3.2.45 \(\int \frac {(a g+b g x) (A+B \log (e (\frac {a+b x}{c+d x})^n))}{(c i+d i x)^2} \, dx\) [145]

Optimal. Leaf size=168 \[ -\frac {A g (a+b x)}{d i^2 (c+d x)}+\frac {B g n (a+b x)}{d i^2 (c+d x)}-\frac {B g (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{d i^2 (c+d x)}-\frac {b g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log \left (\frac {b c-a d}{b (c+d x)}\right )}{d^2 i^2}-\frac {b B g n \text {Li}_2\left (\frac {d (a+b x)}{b (c+d x)}\right )}{d^2 i^2} \]

[Out]

-A*g*(b*x+a)/d/i^2/(d*x+c)+B*g*n*(b*x+a)/d/i^2/(d*x+c)-B*g*(b*x+a)*ln(e*((b*x+a)/(d*x+c))^n)/d/i^2/(d*x+c)-b*g
*(A+B*ln(e*((b*x+a)/(d*x+c))^n))*ln((-a*d+b*c)/b/(d*x+c))/d^2/i^2-b*B*g*n*polylog(2,d*(b*x+a)/b/(d*x+c))/d^2/i
^2

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Rubi [A]
time = 0.11, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {2561, 45, 2393, 2332, 2354, 2438} \begin {gather*} -\frac {b B g n \text {PolyLog}\left (2,\frac {d (a+b x)}{b (c+d x)}\right )}{d^2 i^2}-\frac {b g \log \left (\frac {b c-a d}{b (c+d x)}\right ) \left (B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )+A\right )}{d^2 i^2}-\frac {A g (a+b x)}{d i^2 (c+d x)}-\frac {B g (a+b x) \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{d i^2 (c+d x)}+\frac {B g n (a+b x)}{d i^2 (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*g*(a + b*x))/(d*i^2*(c + d*x))) + (B*g*n*(a + b*x))/(d*i^2*(c + d*x)) - (B*g*(a + b*x)*Log[e*((a + b*x)/(
c + d*x))^n])/(d*i^2*(c + d*x)) - (b*g*(A + B*Log[e*((a + b*x)/(c + d*x))^n])*Log[(b*c - a*d)/(b*(c + d*x))])/
(d^2*i^2) - (b*B*g*n*PolyLog[2, (d*(a + b*x))/(b*(c + d*x))])/(d^2*i^2)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2354

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[Log[1 + e*(x/d)]*((a +
b*Log[c*x^n])^p/e), x] - Dist[b*n*(p/e), Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2393

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2561

Int[((A_.) + Log[(e_.)*(((a_.) + (b_.)*(x_))/((c_.) + (d_.)*(x_)))^(n_.)]*(B_.))^(p_.)*((f_.) + (g_.)*(x_))^(m
_.)*((h_.) + (i_.)*(x_))^(q_.), x_Symbol] :> Dist[(b*c - a*d)^(m + q + 1)*(g/b)^m*(i/d)^q, Subst[Int[x^m*((A +
B*Log[e*x^n])^p/(b - d*x)^(m + q + 2)), x], x, (a + b*x)/(c + d*x)], x] /; FreeQ[{a, b, c, d, e, f, g, h, i,
A, B, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[b*f - a*g, 0] && EqQ[d*h - c*i, 0] && IntegersQ[m, q]

Rubi steps

\begin {align*} \int \frac {(a g+b g x) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{(145 c+145 d x)^2} \, dx &=\int \left (\frac {(-b c+a d) g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21025 d (c+d x)^2}+\frac {b g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21025 d (c+d x)}\right ) \, dx\\ &=\frac {(b g) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{c+d x} \, dx}{21025 d}-\frac {((b c-a d) g) \int \frac {A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )}{(c+d x)^2} \, dx}{21025 d}\\ &=\frac {(b c-a d) g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21025 d^2 (c+d x)}+\frac {b g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21025 d^2}-\frac {(b B g n) \int \frac {(c+d x) \left (-\frac {d (a+b x)}{(c+d x)^2}+\frac {b}{c+d x}\right ) \log (c+d x)}{a+b x} \, dx}{21025 d^2}-\frac {(B (b c-a d) g n) \int \frac {b c-a d}{(a+b x) (c+d x)^2} \, dx}{21025 d^2}\\ &=\frac {(b c-a d) g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21025 d^2 (c+d x)}+\frac {b g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21025 d^2}-\frac {(b B g n) \int \left (\frac {b \log (c+d x)}{a+b x}-\frac {d \log (c+d x)}{c+d x}\right ) \, dx}{21025 d^2}-\frac {\left (B (b c-a d)^2 g n\right ) \int \frac {1}{(a+b x) (c+d x)^2} \, dx}{21025 d^2}\\ &=\frac {(b c-a d) g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21025 d^2 (c+d x)}+\frac {b g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21025 d^2}-\frac {\left (b^2 B g n\right ) \int \frac {\log (c+d x)}{a+b x} \, dx}{21025 d^2}+\frac {(b B g n) \int \frac {\log (c+d x)}{c+d x} \, dx}{21025 d}-\frac {\left (B (b c-a d)^2 g n\right ) \int \left (\frac {b^2}{(b c-a d)^2 (a+b x)}-\frac {d}{(b c-a d) (c+d x)^2}-\frac {b d}{(b c-a d)^2 (c+d x)}\right ) \, dx}{21025 d^2}\\ &=-\frac {B (b c-a d) g n}{21025 d^2 (c+d x)}-\frac {b B g n \log (a+b x)}{21025 d^2}+\frac {(b c-a d) g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21025 d^2 (c+d x)}+\frac {b B g n \log (c+d x)}{21025 d^2}-\frac {b B g n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{21025 d^2}+\frac {b g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21025 d^2}+\frac {(b B g n) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,c+d x\right )}{21025 d^2}+\frac {(b B g n) \int \frac {\log \left (\frac {d (a+b x)}{-b c+a d}\right )}{c+d x} \, dx}{21025 d}\\ &=-\frac {B (b c-a d) g n}{21025 d^2 (c+d x)}-\frac {b B g n \log (a+b x)}{21025 d^2}+\frac {(b c-a d) g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21025 d^2 (c+d x)}+\frac {b B g n \log (c+d x)}{21025 d^2}-\frac {b B g n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{21025 d^2}+\frac {b g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21025 d^2}+\frac {b B g n \log ^2(c+d x)}{42050 d^2}+\frac {(b B g n) \text {Subst}\left (\int \frac {\log \left (1+\frac {b x}{-b c+a d}\right )}{x} \, dx,x,c+d x\right )}{21025 d^2}\\ &=-\frac {B (b c-a d) g n}{21025 d^2 (c+d x)}-\frac {b B g n \log (a+b x)}{21025 d^2}+\frac {(b c-a d) g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{21025 d^2 (c+d x)}+\frac {b B g n \log (c+d x)}{21025 d^2}-\frac {b B g n \log \left (-\frac {d (a+b x)}{b c-a d}\right ) \log (c+d x)}{21025 d^2}+\frac {b g \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)}{21025 d^2}+\frac {b B g n \log ^2(c+d x)}{42050 d^2}-\frac {b B g n \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )}{21025 d^2}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 183, normalized size = 1.09 \begin {gather*} \frac {g \left (\frac {2 (b c-a d) \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right )}{c+d x}+2 b \left (A+B \log \left (e \left (\frac {a+b x}{c+d x}\right )^n\right )\right ) \log (c+d x)-2 B n \left (\frac {b c-a d}{c+d x}+b \log (a+b x)-b \log (c+d x)\right )-b B n \left (\left (2 \log \left (\frac {d (a+b x)}{-b c+a d}\right )-\log (c+d x)\right ) \log (c+d x)+2 \text {Li}_2\left (\frac {b (c+d x)}{b c-a d}\right )\right )\right )}{2 d^2 i^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(g*((2*(b*c - a*d)*(A + B*Log[e*((a + b*x)/(c + d*x))^n]))/(c + d*x) + 2*b*(A + B*Log[e*((a + b*x)/(c + d*x))^
n])*Log[c + d*x] - 2*B*n*((b*c - a*d)/(c + d*x) + b*Log[a + b*x] - b*Log[c + d*x]) - b*B*n*((2*Log[(d*(a + b*x
))/(-(b*c) + a*d)] - Log[c + d*x])*Log[c + d*x] + 2*PolyLog[2, (b*(c + d*x))/(b*c - a*d)])))/(2*d^2*i^2)

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (b g x +a g \right ) \left (A +B \ln \left (e \left (\frac {b x +a}{d x +c}\right )^{n}\right )\right )}{\left (d i x +c i \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*g*x+a*g)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^2,x)

[Out]

int((b*g*x+a*g)*(A+B*ln(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^2,x)

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

[Out]

-B*a*g*n*(b*log(b*x + a)/(b*c*d - a*d^2) - b*log(d*x + c)/(b*c*d - a*d^2) + 1/(d^2*x + c*d)) + 1/2*B*b*g*((2*(
d*n*x + c*n)*log(b*x + a)*log(d*x + c) - (d*n*x + c*n)*log(d*x + c)^2 - 2*((d*x + c)*log(d*x + c) + c)*log((b*
x + a)^n) + 2*((d*x + c)*log(d*x + c) + c)*log((d*x + c)^n))/(d^3*x + c*d^2) - 2*integrate((b*d^2*x^2 - b*c^2*
n + a*c*d*n + a*d^2*x + (b*d^2*n*x^2 + a*c*d*n + (b*c*d*n + a*d^2*n)*x)*log(b*x + a))/(b*d^4*x^3 + a*c^2*d^2 +
(2*b*c*d^3 + a*d^4)*x^2 + (b*c^2*d^2 + 2*a*c*d^3)*x), x)) - A*b*g*(c/(d^3*x + c*d^2) + log(d*x + c)/d^2) + B*
a*g*log((b*x/(d*x + c) + a/(d*x + c))^n*e)/(d^2*x + c*d) + A*a*g/(d^2*x + c*d)

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Fricas [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((b*g*x+a*g)*(A+B*log(e*((b*x+a)/(d*x+c))^n))/(d*i*x+c*i)^2,x, algorithm="fricas")

[Out]

integral(-((A + B)*b*g*x + (A + B)*a*g + (B*b*g*n*x + B*a*g*n)*log((b*x + a)/(d*x + c)))/(d^2*x^2 + 2*c*d*x +
c^2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 866 vs. \(2 (151) = 302\).
time = 126.45, size = 866, normalized size = 5.15 \begin {gather*} \frac {1}{2} \, {\left (\frac {{\left (B b^{4} c^{3} g n - 3 \, B a b^{3} c^{2} d g n - \frac {2 \, {\left (b x + a\right )} B b^{3} c^{3} d g n}{d x + c} + 3 \, B a^{2} b^{2} c d^{2} g n + \frac {6 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} g n}{d x + c} - B a^{3} b d^{3} g n - \frac {6 \, {\left (b x + a\right )} B a^{2} b c d^{3} g n}{d x + c} + \frac {2 \, {\left (b x + a\right )} B a^{3} d^{4} g n}{d x + c}\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b^{2} d^{2} - \frac {2 \, {\left (b x + a\right )} b d^{3}}{d x + c} + \frac {{\left (b x + a\right )}^{2} d^{4}}{{\left (d x + c\right )}^{2}}} + \frac {B b^{4} c^{3} g n - 3 \, B a b^{3} c^{2} d g n - \frac {{\left (b x + a\right )} B b^{3} c^{3} d g n}{d x + c} + 3 \, B a^{2} b^{2} c d^{2} g n + \frac {3 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} g n}{d x + c} - B a^{3} b d^{3} g n - \frac {3 \, {\left (b x + a\right )} B a^{2} b c d^{3} g n}{d x + c} + \frac {{\left (b x + a\right )} B a^{3} d^{4} g n}{d x + c} + A b^{4} c^{3} g + B b^{4} c^{3} g - 3 \, A a b^{3} c^{2} d g - 3 \, B a b^{3} c^{2} d g - \frac {2 \, {\left (b x + a\right )} A b^{3} c^{3} d g}{d x + c} - \frac {2 \, {\left (b x + a\right )} B b^{3} c^{3} d g}{d x + c} + 3 \, A a^{2} b^{2} c d^{2} g + 3 \, B a^{2} b^{2} c d^{2} g + \frac {6 \, {\left (b x + a\right )} A a b^{2} c^{2} d^{2} g}{d x + c} + \frac {6 \, {\left (b x + a\right )} B a b^{2} c^{2} d^{2} g}{d x + c} - A a^{3} b d^{3} g - B a^{3} b d^{3} g - \frac {6 \, {\left (b x + a\right )} A a^{2} b c d^{3} g}{d x + c} - \frac {6 \, {\left (b x + a\right )} B a^{2} b c d^{3} g}{d x + c} + \frac {2 \, {\left (b x + a\right )} A a^{3} d^{4} g}{d x + c} + \frac {2 \, {\left (b x + a\right )} B a^{3} d^{4} g}{d x + c}}{b^{2} d^{2} - \frac {2 \, {\left (b x + a\right )} b d^{3}}{d x + c} + \frac {{\left (b x + a\right )}^{2} d^{4}}{{\left (d x + c\right )}^{2}}} + \frac {{\left (B b^{3} c^{3} g n - 3 \, B a b^{2} c^{2} d g n + 3 \, B a^{2} b c d^{2} g n - B a^{3} d^{3} g n\right )} \log \left (-b + \frac {{\left (b x + a\right )} d}{d x + c}\right )}{b d^{2}} - \frac {{\left (B b^{3} c^{3} g n - 3 \, B a b^{2} c^{2} d g n + 3 \, B a^{2} b c d^{2} g n - B a^{3} d^{3} g n\right )} \log \left (\frac {b x + a}{d x + c}\right )}{b d^{2}}\right )} {\left (\frac {b c}{{\left (b c - a d\right )}^{2}} - \frac {a d}{{\left (b c - a d\right )}^{2}}\right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((B*b^4*c^3*g*n - 3*B*a*b^3*c^2*d*g*n - 2*(b*x + a)*B*b^3*c^3*d*g*n/(d*x + c) + 3*B*a^2*b^2*c*d^2*g*n + 6*
(b*x + a)*B*a*b^2*c^2*d^2*g*n/(d*x + c) - B*a^3*b*d^3*g*n - 6*(b*x + a)*B*a^2*b*c*d^3*g*n/(d*x + c) + 2*(b*x +
a)*B*a^3*d^4*g*n/(d*x + c))*log((b*x + a)/(d*x + c))/(b^2*d^2 - 2*(b*x + a)*b*d^3/(d*x + c) + (b*x + a)^2*d^4
/(d*x + c)^2) + (B*b^4*c^3*g*n - 3*B*a*b^3*c^2*d*g*n - (b*x + a)*B*b^3*c^3*d*g*n/(d*x + c) + 3*B*a^2*b^2*c*d^2
*g*n + 3*(b*x + a)*B*a*b^2*c^2*d^2*g*n/(d*x + c) - B*a^3*b*d^3*g*n - 3*(b*x + a)*B*a^2*b*c*d^3*g*n/(d*x + c) +
(b*x + a)*B*a^3*d^4*g*n/(d*x + c) + A*b^4*c^3*g + B*b^4*c^3*g - 3*A*a*b^3*c^2*d*g - 3*B*a*b^3*c^2*d*g - 2*(b*
x + a)*A*b^3*c^3*d*g/(d*x + c) - 2*(b*x + a)*B*b^3*c^3*d*g/(d*x + c) + 3*A*a^2*b^2*c*d^2*g + 3*B*a^2*b^2*c*d^2
*g + 6*(b*x + a)*A*a*b^2*c^2*d^2*g/(d*x + c) + 6*(b*x + a)*B*a*b^2*c^2*d^2*g/(d*x + c) - A*a^3*b*d^3*g - B*a^3
*b*d^3*g - 6*(b*x + a)*A*a^2*b*c*d^3*g/(d*x + c) - 6*(b*x + a)*B*a^2*b*c*d^3*g/(d*x + c) + 2*(b*x + a)*A*a^3*d
^4*g/(d*x + c) + 2*(b*x + a)*B*a^3*d^4*g/(d*x + c))/(b^2*d^2 - 2*(b*x + a)*b*d^3/(d*x + c) + (b*x + a)^2*d^4/(
d*x + c)^2) + (B*b^3*c^3*g*n - 3*B*a*b^2*c^2*d*g*n + 3*B*a^2*b*c*d^2*g*n - B*a^3*d^3*g*n)*log(-b + (b*x + a)*d
/(d*x + c))/(b*d^2) - (B*b^3*c^3*g*n - 3*B*a*b^2*c^2*d*g*n + 3*B*a^2*b*c*d^2*g*n - B*a^3*d^3*g*n)*log((b*x + a
)/(d*x + c))/(b*d^2))*(b*c/(b*c - a*d)^2 - a*d/(b*c - a*d)^2)^2

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a\,g+b\,g\,x\right )\,\left (A+B\,\ln \left (e\,{\left (\frac {a+b\,x}{c+d\,x}\right )}^n\right )\right )}{{\left (c\,i+d\,i\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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