3.1068 \(\int \frac {2-5 x}{\sqrt {x} (2+5 x+3 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=151 \[ -\frac {30 \sqrt {x} (3 x+2)}{\sqrt {3 x^2+5 x+2}}+\frac {2 \sqrt {x} (45 x+38)}{\sqrt {3 x^2+5 x+2}}-\frac {37 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+\frac {30 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}} \]

[Out]

-30*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+2*(38+45*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+30*(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)-37*(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 = 0.09, antiderivative size = 151, 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 = {822, 839, 1189, 1100, 1136} \[ -\frac {30 \sqrt {x} (3 x+2)}{\sqrt {3 x^2+5 x+2}}+\frac {2 \sqrt {x} (45 x+38)}{\sqrt {3 x^2+5 x+2}}-\frac {37 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+\frac {30 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*Sqrt[x]*(2 + 3*x))/Sqrt[2 + 5*x + 3*x^2] + (2*Sqrt[x]*(38 + 45*x))/Sqrt[2 + 5*x + 3*x^2] + (30*Sqrt[2]*(1
+ x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2] - (37*Sqrt[2]*(1 + x)*Sq
rt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/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 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}{\sqrt {x} \left (2+5 x+3 x^2\right )^{3/2}} \, dx &=\frac {2 \sqrt {x} (38+45 x)}{\sqrt {2+5 x+3 x^2}}-\int \frac {37+45 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 \sqrt {x} (38+45 x)}{\sqrt {2+5 x+3 x^2}}-2 \operatorname {Subst}\left (\int \frac {37+45 x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 \sqrt {x} (38+45 x)}{\sqrt {2+5 x+3 x^2}}-74 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )-90 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {30 \sqrt {x} (2+3 x)}{\sqrt {2+5 x+3 x^2}}+\frac {2 \sqrt {x} (38+45 x)}{\sqrt {2+5 x+3 x^2}}+\frac {30 \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 {37 \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*}

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Mathematica [C]  time = 0.16, size = 137, normalized size = 0.91 \[ \frac {-7 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-30 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{3/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-74 x-60}{\sqrt {x} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-60 - 74*x - (30*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/
2] - (7*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(Sqrt[
x]*Sqrt[2 + 5*x + 3*x^2])

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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )} \sqrt {x}}{9 \, x^{5} + 30 \, x^{4} + 37 \, x^{3} + 20 \, x^{2} + 4 \, x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.09, size = 107, normalized size = 0.71 \[ \frac {270 x^{2}+228 x -15 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+8 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{2}+5 x +2}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2-5*x)/(3*x^2+5*x+2)^(3/2)/x^(1/2),x)

[Out]

1/3*(8*(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))-15*(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))+270*x^2+228*x)/(3*x^2+5*x+2)^(1/2)/x^(
1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x - 2)/(x^(1/2)*(5*x + 3*x^2 + 2)^(3/2)),x)

[Out]

-int((5*x - 2)/(x^(1/2)*(5*x + 3*x^2 + 2)^(3/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)/(3*x**2+5*x+2)**(3/2)/x**(1/2),x)

[Out]

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

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