3.1464 \(\int \frac{\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=222 \[ -\frac{2 a^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}+\frac{4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}-\frac{a^4 \cos (c+d x)}{b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)}-\frac{x}{b^2} \]

[Out]

-(x/b^2) - (2*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(5/2)*d) + (4*a^3*(a^2 -
2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(5/2)*d) + Cos[c + d*x]/(2*(a + b)^2
*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/(2*(a - b)^2*d*(1 + Sin[c + d*x])) - (a^4*Cos[c + d*x])/(b*(a^2 - b^2)^2
*d*(a + b*Sin[c + d*x]))

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Rubi [A]  time = 0.364495, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {2897, 2648, 2664, 12, 2660, 618, 204} \[ -\frac{2 a^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}+\frac{4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \left (a^2-b^2\right )^{5/2}}-\frac{a^4 \cos (c+d x)}{b d \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}+\frac{\cos (c+d x)}{2 d (a+b)^2 (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 d (a-b)^2 (\sin (c+d x)+1)}-\frac{x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]

[Out]

-(x/b^2) - (2*a^5*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(5/2)*d) + (4*a^3*(a^2 -
2*b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(5/2)*d) + Cos[c + d*x]/(2*(a + b)^2
*d*(1 - Sin[c + d*x])) - Cos[c + d*x]/(2*(a - b)^2*d*(1 + Sin[c + d*x])) - (a^4*Cos[c + d*x])/(b*(a^2 - b^2)^2
*d*(a + b*Sin[c + d*x]))

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

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]

Rule 2664

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{\sin ^2(c+d x) \tan ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (-\frac{1}{b^2}-\frac{1}{2 (a+b)^2 (-1+\sin (c+d x))}+\frac{1}{2 (a-b)^2 (1+\sin (c+d x))}+\frac{a^4}{b^2 \left (-a^2+b^2\right ) (a+b \sin (c+d x))^2}+\frac{2 \left (a^5-2 a^3 b^2\right )}{b^2 \left (a^2-b^2\right )^2 (a+b \sin (c+d x))}\right ) \, dx\\ &=-\frac{x}{b^2}+\frac{\int \frac{1}{1+\sin (c+d x)} \, dx}{2 (a-b)^2}-\frac{\int \frac{1}{-1+\sin (c+d x)} \, dx}{2 (a+b)^2}+\frac{\left (2 a^3 \left (a^2-2 b^2\right )\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^2}-\frac{a^4 \int \frac{1}{(a+b \sin (c+d x))^2} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=-\frac{x}{b^2}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac{a^4 \int \frac{a}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^2}+\frac{\left (4 a^3 \left (a^2-2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac{x}{b^2}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac{a^5 \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^2 \left (a^2-b^2\right )^2}-\frac{\left (8 a^3 \left (a^2-2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac{x}{b^2}+\frac{4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}-\frac{\left (2 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac{x}{b^2}+\frac{4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}+\frac{\left (4 a^5\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (c+d x)\right )\right )}{b^2 \left (a^2-b^2\right )^2 d}\\ &=-\frac{x}{b^2}-\frac{2 a^5 \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} d}+\frac{4 a^3 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2} d}+\frac{\cos (c+d x)}{2 (a+b)^2 d (1-\sin (c+d x))}-\frac{\cos (c+d x)}{2 (a-b)^2 d (1+\sin (c+d x))}-\frac{a^4 \cos (c+d x)}{b \left (a^2-b^2\right )^2 d (a+b \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 2.03369, size = 236, normalized size = 1.06 \[ \frac{-\frac{-2 a^2 b^2 (c+d x)+a^4 (c+d x)+2 a b^3+b^4 (c+d x)}{\left (b^3-a^2 b\right )^2}+\frac{2 a^3 \left (a^2-4 b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 \left (a^2-b^2\right )^{5/2}}-\frac{a^4 \cos (c+d x)}{b (a-b)^2 (a+b)^2 (a+b \sin (c+d x))}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]

[Out]

(-((2*a*b^3 + a^4*(c + d*x) - 2*a^2*b^2*(c + d*x) + b^4*(c + d*x))/(-(a^2*b) + b^3)^2) + (2*a^3*(a^2 - 4*b^2)*
ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^2*(a^2 - b^2)^(5/2)) + Sin[(c + d*x)/2]/((a + b)^2*(Cos[(
c + d*x)/2] - Sin[(c + d*x)/2])) + Sin[(c + d*x)/2]/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) - (a^4*C
os[c + d*x])/((a - b)^2*b*(a + b)^2*(a + b*Sin[c + d*x])))/d

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Maple [A]  time = 0.131, size = 303, normalized size = 1.4 \begin{align*} -{\frac{1}{d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{b}^{2}}}-2\,{\frac{{a}^{3}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}-2\,{\frac{{a}^{4}}{bd \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2} \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+2\,{\frac{{a}^{5}}{d{b}^{2} \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-8\,{\frac{{a}^{3}}{d \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{d \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^2,x)

[Out]

-1/d/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)-2/d/b^2*arctan(tan(1/2*d*x+1/2*c))-2/d*a^3/(a-b)^2/(a+b)^2/(tan(1/2*d*x+1/
2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*tan(1/2*d*x+1/2*c)-2/d*a^4/b/(a-b)^2/(a+b)^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(
1/2*d*x+1/2*c)*b+a)+2/d*a^5/b^2/(a-b)^2/(a+b)^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b
^2)^(1/2))-8/d*a^3/(a-b)^2/(a+b)^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-1/
d/(a-b)^2/(tan(1/2*d*x+1/2*c)+1)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01788, size = 1577, normalized size = 7.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[-1/2*(2*a^4*b^3 - 4*a^2*b^5 + 2*b^7 + 2*(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*x*cos(d*x + c) + 2*(a^6*b - b
^7)*cos(d*x + c)^2 - ((a^5*b - 4*a^3*b^3)*cos(d*x + c)*sin(d*x + c) + (a^6 - 4*a^4*b^2)*cos(d*x + c))*sqrt(-a^
2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c)
+ b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 2*(a^5*b^2 - 2*a^
3*b^4 + a*b^6 - (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*x*cos(d*x + c))*sin(d*x + c))/((a^6*b^3 - 3*a^4*b^5 +
3*a^2*b^7 - b^9)*d*cos(d*x + c)*sin(d*x + c) + (a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*d*cos(d*x + c)), -(a^
4*b^3 - 2*a^2*b^5 + b^7 + (a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*x*cos(d*x + c) + (a^6*b - b^7)*cos(d*x + c)^
2 + ((a^5*b - 4*a^3*b^3)*cos(d*x + c)*sin(d*x + c) + (a^6 - 4*a^4*b^2)*cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(
a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - (a^5*b^2 - 2*a^3*b^4 + a*b^6 - (a^6*b - 3*a^4*b^3 + 3*a^
2*b^5 - b^7)*d*x*cos(d*x + c))*sin(d*x + c))/((a^6*b^3 - 3*a^4*b^5 + 3*a^2*b^7 - b^9)*d*cos(d*x + c)*sin(d*x +
c) + (a^7*b^2 - 3*a^5*b^4 + 3*a^3*b^6 - a*b^8)*d*cos(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(sec(d*x+c)**2*sin(d*x+c)**4/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.218, size = 356, normalized size = 1.6 \begin{align*} \frac{\frac{2 \,{\left (a^{5} - 4 \, a^{3} b^{2}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} \sqrt{a^{2} - b^{2}}} - \frac{2 \,{\left (2 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{4} - 2 \, a^{2} b^{2}\right )}}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}} - \frac{d x + c}{b^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

(2*(a^5 - 4*a^3*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 -
b^2)))/((a^4*b^2 - 2*a^2*b^4 + b^6)*sqrt(a^2 - b^2)) - 2*(2*a^3*b*tan(1/2*d*x + 1/2*c)^3 + a*b^3*tan(1/2*d*x
+ 1/2*c)^3 + a^4*tan(1/2*d*x + 1/2*c)^2 + 2*b^4*tan(1/2*d*x + 1/2*c)^2 - 3*a*b^3*tan(1/2*d*x + 1/2*c) - a^4 -
2*a^2*b^2)/((a^4*b - 2*a^2*b^3 + b^5)*(a*tan(1/2*d*x + 1/2*c)^4 + 2*b*tan(1/2*d*x + 1/2*c)^3 - 2*b*tan(1/2*d*x
+ 1/2*c) - a)) - (d*x + c)/b^2)/d