3.403 \(\int \frac{\sec ^2(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=169 \[ \frac{(19 B-75 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(B-9 C) \tan (c+d x)}{4 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac{(5 B-13 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

((19*B - 75*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + ((B
- C)*Sec[c + d*x]^2*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((5*B - 13*C)*Tan[c + d*x])/(16*a*d*(a +
a*Sec[c + d*x])^(3/2)) - ((B - 9*C)*Tan[c + d*x])/(4*a^2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.564363, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 42, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4072, 4019, 4008, 4001, 3795, 203} \[ \frac{(19 B-75 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(B-9 C) \tan (c+d x)}{4 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(B-C) \tan (c+d x) \sec ^2(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}-\frac{(5 B-13 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((19*B - 75*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + ((B
- C)*Sec[c + d*x]^2*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((5*B - 13*C)*Tan[c + d*x])/(16*a*d*(a +
a*Sec[c + d*x])^(3/2)) - ((B - 9*C)*Tan[c + d*x])/(4*a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 4072

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

Rule 4019

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

Rule 4008

Int[csc[(e_.) + (f_.)*(x_)]^2*(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)/(b*f*(2*m + 1)), x] + Dist[1/(b^2*(2*
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b*B*(2*m + 1)*Csc[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rule 4001

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

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^2(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=\int \frac{\sec ^3(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\\ &=\frac{(B-C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\sec ^2(c+d x) \left (2 a (B-C)-\frac{1}{2} a (B-9 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(B-C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(5 B-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{\int \frac{\sec (c+d x) \left (-\frac{3}{4} a^2 (5 B-13 C)+a^2 (B-9 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(B-C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(5 B-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(B-9 C) \tan (c+d x)}{4 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(19 B-75 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=\frac{(B-C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(5 B-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(B-9 C) \tan (c+d x)}{4 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(19 B-75 C) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=\frac{(19 B-75 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(B-C) \sec ^2(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(5 B-13 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(B-9 C) \tan (c+d x)}{4 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.47091, size = 144, normalized size = 0.85 \[ \frac{\tan (c+d x) \left (\sqrt{1-\sec (c+d x)} \left ((85 C-13 B) \sec (c+d x)-9 B+32 C \sec ^2(c+d x)+49 C\right )+2 \sqrt{2} (19 B-75 C) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{16 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^2*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((2*Sqrt[2]*(19*B - 75*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2 + Sqrt[1 -
Sec[c + d*x]]*(-9*B + 49*C + (-13*B + 85*C)*Sec[c + d*x] + 32*C*Sec[c + d*x]^2))*Tan[c + d*x])/(16*d*Sqrt[1 -
Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.281, size = 597, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/32/d/a^3*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(19*B*cos(d*x+c)^2*sin(d*x+c)*ln(((-2*cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-75*C*cos(d
*x+c)^2*sin(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)+38*B*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x
+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-150*C*sin(d*x+c)*cos(d*x+c)*ln(((-2*cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+18*B*cos(d*x+c)^3+19*
B*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*sin(d*x+c)-98*C*cos(d*x+c)^3-75*C*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin
(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+8*B*cos(d*x+c)^2-72*C*cos(d*x+c)^2-26*B*cos(d*x+c)+10
6*C*cos(d*x+c)+64*C)/sin(d*x+c)^5

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Maxima [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(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.627088, size = 1273, normalized size = 7.53 \begin{align*} \left [\frac{\sqrt{2}{\left ({\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right ) + 19 \, B - 75 \, C\right )} \sqrt{-a} \log \left (-\frac{2 \, \sqrt{2} \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \,{\left ({\left (9 \, B - 49 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (13 \, B - 85 \, C\right )} \cos \left (d x + c\right ) - 32 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac{\sqrt{2}{\left ({\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (19 \, B - 75 \, C\right )} \cos \left (d x + c\right ) + 19 \, B - 75 \, C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right ) + 2 \,{\left ({\left (9 \, B - 49 \, C\right )} \cos \left (d x + c\right )^{2} +{\left (13 \, B - 85 \, C\right )} \cos \left (d x + c\right ) - 32 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/64*(sqrt(2)*((19*B - 75*C)*cos(d*x + c)^3 + 3*(19*B - 75*C)*cos(d*x + c)^2 + 3*(19*B - 75*C)*cos(d*x + c) +
19*B - 75*C)*sqrt(-a)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x +
c) - 3*a*cos(d*x + c)^2 - 2*a*cos(d*x + c) + a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((9*B - 49*C)*cos(
d*x + c)^2 + (13*B - 85*C)*cos(d*x + c) - 32*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*c
os(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/32*(sqrt(2)*((19*B - 75*C)*cos(d*x
+ c)^3 + 3*(19*B - 75*C)*cos(d*x + c)^2 + 3*(19*B - 75*C)*cos(d*x + c) + 19*B - 75*C)*sqrt(a)*arctan(sqrt(2)*s
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*((9*B - 49*C)*cos(d*x + c)^2 +
(13*B - 85*C)*cos(d*x + c) - 32*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^
3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (B + C \sec{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)**3/(a*(sec(c + d*x) + 1))**(5/2), x)

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Giac [A]  time = 9.55333, size = 390, normalized size = 2.31 \begin{align*} -\frac{\frac{{\left ({\left (\frac{2 \,{\left (\sqrt{2} B a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - \sqrt{2} C a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{8}} + \frac{9 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - 17 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{11 \, \sqrt{2} B a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right ) - 83 \, \sqrt{2} C a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}{a^{8}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} - \frac{{\left (19 \, \sqrt{2} B - 75 \, \sqrt{2} C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/32*(((2*(sqrt(2)*B*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - sqrt(2)*C*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan
(1/2*d*x + 1/2*c)^2/a^8 + (9*sqrt(2)*B*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 17*sqrt(2)*C*a^6*sgn(tan(1/2*d*x
+ 1/2*c)^2 - 1))/a^8)*tan(1/2*d*x + 1/2*c)^2 - (11*sqrt(2)*B*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 83*sqrt(2)*
C*a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))/a^8)*tan(1/2*d*x + 1/2*c)/sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a) - (19*sqr
t(2)*B - 75*sqrt(2)*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a
)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d