\(\int (a+b \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [535]

Optimal result

Integrand size = 33, antiderivative size = 109 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x+\frac {2 a A b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b (A-C) \sin (c+d x)}{d}-\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d} \]

[Out]

1/2*(2*A*b^2+(2*a^2+b^2)*C)*x+2*a*A*b*arctanh(sin(d*x+c))/d-2*a*b*(A-C)*sin(d*x+c)/d-1/2*b^2*(2*A-C)*cos(d*x+c
)*sin(d*x+c)/d+A*(a+b*cos(d*x+c))^2*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {3127, 3112, 3102, 2814, 3855} \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} x \left (C \left (2 a^2+b^2\right )+2 A b^2\right )+\frac {2 a A b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b (A-C) \sin (c+d x)}{d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^2}{d}-\frac {b^2 (2 A-C) \sin (c+d x) \cos (c+d x)}{2 d} \]

[In]

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

[Out]

((2*A*b^2 + (2*a^2 + b^2)*C)*x)/2 + (2*a*A*b*ArcTanh[Sin[c + d*x]])/d - (2*a*b*(A - C)*Sin[c + d*x])/d - (b^2*
(2*A - C)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (A*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/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 3102

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

Rule 3112

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a +
b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*
c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rule 3127

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

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}& = \frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+\int (a+b \cos (c+d x)) \left (2 A b+a C \cos (c+d x)-b (2 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+\frac {1}{2} \int \left (4 a A b+\left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)-4 a b (A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = -\frac {2 a b (A-C) \sin (c+d x)}{d}-\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+\frac {1}{2} \int \left (4 a A b+\left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {1}{2} \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x-\frac {2 a b (A-C) \sin (c+d x)}{d}-\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d}+(2 a A b) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) x+\frac {2 a A b \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a b (A-C) \sin (c+d x)}{d}-\frac {b^2 (2 A-C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^2 \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.40 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.21 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) (c+d x)-8 a A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 a A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 a b C \sin (c+d x)+\left (4 a^2 A+b^2 C+b^2 C \cos (2 (c+d x))\right ) \tan (c+d x)}{4 d} \]

[In]

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

[Out]

(2*(2*A*b^2 + (2*a^2 + b^2)*C)*(c + d*x) - 8*a*A*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 8*a*A*b*Log[Cos[
(c + d*x)/2] + Sin[(c + d*x)/2]] + 8*a*b*C*Sin[c + d*x] + (4*a^2*A + b^2*C + b^2*C*Cos[2*(c + d*x)])*Tan[c + d
*x])/(4*d)

Maple [A] (verified)

Time = 4.54 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {A \,a^{2} \tan \left (d x +c \right )+a^{2} C \left (d x +c \right )+2 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b \sin \left (d x +c \right ) C +A \,b^{2} \left (d x +c \right )+b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(94\)
default	\(\frac {A \,a^{2} \tan \left (d x +c \right )+a^{2} C \left (d x +c \right )+2 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a b \sin \left (d x +c \right ) C +A \,b^{2} \left (d x +c \right )+b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(94\)
parts	\(\frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{2}+a^{2} C \right ) \left (d x +c \right )}{d}+\frac {b^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 \sin \left (d x +c \right ) C a b}{d}\)	\(102\)
parallelrisch	\(\frac {-16 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a b \cos \left (d x +c \right )+16 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a b \cos \left (d x +c \right )+8 a b \sin \left (2 d x +2 c \right ) C +C \sin \left (3 d x +3 c \right ) b^{2}+8 x d \left (\left (A +\frac {C}{2}\right ) b^{2}+a^{2} C \right ) \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (8 A \,a^{2}+b^{2} C \right )}{8 d \cos \left (d x +c \right )}\)	\(133\)
risch	\(x A \,b^{2}+a^{2} C x +\frac {b^{2} C x}{2}-\frac {i b^{2} C \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i C a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {i C a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i b^{2} C \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i A \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 A a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 A a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(160\)
norman	\(\frac {\left (-A \,b^{2}-a^{2} C -\frac {1}{2} b^{2} C \right ) x +\left (-3 A \,b^{2}-3 a^{2} C -\frac {3}{2} b^{2} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,b^{2}+a^{2} C +\frac {1}{2} b^{2} C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,b^{2}+3 a^{2} C +\frac {3}{2} b^{2} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A \,b^{2}-2 a^{2} C -b^{2} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 A \,b^{2}+2 a^{2} C +b^{2} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 \left (6 A \,a^{2}-b^{2} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 A \,a^{2}-4 C a b +b^{2} C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 A \,a^{2}+4 C a b +b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {8 a \left (a A -C b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a \left (a A +C b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 A a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 A a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(394\)

[In]

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

[Out]

1/d*(A*a^2*tan(d*x+c)+a^2*C*(d*x+c)+2*A*a*b*ln(sec(d*x+c)+tan(d*x+c))+2*a*b*sin(d*x+c)*C+A*b^2*(d*x+c)+b^2*C*(
1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.09 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2 \, A a b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, A a b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{2} + {\left (2 \, A + C\right )} b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (C b^{2} \cos \left (d x + c\right )^{2} + 4 \, C a b \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((A + C*cos(c + d*x)**2)*(a + b*cos(c + d*x))**2*sec(c + d*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.91 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C a^{2} + 4 \, {\left (d x + c\right )} A b^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2} + 4 \, A a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a b \sin \left (d x + c\right ) + 4 \, A a^{2} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.61 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {4 \, A a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, A a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + {\left (2 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

1/2*(4*A*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 4*A*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 4*A*a^2*tan(1/2
*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + (2*C*a^2 + 2*A*b^2 + C*b^2)*(d*x + c) + 2*(4*C*a*b*tan(1/2*d*x +
1/2*c)^3 - C*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*C*a*b*tan(1/2*d*x + 1/2*c) + C*b^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 = 1.93 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.77 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {2\,C\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {C\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}-\frac {A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}}{d}-\frac {A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}}{d} \]

[In]

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

[Out]

(A*a^2*sin(c + d*x))/(d*cos(c + d*x)) - (C*a^2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d - (C*b^
2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i)/d - (A*b^2*atanh((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*
x)/2))*2i)/d + (2*C*a*b*sin(c + d*x))/d - (A*a*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*4i)/d + (C*b
^2*cos(c + d*x)*sin(c + d*x))/(2*d)