3.98 \(\int \frac{\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=170 \[ -\frac{4 (3 A-2 B) \sin ^3(c+d x)}{3 a^2 d}+\frac{4 (3 A-2 B) \sin (c+d x)}{a^2 d}-\frac{(10 A-7 B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{(10 A-7 B) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac{x (10 A-7 B)}{2 a^2}-\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-((10*A - 7*B)*x)/(2*a^2) + (4*(3*A - 2*B)*Sin[c + d*x])/(a^2*d) - ((10*A - 7*B)*Cos[c + d*x]*Sin[c + d*x])/(2
*a^2*d) - ((10*A - 7*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B)*Cos[c + d*x]^2*Si
n[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) - (4*(3*A - 2*B)*Sin[c + d*x]^3)/(3*a^2*d)

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Rubi [A]  time = 0.319436, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4020, 3787, 2633, 2635, 8} \[ -\frac{4 (3 A-2 B) \sin ^3(c+d x)}{3 a^2 d}+\frac{4 (3 A-2 B) \sin (c+d x)}{a^2 d}-\frac{(10 A-7 B) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac{(10 A-7 B) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac{x (10 A-7 B)}{2 a^2}-\frac{(A-B) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

-((10*A - 7*B)*x)/(2*a^2) + (4*(3*A - 2*B)*Sin[c + d*x])/(a^2*d) - ((10*A - 7*B)*Cos[c + d*x]*Sin[c + d*x])/(2
*a^2*d) - ((10*A - 7*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A - B)*Cos[c + d*x]^2*Si
n[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) - (4*(3*A - 2*B)*Sin[c + d*x]^3)/(3*a^2*d)

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 2633

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

Rule 2635

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

Rule 8

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

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx &=-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \frac{\cos ^3(c+d x) (3 a (2 A-B)-4 a (A-B) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac{(10 A-7 B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac{\int \cos ^3(c+d x) \left (12 a^2 (3 A-2 B)-3 a^2 (10 A-7 B) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac{(10 A-7 B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(10 A-7 B) \int \cos ^2(c+d x) \, dx}{a^2}+\frac{(4 (3 A-2 B)) \int \cos ^3(c+d x) \, dx}{a^2}\\ &=-\frac{(10 A-7 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac{(10 A-7 B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{(10 A-7 B) \int 1 \, dx}{2 a^2}-\frac{(4 (3 A-2 B)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=-\frac{(10 A-7 B) x}{2 a^2}+\frac{4 (3 A-2 B) \sin (c+d x)}{a^2 d}-\frac{(10 A-7 B) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac{(10 A-7 B) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac{(A-B) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac{4 (3 A-2 B) \sin ^3(c+d x)}{3 a^2 d}\\ \end{align*}

Mathematica [B]  time = 0.713892, size = 369, normalized size = 2.17 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-36 d x (10 A-7 B) \cos \left (c+\frac{d x}{2}\right )-36 d x (10 A-7 B) \cos \left (\frac{d x}{2}\right )-156 A \sin \left (c+\frac{d x}{2}\right )+342 A \sin \left (c+\frac{3 d x}{2}\right )+118 A \sin \left (2 c+\frac{3 d x}{2}\right )+30 A \sin \left (2 c+\frac{5 d x}{2}\right )+30 A \sin \left (3 c+\frac{5 d x}{2}\right )-3 A \sin \left (3 c+\frac{7 d x}{2}\right )-3 A \sin \left (4 c+\frac{7 d x}{2}\right )+A \sin \left (4 c+\frac{9 d x}{2}\right )+A \sin \left (5 c+\frac{9 d x}{2}\right )-120 A d x \cos \left (c+\frac{3 d x}{2}\right )-120 A d x \cos \left (2 c+\frac{3 d x}{2}\right )+516 A \sin \left (\frac{d x}{2}\right )+147 B \sin \left (c+\frac{d x}{2}\right )-239 B \sin \left (c+\frac{3 d x}{2}\right )-63 B \sin \left (2 c+\frac{3 d x}{2}\right )-15 B \sin \left (2 c+\frac{5 d x}{2}\right )-15 B \sin \left (3 c+\frac{5 d x}{2}\right )+3 B \sin \left (3 c+\frac{7 d x}{2}\right )+3 B \sin \left (4 c+\frac{7 d x}{2}\right )+84 B d x \cos \left (c+\frac{3 d x}{2}\right )+84 B d x \cos \left (2 c+\frac{3 d x}{2}\right )-381 B \sin \left (\frac{d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^2,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(10*A - 7*B)*d*x*Cos[(d*x)/2] - 36*(10*A - 7*B)*d*x*Cos[c + (d*x)/2] - 120*A*d
*x*Cos[c + (3*d*x)/2] + 84*B*d*x*Cos[c + (3*d*x)/2] - 120*A*d*x*Cos[2*c + (3*d*x)/2] + 84*B*d*x*Cos[2*c + (3*d
*x)/2] + 516*A*Sin[(d*x)/2] - 381*B*Sin[(d*x)/2] - 156*A*Sin[c + (d*x)/2] + 147*B*Sin[c + (d*x)/2] + 342*A*Sin
[c + (3*d*x)/2] - 239*B*Sin[c + (3*d*x)/2] + 118*A*Sin[2*c + (3*d*x)/2] - 63*B*Sin[2*c + (3*d*x)/2] + 30*A*Sin
[2*c + (5*d*x)/2] - 15*B*Sin[2*c + (5*d*x)/2] + 30*A*Sin[3*c + (5*d*x)/2] - 15*B*Sin[3*c + (5*d*x)/2] - 3*A*Si
n[3*c + (7*d*x)/2] + 3*B*Sin[3*c + (7*d*x)/2] - 3*A*Sin[4*c + (7*d*x)/2] + 3*B*Sin[4*c + (7*d*x)/2] + A*Sin[4*
c + (9*d*x)/2] + A*Sin[5*c + (9*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

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Maple [B]  time = 0.11, size = 322, normalized size = 1.9 \begin{align*} -{\frac{A}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{B}{6\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{9\,A}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{7\,B}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+10\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}A}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}B}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{40\,A}{3\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-3}}-8\,{\frac{B \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+6\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-3\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-10\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}+7\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x)

[Out]

-1/6/d/a^2*A*tan(1/2*d*x+1/2*c)^3+1/6/d/a^2*B*tan(1/2*d*x+1/2*c)^3+9/2/d/a^2*A*tan(1/2*d*x+1/2*c)-7/2/d/a^2*B*
tan(1/2*d*x+1/2*c)+10/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A-5/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)
^3*tan(1/2*d*x+1/2*c)^5*B+40/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*A*tan(1/2*d*x+1/2*c)^3-8/d/a^2/(1+tan(1/2*d*x+
1/2*c)^2)^3*B*tan(1/2*d*x+1/2*c)^3+6/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*A*tan(1/2*d*x+1/2*c)-3/d/a^2/(1+tan(1/2*
d*x+1/2*c)^2)^3*B*tan(1/2*d*x+1/2*c)-10/d/a^2*A*arctan(tan(1/2*d*x+1/2*c))+7/d/a^2*arctan(tan(1/2*d*x+1/2*c))*
B

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Maxima [B]  time = 1.53032, size = 502, normalized size = 2.95 \begin{align*} \frac{A{\left (\frac{4 \,{\left (\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{\frac{27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{60 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - B{\left (\frac{6 \,{\left (\frac{3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac{2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac{42 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(A*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 60*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(c
os(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arct
an(sin(d*x + c)/(cos(d*x + c) + 1))/a^2))/d

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Fricas [A]  time = 0.480637, size = 389, normalized size = 2.29 \begin{align*} -\frac{3 \,{\left (10 \, A - 7 \, B\right )} d x \cos \left (d x + c\right )^{2} + 6 \,{\left (10 \, A - 7 \, B\right )} d x \cos \left (d x + c\right ) + 3 \,{\left (10 \, A - 7 \, B\right )} d x -{\left (2 \, A \cos \left (d x + c\right )^{4} -{\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \,{\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} +{\left (66 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 48 \, A - 32 \, B\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/6*(3*(10*A - 7*B)*d*x*cos(d*x + c)^2 + 6*(10*A - 7*B)*d*x*cos(d*x + c) + 3*(10*A - 7*B)*d*x - (2*A*cos(d*x
+ c)^4 - (2*A - 3*B)*cos(d*x + c)^3 + 6*(2*A - B)*cos(d*x + c)^2 + (66*A - 43*B)*cos(d*x + c) + 48*A - 32*B)*s
in(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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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(cos(d*x+c)**3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.29931, size = 259, normalized size = 1.52 \begin{align*} -\frac{\frac{3 \,{\left (d x + c\right )}{\left (10 \, A - 7 \, B\right )}}{a^{2}} - \frac{2 \,{\left (30 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 9 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac{A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 27 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 21 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{6}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(3*(d*x + c)*(10*A - 7*B)/a^2 - 2*(30*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 + 40*A*tan(1
/2*d*x + 1/2*c)^3 - 24*B*tan(1/2*d*x + 1/2*c)^3 + 18*A*tan(1/2*d*x + 1/2*c) - 9*B*tan(1/2*d*x + 1/2*c))/((tan(
1/2*d*x + 1/2*c)^2 + 1)^3*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 - 27*A*a^4*tan(1
/2*d*x + 1/2*c) + 21*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d