3.242 \(\int \frac{\csc ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=138 \[ \frac{4 A \cot (c+d x)}{a^3 d}+\frac{164 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}+\frac{29 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}+\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac{19 A \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{A \cot (c+d x) \csc (c+d x)}{2 a^3 d} \]

[Out]

(-19*A*ArcTanh[Cos[c + d*x]])/(2*a^3*d) + (4*A*Cot[c + d*x])/(a^3*d) - (A*Cot[c + d*x]*Csc[c + d*x])/(2*a^3*d)
+ (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) + (29*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x])^2) + (
164*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x]))

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Rubi [A]  time = 0.22469, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {2966, 3770, 3767, 8, 3768, 2650, 2648} \[ \frac{4 A \cot (c+d x)}{a^3 d}+\frac{164 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)}+\frac{29 A \cos (c+d x)}{15 a^3 d (\sin (c+d x)+1)^2}+\frac{2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac{19 A \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}-\frac{A \cot (c+d x) \csc (c+d x)}{2 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^3*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(-19*A*ArcTanh[Cos[c + d*x]])/(2*a^3*d) + (4*A*Cot[c + d*x])/(a^3*d) - (A*Cot[c + d*x]*Csc[c + d*x])/(2*a^3*d)
+ (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) + (29*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x])^2) + (
164*A*Cos[c + d*x])/(15*a^3*d*(1 + Sin[c + d*x]))

Rule 2966

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\csc ^3(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=\int \left (\frac{9 A \csc (c+d x)}{a^3}-\frac{4 A \csc ^2(c+d x)}{a^3}+\frac{A \csc ^3(c+d x)}{a^3}-\frac{2 A}{a^3 (1+\sin (c+d x))^3}-\frac{5 A}{a^3 (1+\sin (c+d x))^2}-\frac{9 A}{a^3 (1+\sin (c+d x))}\right ) \, dx\\ &=\frac{A \int \csc ^3(c+d x) \, dx}{a^3}-\frac{(2 A) \int \frac{1}{(1+\sin (c+d x))^3} \, dx}{a^3}-\frac{(4 A) \int \csc ^2(c+d x) \, dx}{a^3}-\frac{(5 A) \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{a^3}+\frac{(9 A) \int \csc (c+d x) \, dx}{a^3}-\frac{(9 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{a^3}\\ &=-\frac{9 A \tanh ^{-1}(\cos (c+d x))}{a^3 d}-\frac{A \cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac{5 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}+\frac{9 A \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac{A \int \csc (c+d x) \, dx}{2 a^3}-\frac{(4 A) \int \frac{1}{(1+\sin (c+d x))^2} \, dx}{5 a^3}-\frac{(5 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{3 a^3}+\frac{(4 A) \operatorname{Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=-\frac{19 A \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{4 A \cot (c+d x)}{a^3 d}-\frac{A \cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac{29 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}+\frac{32 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))}-\frac{(4 A) \int \frac{1}{1+\sin (c+d x)} \, dx}{15 a^3}\\ &=-\frac{19 A \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{4 A \cot (c+d x)}{a^3 d}-\frac{A \cot (c+d x) \csc (c+d x)}{2 a^3 d}+\frac{2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac{29 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))^2}+\frac{164 A \cos (c+d x)}{15 a^3 d (1+\sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 4.23625, size = 245, normalized size = 1.78 \[ \frac{A \left (-240 \tan \left (\frac{1}{2} (c+d x)\right )+240 \cot \left (\frac{1}{2} (c+d x)\right )-15 \csc ^2\left (\frac{1}{2} (c+d x)\right )+15 \sec ^2\left (\frac{1}{2} (c+d x)\right )+1140 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )-1140 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-\frac{2624 \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{232}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{464 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{48}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{96 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5}\right )}{120 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^3*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(A*(240*Cot[(c + d*x)/2] - 15*Csc[(c + d*x)/2]^2 - 1140*Log[Cos[(c + d*x)/2]] + 1140*Log[Sin[(c + d*x)/2]] + 1
5*Sec[(c + d*x)/2]^2 - (96*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5 + 48/(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^4 - (464*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + 232/(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^2 - (2624*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - 240*Tan[(c + d*x)/2]))/
(120*a^3*d)

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Maple [A]  time = 0.197, size = 209, normalized size = 1.5 \begin{align*}{\frac{A}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3}}}+{\frac{16\,A}{5\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}-8\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{4}}}+{\frac{52\,A}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-18\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}+32\,{\frac{A}{d{a}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-{\frac{A}{8\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+2\,{\frac{A}{d{a}^{3}\tan \left ( 1/2\,dx+c/2 \right ) }}+{\frac{19\,A}{2\,d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x)

[Out]

1/8/d*A/a^3*tan(1/2*d*x+1/2*c)^2-2/d*A/a^3*tan(1/2*d*x+1/2*c)+16/5/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^5-8/d*A/a^3/
(tan(1/2*d*x+1/2*c)+1)^4+52/3/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^3-18/d*A/a^3/(tan(1/2*d*x+1/2*c)+1)^2+32/d*A/a^3/
(tan(1/2*d*x+1/2*c)+1)-1/8/d*A/a^3/tan(1/2*d*x+1/2*c)^2+2/d*A/a^3/tan(1/2*d*x+1/2*c)+19/2/d*A/a^3*ln(tan(1/2*d
*x+1/2*c))

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Maxima [B]  time = 1.02398, size = 840, normalized size = 6.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/120*(12*A*((121*sin(d*x + c)/(cos(d*x + c) + 1) + 410*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 610*sin(d*x + c)
^3/(cos(d*x + c) + 1)^3 + 425*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 125*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 +
5)/(a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(c
os(d*x + c) + 1)^3 + 10*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 5*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 +
a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + 30*log(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 - 5*sin(d*x + c)/(a^3*(
cos(d*x + c) + 1))) + A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 2782*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 941
0*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 13645*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 9285*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5 + 2580*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 15)/(a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5*a
^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 10*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 10*a^3*sin(d*x + c)^5/(c
os(d*x + c) + 1)^5 + 5*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7) - 15
*(12*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^2/(cos(d*x + c) + 1)^2)/a^3 + 780*log(sin(d*x + c)/(cos(d*
x + c) + 1))/a^3))/d

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Fricas [B]  time = 2.04622, size = 1327, normalized size = 9.62 \begin{align*} \frac{896 \, A \cos \left (d x + c\right )^{5} - 1222 \, A \cos \left (d x + c\right )^{4} - 3218 \, A \cos \left (d x + c\right )^{3} + 1168 \, A \cos \left (d x + c\right )^{2} + 2292 \, A \cos \left (d x + c\right ) - 285 \,{\left (A \cos \left (d x + c\right )^{5} + 3 \, A \cos \left (d x + c\right )^{4} - 3 \, A \cos \left (d x + c\right )^{3} - 7 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) +{\left (A \cos \left (d x + c\right )^{4} - 2 \, A \cos \left (d x + c\right )^{3} - 5 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + 4 \, A\right )} \sin \left (d x + c\right ) + 4 \, A\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) + 285 \,{\left (A \cos \left (d x + c\right )^{5} + 3 \, A \cos \left (d x + c\right )^{4} - 3 \, A \cos \left (d x + c\right )^{3} - 7 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) +{\left (A \cos \left (d x + c\right )^{4} - 2 \, A \cos \left (d x + c\right )^{3} - 5 \, A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + 4 \, A\right )} \sin \left (d x + c\right ) + 4 \, A\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) - 2 \,{\left (448 \, A \cos \left (d x + c\right )^{4} + 1059 \, A \cos \left (d x + c\right )^{3} - 550 \, A \cos \left (d x + c\right )^{2} - 1134 \, A \cos \left (d x + c\right ) + 12 \, A\right )} \sin \left (d x + c\right ) + 24 \, A}{60 \,{\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} - 3 \, a^{3} d \cos \left (d x + c\right )^{3} - 7 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + 4 \, a^{3} d +{\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{3} - 5 \, a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(896*A*cos(d*x + c)^5 - 1222*A*cos(d*x + c)^4 - 3218*A*cos(d*x + c)^3 + 1168*A*cos(d*x + c)^2 + 2292*A*co
s(d*x + c) - 285*(A*cos(d*x + c)^5 + 3*A*cos(d*x + c)^4 - 3*A*cos(d*x + c)^3 - 7*A*cos(d*x + c)^2 + 2*A*cos(d*
x + c) + (A*cos(d*x + c)^4 - 2*A*cos(d*x + c)^3 - 5*A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + 4*A)*sin(d*x + c) +
4*A)*log(1/2*cos(d*x + c) + 1/2) + 285*(A*cos(d*x + c)^5 + 3*A*cos(d*x + c)^4 - 3*A*cos(d*x + c)^3 - 7*A*cos(d
*x + c)^2 + 2*A*cos(d*x + c) + (A*cos(d*x + c)^4 - 2*A*cos(d*x + c)^3 - 5*A*cos(d*x + c)^2 + 2*A*cos(d*x + c)
+ 4*A)*sin(d*x + c) + 4*A)*log(-1/2*cos(d*x + c) + 1/2) - 2*(448*A*cos(d*x + c)^4 + 1059*A*cos(d*x + c)^3 - 55
0*A*cos(d*x + c)^2 - 1134*A*cos(d*x + c) + 12*A)*sin(d*x + c) + 24*A)/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x
+ c)^4 - 3*a^3*d*cos(d*x + c)^3 - 7*a^3*d*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c) + 4*a^3*d + (a^3*d*cos(d*x + c
)^4 - 2*a^3*d*cos(d*x + c)^3 - 5*a^3*d*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c) + 4*a^3*d)*sin(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.17636, size = 243, normalized size = 1.76 \begin{align*} \frac{\frac{1140 \, A \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{15 \,{\left (114 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 16 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A\right )}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}} + \frac{15 \,{\left (A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 16 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a^{6}} + \frac{16 \,{\left (240 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 825 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 1165 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 755 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 199 \, A\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/120*(1140*A*log(abs(tan(1/2*d*x + 1/2*c)))/a^3 - 15*(114*A*tan(1/2*d*x + 1/2*c)^2 - 16*A*tan(1/2*d*x + 1/2*c
) + A)/(a^3*tan(1/2*d*x + 1/2*c)^2) + 15*(A*a^3*tan(1/2*d*x + 1/2*c)^2 - 16*A*a^3*tan(1/2*d*x + 1/2*c))/a^6 +
16*(240*A*tan(1/2*d*x + 1/2*c)^4 + 825*A*tan(1/2*d*x + 1/2*c)^3 + 1165*A*tan(1/2*d*x + 1/2*c)^2 + 755*A*tan(1/
2*d*x + 1/2*c) + 199*A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d