3.30 \(\int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=150 \[ -\frac {4 a^4 (5 A+4 B) \sin ^3(c+d x)}{15 d}+\frac {8 a^4 (5 A+4 B) \sin (c+d x)}{5 d}+\frac {a^4 (5 A+4 B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {27 a^4 (5 A+4 B) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {7}{8} a^4 x (5 A+4 B)+\frac {B \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d} \]

[Out]

7/8*a^4*(5*A+4*B)*x+8/5*a^4*(5*A+4*B)*sin(d*x+c)/d+27/40*a^4*(5*A+4*B)*cos(d*x+c)*sin(d*x+c)/d+1/20*a^4*(5*A+4
*B)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*B*(a+a*cos(d*x+c))^4*sin(d*x+c)/d-4/15*a^4*(5*A+4*B)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.14, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2751, 2645, 2637, 2635, 8, 2633} \[ -\frac {4 a^4 (5 A+4 B) \sin ^3(c+d x)}{15 d}+\frac {8 a^4 (5 A+4 B) \sin (c+d x)}{5 d}+\frac {a^4 (5 A+4 B) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {27 a^4 (5 A+4 B) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {7}{8} a^4 x (5 A+4 B)+\frac {B \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x]),x]

[Out]

(7*a^4*(5*A + 4*B)*x)/8 + (8*a^4*(5*A + 4*B)*Sin[c + d*x])/(5*d) + (27*a^4*(5*A + 4*B)*Cos[c + d*x]*Sin[c + d*
x])/(40*d) + (a^4*(5*A + 4*B)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (B*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(5
*d) - (4*a^4*(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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2645

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

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^4 (A+B \cos (c+d x)) \, dx &=\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {1}{5} (5 A+4 B) \int (a+a \cos (c+d x))^4 \, dx\\ &=\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {1}{5} (5 A+4 B) \int \left (a^4+4 a^4 \cos (c+d x)+6 a^4 \cos ^2(c+d x)+4 a^4 \cos ^3(c+d x)+a^4 \cos ^4(c+d x)\right ) \, dx\\ &=\frac {1}{5} a^4 (5 A+4 B) x+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {1}{5} \left (a^4 (5 A+4 B)\right ) \int \cos ^4(c+d x) \, dx+\frac {1}{5} \left (4 a^4 (5 A+4 B)\right ) \int \cos (c+d x) \, dx+\frac {1}{5} \left (4 a^4 (5 A+4 B)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{5} \left (6 a^4 (5 A+4 B)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {1}{5} a^4 (5 A+4 B) x+\frac {4 a^4 (5 A+4 B) \sin (c+d x)}{5 d}+\frac {3 a^4 (5 A+4 B) \cos (c+d x) \sin (c+d x)}{5 d}+\frac {a^4 (5 A+4 B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {1}{20} \left (3 a^4 (5 A+4 B)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{5} \left (3 a^4 (5 A+4 B)\right ) \int 1 \, dx-\frac {\left (4 a^4 (5 A+4 B)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {4}{5} a^4 (5 A+4 B) x+\frac {8 a^4 (5 A+4 B) \sin (c+d x)}{5 d}+\frac {27 a^4 (5 A+4 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a^4 (5 A+4 B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}-\frac {4 a^4 (5 A+4 B) \sin ^3(c+d x)}{15 d}+\frac {1}{40} \left (3 a^4 (5 A+4 B)\right ) \int 1 \, dx\\ &=\frac {7}{8} a^4 (5 A+4 B) x+\frac {8 a^4 (5 A+4 B) \sin (c+d x)}{5 d}+\frac {27 a^4 (5 A+4 B) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a^4 (5 A+4 B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {B (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}-\frac {4 a^4 (5 A+4 B) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 108, normalized size = 0.72 \[ \frac {a^4 (420 (8 A+7 B) \sin (c+d x)+120 (7 A+8 B) \sin (2 (c+d x))+160 A \sin (3 (c+d x))+15 A \sin (4 (c+d x))+2100 A d x+290 B \sin (3 (c+d x))+60 B \sin (4 (c+d x))+6 B \sin (5 (c+d x))+1680 B d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^4*(A + B*Cos[c + d*x]),x]

[Out]

(a^4*(2100*A*d*x + 1680*B*d*x + 420*(8*A + 7*B)*Sin[c + d*x] + 120*(7*A + 8*B)*Sin[2*(c + d*x)] + 160*A*Sin[3*
(c + d*x)] + 290*B*Sin[3*(c + d*x)] + 15*A*Sin[4*(c + d*x)] + 60*B*Sin[4*(c + d*x)] + 6*B*Sin[5*(c + d*x)]))/(
480*d)

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fricas [A]  time = 0.62, size = 110, normalized size = 0.73 \[ \frac {105 \, {\left (5 \, A + 4 \, B\right )} a^{4} d x + {\left (24 \, B a^{4} \cos \left (d x + c\right )^{4} + 30 \, {\left (A + 4 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} + 16 \, {\left (10 \, A + 17 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \, {\left (27 \, A + 28 \, B\right )} a^{4} \cos \left (d x + c\right ) + 8 \, {\left (100 \, A + 83 \, B\right )} a^{4}\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(105*(5*A + 4*B)*a^4*d*x + (24*B*a^4*cos(d*x + c)^4 + 30*(A + 4*B)*a^4*cos(d*x + c)^3 + 16*(10*A + 17*B)
*a^4*cos(d*x + c)^2 + 15*(27*A + 28*B)*a^4*cos(d*x + c) + 8*(100*A + 83*B)*a^4)*sin(d*x + c))/d

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giac [A]  time = 1.39, size = 139, normalized size = 0.93 \[ \frac {B a^{4} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {7}{8} \, {\left (5 \, A a^{4} + 4 \, B a^{4}\right )} x + \frac {{\left (A a^{4} + 4 \, B a^{4}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (16 \, A a^{4} + 29 \, B a^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (7 \, A a^{4} + 8 \, B a^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {7 \, {\left (8 \, A a^{4} + 7 \, B a^{4}\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

1/80*B*a^4*sin(5*d*x + 5*c)/d + 7/8*(5*A*a^4 + 4*B*a^4)*x + 1/32*(A*a^4 + 4*B*a^4)*sin(4*d*x + 4*c)/d + 1/48*(
16*A*a^4 + 29*B*a^4)*sin(3*d*x + 3*c)/d + 1/4*(7*A*a^4 + 8*B*a^4)*sin(2*d*x + 2*c)/d + 7/8*(8*A*a^4 + 7*B*a^4)
*sin(d*x + c)/d

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maple [A]  time = 0.06, size = 248, normalized size = 1.65 \[ \frac {\frac {a^{4} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+A \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+4 a^{4} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 A \,a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 a^{4} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+6 A \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{4} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 A \,a^{4} \sin \left (d x +c \right )+a^{4} B \sin \left (d x +c \right )+A \,a^{4} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)),x)

[Out]

1/d*(1/5*a^4*B*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+A*a^4*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x
+c)+3/8*d*x+3/8*c)+4*a^4*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)+4/3*A*a^4*(2+cos(d*x+c
)^2)*sin(d*x+c)+2*a^4*B*(2+cos(d*x+c)^2)*sin(d*x+c)+6*A*a^4*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+4*a^4*B*
(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+4*A*a^4*sin(d*x+c)+a^4*B*sin(d*x+c)+A*a^4*(d*x+c))

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maxima [A]  time = 0.84, size = 236, normalized size = 1.57 \[ -\frac {640 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 720 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 480 \, {\left (d x + c\right )} A a^{4} - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{4} + 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} - 60 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 480 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 1920 \, A a^{4} \sin \left (d x + c\right ) - 480 \, B a^{4} \sin \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

-1/480*(640*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c
))*A*a^4 - 720*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 480*(d*x + c)*A*a^4 - 32*(3*sin(d*x + c)^5 - 10*sin(d*
x + c)^3 + 15*sin(d*x + c))*B*a^4 + 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4 - 60*(12*d*x + 12*c + sin(4*d*
x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^4 - 480*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^4 - 1920*A*a^4*sin(d*x + c) -
480*B*a^4*sin(d*x + c))/d

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mupad [B]  time = 1.56, size = 278, normalized size = 1.85 \[ \frac {\left (\frac {35\,A\,a^4}{4}+7\,B\,a^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {245\,A\,a^4}{6}+\frac {98\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {224\,A\,a^4}{3}+\frac {896\,B\,a^4}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {395\,A\,a^4}{6}+\frac {158\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {93\,A\,a^4}{4}+25\,B\,a^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {7\,a^4\,\left (5\,A+4\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d}+\frac {7\,a^4\,\mathrm {atan}\left (\frac {7\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,A+4\,B\right )}{4\,\left (\frac {35\,A\,a^4}{4}+7\,B\,a^4\right )}\right )\,\left (5\,A+4\,B\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^4,x)

[Out]

(tan(c/2 + (d*x)/2)*((93*A*a^4)/4 + 25*B*a^4) + tan(c/2 + (d*x)/2)^9*((35*A*a^4)/4 + 7*B*a^4) + tan(c/2 + (d*x
)/2)^7*((245*A*a^4)/6 + (98*B*a^4)/3) + tan(c/2 + (d*x)/2)^3*((395*A*a^4)/6 + (158*B*a^4)/3) + tan(c/2 + (d*x)
/2)^5*((224*A*a^4)/3 + (896*B*a^4)/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)) - (7*a^4*(5*A + 4*B)*(atan(tan(c/2 + (d*x)/2))
- (d*x)/2))/(4*d) + (7*a^4*atan((7*a^4*tan(c/2 + (d*x)/2)*(5*A + 4*B))/(4*((35*A*a^4)/4 + 7*B*a^4)))*(5*A + 4
*B))/(4*d)

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sympy [A]  time = 3.02, size = 544, normalized size = 3.63 \[ \begin {cases} \frac {3 A a^{4} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + 3 A a^{4} x \sin ^{2}{\left (c + d x \right )} + \frac {3 A a^{4} x \cos ^{4}{\left (c + d x \right )}}{8} + 3 A a^{4} x \cos ^{2}{\left (c + d x \right )} + A a^{4} x + \frac {3 A a^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {8 A a^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 A a^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {4 A a^{4} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 A a^{4} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {4 A a^{4} \sin {\left (c + d x \right )}}{d} + \frac {3 B a^{4} x \sin ^{4}{\left (c + d x \right )}}{2} + 3 B a^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} + 2 B a^{4} x \sin ^{2}{\left (c + d x \right )} + \frac {3 B a^{4} x \cos ^{4}{\left (c + d x \right )}}{2} + 2 B a^{4} x \cos ^{2}{\left (c + d x \right )} + \frac {8 B a^{4} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 B a^{4} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 B a^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {4 B a^{4} \sin ^{3}{\left (c + d x \right )}}{d} + \frac {B a^{4} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 B a^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{2 d} + \frac {6 B a^{4} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {2 B a^{4} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {B a^{4} \sin {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + B \cos {\relax (c )}\right ) \left (a \cos {\relax (c )} + a\right )^{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**4*(A+B*cos(d*x+c)),x)

[Out]

Piecewise((3*A*a**4*x*sin(c + d*x)**4/8 + 3*A*a**4*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*a**4*x*sin(c + d*
x)**2 + 3*A*a**4*x*cos(c + d*x)**4/8 + 3*A*a**4*x*cos(c + d*x)**2 + A*a**4*x + 3*A*a**4*sin(c + d*x)**3*cos(c
+ d*x)/(8*d) + 8*A*a**4*sin(c + d*x)**3/(3*d) + 5*A*a**4*sin(c + d*x)*cos(c + d*x)**3/(8*d) + 4*A*a**4*sin(c +
d*x)*cos(c + d*x)**2/d + 3*A*a**4*sin(c + d*x)*cos(c + d*x)/d + 4*A*a**4*sin(c + d*x)/d + 3*B*a**4*x*sin(c +
d*x)**4/2 + 3*B*a**4*x*sin(c + d*x)**2*cos(c + d*x)**2 + 2*B*a**4*x*sin(c + d*x)**2 + 3*B*a**4*x*cos(c + d*x)*
*4/2 + 2*B*a**4*x*cos(c + d*x)**2 + 8*B*a**4*sin(c + d*x)**5/(15*d) + 4*B*a**4*sin(c + d*x)**3*cos(c + d*x)**2
/(3*d) + 3*B*a**4*sin(c + d*x)**3*cos(c + d*x)/(2*d) + 4*B*a**4*sin(c + d*x)**3/d + B*a**4*sin(c + d*x)*cos(c
+ d*x)**4/d + 5*B*a**4*sin(c + d*x)*cos(c + d*x)**3/(2*d) + 6*B*a**4*sin(c + d*x)*cos(c + d*x)**2/d + 2*B*a**4
*sin(c + d*x)*cos(c + d*x)/d + B*a**4*sin(c + d*x)/d, Ne(d, 0)), (x*(A + B*cos(c))*(a*cos(c) + a)**4, True))

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