\(\int (a+a \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [240]

Optimal result

Integrand size = 40, antiderivative size = 88 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=a^2 C x+\frac {a^2 (3 B+4 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (3 B+2 C) \tan (c+d x)}{2 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

a^2*C*x+1/2*a^2*(3*B+4*C)*arctanh(sin(d*x+c))/d+1/2*a^2*(3*B+2*C)*tan(d*x+c)/d+1/2*B*(a^2+a^2*cos(d*x+c))*sec(
d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3108, 3054, 3047, 3100, 2814, 3855} \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a^2 (3 B+4 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (3 B+2 C) \tan (c+d x)}{2 d}+\frac {B \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}+a^2 C x \]

[In]

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

[Out]

a^2*C*x + (a^2*(3*B + 4*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a^2*(3*B + 2*C)*Tan[c + d*x])/(2*d) + (B*(a^2 + a^2
*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 2814

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

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 3054

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

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3108

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rule 3855

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

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

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.60 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a^2 (2 C d x+(3 B+4 C) \text {arctanh}(\sin (c+d x))+(4 B+2 C+B \sec (c+d x)) \tan (c+d x))}{2 d} \]

[In]

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

[Out]

(a^2*(2*C*d*x + (3*B + 4*C)*ArcTanh[Sin[c + d*x]] + (4*B + 2*C + B*Sec[c + d*x])*Tan[c + d*x]))/(2*d)

Maple [A] (verified)

Time = 6.50 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.20

method	result	size
parts	\(\frac {\left (B \,a^{2}+2 a^{2} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (2 B \,a^{2}+a^{2} C \right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a^{2} C \left (d x +c \right )}{d}\)	\(106\)
derivativedivides	\(\frac {a^{2} C \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B \,a^{2} \tan \left (d x +c \right )+a^{2} C \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(114\)
default	\(\frac {a^{2} C \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B \,a^{2} \tan \left (d x +c \right )+a^{2} C \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(114\)
parallelrisch	\(-\frac {3 a^{2} \left (\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B +\frac {4 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B +\frac {4 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 d x C \cos \left (2 d x +2 c \right )}{3}+\frac {2 \left (-2 B -C \right ) \sin \left (2 d x +2 c \right )}{3}-\frac {2 d x C}{3}-\frac {2 B \sin \left (d x +c \right )}{3}\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(128\)
risch	\(a^{2} C x -\frac {i a^{2} \left (B \,{\mathrm e}^{3 i \left (d x +c \right )}-4 B \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C \,{\mathrm e}^{2 i \left (d x +c \right )}-B \,{\mathrm e}^{i \left (d x +c \right )}-4 B -2 C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\)	\(172\)
norman	\(\frac {a^{2} C x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} C x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{2} \left (9 B +2 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-a^{2} C x -a^{2} C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{2} C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{2} C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{2} C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{2} C x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {a^{2} \left (B -2 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (B +C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{2} \left (2 B +C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (3 B +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (3 B +2 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (5 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a^{2} \left (3 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (3 B +4 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(388\)

[In]

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

[Out]

(B*a^2+2*C*a^2)/d*ln(sec(d*x+c)+tan(d*x+c))+(2*B*a^2+C*a^2)/d*tan(d*x+c)+B*a^2/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/
2*ln(sec(d*x+c)+tan(d*x+c)))+a^2*C/d*(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.35 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, C a^{2} d x \cos \left (d x + c\right )^{2} + {\left (3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) + B a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

1/4*(4*C*a^2*d*x*cos(d*x + c)^2 + (3*B + 4*C)*a^2*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (3*B + 4*C)*a^2*cos(d
*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*(2*B + C)*a^2*cos(d*x + c) + B*a^2)*sin(d*x + c))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.61 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C a^{2} - B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, B a^{2} \tan \left (d x + c\right ) + 4 \, C a^{2} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*(4*(d*x + c)*C*a^2 - B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c)
- 1)) + 2*B*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*
x + c) - 1)) + 8*B*a^2*tan(d*x + c) + 4*C*a^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.75 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} C a^{2} + {\left (3 \, B a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (3 \, B a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(2*(d*x + c)*C*a^2 + (3*B*a^2 + 4*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (3*B*a^2 + 4*C*a^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 2*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 5*B*a^2*tan(
1/2*d*x + 1/2*c) - 2*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

Mupad [B] (verification not implemented)

Time = 1.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.84 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

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