3.11.65 \(\int \frac {(2-5 x) x^{5/2}}{(2+5 x+3 x^2)^{3/2}} \, dx\) [1065]

Optimal. Leaf size=182 \[ \frac {1804 \sqrt {x} (2+3 x)}{81 \sqrt {2+5 x+3 x^2}}+\frac {2 x^{3/2} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {580}{27} \sqrt {x} \sqrt {2+5 x+3 x^2}-\frac {1804 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {2+5 x+3 x^2}}+\frac {580 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{27 \sqrt {2+5 x+3 x^2}} \]

[Out]

2/3*x^(3/2)*(74+95*x)/(3*x^2+5*x+2)^(1/2)+1804/81*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)-1804/81*(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)+580
/27*(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)-580/27*x^(1/2)*(3*x^2+5*x+2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 182, 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 = {832, 846, 853, 1203, 1114, 1150} \begin {gather*} \frac {580 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\text {ArcTan}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{27 \sqrt {3 x^2+5 x+2}}-\frac {1804 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\text {ArcTan}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {3 x^2+5 x+2}}-\frac {580}{27} \sqrt {3 x^2+5 x+2} \sqrt {x}+\frac {1804 (3 x+2) \sqrt {x}}{81 \sqrt {3 x^2+5 x+2}}+\frac {2 (95 x+74) x^{3/2}}{3 \sqrt {3 x^2+5 x+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1804*Sqrt[x]*(2 + 3*x))/(81*Sqrt[2 + 5*x + 3*x^2]) + (2*x^(3/2)*(74 + 95*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (580
*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])/27 - (1804*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]],
-1/2])/(81*Sqrt[2 + 5*x + 3*x^2]) + (580*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1
/2])/(27*Sqrt[2 + 5*x + 3*x^2])

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(-(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] &&
NeQ[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 846

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 853

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 1114

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)]/(2*a*Rt[(b - q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*Elli
pticF[ArcTan[Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^
2 - 4*a*c, 0]

Rule 1150

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)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan[Rt[(b - q)/(2*a), 2]*x], -2*(
q/(b - q))], x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1203

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^{5/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx &=\frac {2 x^{3/2} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}+\frac {2}{3} \int \frac {(-111-145 x) \sqrt {x}}{\sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 x^{3/2} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {580}{27} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {4}{27} \int \frac {145+\frac {451 x}{2}}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {2 x^{3/2} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {580}{27} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {8}{27} \text {Subst}\left (\int \frac {145+\frac {451 x^2}{2}}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 x^{3/2} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {580}{27} \sqrt {x} \sqrt {2+5 x+3 x^2}+\frac {1160}{27} \text {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )+\frac {1804}{27} \text {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {1804 \sqrt {x} (2+3 x)}{81 \sqrt {2+5 x+3 x^2}}+\frac {2 x^{3/2} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {580}{27} \sqrt {x} \sqrt {2+5 x+3 x^2}-\frac {1804 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {2+5 x+3 x^2}}+\frac {580 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{27 \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 20.14, size = 150, normalized size = 0.82 \begin {gather*} \frac {3608+5540 x+708 x^2-90 x^3+1804 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-64 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )}{81 \sqrt {x} \sqrt {2+5 x+3 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3608 + 5540*x + 708*x^2 - 90*x^3 + (1804*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSin
h[Sqrt[2/3]/Sqrt[x]], 3/2] - (64*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[2/
3]/Sqrt[x]], 3/2])/(81*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])

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Maple [A]
time = 0.77, size = 112, normalized size = 0.62

method	result	size
default	\(-\frac {2 \left (483 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-451 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+135 x^{3}+7056 x^{2}+5220 x \right )}{243 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(112\)
elliptic	\(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {2 x \left (\frac {190}{27}+\frac {253 x}{27}\right ) \sqrt {3}}{\sqrt {x \left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )}}-\frac {10 \sqrt {3 x^{3}+5 x^{2}+2 x}}{27}+\frac {580 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {-6 x}\, \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{81 \sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {902 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {-6 x}\, \left (\frac {\EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-\EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{81 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\)	\(200\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/243*(483*(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))-451*(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))+135*x^3+7056*x^2+5220*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.36, size = 87, normalized size = 0.48 \begin {gather*} \frac {2 \, {\left (710 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 8118 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 27 \, {\left (15 \, x^{2} + 784 \, x + 580\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}\right )}}{729 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/729*(710*sqrt(3)*(3*x^2 + 5*x + 2)*weierstrassPInverse(28/27, 80/729, x + 5/9) - 8118*sqrt(3)*(3*x^2 + 5*x +
2)*weierstrassZeta(28/27, 80/729, weierstrassPInverse(28/27, 80/729, x + 5/9)) - 27*(15*x^2 + 784*x + 580)*sq
rt(3*x^2 + 5*x + 2)*sqrt(x))/(3*x^2 + 5*x + 2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \left (- \frac {2 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}}\right )\, dx - \int \frac {5 x^{\frac {7}{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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(-2*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) - Integral(5*x**(7/2)/(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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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