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

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

[Out]

-2/3*(30+37*x)*x^(1/2)/(3*x^2+5*x+2)^(3/2)-198*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)+2*(250+297*x)*x^(1/2)/(3*x^
2+5*x+2)^(1/2)+198*(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)-245*(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 = 0.11, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, 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 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 )^{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*}

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Mathematica [C]  time = 0.30, size = 165, normalized size = 0.92 \[ -\frac {47 i \sqrt {\frac {2}{x}+2} \sqrt {\frac {2}{x}+3} x F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )}{\sqrt {3 x^2+5 x+2}}-\frac {198 i \sqrt {\frac {2}{x}+2} \sqrt {\frac {2}{x}+3} x E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )}{\sqrt {3 x^2+5 x+2}}-\frac {2 \left (2205 x^3+5494 x^2+4470 x+1188\right )}{3 \sqrt {x} \left (3 x^2+5 x+2\right )^{3/2}} \]

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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fricas [F]  time = 0.96, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {3 \, x^{2} + 5 \, x + 2} {\left (5 \, x - 2\right )} \sqrt {x}}{27 \, x^{6} + 135 \, x^{5} + 279 \, x^{4} + 305 \, x^{3} + 186 \, x^{2} + 60 \, x + 8}, x\right ) \]

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

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)

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maple [A]  time = 0.09, size = 297, normalized size = 1.66 \[ \frac {\left (5346 x^{4}+13410 x^{3}-297 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{2} \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+156 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{2} \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+10990 x^{2}-495 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+260 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+2940 x -198 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+104 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right ) \sqrt {3 x^{2}+5 x +2}}{3 \left (x +1\right )^{2} \left (3 x +2\right )^{2} \sqrt {x}} \]

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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