\(\int \frac {a+b \log (c (d+e x)^n)}{x^2 (f+g x^2)^2} \, dx\) [274]

Optimal result

Integrand size = 27, antiderivative size = 560 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\frac {b e n \log (x)}{d f^2}-\frac {b e n \log (d+e x)}{d f^2}-\frac {b e \sqrt {g} n \log (d+e x)}{4 f^2 \left (e \sqrt {-f}+d \sqrt {g}\right )}-\frac {b e \sqrt {g} n \log (d+e x)}{4 f \left (e (-f)^{3/2}+d f \sqrt {g}\right )}-\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x}+\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f^2 \left (\sqrt {-f}-\sqrt {g} x\right )}-\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f^2 \left (\sqrt {-f}+\sqrt {g} x\right )}+\frac {b e \sqrt {g} n \log \left (\sqrt {-f}-\sqrt {g} x\right )}{4 f^2 \left (e \sqrt {-f}+d \sqrt {g}\right )}-\frac {3 \sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{4 (-f)^{5/2}}+\frac {b e \sqrt {g} n \log \left (\sqrt {-f}+\sqrt {g} x\right )}{4 f \left (e (-f)^{3/2}+d f \sqrt {g}\right )}+\frac {3 \sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{5/2}}+\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{5/2}}-\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{4 (-f)^{5/2}} \] Output:

b*e*n*ln(x)/d/f^2-b*e*n*ln(e*x+d)/d/f^2-1/4*b*e*g^(1/2)*n*ln(e*x+d)/f^2/(e 
*(-f)^(1/2)+d*g^(1/2))-1/4*b*e*g^(1/2)*n*ln(e*x+d)/f/(e*(-f)^(3/2)+d*f*g^( 
1/2))-(a+b*ln(c*(e*x+d)^n))/f^2/x+1/4*g^(1/2)*(a+b*ln(c*(e*x+d)^n))/f^2/(( 
-f)^(1/2)-g^(1/2)*x)-1/4*g^(1/2)*(a+b*ln(c*(e*x+d)^n))/f^2/((-f)^(1/2)+g^( 
1/2)*x)+1/4*b*e*g^(1/2)*n*ln((-f)^(1/2)-g^(1/2)*x)/f^2/(e*(-f)^(1/2)+d*g^( 
1/2))-3/4*g^(1/2)*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)-g^(1/2)*x)/(e*(-f 
)^(1/2)+d*g^(1/2)))/(-f)^(5/2)+1/4*b*e*g^(1/2)*n*ln((-f)^(1/2)+g^(1/2)*x)/ 
f/(e*(-f)^(3/2)+d*f*g^(1/2))+3/4*g^(1/2)*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^ 
(1/2)+g^(1/2)*x)/(e*(-f)^(1/2)-d*g^(1/2)))/(-f)^(5/2)+3/4*b*g^(1/2)*n*poly 
log(2,-g^(1/2)*(e*x+d)/(e*(-f)^(1/2)-d*g^(1/2)))/(-f)^(5/2)-3/4*b*g^(1/2)* 
n*polylog(2,g^(1/2)*(e*x+d)/(e*(-f)^(1/2)+d*g^(1/2)))/(-f)^(5/2)
 

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 475, normalized size of antiderivative = 0.85 \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\frac {1}{4} \left (\frac {4 b e n (\log (x)-\log (d+e x))}{d f^2}-\frac {4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 x}+\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (\sqrt {-f}-\sqrt {g} x\right )}-\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (\sqrt {-f}+\sqrt {g} x\right )}+\frac {b e \sqrt {g} n \left (-\log (d+e x)+\log \left (\sqrt {-f}-\sqrt {g} x\right )\right )}{f^2 \left (e \sqrt {-f}+d \sqrt {g}\right )}-\frac {3 \sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2}}+\frac {b e \sqrt {g} n \left (\log (d+e x)-\log \left (\sqrt {-f}+\sqrt {g} x\right )\right )}{f^2 \left (e \sqrt {-f}-d \sqrt {g}\right )}+\frac {3 \sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{5/2}}+\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{(-f)^{5/2}}-\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{(-f)^{5/2}}\right ) \] Input:

Integrate[(a + b*Log[c*(d + e*x)^n])/(x^2*(f + g*x^2)^2),x]
 

Output:

((4*b*e*n*(Log[x] - Log[d + e*x]))/(d*f^2) - (4*(a + b*Log[c*(d + e*x)^n]) 
)/(f^2*x) + (Sqrt[g]*(a + b*Log[c*(d + e*x)^n]))/(f^2*(Sqrt[-f] - Sqrt[g]* 
x)) - (Sqrt[g]*(a + b*Log[c*(d + e*x)^n]))/(f^2*(Sqrt[-f] + Sqrt[g]*x)) + 
(b*e*Sqrt[g]*n*(-Log[d + e*x] + Log[Sqrt[-f] - Sqrt[g]*x]))/(f^2*(e*Sqrt[- 
f] + d*Sqrt[g])) - (3*Sqrt[g]*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] 
- Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(-f)^(5/2) + (b*e*Sqrt[g]*n*(Log[ 
d + e*x] - Log[Sqrt[-f] + Sqrt[g]*x]))/(f^2*(e*Sqrt[-f] - d*Sqrt[g])) + (3 
*Sqrt[g]*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt 
[-f] - d*Sqrt[g])])/(-f)^(5/2) + (3*b*Sqrt[g]*n*PolyLog[2, -((Sqrt[g]*(d + 
 e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/(-f)^(5/2) - (3*b*Sqrt[g]*n*PolyLog[2, 
(Sqrt[g]*(d + e*x))/(e*Sqrt[-f] + d*Sqrt[g])])/(-f)^(5/2))/4
 

Rubi [A] (verified)

Time = 1.93 (sec) , antiderivative size = 560, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2863, 2009}

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.

\(\displaystyle \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx\)

\(\Big \downarrow \) 2863

\(\displaystyle \int \left (-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f^2 \left (f+g x^2\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x^2}-\frac {g \left (a+b \log \left (c (d+e x)^n\right )\right )}{f \left (f+g x^2\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f^2 \left (\sqrt {-f}-\sqrt {g} x\right )}-\frac {\sqrt {g} \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 f^2 \left (\sqrt {-f}+\sqrt {g} x\right )}-\frac {a+b \log \left (c (d+e x)^n\right )}{f^2 x}-\frac {3 \sqrt {g} \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 (-f)^{5/2}}+\frac {3 \sqrt {g} \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 (-f)^{5/2}}-\frac {b e \sqrt {g} n \log (d+e x)}{4 f^2 \left (d \sqrt {g}+e \sqrt {-f}\right )}+\frac {b e \sqrt {g} n \log \left (\sqrt {-f}-\sqrt {g} x\right )}{4 f^2 \left (d \sqrt {g}+e \sqrt {-f}\right )}+\frac {b e n \log (x)}{d f^2}-\frac {b e n \log (d+e x)}{d f^2}+\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{4 (-f)^{5/2}}-\frac {3 b \sqrt {g} n \operatorname {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{4 (-f)^{5/2}}-\frac {b e \sqrt {g} n \log (d+e x)}{4 f \left (d f \sqrt {g}+e (-f)^{3/2}\right )}+\frac {b e \sqrt {g} n \log \left (\sqrt {-f}+\sqrt {g} x\right )}{4 f \left (d f \sqrt {g}+e (-f)^{3/2}\right )}\)

Input:

Int[(a + b*Log[c*(d + e*x)^n])/(x^2*(f + g*x^2)^2),x]
 

Output:

(b*e*n*Log[x])/(d*f^2) - (b*e*n*Log[d + e*x])/(d*f^2) - (b*e*Sqrt[g]*n*Log 
[d + e*x])/(4*f^2*(e*Sqrt[-f] + d*Sqrt[g])) - (b*e*Sqrt[g]*n*Log[d + e*x]) 
/(4*f*(e*(-f)^(3/2) + d*f*Sqrt[g])) - (a + b*Log[c*(d + e*x)^n])/(f^2*x) + 
 (Sqrt[g]*(a + b*Log[c*(d + e*x)^n]))/(4*f^2*(Sqrt[-f] - Sqrt[g]*x)) - (Sq 
rt[g]*(a + b*Log[c*(d + e*x)^n]))/(4*f^2*(Sqrt[-f] + Sqrt[g]*x)) + (b*e*Sq 
rt[g]*n*Log[Sqrt[-f] - Sqrt[g]*x])/(4*f^2*(e*Sqrt[-f] + d*Sqrt[g])) - (3*S 
qrt[g]*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[- 
f] + d*Sqrt[g])])/(4*(-f)^(5/2)) + (b*e*Sqrt[g]*n*Log[Sqrt[-f] + Sqrt[g]*x 
])/(4*f*(e*(-f)^(3/2) + d*f*Sqrt[g])) + (3*Sqrt[g]*(a + b*Log[c*(d + e*x)^ 
n])*Log[(e*(Sqrt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/(4*(-f)^(5/2 
)) + (3*b*Sqrt[g]*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[ 
g]))])/(4*(-f)^(5/2)) - (3*b*Sqrt[g]*n*PolyLog[2, (Sqrt[g]*(d + e*x))/(e*S 
qrt[-f] + d*Sqrt[g])])/(4*(-f)^(5/2))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2863

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_)) 
^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a 
 + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]
 

Maple [C] (warning: unable to verify)

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

Time = 4.82 (sec) , antiderivative size = 1619, normalized size of antiderivative = 2.89

method	result	size
risch	\(\text {Expression too large to display}\)	\(1619\)

Input:

int((a+b*ln(c*(e*x+d)^n))/x^2/(g*x^2+f)^2,x,method=_RETURNVERBOSE)
 

Output:

1/4*b*e*n/f^2*g/(d^2*g+e^2*f)*d*ln(g*(e*x+d)^2-2*(e*x+d)*d*g+d^2*g+f*e^2)+ 
1/2*b*e^2*n/f*g/(d^2*g+e^2*f)/(g*f)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e 
/(g*f)^(1/2))+1/2*b*e^2/f^2*g*x/(e^2*g*x^2+e^2*f)*n*ln(e*x+d)-3/2*b/f^2*g/ 
(g*f)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e/(g*f)^(1/2))*ln((e*x+d)^n)-3/ 
4*b*n/f^2*g/(-g*f)^(1/2)*dilog((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1 
/2)+d*g))+3/4*b*n/f^2*g/(-g*f)^(1/2)*dilog((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/ 
(e*(-g*f)^(1/2)-d*g))+b*e*n/f^2/d*ln(e*x)-1/2*b*e^2/f^2*g*x/(e^2*g*x^2+e^2 
*f)*ln((e*x+d)^n)+3/2*b/f^2*g/(g*f)^(1/2)*arctan(1/2*(2*g*(e*x+d)-2*d*g)/e 
/(g*f)^(1/2))*n*ln(e*x+d)-1/2*b*n/f^2*g*ln(e*x+d)/(-g*f)^(1/2)*ln((e*(-g*f 
)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))+1/2*b*n/f^2*g*ln(e*x+d)/(-g*f 
)^(1/2)*ln((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))-b*ln((e*x+ 
d)^n)/f^2/x-1/4*b*e^2*n/f^2*g^3*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/ 
(-g*f)^(1/2)*ln((e*(-g*f)^(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))*x^2*d 
^2+1/4*b*e^4*n/f*g^2*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-g*f)^(1/2 
)*ln((e*(-g*f)^(1/2)+g*(e*x+d)-d*g)/(e*(-g*f)^(1/2)-d*g))*x^2-1/4*b*e^4*n/ 
f*g^2*ln(e*x+d)/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-g*f)^(1/2)*ln((e*(-g*f)^ 
(1/2)-g*(e*x+d)+d*g)/(e*(-g*f)^(1/2)+d*g))*x^2-1/4*b*e^2*n/f*g^2*ln(e*x+d) 
/(d^2*g+e^2*f)/(e^2*g*x^2+e^2*f)/(-g*f)^(1/2)*ln((e*(-g*f)^(1/2)-g*(e*x+d) 
+d*g)/(e*(-g*f)^(1/2)+d*g))*d^2+(1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e* 
x+d)^n)^2-1/2*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)*csgn(I*c)-1/...
 

Fricas [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x^{2}} \,d x } \] Input:

integrate((a+b*log(c*(e*x+d)^n))/x^2/(g*x^2+f)^2,x, algorithm="fricas")
 

Output:

integral((b*log((e*x + d)^n*c) + a)/(g^2*x^6 + 2*f*g*x^4 + f^2*x^2), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((a+b*ln(c*(e*x+d)**n))/x**2/(g*x**2+f)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x^{2}} \,d x } \] Input:

integrate((a+b*log(c*(e*x+d)^n))/x^2/(g*x^2+f)^2,x, algorithm="maxima")
 

Output:

-1/2*a*((3*g*x^2 + 2*f)/(f^2*g*x^3 + f^3*x) + 3*g*arctan(g*x/sqrt(f*g))/(s 
qrt(f*g)*f^2)) + b*integrate((log((e*x + d)^n) + log(c))/(g^2*x^6 + 2*f*g* 
x^4 + f^2*x^2), x)
 

Giac [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\int { \frac {b \log \left ({\left (e x + d\right )}^{n} c\right ) + a}{{\left (g x^{2} + f\right )}^{2} x^{2}} \,d x } \] Input:

integrate((a+b*log(c*(e*x+d)^n))/x^2/(g*x^2+f)^2,x, algorithm="giac")
 

Output:

integrate((b*log((e*x + d)^n*c) + a)/((g*x^2 + f)^2*x^2), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{x^2\,{\left (g\,x^2+f\right )}^2} \,d x \] Input:

int((a + b*log(c*(d + e*x)^n))/(x^2*(f + g*x^2)^2),x)
 

Output:

int((a + b*log(c*(d + e*x)^n))/(x^2*(f + g*x^2)^2), x)
 

Reduce [F]

\[ \int \frac {a+b \log \left (c (d+e x)^n\right )}{x^2 \left (f+g x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((a+b*log(c*(e*x+d)^n))/x^2/(g*x^2+f)^2,x)
 

Output:

( - 3*sqrt(g)*sqrt(f)*atan((g*x)/(sqrt(g)*sqrt(f)))*a*d**3*f*g*x - 3*sqrt( 
g)*sqrt(f)*atan((g*x)/(sqrt(g)*sqrt(f)))*a*d**3*g**2*x**3 - 3*sqrt(g)*sqrt 
(f)*atan((g*x)/(sqrt(g)*sqrt(f)))*a*d*e**2*f**2*x - 3*sqrt(g)*sqrt(f)*atan 
((g*x)/(sqrt(g)*sqrt(f)))*a*d*e**2*f*g*x**3 - 2*sqrt(g)*sqrt(f)*atan((g*x) 
/(sqrt(g)*sqrt(f)))*b*d*e**2*f**2*n*x - 2*sqrt(g)*sqrt(f)*atan((g*x)/(sqrt 
(g)*sqrt(f)))*b*d*e**2*f*g*n*x**3 - 6*int(log((d + e*x)**n*c)/(f**2 + 2*f* 
g*x**2 + g**2*x**4),x)*b*d**3*f**3*g**2*x - 6*int(log((d + e*x)**n*c)/(f** 
2 + 2*f*g*x**2 + g**2*x**4),x)*b*d**3*f**2*g**3*x**3 - 6*int(log((d + e*x) 
**n*c)/(f**2 + 2*f*g*x**2 + g**2*x**4),x)*b*d*e**2*f**4*g*x - 6*int(log((d 
 + e*x)**n*c)/(f**2 + 2*f*g*x**2 + g**2*x**4),x)*b*d*e**2*f**3*g**2*x**3 - 
 2*log(d + e*x)*b*e**3*f**3*n*x - 2*log(d + e*x)*b*e**3*f**2*g*n*x**3 - lo 
g(f + g*x**2)*b*d**2*e*f**2*g*n*x - log(f + g*x**2)*b*d**2*e*f*g**2*n*x**3 
 - 2*log((d + e*x)**n*c)*b*d**3*f**2*g - 2*log((d + e*x)**n*c)*b*d*e**2*f* 
*3 + 2*log(x)*b*d**2*e*f**2*g*n*x + 2*log(x)*b*d**2*e*f*g**2*n*x**3 + 2*lo 
g(x)*b*e**3*f**3*n*x + 2*log(x)*b*e**3*f**2*g*n*x**3 - 2*a*d**3*f**2*g - 3 
*a*d**3*f*g**2*x**2 - 2*a*d*e**2*f**3 - 3*a*d*e**2*f**2*g*x**2)/(2*d*f**3* 
x*(d**2*f*g + d**2*g**2*x**2 + e**2*f**2 + e**2*f*g*x**2))