3.1061 \(\int \frac {2-5 x}{x^{3/2} \sqrt {2+5 x+3 x^2}} \, dx\)

Optimal. Leaf size=146 \[ \frac {2 \sqrt {x} (3 x+2)}{\sqrt {3 x^2+5 x+2}}-\frac {2 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {5 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}-\frac {2 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 146, 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 = {834, 839, 1189, 1100, 1136} \[ \frac {2 \sqrt {x} (3 x+2)}{\sqrt {3 x^2+5 x+2}}-\frac {2 \sqrt {3 x^2+5 x+2}}{\sqrt {x}}-\frac {5 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}-\frac {2 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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} \sqrt {2+5 x+3 x^2}} \, dx &=-\frac {2 \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-\int \frac {5-3 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=-\frac {2 \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-2 \operatorname {Subst}\left (\int \frac {5-3 x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {2 \sqrt {2+5 x+3 x^2}}{\sqrt {x}}+6 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )-10 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 \sqrt {x} (2+3 x)}{\sqrt {2+5 x+3 x^2}}-\frac {2 \sqrt {2+5 x+3 x^2}}{\sqrt {x}}-\frac {2 \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 {5 \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 = 0.14, size = 90, normalized size = 0.62 \[ \frac {i \sqrt {\frac {2}{x}+2} \sqrt {\frac {2}{x}+3} x \left (2 E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-7 F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )\right )}{\sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )} \sqrt {x}}{3 \, x^{4} + 5 \, x^{3} + 2 \, x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(3*x^2 + 5*x + 2)*(5*x - 2)*sqrt(x)/(3*x^4 + 5*x^3 + 2*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 108, normalized size = 0.74 \[ -\frac {18 x^{2}+30 x -\sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+8 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+12}{3 \sqrt {3 x^{2}+5 x +2}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2-5*x)/x^(3/2)/(3*x^2+5*x+2)^(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))-(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))+18*x^2+30*x+12)/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 \[ -\int \frac {5 \, x - 2}{\sqrt {3 \, x^{2} + 5 \, x + 2} x^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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