrt{3 x^2+5 x+2}}{\sqrt{x}}-\frac{2 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}} \]

[Out]

(2*Sqrt[x]*(2 + 3*x))/Sqrt[2 + 5*x + 3*x^2] - (2*Sqrt[2 + 5*x + 3*x^2])/Sqrt[x] - (2*Sqrt[2]*(1 + x)*Sqrt[(2 +
3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2] - (5*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 = 0.0915021, antiderivative size = 146, 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 = {834, 839, 1189, 1100, 1136} \[ \frac{2 \sqrt{x} (3 x+2)}{\sqrt{3 x^2+5 x+2}}-\frac{2 \sqrt{3 x^2+5 x+2}}{\sqrt{x}}-\frac{5 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}-\frac{2 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*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] && LtQ[m, -1] && (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^{3/2} \sqrt{2+5 x+3 x^2}} \, dx &=-\frac{2 \sqrt{2+5 x+3 x^2}}{\sqrt{x}}-\int \frac{5-3 x}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{2 \sqrt{2+5 x+3 x^2}}{\sqrt{x}}-2 \operatorname{Subst}\left (\int \frac{5-3 x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \sqrt{2+5 x+3 x^2}}{\sqrt{x}}+6 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )-10 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 \sqrt{x} (2+3 x)}{\sqrt{2+5 x+3 x^2}}-\frac{2 \sqrt{2+5 x+3 x^2}}{\sqrt{x}}-\frac{2 \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{5 \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 = 0.142375, size = 90, normalized size = 0.62 \[ \frac{i \sqrt{\frac{2}{x}+2} \sqrt{\frac{2}{x}+3} x \left (2 E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )-7 \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )\right )}{\sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*Sqrt[2 + 2/x]*Sqrt[3 + 2/x]*x*(2*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - 7*EllipticF[I*ArcSinh[Sqrt[
2/3]/Sqrt[x]], 3/2]))/Sqrt[2 + 5*x + 3*x^2]

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Maple [A]  time = 0.016, size = 108, normalized size = 0.7 \begin{align*} -{\frac{1}{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 ) -\sqrt{6\,x+4}\sqrt{3+3\,x}\sqrt{6}\sqrt{-x}{\it EllipticE} \left ({\frac{1}{2}\sqrt{6\,x+4}},i\sqrt{2} \right ) +18\,{x}^{2}+30\,x+12 \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^(3/2)/(3*x^2+5*x+2)^(1/2),x)

[Out]

-1/3*(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*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))+18*x^2+30*x+12)/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{5 \, x - 2}{\sqrt{3 \, x^{2} + 5 \, x + 2} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((5*x - 2)/(sqrt(3*x^2 + 5*x + 2)*x^(3/2)), 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 )} \sqrt{x}}{3 \, x^{4} + 5 \, x^{3} + 2 \, x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)/x^(3/2)/(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^4 + 5*x^3 + 2*x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{2}{x^{\frac{3}{2}} \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**(3/2)/(3*x**2+5*x+2)**(1/2),x)

[Out]

-Integral(-2/(x**(3/2)*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} x^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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