3.1.1 \(\int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx\) [1]

Optimal. Leaf size=125 \[ \frac {3}{8} a (A+B) x+\frac {a (5 A+4 B) \sin (c+d x)}{5 d}+\frac {3 a (A+B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (5 A+4 B) \sin ^3(c+d x)}{15 d} \]

[Out]

3/8*a*(A+B)*x+1/5*a*(5*A+4*B)*sin(d*x+c)/d+3/8*a*(A+B)*cos(d*x+c)*sin(d*x+c)/d+1/4*a*(A+B)*cos(d*x+c)^3*sin(d*
x+c)/d+1/5*a*B*cos(d*x+c)^4*sin(d*x+c)/d-1/15*a*(5*A+4*B)*sin(d*x+c)^3/d

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Rubi [A]
time = 0.11, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3047, 3102, 2827, 2713, 2715, 8} \begin {gather*} -\frac {a (5 A+4 B) \sin ^3(c+d x)}{15 d}+\frac {a (5 A+4 B) \sin (c+d x)}{5 d}+\frac {a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 a (A+B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3}{8} a x (A+B)+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(A + B)*x)/8 + (a*(5*A + 4*B)*Sin[c + d*x])/(5*d) + (3*a*(A + B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*(A
+ B)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*B*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - (a*(5*A + 4*B)*Sin[c + d*
x]^3)/(15*d)

Rule 8

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

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx &=\int \cos ^3(c+d x) \left (a A+(a A+a B) \cos (c+d x)+a B \cos ^2(c+d x)\right ) \, dx\\ &=\frac {a B \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^3(c+d x) (a (5 A+4 B)+5 a (A+B) \cos (c+d x)) \, dx\\ &=\frac {a B \cos ^4(c+d x) \sin (c+d x)}{5 d}+(a (A+B)) \int \cos ^4(c+d x) \, dx+\frac {1}{5} (a (5 A+4 B)) \int \cos ^3(c+d x) \, dx\\ &=\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a B \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac {1}{4} (3 a (A+B)) \int \cos ^2(c+d x) \, dx-\frac {(a (5 A+4 B)) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {a (5 A+4 B) \sin (c+d x)}{5 d}+\frac {3 a (A+B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (5 A+4 B) \sin ^3(c+d x)}{15 d}+\frac {1}{8} (3 a (A+B)) \int 1 \, dx\\ &=\frac {3}{8} a (A+B) x+\frac {a (5 A+4 B) \sin (c+d x)}{5 d}+\frac {3 a (A+B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (5 A+4 B) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]
time = 0.29, size = 110, normalized size = 0.88 \begin {gather*} \frac {a \left (180 A c+180 B c+180 A d x+180 B d x+60 (8 A+5 B) \sin (c+d x)-160 A \sin ^3(c+d x)+120 (A+B) \sin (2 (c+d x))+50 B \sin (3 (c+d x))+15 A \sin (4 (c+d x))+15 B \sin (4 (c+d x))+6 B \sin (5 (c+d x))\right )}{480 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(180*A*c + 180*B*c + 180*A*d*x + 180*B*d*x + 60*(8*A + 5*B)*Sin[c + d*x] - 160*A*Sin[c + d*x]^3 + 120*(A +
B)*Sin[2*(c + d*x)] + 50*B*Sin[3*(c + d*x)] + 15*A*Sin[4*(c + d*x)] + 15*B*Sin[4*(c + d*x)] + 6*B*Sin[5*(c + d
*x)]))/(480*d)

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Maple [A]
time = 0.17, size = 128, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {\frac {a B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(128\)
default	\(\frac {\frac {a B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a A \left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(128\)
risch	\(\frac {3 a x A}{8}+\frac {3 a B x}{8}+\frac {3 \sin \left (d x +c \right ) a A}{4 d}+\frac {5 a B \sin \left (d x +c \right )}{8 d}+\frac {a B \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) a A}{32 d}+\frac {\sin \left (4 d x +4 c \right ) a B}{32 d}+\frac {\sin \left (3 d x +3 c \right ) a A}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) a B}{48 d}+\frac {\sin \left (2 d x +2 c \right ) a A}{4 d}+\frac {a B \sin \left (2 d x +2 c \right )}{4 d}\)	\(150\)
norman	\(\frac {\frac {3 a \left (A +B \right ) x}{8}+\frac {15 a \left (A +B \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {15 a \left (A +B \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a \left (A +B \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {15 a \left (A +B \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {3 a \left (A +B \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {4 a \left (25 A +29 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {a \left (29 A +13 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {a \left (35 A +19 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {13 \left (A +B \right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {3 \left (A +B \right ) a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(225\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d*(1/5*a*B*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a*A*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+
3/8*d*x+3/8*c)+a*B*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/3*a*A*(cos(d*x+c)^2+2)*sin(d
*x+c))

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Maxima [A]
time = 0.26, size = 124, normalized size = 0.99 \begin {gather*} -\frac {160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a}{480 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))
*A*a - 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a - 15*(12*d*x + 12*c + sin(4*d*x + 4*c)
+ 8*sin(2*d*x + 2*c))*B*a)/d

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Fricas [A]
time = 0.41, size = 88, normalized size = 0.70 \begin {gather*} \frac {45 \, {\left (A + B\right )} a d x + {\left (24 \, B a \cos \left (d x + c\right )^{4} + 30 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{2} + 45 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 16 \, {\left (5 \, A + 4 \, B\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(45*(A + B)*a*d*x + (24*B*a*cos(d*x + c)^4 + 30*(A + B)*a*cos(d*x + c)^3 + 8*(5*A + 4*B)*a*cos(d*x + c)^
2 + 45*(A + B)*a*cos(d*x + c) + 16*(5*A + 4*B)*a)*sin(d*x + c))/d

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (117) = 234\).
time = 0.34, size = 333, normalized size = 2.66 \begin {gather*} \begin {cases} \frac {3 A a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 A a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {A a \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 B a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 B a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {8 B a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 B a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {B a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 B a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right ) \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)),x)

[Out]

Piecewise((3*A*a*x*sin(c + d*x)**4/8 + 3*A*a*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*a*x*cos(c + d*x)**4/8 +
3*A*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*A*a*sin(c + d*x)**3/(3*d) + 5*A*a*sin(c + d*x)*cos(c + d*x)**3/(
8*d) + A*a*sin(c + d*x)*cos(c + d*x)**2/d + 3*B*a*x*sin(c + d*x)**4/8 + 3*B*a*x*sin(c + d*x)**2*cos(c + d*x)**
2/4 + 3*B*a*x*cos(c + d*x)**4/8 + 8*B*a*sin(c + d*x)**5/(15*d) + 4*B*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) +
3*B*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + B*a*sin(c + d*x)*cos(c + d*x)**4/d + 5*B*a*sin(c + d*x)*cos(c + d*
x)**3/(8*d), Ne(d, 0)), (x*(A + B*cos(c))*(a*cos(c) + a)*cos(c)**3, True))

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Giac [A]
time = 0.46, size = 112, normalized size = 0.90 \begin {gather*} \frac {3}{8} \, {\left (A a + B a\right )} x + \frac {B a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {{\left (A a + B a\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (4 \, A a + 5 \, B a\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a + B a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (6 \, A a + 5 \, B a\right )} \sin \left (d x + c\right )}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(A*a + B*a)*x + 1/80*B*a*sin(5*d*x + 5*c)/d + 1/32*(A*a + B*a)*sin(4*d*x + 4*c)/d + 1/48*(4*A*a + 5*B*a)*s
in(3*d*x + 3*c)/d + 1/4*(A*a + B*a)*sin(2*d*x + 2*c)/d + 1/8*(6*A*a + 5*B*a)*sin(d*x + c)/d

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Mupad [B]
time = 1.55, size = 236, normalized size = 1.89 \begin {gather*} \frac {\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {29\,A\,a}{6}+\frac {13\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,a}{3}+\frac {116\,B\,a}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {35\,A\,a}{6}+\frac {19\,B\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+\frac {13\,B\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,a\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,\left (A+B\right )}{4\,d}+\frac {3\,a\,\mathrm {atan}\left (\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{4\,\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}\right )}\right )\,\left (A+B\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((13*A*a)/4 + (13*B*a)/4) + tan(c/2 + (d*x)/2)^9*((3*A*a)/4 + (3*B*a)/4) + tan(c/2 + (d*x)
/2)^7*((29*A*a)/6 + (13*B*a)/6) + tan(c/2 + (d*x)/2)^3*((35*A*a)/6 + (19*B*a)/6) + tan(c/2 + (d*x)/2)^5*((20*A
*a)/3 + (116*B*a)/15))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(
c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (3*a*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(A + B))/(4*d) + (3
*a*atan((3*a*tan(c/2 + (d*x)/2)*(A + B))/(4*((3*A*a)/4 + (3*B*a)/4)))*(A + B))/(4*d)

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