3.1078 \(\int \frac{(2-5 x) \sqrt{x}}{(2+5 x+3 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=179 \[ -\frac{245 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} \text{EllipticF}\left (\tan ^{-1}\left (\sqrt{x}\right ),-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}-\frac{198 \sqrt{x} (3 x+2)}{\sqrt{3 x^2+5 x+2}}+\frac{2 \sqrt{x} (297 x+250)}{\sqrt{3 x^2+5 x+2}}-\frac{2 \sqrt{x} (37 x+30)}{3 \left (3 x^2+5 x+2\right )^{3/2}}+\frac{198 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}} \]

[Out]

(-2*Sqrt[x]*(30 + 37*x))/(3*(2 + 5*x + 3*x^2)^(3/2)) - (198*Sqrt[x]*(2 + 3*x))/Sqrt[2 + 5*x + 3*x^2] + (2*Sqrt
[x]*(250 + 297*x))/Sqrt[2 + 5*x + 3*x^2] + (198*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[
x]], -1/2])/Sqrt[2 + 5*x + 3*x^2] - (245*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 = 0.112693, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {820, 822, 839, 1189, 1100, 1136} \[ -\frac{198 \sqrt{x} (3 x+2)}{\sqrt{3 x^2+5 x+2}}+\frac{2 \sqrt{x} (297 x+250)}{\sqrt{3 x^2+5 x+2}}-\frac{2 \sqrt{x} (37 x+30)}{3 \left (3 x^2+5 x+2\right )^{3/2}}-\frac{245 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}}+\frac{198 \sqrt{2} (x+1) \sqrt{\frac{3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt{x}\right )|-\frac{1}{2}\right )}{\sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[x]*(30 + 37*x))/(3*(2 + 5*x + 3*x^2)^(3/2)) - (198*Sqrt[x]*(2 + 3*x))/Sqrt[2 + 5*x + 3*x^2] + (2*Sqrt
[x]*(250 + 297*x))/Sqrt[2 + 5*x + 3*x^2] + (198*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[
x]], -1/2])/Sqrt[2 + 5*x + 3*x^2] - (245*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1
/2])/Sqrt[2 + 5*x + 3*x^2]

Rule 820

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*(f*b - 2*a*g + (2*c*f - b*g)*x))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/
((p + 1)*(b^2 - 4*a*c)), Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g*(2*a*e*m + b*d*(2*p + 3)) - f*
(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)*(m + 2*p + 3)*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] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p]
|| IntegersQ[2*m, 2*p])

Rule 822

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2*a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x
)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*
c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2
*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d*m + b*e*m) - b*d*(3*c*d -
b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b,
c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -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) \sqrt{x}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx &=-\frac{2 \sqrt{x} (30+37 x)}{3 \left (2+5 x+3 x^2\right )^{3/2}}-\frac{2}{3} \int \frac{-15+\frac{111 x}{2}}{\sqrt{x} \left (2+5 x+3 x^2\right )^{3/2}} \, dx\\ &=-\frac{2 \sqrt{x} (30+37 x)}{3 \left (2+5 x+3 x^2\right )^{3/2}}+\frac{2 \sqrt{x} (250+297 x)}{\sqrt{2+5 x+3 x^2}}+\frac{2}{3} \int \frac{-\frac{735}{2}-\frac{891 x}{2}}{\sqrt{x} \sqrt{2+5 x+3 x^2}} \, dx\\ &=-\frac{2 \sqrt{x} (30+37 x)}{3 \left (2+5 x+3 x^2\right )^{3/2}}+\frac{2 \sqrt{x} (250+297 x)}{\sqrt{2+5 x+3 x^2}}+\frac{4}{3} \operatorname{Subst}\left (\int \frac{-\frac{735}{2}-\frac{891 x^2}{2}}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \sqrt{x} (30+37 x)}{3 \left (2+5 x+3 x^2\right )^{3/2}}+\frac{2 \sqrt{x} (250+297 x)}{\sqrt{2+5 x+3 x^2}}-490 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )-594 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{2+5 x^2+3 x^4}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{2 \sqrt{x} (30+37 x)}{3 \left (2+5 x+3 x^2\right )^{3/2}}-\frac{198 \sqrt{x} (2+3 x)}{\sqrt{2+5 x+3 x^2}}+\frac{2 \sqrt{x} (250+297 x)}{\sqrt{2+5 x+3 x^2}}+\frac{198 \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{245 \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 = 0.306502, size = 165, normalized size = 0.92 \[ -\frac{47 i \sqrt{\frac{2}{x}+2} \sqrt{\frac{2}{x}+3} x \text{EllipticF}\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right ),\frac{3}{2}\right )}{\sqrt{3 x^2+5 x+2}}-\frac{2 \left (2205 x^3+5494 x^2+4470 x+1188\right )}{3 \sqrt{x} \left (3 x^2+5 x+2\right )^{3/2}}-\frac{198 i \sqrt{\frac{2}{x}+2} \sqrt{\frac{2}{x}+3} x E\left (i \sinh ^{-1}\left (\frac{\sqrt{\frac{2}{3}}}{\sqrt{x}}\right )|\frac{3}{2}\right )}{\sqrt{3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(1188 + 4470*x + 5494*x^2 + 2205*x^3))/(3*Sqrt[x]*(2 + 5*x + 3*x^2)^(3/2)) - ((198*I)*Sqrt[2 + 2/x]*Sqrt[3
+ 2/x]*x*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/Sqrt[2 + 5*x + 3*x^2] - ((47*I)*Sqrt[2 + 2/x]*Sqrt[3 +
2/x]*x*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/Sqrt[2 + 5*x + 3*x^2]

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Maple [A]  time = 0.022, size = 297, normalized size = 1.7 \begin{align*}{\frac{1}{3\, \left ( 1+x \right ) ^{2} \left ( 2+3\,x \right ) ^{2}} \left ( 156\,\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}^{2}-297\,\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}^{2}+260\,\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-495\,\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+104\,\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 ) -198\,\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 ) +5346\,{x}^{4}+13410\,{x}^{3}+10990\,{x}^{2}+2940\,x \right ) \sqrt{3\,{x}^{2}+5\,x+2}{\frac{1}{\sqrt{x}}}} \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)^(5/2),x)

[Out]

1/3*(156*(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^2-297*(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^2+260*(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-495*(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+104*(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))-198*(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))+5346*x^4+13410*x^3+10990*x^2+2940*x)*(3*x^2+5*x+2)^(1/2)/x^(1/2)/(1+x)^2/(2+3*x)^2

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

[Out]

-integrate((5*x - 2)*sqrt(x)/(3*x^2 + 5*x + 2)^(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 )} \sqrt{x}}{27 \, x^{6} + 135 \, x^{5} + 279 \, x^{4} + 305 \, x^{3} + 186 \, x^{2} + 60 \, x + 8}, 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)^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(3*x^2 + 5*x + 2)*(5*x - 2)*sqrt(x)/(27*x^6 + 135*x^5 + 279*x^4 + 305*x^3 + 186*x^2 + 60*x + 8),
x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int - \frac{2 \sqrt{x}}{9 x^{4} \sqrt{3 x^{2} + 5 x + 2} + 30 x^{3} \sqrt{3 x^{2} + 5 x + 2} + 37 x^{2} \sqrt{3 x^{2} + 5 x + 2} + 20 x \sqrt{3 x^{2} + 5 x + 2} + 4 \sqrt{3 x^{2} + 5 x + 2}}\, dx - \int \frac{5 x^{\frac{3}{2}}}{9 x^{4} \sqrt{3 x^{2} + 5 x + 2} + 30 x^{3} \sqrt{3 x^{2} + 5 x + 2} + 37 x^{2} \sqrt{3 x^{2} + 5 x + 2} + 20 x \sqrt{3 x^{2} + 5 x + 2} + 4 \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)**(5/2),x)

[Out]

-Integral(-2*sqrt(x)/(9*x**4*sqrt(3*x**2 + 5*x + 2) + 30*x**3*sqrt(3*x**2 + 5*x + 2) + 37*x**2*sqrt(3*x**2 + 5
*x + 2) + 20*x*sqrt(3*x**2 + 5*x + 2) + 4*sqrt(3*x**2 + 5*x + 2)), x) - Integral(5*x**(3/2)/(9*x**4*sqrt(3*x**
2 + 5*x + 2) + 30*x**3*sqrt(3*x**2 + 5*x + 2) + 37*x**2*sqrt(3*x**2 + 5*x + 2) + 20*x*sqrt(3*x**2 + 5*x + 2) +
4*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}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac{5}{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)^(5/2),x, algorithm="giac")

[Out]

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