\(\int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [328]

Optimal result

Integrand size = 40, antiderivative size = 125 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a^3 (5 B+7 C) x+\frac {a^3 C \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \]

[Out]

1/2*a^3*(5*B+7*C)*x+a^3*C*arctanh(sin(d*x+c))/d+5/2*a^3*(B+C)*sin(d*x+c)/d+1/3*a*B*cos(d*x+c)^2*(a+a*sec(d*x+c
))^2*sin(d*x+c)/d+1/6*(5*B+3*C)*cos(d*x+c)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4157, 4102, 4081, 3855} \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 C \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {(5 B+3 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{6 d}+\frac {1}{2} a^3 x (5 B+7 C)+\frac {a B \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(5*B + 7*C)*x)/2 + (a^3*C*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(B + C)*Sin[c + d*x])/(2*d) + (a*B*Cos[c + d*
x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + ((5*B + 3*C)*Cos[c + d*x]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d
*x])/(6*d)

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 4102

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[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 ^3(c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx \\ & = \frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 (a (5 B+3 C)+3 a C \sec (c+d x)) \, dx \\ & = \frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^2 (B+C)+6 a^2 C \sec (c+d x)\right ) \, dx \\ & = \frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac {1}{6} \int \left (-3 a^3 (5 B+7 C)-6 a^3 C \sec (c+d x)\right ) \, dx \\ & = \frac {1}{2} a^3 (5 B+7 C) x+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^3 C\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^3 (5 B+7 C) x+\frac {a^3 C \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {a B \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(5 B+3 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (30 B d x+42 C d x-12 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 (5 B+4 C) \sin (c+d x)+3 (3 B+C) \sin (2 (c+d x))+B \sin (3 (c+d x))\right )}{12 d} \]

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(30*B*d*x + 42*C*d*x - 12*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*C*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] + 9*(5*B + 4*C)*Sin[c + d*x] + 3*(3*B + C)*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(12*d)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {5 a^{3} \left (-\frac {2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{5}+\frac {2 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{5}+\frac {\left (3 B +C \right ) \sin \left (2 d x +2 c \right )}{10}+\frac {B \sin \left (3 d x +3 c \right )}{30}+3 \left (\frac {B}{2}+\frac {2 C}{5}\right ) \sin \left (d x +c \right )+d x \left (B +\frac {7 C}{5}\right )\right )}{2 d}\)	\(91\)
derivativedivides	\(\frac {B \,a^{3} \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \sin \left (d x +c \right )+3 a^{3} C \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \sin \left (d x +c \right )+\frac {B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{3} C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(147\)
default	\(\frac {B \,a^{3} \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \sin \left (d x +c \right )+3 a^{3} C \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \sin \left (d x +c \right )+\frac {B \,a^{3} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{3} C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(147\)
risch	\(\frac {5 a^{3} B x}{2}+\frac {7 a^{3} x C}{2}-\frac {15 i B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{3} C}{2 d}+\frac {15 i B \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} C}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {B \,a^{3} \sin \left (3 d x +3 c \right )}{12 d}+\frac {3 \sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} C}{4 d}\)	\(189\)
norman	\(\frac {\left (\frac {5}{2} B \,a^{3}+\frac {7}{2} a^{3} C \right ) x +\left (\frac {5}{2} B \,a^{3}+\frac {7}{2} a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (-10 B \,a^{3}-14 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-10 B \,a^{3}-14 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (15 B \,a^{3}+21 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {a^{3} \left (11 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {19 a^{3} \left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}+\frac {5 a^{3} \left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-\frac {a^{3} \left (5 B +9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}+\frac {a^{3} \left (11 B +39 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}-\frac {a^{3} \left (13 B +17 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{3} \left (59 B +27 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {a^{3} \left (101 B +69 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}+\frac {a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(391\)

[In]

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

5/2*a^3*(-2/5*C*ln(tan(1/2*d*x+1/2*c)-1)+2/5*C*ln(tan(1/2*d*x+1/2*c)+1)+1/10*(3*B+C)*sin(2*d*x+2*c)+1/30*B*sin
(3*d*x+3*c)+3*(1/2*B+2/5*C)*sin(d*x+c)+d*x*(B+7/5*C))/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.82 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (5 \, B + 7 \, C\right )} a^{3} d x + 3 \, C a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + 2 \, {\left (11 \, B + 9 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(3*(5*B + 7*C)*a^3*d*x + 3*C*a^3*log(sin(d*x + c) + 1) - 3*C*a^3*log(-sin(d*x + c) + 1) + (2*B*a^3*cos(d*x
+ c)^2 + 3*(3*B + C)*a^3*cos(d*x + c) + 2*(11*B + 9*C)*a^3)*sin(d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))**3*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.18 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 12 \, {\left (d x + c\right )} B a^{3} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 36 \, {\left (d x + c\right )} C a^{3} - 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} \sin \left (d x + c\right ) - 36 \, C a^{3} \sin \left (d x + c\right )}{12 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 12*(d*x + c)*B*a
^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 - 36*(d*x + c)*C*a^3 - 6*C*a^3*(log(sin(d*x + c) + 1) - log(sin(
d*x + c) - 1)) - 36*B*a^3*sin(d*x + c) - 36*C*a^3*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.44 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (5 \, B a^{3} + 7 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(6*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(5*B*a^3 + 7*
C*a^3)*(d*x + c) + 2*(15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^3*tan(1/2*d*x
+ 1/2*c)^3 + 36*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*B*a^3*tan(1/2*d*x + 1/2*c) + 21*C*a^3*tan(1/2*d*x + 1/2*c))
/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

Mupad [B] (verification not implemented)

Time = 16.57 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.42 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15\,B\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

[In]

int(cos(c + d*x)^4*(B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3,x)

[Out]

(15*B*a^3*sin(c + d*x))/(4*d) + (3*C*a^3*sin(c + d*x))/d + (5*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (7*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (
d*x)/2)))/d + (3*B*a^3*sin(2*c + 2*d*x))/(4*d) + (B*a^3*sin(3*c + 3*d*x))/(12*d) + (C*a^3*sin(2*c + 2*d*x))/(4
*d)