\(\int \cos ^6(c+d x) (a+b \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [774]

Optimal result

Integrand size = 38, antiderivative size = 136 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3}{8} (b B+a C) x+\frac {(4 a B+5 b C) \sin (c+d x)}{5 d}+\frac {3 (b B+a C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {(b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {(4 a B+5 b C) \sin ^3(c+d x)}{15 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4157, 4081, 3872, 2715, 8, 2713} \[ \int \cos ^6(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {(4 a B+5 b C) \sin ^3(c+d x)}{15 d}+\frac {(4 a B+5 b C) \sin (c+d x)}{5 d}+\frac {(a C+b B) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 (a C+b B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3}{8} x (a C+b B)+\frac {a B \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

[In]

Int[Cos[c + d*x]^6*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(3*(b*B + a*C)*x)/8 + ((4*a*B + 5*b*C)*Sin[c + d*x])/(5*d) + (3*(b*B + a*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) +
((b*B + a*C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*B*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - ((4*a*B + 5*b*C)*
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 3872

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 4081

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

Rule 4157

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]

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.65 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {480 (a B+b C) \sin (c+d x)-160 (2 a B+b C) \sin ^3(c+d x)+96 a B \sin ^5(c+d x)+15 (b B+a C) (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{480 d} \]

[In]

Integrate[Cos[c + d*x]^6*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(480*(a*B + b*C)*Sin[c + d*x] - 160*(2*a*B + b*C)*Sin[c + d*x]^3 + 96*a*B*Sin[c + d*x]^5 + 15*(b*B + a*C)*(12*
(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(480*d)

Maple [A] (verified)

Time = 0.74 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(\frac {120 \left (B b +C a \right ) \sin \left (2 d x +2 c \right )+10 \left (5 a B +4 C b \right ) \sin \left (3 d x +3 c \right )+15 \left (B b +C a \right ) \sin \left (4 d x +4 c \right )+6 a B \sin \left (5 d x +5 c \right )+60 \left (5 a B +6 C b \right ) \sin \left (d x +c \right )+180 \left (B b +C a \right ) x d}{480 d}\)	\(104\)
derivativedivides	\(\frac {\frac {a B \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+B b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C a \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(128\)
default	\(\frac {\frac {a B \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+B b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C a \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {C b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(128\)
risch	\(\frac {3 B b x}{8}+\frac {3 a x C}{8}+\frac {5 a B \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) C b}{4 d}+\frac {a B \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) B b}{32 d}+\frac {\sin \left (4 d x +4 c \right ) C a}{32 d}+\frac {5 a B \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) C b}{12 d}+\frac {\sin \left (2 d x +2 c \right ) B b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a}{4 d}\)	\(150\)
norman	\(\frac {\left (\frac {3 B b}{8}+\frac {3 C a}{8}\right ) x +\left (-\frac {15 B b}{4}-\frac {15 C a}{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-\frac {3 B b}{2}-\frac {3 C a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {3 B b}{2}-\frac {3 C a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {3 B b}{2}+\frac {3 C a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3 B b}{2}+\frac {3 C a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {3 B b}{2}+\frac {3 C a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {3 B b}{2}+\frac {3 C a}{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {3 B b}{8}+\frac {3 C a}{8}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\frac {\left (8 a B -9 B b -9 C a +40 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {\left (8 a B -5 B b -5 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}+\frac {\left (8 a B +5 B b +5 C a +8 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (8 a B +9 B b +9 C a +40 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 d}+\frac {\left (184 a B -105 B b -105 C a -40 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 d}+\frac {\left (184 a B +105 B b +105 C a -40 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{60 d}-\frac {\left (344 a B -15 B b -15 C a +280 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{60 d}-\frac {\left (344 a B +15 B b +15 C a +280 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\)	\(482\)

[In]

int(cos(d*x+c)^6*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/480*(120*(B*b+C*a)*sin(2*d*x+2*c)+10*(5*B*a+4*C*b)*sin(3*d*x+3*c)+15*(B*b+C*a)*sin(4*d*x+4*c)+6*a*B*sin(5*d*
x+5*c)+60*(5*B*a+6*C*b)*sin(d*x+c)+180*(B*b+C*a)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.71 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, {\left (C a + B b\right )} d x + {\left (24 \, B a \cos \left (d x + c\right )^{4} + 30 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, B a + 5 \, C b\right )} \cos \left (d x + c\right )^{2} + 64 \, B a + 80 \, C b + 45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(45*(C*a + B*b)*d*x + (24*B*a*cos(d*x + c)^4 + 30*(C*a + B*b)*cos(d*x + c)^3 + 8*(4*B*a + 5*C*b)*cos(d*x
+ c)^2 + 64*B*a + 80*C*b + 45*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b}{480 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/480*(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))*C*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*b - 160*(sin(d*x + c)
^3 - 3*sin(d*x + c))*C*b)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (124) = 248\).

Time = 0.29 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.21 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {45 \, {\left (C a + B b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(45*(C*a + B*b)*(d*x + c) + 2*(120*B*a*tan(1/2*d*x + 1/2*c)^9 - 75*C*a*tan(1/2*d*x + 1/2*c)^9 - 75*B*b*t
an(1/2*d*x + 1/2*c)^9 + 120*C*b*tan(1/2*d*x + 1/2*c)^9 + 160*B*a*tan(1/2*d*x + 1/2*c)^7 - 30*C*a*tan(1/2*d*x +
1/2*c)^7 - 30*B*b*tan(1/2*d*x + 1/2*c)^7 + 320*C*b*tan(1/2*d*x + 1/2*c)^7 + 464*B*a*tan(1/2*d*x + 1/2*c)^5 +
400*C*b*tan(1/2*d*x + 1/2*c)^5 + 160*B*a*tan(1/2*d*x + 1/2*c)^3 + 30*C*a*tan(1/2*d*x + 1/2*c)^3 + 30*B*b*tan(1
/2*d*x + 1/2*c)^3 + 320*C*b*tan(1/2*d*x + 1/2*c)^3 + 120*B*a*tan(1/2*d*x + 1/2*c) + 75*C*a*tan(1/2*d*x + 1/2*c
) + 75*B*b*tan(1/2*d*x + 1/2*c) + 120*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

Mupad [B] (verification not implemented)

Time = 17.12 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.10 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,B\,b\,x}{8}+\frac {3\,C\,a\,x}{8}+\frac {5\,B\,a\,\sin \left (c+d\,x\right )}{8\,d}+\frac {3\,C\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {5\,B\,a\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {B\,a\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {B\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

[In]

int(cos(c + d*x)^6*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x)),x)

[Out]

(3*B*b*x)/8 + (3*C*a*x)/8 + (5*B*a*sin(c + d*x))/(8*d) + (3*C*b*sin(c + d*x))/(4*d) + (5*B*a*sin(3*c + 3*d*x))
/(48*d) + (B*a*sin(5*c + 5*d*x))/(80*d) + (B*b*sin(2*c + 2*d*x))/(4*d) + (C*a*sin(2*c + 2*d*x))/(4*d) + (B*b*s
in(4*c + 4*d*x))/(32*d) + (C*a*sin(4*c + 4*d*x))/(32*d) + (C*b*sin(3*c + 3*d*x))/(12*d)