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

Optimal. Leaf size=136 \[ \frac {3}{8} (b B+a C) x+\frac {(4 a B+5 b C) \sin (c+d x)}{5 d}+\frac {3 (b B+a C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {(b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {(4 a B+5 b C) \sin ^3(c+d x)}{15 d} \]

[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

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Rubi [A]
time = 0.15, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4157, 4081, 3872, 2715, 8, 2713} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[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*} \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 &=\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*}

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Mathematica [A]
time = 0.29, size = 124, normalized size = 0.91 \begin {gather*} \frac {180 b B c+180 a c C+180 b B d x+180 a C d x+60 (5 a B+8 b C) \sin (c+d x)-160 b C \sin ^3(c+d x)+120 (b B+a C) \sin (2 (c+d x))+50 a B \sin (3 (c+d x))+15 b B \sin (4 (c+d x))+15 a C \sin (4 (c+d x))+6 a B \sin (5 (c+d x))}{480 d} \end {gather*}

Antiderivative was successfully verified.

[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]

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

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Maple [A]
time = 0.13, size = 128, normalized size = 0.94

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

Verification of antiderivative is not currently implemented for this CAS.

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

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Maxima [A]
time = 0.30, size = 124, normalized size = 0.91 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]
time = 2.80, size = 96, normalized size = 0.71 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (124) = 248\).
time = 0.49, size = 300, normalized size = 2.21 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Mupad [B]
time = 3.97, size = 149, normalized size = 1.10 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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