\(\int \frac {(4+3 x)^3}{(5-2 x)^{3/2} \sqrt {2+3 x+x^2}} \, dx\) [586]

Optimal result

Integrand size = 29, antiderivative size = 135 \[ \int \frac {(4+3 x)^3}{(5-2 x)^{3/2} \sqrt {2+3 x+x^2}} \, dx=\frac {12167 \sqrt {2+3 x+x^2}}{126 \sqrt {5-2 x}}+\frac {9}{2} \sqrt {5-2 x} \sqrt {2+3 x+x^2}+\frac {16573 \sqrt {-2-3 x-x^2} E\left (\arcsin \left (\sqrt {2+x}\right )|\frac {2}{9}\right )}{42 \sqrt {2+3 x+x^2}}-\frac {699 \sqrt {-2-3 x-x^2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {2+x}\right ),\frac {2}{9}\right )}{2 \sqrt {2+3 x+x^2}} \] Output:

12167/126*(x^2+3*x+2)^(1/2)/(5-2*x)^(1/2)+9/2*(5-2*x)^(1/2)*(x^2+3*x+2)^(1 
/2)+16573/42*(-x^2-3*x-2)^(1/2)*EllipticE((2+x)^(1/2),1/3*2^(1/2))/(x^2+3* 
x+2)^(1/2)-699/2*(-x^2-3*x-2)^(1/2)*EllipticF((2+x)^(1/2),1/3*2^(1/2))/(x^ 
2+3*x+2)^(1/2)
 

Mathematica [A] (verified)

Time = 32.80 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.15 \[ \int \frac {(4+3 x)^3}{(5-2 x)^{3/2} \sqrt {2+3 x+x^2}} \, dx=\frac {-378 \left (32+50 x+19 x^2+x^3\right )-16573 (5-2 x)^{3/2} \sqrt {\frac {1+x}{-5+2 x}} \sqrt {\frac {2+x}{-5+2 x}} E\left (\arcsin \left (\frac {3}{\sqrt {5-2 x}}\right )|\frac {7}{9}\right )+1894 (5-2 x)^{3/2} \sqrt {\frac {1+x}{-5+2 x}} \sqrt {\frac {2+x}{-5+2 x}} \operatorname {EllipticF}\left (\arcsin \left (\frac {3}{\sqrt {5-2 x}}\right ),\frac {7}{9}\right )}{42 \sqrt {5-2 x} \sqrt {2+3 x+x^2}} \] Input:

Integrate[(4 + 3*x)^3/((5 - 2*x)^(3/2)*Sqrt[2 + 3*x + x^2]),x]
 

Output:

(-378*(32 + 50*x + 19*x^2 + x^3) - 16573*(5 - 2*x)^(3/2)*Sqrt[(1 + x)/(-5 
+ 2*x)]*Sqrt[(2 + x)/(-5 + 2*x)]*EllipticE[ArcSin[3/Sqrt[5 - 2*x]], 7/9] + 
 1894*(5 - 2*x)^(3/2)*Sqrt[(1 + x)/(-5 + 2*x)]*Sqrt[(2 + x)/(-5 + 2*x)]*El 
lipticF[ArcSin[3/Sqrt[5 - 2*x]], 7/9])/(42*Sqrt[5 - 2*x]*Sqrt[2 + 3*x + x^ 
2])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {1290, 27, 2184, 27, 1269, 1172, 27, 321, 327}

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.

\(\displaystyle \int \frac {(3 x+4)^3}{(5-2 x)^{3/2} \sqrt {x^2+3 x+2}} \, dx\)

\(\Big \downarrow \) 1290

\(\displaystyle \frac {12167 \sqrt {x^2+3 x+2}}{126 \sqrt {5-2 x}}-\frac {2}{63} \int \frac {3402 x^2+34280 x+43009}{8 \sqrt {5-2 x} \sqrt {x^2+3 x+2}}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {12167 \sqrt {x^2+3 x+2}}{126 \sqrt {5-2 x}}-\frac {1}{252} \int \frac {3402 x^2+34280 x+43009}{\sqrt {5-2 x} \sqrt {x^2+3 x+2}}dx\)

\(\Big \downarrow \) 2184

\(\displaystyle \frac {1}{252} \left (1134 \sqrt {5-2 x} \sqrt {x^2+3 x+2}-\frac {1}{6} \int \frac {12 (16573 x+24623)}{\sqrt {5-2 x} \sqrt {x^2+3 x+2}}dx\right )+\frac {12167 \sqrt {x^2+3 x+2}}{126 \sqrt {5-2 x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{252} \left (1134 \sqrt {5-2 x} \sqrt {x^2+3 x+2}-2 \int \frac {16573 x+24623}{\sqrt {5-2 x} \sqrt {x^2+3 x+2}}dx\right )+\frac {12167 \sqrt {x^2+3 x+2}}{126 \sqrt {5-2 x}}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {1}{252} \left (1134 \sqrt {5-2 x} \sqrt {x^2+3 x+2}-2 \left (\frac {132111}{2} \int \frac {1}{\sqrt {5-2 x} \sqrt {x^2+3 x+2}}dx-\frac {16573}{2} \int \frac {\sqrt {5-2 x}}{\sqrt {x^2+3 x+2}}dx\right )\right )+\frac {12167 \sqrt {x^2+3 x+2}}{126 \sqrt {5-2 x}}\)

\(\Big \downarrow \) 1172

\(\displaystyle \frac {1}{252} \left (1134 \sqrt {5-2 x} \sqrt {x^2+3 x+2}-2 \left (\frac {44037 \sqrt {-x^2-3 x-2} \int \frac {3}{\sqrt {-x-1} \sqrt {9-2 (x+2)}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}-\frac {49719 \sqrt {-x^2-3 x-2} \int \frac {\sqrt {9-2 (x+2)}}{3 \sqrt {-x-1}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}\right )\right )+\frac {12167 \sqrt {x^2+3 x+2}}{126 \sqrt {5-2 x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{252} \left (1134 \sqrt {5-2 x} \sqrt {x^2+3 x+2}-2 \left (\frac {132111 \sqrt {-x^2-3 x-2} \int \frac {1}{\sqrt {-x-1} \sqrt {9-2 (x+2)}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}-\frac {16573 \sqrt {-x^2-3 x-2} \int \frac {\sqrt {9-2 (x+2)}}{\sqrt {-x-1}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}\right )\right )+\frac {12167 \sqrt {x^2+3 x+2}}{126 \sqrt {5-2 x}}\)

\(\Big \downarrow \) 321

\(\displaystyle \frac {1}{252} \left (1134 \sqrt {5-2 x} \sqrt {x^2+3 x+2}-2 \left (\frac {44037 \sqrt {-x^2-3 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x+2}\right ),\frac {2}{9}\right )}{\sqrt {x^2+3 x+2}}-\frac {16573 \sqrt {-x^2-3 x-2} \int \frac {\sqrt {9-2 (x+2)}}{\sqrt {-x-1}}d\sqrt {x+2}}{\sqrt {x^2+3 x+2}}\right )\right )+\frac {12167 \sqrt {x^2+3 x+2}}{126 \sqrt {5-2 x}}\)

\(\Big \downarrow \) 327

\(\displaystyle \frac {1}{252} \left (1134 \sqrt {5-2 x} \sqrt {x^2+3 x+2}-2 \left (\frac {44037 \sqrt {-x^2-3 x-2} \operatorname {EllipticF}\left (\arcsin \left (\sqrt {x+2}\right ),\frac {2}{9}\right )}{\sqrt {x^2+3 x+2}}-\frac {49719 \sqrt {-x^2-3 x-2} E\left (\arcsin \left (\sqrt {x+2}\right )|\frac {2}{9}\right )}{\sqrt {x^2+3 x+2}}\right )\right )+\frac {12167 \sqrt {x^2+3 x+2}}{126 \sqrt {5-2 x}}\)

Input:

Int[(4 + 3*x)^3/((5 - 2*x)^(3/2)*Sqrt[2 + 3*x + x^2]),x]
 

Output:

(12167*Sqrt[2 + 3*x + x^2])/(126*Sqrt[5 - 2*x]) + (1134*Sqrt[5 - 2*x]*Sqrt 
[2 + 3*x + x^2] - 2*((-49719*Sqrt[-2 - 3*x - x^2]*EllipticE[ArcSin[Sqrt[2 
+ x]], 2/9])/Sqrt[2 + 3*x + x^2] + (44037*Sqrt[-2 - 3*x - x^2]*EllipticF[A 
rcSin[Sqrt[2 + x]], 2/9])/Sqrt[2 + 3*x + x^2]))/252
 

Defintions of rubi rules used

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 321

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c 
/(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 
0] &&  !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
 

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 1172

Int[((d_.) + (e_.)*(x_))^(m_)/Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Sy 
mbol] :> Simp[2*Rt[b^2 - 4*a*c, 2]*(d + e*x)^m*(Sqrt[(-c)*((a + b*x + c*x^2 
)/(b^2 - 4*a*c))]/(c*Sqrt[a + b*x + c*x^2]*(2*c*((d + e*x)/(2*c*d - b*e - e 
*Rt[b^2 - 4*a*c, 2])))^m))   Subst[Int[(1 + 2*e*Rt[b^2 - 4*a*c, 2]*(x^2/(2* 
c*d - b*e - e*Rt[b^2 - 4*a*c, 2])))^m/Sqrt[1 - x^2], x], x, Sqrt[(b + Rt[b^ 
2 - 4*a*c, 2] + 2*c*x)/(2*Rt[b^2 - 4*a*c, 2])]], x] /; FreeQ[{a, b, c, d, e 
}, x] && EqQ[m^2, 1/4]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 1290

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(f + g*x)^ 
n, d + e*x, x], R = PolynomialRemainder[(f + g*x)^n, d + e*x, x]}, Simp[(e* 
R*(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] + Simp[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*ExpandToSum[(m + 1)*(c*d^2 - b*d*e + a*e^2)*Q + c*d*R 
*(m + 1) - b*e*R*(m + p + 2) - c*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[ 
{a, b, c, d, e, f, g, p}, x] && IGtQ[n, 1] && LtQ[m, -1]
 

rule 2184

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S 
imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q 
+ 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + 
b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 
1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q - 1) - c 
*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[ 
q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && Pol 
yQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !(IGt 
Q[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.93

method	result	size
default	\(-\frac {\sqrt {5-2 x}\, \sqrt {x^{2}+3 x +2}\, \left (2300 \sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \operatorname {EllipticF}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )+16573 \sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \operatorname {EllipticE}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )-2268 x^{3}+23200 x^{2}+85476 x +60008\right )}{252 \left (2 x^{3}+x^{2}-11 x -10\right )}\)	\(126\)
elliptic	\(\frac {\sqrt {-\left (-5+2 x \right ) \left (x^{2}+3 x +2\right )}\, \left (-\frac {12167 \left (-2 x^{2}-6 x -4\right )}{252 \sqrt {\left (x -\frac {5}{2}\right ) \left (-2 x^{2}-6 x -4\right )}}+\frac {24623 \sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \operatorname {EllipticF}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )}{882 \sqrt {-2 x^{3}-x^{2}+11 x +10}}+\frac {16573 \sqrt {5-2 x}\, \sqrt {14 x +14}\, \sqrt {2 x +4}\, \left (\frac {7 \operatorname {EllipticE}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )}{2}-\operatorname {EllipticF}\left (\frac {\sqrt {5-2 x}}{3}, \frac {3 \sqrt {7}}{7}\right )\right )}{882 \sqrt {-2 x^{3}-x^{2}+11 x +10}}+\frac {9 \sqrt {-2 x^{3}-x^{2}+11 x +10}}{2}\right )}{\sqrt {5-2 x}\, \sqrt {x^{2}+3 x +2}}\)	\(214\)

Input:

int((3*x+4)^3/(5-2*x)^(3/2)/(x^2+3*x+2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/252*(5-2*x)^(1/2)*(x^2+3*x+2)^(1/2)*(2300*(5-2*x)^(1/2)*(14*x+14)^(1/2) 
*(2*x+4)^(1/2)*EllipticF(1/3*(5-2*x)^(1/2),3/7*7^(1/2))+16573*(5-2*x)^(1/2 
)*(14*x+14)^(1/2)*(2*x+4)^(1/2)*EllipticE(1/3*(5-2*x)^(1/2),3/7*7^(1/2))-2 
268*x^3+23200*x^2+85476*x+60008)/(2*x^3+x^2-11*x-10)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.51 \[ \int \frac {(4+3 x)^3}{(5-2 x)^{3/2} \sqrt {2+3 x+x^2}} \, dx=\frac {131165 \, \sqrt {-2} {\left (2 \, x - 5\right )} {\rm weierstrassPInverse}\left (\frac {67}{3}, \frac {440}{27}, x + \frac {1}{6}\right ) - 99438 \, \sqrt {-2} {\left (2 \, x - 5\right )} {\rm weierstrassZeta}\left (\frac {67}{3}, \frac {440}{27}, {\rm weierstrassPInverse}\left (\frac {67}{3}, \frac {440}{27}, x + \frac {1}{6}\right )\right ) + 12 \, \sqrt {x^{2} + 3 \, x + 2} {\left (567 \, x - 7501\right )} \sqrt {-2 \, x + 5}}{756 \, {\left (2 \, x - 5\right )}} \] Input:

integrate((4+3*x)^3/(5-2*x)^(3/2)/(x^2+3*x+2)^(1/2),x, algorithm="fricas")
 

Output:

1/756*(131165*sqrt(-2)*(2*x - 5)*weierstrassPInverse(67/3, 440/27, x + 1/6 
) - 99438*sqrt(-2)*(2*x - 5)*weierstrassZeta(67/3, 440/27, weierstrassPInv 
erse(67/3, 440/27, x + 1/6)) + 12*sqrt(x^2 + 3*x + 2)*(567*x - 7501)*sqrt( 
-2*x + 5))/(2*x - 5)
 

Sympy [F]

\[ \int \frac {(4+3 x)^3}{(5-2 x)^{3/2} \sqrt {2+3 x+x^2}} \, dx=\int \frac {\left (3 x + 4\right )^{3}}{\sqrt {\left (x + 1\right ) \left (x + 2\right )} \left (5 - 2 x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((4+3*x)**3/(5-2*x)**(3/2)/(x**2+3*x+2)**(1/2),x)
 

Output:

Integral((3*x + 4)**3/(sqrt((x + 1)*(x + 2))*(5 - 2*x)**(3/2)), x)
 

Maxima [F]

\[ \int \frac {(4+3 x)^3}{(5-2 x)^{3/2} \sqrt {2+3 x+x^2}} \, dx=\int { \frac {{\left (3 \, x + 4\right )}^{3}}{\sqrt {x^{2} + 3 \, x + 2} {\left (-2 \, x + 5\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((4+3*x)^3/(5-2*x)^(3/2)/(x^2+3*x+2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {(4+3 x)^3}{(5-2 x)^{3/2} \sqrt {2+3 x+x^2}} \, dx=\int { \frac {{\left (3 \, x + 4\right )}^{3}}{\sqrt {x^{2} + 3 \, x + 2} {\left (-2 \, x + 5\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((4+3*x)^3/(5-2*x)^(3/2)/(x^2+3*x+2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(4+3 x)^3}{(5-2 x)^{3/2} \sqrt {2+3 x+x^2}} \, dx=\int \frac {{\left (3\,x+4\right )}^3}{{\left (5-2\,x\right )}^{3/2}\,\sqrt {x^2+3\,x+2}} \,d x \] Input:

int((3*x + 4)^3/((5 - 2*x)^(3/2)*(3*x + x^2 + 2)^(1/2)),x)
 

Output:

int((3*x + 4)^3/((5 - 2*x)^(3/2)*(3*x + x^2 + 2)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {(4+3 x)^3}{(5-2 x)^{3/2} \sqrt {2+3 x+x^2}} \, dx=\frac {180 \sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}\, x -1314 \sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}+8388 \left (\int \frac {\sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}\, x^{2}}{4 x^{4}-8 x^{3}-27 x^{2}+35 x +50}d x \right ) x -20970 \left (\int \frac {\sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}\, x^{2}}{4 x^{4}-8 x^{3}-27 x^{2}+35 x +50}d x \right )-18806 \left (\int \frac {\sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}}{4 x^{4}-8 x^{3}-27 x^{2}+35 x +50}d x \right ) x +47015 \left (\int \frac {\sqrt {-2 x +5}\, \sqrt {x^{2}+3 x +2}}{4 x^{4}-8 x^{3}-27 x^{2}+35 x +50}d x \right )}{40 x -100} \] Input:

int((4+3*x)^3/(5-2*x)^(3/2)/(x^2+3*x+2)^(1/2),x)
 

Output:

(180*sqrt( - 2*x + 5)*sqrt(x**2 + 3*x + 2)*x - 1314*sqrt( - 2*x + 5)*sqrt( 
x**2 + 3*x + 2) + 8388*int((sqrt( - 2*x + 5)*sqrt(x**2 + 3*x + 2)*x**2)/(4 
*x**4 - 8*x**3 - 27*x**2 + 35*x + 50),x)*x - 20970*int((sqrt( - 2*x + 5)*s 
qrt(x**2 + 3*x + 2)*x**2)/(4*x**4 - 8*x**3 - 27*x**2 + 35*x + 50),x) - 188 
06*int((sqrt( - 2*x + 5)*sqrt(x**2 + 3*x + 2))/(4*x**4 - 8*x**3 - 27*x**2 
+ 35*x + 50),x)*x + 47015*int((sqrt( - 2*x + 5)*sqrt(x**2 + 3*x + 2))/(4*x 
**4 - 8*x**3 - 27*x**2 + 35*x + 50),x))/(20*(2*x - 5))