Integrand size = 40, antiderivative size = 80 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=b^2 C x+\frac {\left (a^2 B+2 b^2 B+4 a b C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (2 b B+a C) \tan (c+d x)}{d}+\frac {a^2 B \sec (c+d x) \tan (c+d x)}{2 d} \]
b^2*C*x+1/2*(B*a^2+2*B*b^2+4*C*a*b)*arctanh(sin(d*x+c))/d+a*(2*B*b+C*a)*ta n(d*x+c)/d+1/2*a^2*B*sec(d*x+c)*tan(d*x+c)/d
Time = 0.29 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.84 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2 b^2 C d x+\left (a^2 B+2 b^2 B+4 a b C\right ) \text {arctanh}(\sin (c+d x))+a (4 b B+2 a C+a B \sec (c+d x)) \tan (c+d x)}{2 d} \]
(2*b^2*C*d*x + (a^2*B + 2*b^2*B + 4*a*b*C)*ArcTanh[Sin[c + d*x]] + a*(4*b* B + 2*a*C + a*B*Sec[c + d*x])*Tan[c + d*x])/(2*d)
Time = 0.76 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.275, Rules used = {3042, 3508, 3042, 3467, 25, 3042, 3500, 3042, 3214, 3042, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^3(c+d x) (a+b \cos (c+d x))^2 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3467 |
| \(\displaystyle \frac {a^2 B \tan (c+d x) \sec (c+d x)}{2 d}-\frac {1}{2} \int -\left (\left (2 b^2 C \cos ^2(c+d x)+\left (B a^2+4 b C a+2 b^2 B\right ) \cos (c+d x)+2 a (2 b B+a C)\right ) \sec ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \int \left (2 b^2 C \cos ^2(c+d x)+\left (B a^2+4 b C a+2 b^2 B\right ) \cos (c+d x)+2 a (2 b B+a C)\right ) \sec ^2(c+d x)dx+\frac {a^2 B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {2 b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (B a^2+4 b C a+2 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a (2 b B+a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{2} \left (\int \left (B a^2+4 b C a+2 b^2 B+2 b^2 C \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 a (a C+2 b B) \tan (c+d x)}{d}\right )+\frac {a^2 B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {B a^2+4 b C a+2 b^2 B+2 b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 a (a C+2 b B) \tan (c+d x)}{d}\right )+\frac {a^2 B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (\left (a^2 B+4 a b C+2 b^2 B\right ) \int \sec (c+d x)dx+\frac {2 a (a C+2 b B) \tan (c+d x)}{d}+2 b^2 C x\right )+\frac {a^2 B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\left (a^2 B+4 a b C+2 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a (a C+2 b B) \tan (c+d x)}{d}+2 b^2 C x\right )+\frac {a^2 B \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {\left (a^2 B+4 a b C+2 b^2 B\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a (a C+2 b B) \tan (c+d x)}{d}+2 b^2 C x\right )+\frac {a^2 B \tan (c+d x) \sec (c+d x)}{2 d}\) |
(a^2*B*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2*b^2*C*x + ((a^2*B + 2*b^2*B + 4*a*b*C)*ArcTanh[Sin[c + d*x]])/d + (2*a*(2*b*B + a*C)*Tan[c + d*x])/d)/2
3.8.83.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(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]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f _.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ (B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 *(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2)) Int[(c + d* Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[ n, -1]
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] + Simp[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]
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] :> Simp[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]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.30
| method | result | size |
| parts | \(\frac {\left (B \,b^{2}+2 C a b \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (2 B a b +a^{2} C \right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {b^{2} C \left (d x +c \right )}{d}\) | \(104\) |
| derivativedivides | \(\frac {a^{2} C \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B a b \tan \left (d x +c \right )+b^{2} C \left (d x +c \right )+B \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(112\) |
| default | \(\frac {a^{2} C \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B a b \tan \left (d x +c \right )+b^{2} C \left (d x +c \right )+B \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(112\) |
| parallelrisch | \(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B \,a^{2}+2 B \,b^{2}+4 C a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B \,a^{2}+2 B \,b^{2}+4 C a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 C \,b^{2} d x \cos \left (2 d x +2 c \right )+\left (4 B a b +2 a^{2} C \right ) \sin \left (2 d x +2 c \right )+2 C \,b^{2} d x +2 B \sin \left (d x +c \right ) a^{2}}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(162\) |
| risch | \(b^{2} C x -\frac {i a \left (B a \,{\mathrm e}^{3 i \left (d x +c \right )}-4 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a C \,{\mathrm e}^{2 i \left (d x +c \right )}-B a \,{\mathrm e}^{i \left (d x +c \right )}-4 B b -2 a C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C a b}{d}+\frac {B \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C a b}{d}\) | \(217\) |
| norman | \(\frac {\frac {a \left (B a -4 B b -2 a C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (5 B a +4 B b +2 a C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+b^{2} C x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+b^{2} C x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b^{2} C x +\frac {8 a \left (2 B b +a C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (B a -2 B b -a C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a \left (B a +2 B b +a C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a \left (B a +4 B b +2 a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \left (5 B a -4 B b -2 a C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-b^{2} C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 b^{2} C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 b^{2} C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 b^{2} C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 b^{2} C x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {\left (B \,a^{2}+2 B \,b^{2}+4 C a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (B \,a^{2}+2 B \,b^{2}+4 C a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(429\) |
(B*b^2+2*C*a*b)/d*ln(sec(d*x+c)+tan(d*x+c))+(2*B*a*b+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)))+b^2*C/d*( d*x+c)
Time = 0.30 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.70 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, C b^{2} d x \cos \left (d x + c\right )^{2} + {\left (B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B a^{2} + 2 \, {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")
1/4*(4*C*b^2*d*x*cos(d*x + c)^2 + (B*a^2 + 4*C*a*b + 2*B*b^2)*cos(d*x + c) ^2*log(sin(d*x + c) + 1) - (B*a^2 + 4*C*a*b + 2*B*b^2)*cos(d*x + c)^2*log( -sin(d*x + c) + 1) + 2*(B*a^2 + 2*(C*a^2 + 2*B*a*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.75 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} C b^{2} - B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{2} \tan \left (d x + c\right ) + 8 \, B a b \tan \left (d x + c\right )}{4 \, d} \]
integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")
1/4*(4*(d*x + c)*C*b^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)) + 4*C*a*b*(log(sin(d*x + c) + 1 ) - log(sin(d*x + c) - 1)) + 2*B*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*C*a^2*tan(d*x + c) + 8*B*a*b*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (76) = 152\).
Time = 0.36 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.38 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2 \, {\left (d x + c\right )} C b^{2} + {\left (B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
1/2*(2*(d*x + c)*C*b^2 + (B*a^2 + 4*C*a*b + 2*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (B*a^2 + 4*C*a*b + 2*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(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 - 4*B*a*b*tan(1/2*d*x + 1/2*c)^3 + B*a^2*tan(1/2*d*x + 1/2*c) + 2*C*a^2*tan( 1/2*d*x + 1/2*c) + 4*B*a*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 2.53 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.20 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2\,\left (\frac {B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+B\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,C\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{2}+B\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]
(2*((B*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + B*b^2*atanh(s in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + C*b^2*atan(sin(c/2 + (d*x)/2)/cos( c/2 + (d*x)/2)) + 2*C*a*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/d + ((C*a^2*sin(2*c + 2*d*x))/2 + (B*a^2*sin(c + d*x))/2 + B*a*b*sin(2*c + 2*d*x))/(d*(cos(2*c + 2*d*x)/2 + 1/2))