3.1080 \(\int \frac {2-5 x}{x^{3/2} (2+5 x+3 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=208 \[ -\frac {838 \sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}+\frac {838 \sqrt {3 x^2+5 x+2}}{3 \sqrt {x}}-\frac {2085 x+1717}{3 \sqrt {x} \sqrt {3 x^2+5 x+2}}+\frac {2 (45 x+38)}{3 \sqrt {x} \left (3 x^2+5 x+2\right )^{3/2}}-\frac {695 (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+\frac {838 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}} \]

[Out]

2/3*(38+45*x)/(3*x^2+5*x+2)^(3/2)/x^(1/2)+1/3*(-1717-2085*x)/x^(1/2)/(3*x^2+5*x+2)^(1/2)-838/3*(2+3*x)*x^(1/2)
/(3*x^2+5*x+2)^(1/2)-695/2*(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticF(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))*2^(1/2)*((
2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)+838/3*(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticE(x^(1/2)/(1+x)^(1/2),1/2*I*
2^(1/2))*2^(1/2)*((2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)+838/3*(3*x^2+5*x+2)^(1/2)/x^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {822, 834, 839, 1189, 1100, 1136} \[ -\frac {838 \sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}+\frac {838 \sqrt {3 x^2+5 x+2}}{3 \sqrt {x}}-\frac {2085 x+1717}{3 \sqrt {x} \sqrt {3 x^2+5 x+2}}+\frac {2 (45 x+38)}{3 \sqrt {x} \left (3 x^2+5 x+2\right )^{3/2}}-\frac {695 (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+\frac {838 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(38 + 45*x))/(3*Sqrt[x]*(2 + 5*x + 3*x^2)^(3/2)) - (838*Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (171
7 + 2085*x)/(3*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2]) + (838*Sqrt[2 + 5*x + 3*x^2])/(3*Sqrt[x]) + (838*Sqrt[2]*(1 + x)
*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2]) - (695*(1 + x)*Sqrt[(2 +
3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2]*Sqrt[2 + 5*x + 3*x^2])

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 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 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]

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]

Rubi steps

\begin {align*} \int \frac {2-5 x}{x^{3/2} \left (2+5 x+3 x^2\right )^{5/2}} \, dx &=\frac {2 (38+45 x)}{3 \sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}}-\frac {1}{3} \int \frac {-41-225 x}{x^{3/2} \left (2+5 x+3 x^2\right )^{3/2}} \, dx\\ &=\frac {2 (38+45 x)}{3 \sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}}-\frac {1717+2085 x}{3 \sqrt {x} \sqrt {2+5 x+3 x^2}}+\frac {1}{3} \int \frac {-838-\frac {2085 x}{2}}{x^{3/2} \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 (38+45 x)}{3 \sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}}-\frac {1717+2085 x}{3 \sqrt {x} \sqrt {2+5 x+3 x^2}}+\frac {838 \sqrt {2+5 x+3 x^2}}{3 \sqrt {x}}-\frac {1}{3} \int \frac {\frac {2085}{2}+1257 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 (38+45 x)}{3 \sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}}-\frac {1717+2085 x}{3 \sqrt {x} \sqrt {2+5 x+3 x^2}}+\frac {838 \sqrt {2+5 x+3 x^2}}{3 \sqrt {x}}-\frac {2}{3} \operatorname {Subst}\left (\int \frac {\frac {2085}{2}+1257 x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 (38+45 x)}{3 \sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}}-\frac {1717+2085 x}{3 \sqrt {x} \sqrt {2+5 x+3 x^2}}+\frac {838 \sqrt {2+5 x+3 x^2}}{3 \sqrt {x}}-695 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )-838 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 (38+45 x)}{3 \sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}}-\frac {838 \sqrt {x} (2+3 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {1717+2085 x}{3 \sqrt {x} \sqrt {2+5 x+3 x^2}}+\frac {838 \sqrt {2+5 x+3 x^2}}{3 \sqrt {x}}+\frac {838 \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 {695 (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {2} \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 167, normalized size = 0.80 \[ \frac {-409 i \sqrt {\frac {2}{x}+2} \sqrt {\frac {2}{x}+3} \left (3 x^2+5 x+2\right ) x^{3/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-1676 i \sqrt {\frac {2}{x}+2} \sqrt {\frac {2}{x}+3} \left (3 x^2+5 x+2\right ) x^{3/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-2 \left (6255 x^3+15576 x^2+12665 x+3358\right )}{6 \sqrt {x} \left (3 x^2+5 x+2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3358 + 12665*x + 15576*x^2 + 6255*x^3) - (1676*I)*Sqrt[2 + 2/x]*Sqrt[3 + 2/x]*x^(3/2)*(2 + 5*x + 3*x^2)*E
llipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (409*I)*Sqrt[2 + 2/x]*Sqrt[3 + 2/x]*x^(3/2)*(2 + 5*x + 3*x^2)*El
lipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(6*Sqrt[x]*(2 + 5*x + 3*x^2)^(3/2))

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )} \sqrt {x}}{27 \, x^{8} + 135 \, x^{7} + 279 \, x^{6} + 305 \, x^{5} + 186 \, x^{4} + 60 \, x^{3} + 8 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)/x^(3/2)/(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^8 + 135*x^7 + 279*x^6 + 305*x^5 + 186*x^4 + 60*x^3 + 8
*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {5 \, x - 2}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}} x^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 298, normalized size = 1.43 \[ \frac {45252 x^{4}+113310 x^{3}-2514 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{2} \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+1287 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{2} \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+92580 x^{2}-4190 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+2145 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+24570 x -1676 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+858 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-36}{18 \left (x +1\right ) \left (3 x +2\right ) \sqrt {3 x^{2}+5 x +2}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(1287*(6*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))*x^2-2514*(6*x
+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*x^2+2145*(6*x+4)^(1/2)*(3*x+
3)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))*x-4190*(6*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*
(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*x+858*(6*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*Ellipti
cF(1/2*(6*x+4)^(1/2),I*2^(1/2))-1676*(6*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2
),I*2^(1/2))+45252*x^4+113310*x^3+92580*x^2+24570*x-36)/(x+1)/(3*x+2)/x^(1/2)/(3*x^2+5*x+2)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {5 \, x - 2}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}} x^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int -\frac {5\,x-2}{x^{3/2}\,{\left (3\,x^2+5\,x+2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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