3.1073 \(\int \frac {(2-5 x) x^{11/2}}{(2+5 x+3 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=233 \[ -\frac {16040}{243} \sqrt {3 x^2+5 x+2} \sqrt {x}+\frac {33608 (3 x+2) \sqrt {x}}{729 \sqrt {3 x^2+5 x+2}}+\frac {16040 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{243 \sqrt {3 x^2+5 x+2}}-\frac {33608 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{729 \sqrt {3 x^2+5 x+2}}+\frac {2 (95 x+74) x^{9/2}}{9 \left (3 x^2+5 x+2\right )^{3/2}}-\frac {8 (905 x+773) x^{5/2}}{27 \sqrt {3 x^2+5 x+2}}+\frac {2348}{27} \sqrt {3 x^2+5 x+2} x^{3/2} \]

[Out]

2/9*x^(9/2)*(74+95*x)/(3*x^2+5*x+2)^(3/2)-8/27*x^(5/2)*(773+905*x)/(3*x^2+5*x+2)^(1/2)+33608/729*(2+3*x)*x^(1/
2)/(3*x^2+5*x+2)^(1/2)-33608/729*(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)+16040/243*(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)+2348/27*x^(3/2)*(3*x^2+5*x+2)^(1/2)-16040
/243*x^(1/2)*(3*x^2+5*x+2)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {818, 832, 839, 1189, 1100, 1136} \[ \frac {2 (95 x+74) x^{9/2}}{9 \left (3 x^2+5 x+2\right )^{3/2}}-\frac {8 (905 x+773) x^{5/2}}{27 \sqrt {3 x^2+5 x+2}}+\frac {2348}{27} \sqrt {3 x^2+5 x+2} x^{3/2}-\frac {16040}{243} \sqrt {3 x^2+5 x+2} \sqrt {x}+\frac {33608 (3 x+2) \sqrt {x}}{729 \sqrt {3 x^2+5 x+2}}+\frac {16040 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{243 \sqrt {3 x^2+5 x+2}}-\frac {33608 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{729 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

Int[((2 - 5*x)*x^(11/2))/(2 + 5*x + 3*x^2)^(5/2),x]

[Out]

(2*x^(9/2)*(74 + 95*x))/(9*(2 + 5*x + 3*x^2)^(3/2)) + (33608*Sqrt[x]*(2 + 3*x))/(729*Sqrt[2 + 5*x + 3*x^2]) -
(8*x^(5/2)*(773 + 905*x))/(27*Sqrt[2 + 5*x + 3*x^2]) - (16040*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])/243 + (2348*x^(3/
2)*Sqrt[2 + 5*x + 3*x^2])/27 - (33608*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2]
)/(729*Sqrt[2 + 5*x + 3*x^2]) + (16040*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2
])/(243*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 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*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] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 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^{11/2}}{\left (2+5 x+3 x^2\right )^{5/2}} \, dx &=\frac {2 x^{9/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {2}{9} \int \frac {(-333-245 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx\\ &=\frac {2 x^{9/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}-\frac {8 x^{5/2} (773+905 x)}{27 \sqrt {2+5 x+3 x^2}}+\frac {4}{27} \int \frac {x^{3/2} \left (3865+\frac {8805 x}{2}\right )}{\sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 x^{9/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}-\frac {8 x^{5/2} (773+905 x)}{27 \sqrt {2+5 x+3 x^2}}+\frac {2348}{27} x^{3/2} \sqrt {2+5 x+3 x^2}+\frac {8}{405} \int \frac {\left (-\frac {26415}{2}-\frac {30075 x}{2}\right ) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 x^{9/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}-\frac {8 x^{5/2} (773+905 x)}{27 \sqrt {2+5 x+3 x^2}}-\frac {16040}{243} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {2348}{27} x^{3/2} \sqrt {2+5 x+3 x^2}+\frac {16 \int \frac {\frac {30075}{2}+\frac {63015 x}{4}}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx}{3645}\\ &=\frac {2 x^{9/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}-\frac {8 x^{5/2} (773+905 x)}{27 \sqrt {2+5 x+3 x^2}}-\frac {16040}{243} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {2348}{27} x^{3/2} \sqrt {2+5 x+3 x^2}+\frac {32 \operatorname {Subst}\left (\int \frac {\frac {30075}{2}+\frac {63015 x^2}{4}}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )}{3645}\\ &=\frac {2 x^{9/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}-\frac {8 x^{5/2} (773+905 x)}{27 \sqrt {2+5 x+3 x^2}}-\frac {16040}{243} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {2348}{27} x^{3/2} \sqrt {2+5 x+3 x^2}+\frac {32080}{243} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )+\frac {33608}{243} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 x^{9/2} (74+95 x)}{9 \left (2+5 x+3 x^2\right )^{3/2}}+\frac {33608 \sqrt {x} (2+3 x)}{729 \sqrt {2+5 x+3 x^2}}-\frac {8 x^{5/2} (773+905 x)}{27 \sqrt {2+5 x+3 x^2}}-\frac {16040}{243} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {2348}{27} x^{3/2} \sqrt {2+5 x+3 x^2}-\frac {33608 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{729 \sqrt {2+5 x+3 x^2}}+\frac {16040 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{243 \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 179, normalized size = 0.77 \[ \frac {-486 x^6+2484 x^5-21276 x^4+161784 x^3+534680 x^2+14512 i \sqrt {\frac {2}{x}+2} \sqrt {\frac {2}{x}+3} \left (3 x^2+5 x+2\right ) x^{3/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+33608 i \sqrt {\frac {2}{x}+2} \sqrt {\frac {2}{x}+3} \left (3 x^2+5 x+2\right ) x^{3/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+479680 x+134432}{729 \sqrt {x} \left (3 x^2+5 x+2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((2 - 5*x)*x^(11/2))/(2 + 5*x + 3*x^2)^(5/2),x]

[Out]

(134432 + 479680*x + 534680*x^2 + 161784*x^3 - 21276*x^4 + 2484*x^5 - 486*x^6 + (33608*I)*Sqrt[2 + 2/x]*Sqrt[3
+ 2/x]*x^(3/2)*(2 + 5*x + 3*x^2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (14512*I)*Sqrt[2 + 2/x]*Sqrt[
3 + 2/x]*x^(3/2)*(2 + 5*x + 3*x^2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(729*Sqrt[x]*(2 + 5*x + 3*x^2
)^(3/2))

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fricas [F]  time = 0.91, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (5 \, x^{6} - 2 \, x^{5}\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}}{27 \, x^{6} + 135 \, x^{5} + 279 \, x^{4} + 305 \, x^{3} + 186 \, x^{2} + 60 \, x + 8}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*x^(11/2)/(3*x^2+5*x+2)^(5/2),x, algorithm="fricas")

[Out]

integral(-(5*x^6 - 2*x^5)*sqrt(3*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 \[ \int -\frac {{\left (5 \, x - 2\right )} x^{\frac {11}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*x^(11/2)/(3*x^2+5*x+2)^(5/2),x, algorithm="giac")

[Out]

integrate(-(5*x - 2)*x^(11/2)/(3*x^2 + 5*x + 2)^(5/2), x)

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maple [A]  time = 0.09, size = 307, normalized size = 1.32 \[ -\frac {2 \left (729 x^{6}-3726 x^{5}+485622 x^{4}+1269684 x^{3}-25206 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{2} \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+3438 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{2} \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+1063224 x^{2}-42010 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+5730 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+288720 x -16804 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+2292 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right ) \sqrt {3 x^{2}+5 x +2}}{2187 \left (x +1\right )^{2} \left (3 x +2\right )^{2} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2-5*x)*x^(11/2)/(3*x^2+5*x+2)^(5/2),x)

[Out]

-2/2187*(3438*(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^2-25206*
(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+5730*(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-42010*(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+729*x^6+2292*(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))-16804*(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))-3726*x^5+485622*x^4+1269684*x^3+1063224*x^2+288720*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 \[ -\int \frac {{\left (5 \, x - 2\right )} x^{\frac {11}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*x^(11/2)/(3*x^2+5*x+2)^(5/2),x, algorithm="maxima")

[Out]

-integrate((5*x - 2)*x^(11/2)/(3*x^2 + 5*x + 2)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ -\int \frac {x^{11/2}\,\left (5\,x-2\right )}{{\left (3\,x^2+5\,x+2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^(11/2)*(5*x - 2))/(5*x + 3*x^2 + 2)^(5/2),x)

[Out]

-int((x^(11/2)*(5*x - 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**(11/2)/(3*x**2+5*x+2)**(5/2),x)

[Out]

Timed out

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