3.1059 \(\int \frac{(2-5 x) \sqrt{x}}{\sqrt{2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=154 \[ \frac{10 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{9 \sqrt{3 x^2+5 x+2}}+\frac{136 \sqrt{x} (3 x+2)}{27 \sqrt{3 x^2+5 x+2}}-\frac{10}{9} \sqrt{x} \sqrt{3 x^2+5 x+2}-\frac{136 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{27 \sqrt{3 x^2+5 x+2}} \]

[Out]

(136*Sqrt[x]*(2 + 3*x))/(27*Sqrt[2 + 5*x + 3*x^2]) - (10*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])/9 - (136*Sqrt[2]*(1 +
x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(27*Sqrt[2 + 5*x + 3*x^2]) + (10*Sqrt[2]*(1 + x)*
Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(9*Sqrt[2 + 5*x + 3*x^2])

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Rubi [A]  time = 0.0986726, antiderivative size = 154, 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 = {832, 839, 1189, 1100, 1136} \[ \frac{136 \sqrt{x} (3 x+2)}{27 \sqrt{3 x^2+5 x+2}}-\frac{10}{9} \sqrt{x} \sqrt{3 x^2+5 x+2}+\frac{10 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{9 \sqrt{3 x^2+5 x+2}}-\frac{136 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{27 \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(136*Sqrt[x]*(2 + 3*x))/(27*Sqrt[2 + 5*x + 3*x^2]) - (10*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])/9 - (136*Sqrt[2]*(1 +
x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(27*Sqrt[2 + 5*x + 3*x^2]) + (10*Sqrt[2]*(1 + x)*
Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(9*Sqrt[2 + 5*x + 3*x^2])

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) \sqrt{x}}{\sqrt{2+5 x+3 x^2}} \, dx &=-\frac{10}{9} \sqrt{x} \sqrt{2+5 x+3 x^2}+\frac{2}{9} \int \frac{5+34 x}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{10}{9} \sqrt{x} \sqrt{2+5 x+3 x^2}+\frac{4}{9} \operatorname{Subst}\left (\int \frac{5+34 x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{10}{9} \sqrt{x} \sqrt{2+5 x+3 x^2}+\frac{20}{9} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )+\frac{136}{9} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=\frac{136 \sqrt{x} (2+3 x)}{27 \sqrt{2+5 x+3 x^2}}-\frac{10}{9} \sqrt{x} \sqrt{2+5 x+3 x^2}-\frac{136 \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{10 \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 = 0.167943, size = 150, normalized size = 0.97 \[ \frac{-106 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )-90 x^3+258 x^2+136 i \sqrt{2} \sqrt{\frac{1}{x}+1} \sqrt{\frac{2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )+620 x+272}{27 \sqrt{x} \sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(272 + 620*x + 258*x^2 - 90*x^3 + (136*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[S
qrt[2/3]/Sqrt[x]], 3/2] - (106*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]
/Sqrt[x]], 3/2])/(27*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]  time = 0.014, size = 112, normalized size = 0.7 \begin{align*} -{\frac{2}{81} \left ( 87\,\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 ) -34\,\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 ) +135\,{x}^{3}+225\,{x}^{2}+90\,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^(1/2)/(3*x^2+5*x+2)^(1/2),x)

[Out]

-2/81/x^(1/2)/(3*x^2+5*x+2)^(1/2)*(87*(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))-34*(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))+135*x^3
+225*x^2+90*x)

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

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