3.1066 \(\int \frac {(2-5 x) x^{3/2}}{(2+5 x+3 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=159 \[ -\frac {200 \sqrt {x} (3 x+2)}{9 \sqrt {3 x^2+5 x+2}}+\frac {2 \sqrt {x} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {74 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}+\frac {200 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{9 \sqrt {3 x^2+5 x+2}} \]

[Out]

-200/9*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+2/3*(74+95*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+200/9*(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)-74/3*(
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)

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Rubi [A]  time = 0.10, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {818, 839, 1189, 1100, 1136} \[ -\frac {200 \sqrt {x} (3 x+2)}{9 \sqrt {3 x^2+5 x+2}}+\frac {2 \sqrt {x} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {74 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}+\frac {200 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{9 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-200*Sqrt[x]*(2 + 3*x))/(9*Sqrt[2 + 5*x + 3*x^2]) + (2*Sqrt[x]*(74 + 95*x))/(3*Sqrt[2 + 5*x + 3*x^2]) + (200*
Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(9*Sqrt[2 + 5*x + 3*x^2]) - (74*Sqrt
[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2])

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 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 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 )^{3/2}} \, dx &=\frac {2 \sqrt {x} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {2}{3} \int \frac {-37-50 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 \sqrt {x} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {4}{3} \operatorname {Subst}\left (\int \frac {-37-50 x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 \sqrt {x} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {148}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )-\frac {200}{3} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {200 \sqrt {x} (2+3 x)}{9 \sqrt {2+5 x+3 x^2}}+\frac {2 \sqrt {x} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {200 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{9 \sqrt {2+5 x+3 x^2}}-\frac {74 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 145, normalized size = 0.91 \[ \frac {-22 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-200 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-30 x^2-556 x-400}{9 \sqrt {x} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-400 - 556*x - 30*x^2 - (200*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/
Sqrt[x]], 3/2] - (22*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]],
3/2])/(9*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (5 \, x^{2} - 2 \, x\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}}{9 \, x^{4} + 30 \, x^{3} + 37 \, x^{2} + 20 \, x + 4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(5*x^2 - 2*x)*sqrt(3*x^2 + 5*x + 2)*sqrt(x)/(9*x^4 + 30*x^3 + 37*x^2 + 20*x + 4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 107, normalized size = 0.67 \[ \frac {\frac {190 x^{2}}{3}+\frac {148 x}{3}-\frac {100 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{27}+\frac {26 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{9}}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/27*(39*(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))-50*(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))+855*x^2+666*x)/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 \[ -\int \frac {{\left (5 \, x - 2\right )} x^{\frac {3}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {x^{3/2}\,\left (5\,x-2\right )}{{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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