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

3.11.65.1 Optimal result
3.11.65.2 Mathematica [C] (verified)
3.11.65.3 Rubi [A] (verified)
3.11.65.4 Maple [A] (verified)
3.11.65.5 Fricas [C] (verification not implemented)
3.11.65.6 Sympy [F]
3.11.65.7 Maxima [F]
3.11.65.8 Giac [F]
3.11.65.9 Mupad [F(-1)]

3.11.65.1 Optimal result

Integrand size = 25, antiderivative size = 182 \[ \int \frac {(2-5 x) x^{5/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {1804 \sqrt {x} (2+3 x)}{81 \sqrt {2+5 x+3 x^2}}+\frac {2 x^{3/2} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {580}{27} \sqrt {x} \sqrt {2+5 x+3 x^2}-\frac {1804 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{81 \sqrt {2+5 x+3 x^2}}+\frac {580 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{27 \sqrt {2+5 x+3 x^2}} \]

output
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)+580/2 
7*(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)
 
3.11.65.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.82 \[ \int \frac {(2-5 x) x^{5/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {3608+5540 x+708 x^2-90 x^3+1804 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-64 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{81 \sqrt {x} \sqrt {2+5 x+3 x^2}} \]

input
Integrate[((2 - 5*x)*x^(5/2))/(2 + 5*x + 3*x^2)^(3/2),x]
 
output
(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*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (64*I)*Sq 
rt[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])
 
3.11.65.3 Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {1233, 25, 1236, 27, 1240, 1503, 1413, 1456}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(2-5 x) x^{5/2}}{\left (3 x^2+5 x+2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1233

\(\displaystyle \frac {2}{3} \int -\frac {\sqrt {x} (145 x+111)}{\sqrt {3 x^2+5 x+2}}dx+\frac {2 (95 x+74) x^{3/2}}{3 \sqrt {3 x^2+5 x+2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 x^{3/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {2}{3} \int \frac {\sqrt {x} (145 x+111)}{\sqrt {3 x^2+5 x+2}}dx\)

\(\Big \downarrow \) 1236

\(\displaystyle \frac {2 x^{3/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {2}{3} \left (\frac {2}{9} \int -\frac {451 x+290}{2 \sqrt {x} \sqrt {3 x^2+5 x+2}}dx+\frac {290}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 x^{3/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {2}{3} \left (\frac {290}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {1}{9} \int \frac {451 x+290}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx\right )\)

\(\Big \downarrow \) 1240

\(\displaystyle \frac {2 x^{3/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {2}{3} \left (\frac {290}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {2}{9} \int \frac {451 x+290}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\)

\(\Big \downarrow \) 1503

\(\displaystyle \frac {2 x^{3/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {2}{3} \left (\frac {290}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {2}{9} \left (290 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+451 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\right )\)

\(\Big \downarrow \) 1413

\(\displaystyle \frac {2 x^{3/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {2}{3} \left (\frac {290}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {2}{9} \left (451 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {145 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}\right )\right )\)

\(\Big \downarrow \) 1456

\(\displaystyle \frac {2 x^{3/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {2}{3} \left (\frac {290}{9} \sqrt {x} \sqrt {3 x^2+5 x+2}-\frac {2}{9} \left (\frac {145 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+451 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )\right )\)

input
Int[((2 - 5*x)*x^(5/2))/(2 + 5*x + 3*x^2)^(3/2),x]
 
output
(2*x^(3/2)*(74 + 95*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (2*((290*Sqrt[x]*Sqrt[ 
2 + 5*x + 3*x^2])/9 - (2*(451*((Sqrt[x]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2 
]) - (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], - 
1/2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (145*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 
+ x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5*x + 3*x^2]))/9))/3
 

3.11.65.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1233
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(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] - Simp[1/(c*( 
p + 1)*(b^2 - 4*a*c))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim 
p[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] && 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 1236
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 
1)/(c*(m + 2*p + 2))), x] + Simp[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] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege 
rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])
 

rule 1240
Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), 
x_Symbol] :> Simp[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]
 

rule 1413
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]))*EllipticF 
[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 1456
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 1503
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[d   Int[1/Sqrt[a + b*x^2 + c*x^4] 
, x], x] + Simp[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]
 
3.11.65.4 Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.62

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

input
int((2-5*x)*x^(5/2)/(3*x^2+5*x+2)^(3/2),x,method=_RETURNVERBOSE)
 
output
-2/243*(483*(6*x+4)^(1/2)*(3+3*x)^(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+3*x)^(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)
 
3.11.65.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.48 \[ \int \frac {(2-5 x) x^{5/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (710 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 8118 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 27 \, {\left (15 \, x^{2} + 784 \, x + 580\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}\right )}}{729 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \]

input
integrate((2-5*x)*x^(5/2)/(3*x^2+5*x+2)^(3/2),x, algorithm="fricas")
 
output
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, wei 
erstrassPInverse(28/27, 80/729, x + 5/9)) - 27*(15*x^2 + 784*x + 580)*sqrt 
(3*x^2 + 5*x + 2)*sqrt(x))/(3*x^2 + 5*x + 2)
 
3.11.65.6 Sympy [F]

\[ \int \frac {(2-5 x) x^{5/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=- \int \left (- \frac {2 x^{\frac {5}{2}}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 x^{\frac {7}{2}}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\, dx \]

input
integrate((2-5*x)*x**(5/2)/(3*x**2+5*x+2)**(3/2),x)
 
output
-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)
 
3.11.65.7 Maxima [F]

\[ \int \frac {(2-5 x) x^{5/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} x^{\frac {5}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \,d x } \]

input
integrate((2-5*x)*x^(5/2)/(3*x^2+5*x+2)^(3/2),x, algorithm="maxima")
 
output
-integrate((5*x - 2)*x^(5/2)/(3*x^2 + 5*x + 2)^(3/2), x)
 
3.11.65.8 Giac [F]

\[ \int \frac {(2-5 x) x^{5/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} x^{\frac {5}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \,d x } \]

input
integrate((2-5*x)*x^(5/2)/(3*x^2+5*x+2)^(3/2),x, algorithm="giac")
 
output
integrate(-(5*x - 2)*x^(5/2)/(3*x^2 + 5*x + 2)^(3/2), x)
 
3.11.65.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(2-5 x) x^{5/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\int \frac {x^{5/2}\,\left (5\,x-2\right )}{{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \]

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