Integrand size = 31, antiderivative size = 109 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=2 a A b x+\frac {\left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {2 a b (A-C) \tan (c+d x)}{d}-\frac {b^2 (2 A-C) \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
2*a*A*b*x+1/2*(2*A*b^2+(2*a^2+b^2)*C)*arctanh(sin(d*x+c))/d+A*(a+b*sec(d*x +c))^2*sin(d*x+c)/d-2*a*b*(A-C)*tan(d*x+c)/d-1/2*b^2*(2*A-C)*sec(d*x+c)*ta n(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(352\) vs. \(2(109)=218\).
Time = 3.26 (sec) , antiderivative size = 352, normalized size of antiderivative = 3.23 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\sec ^2(c+d x) \left (4 a A b c+4 a A b d x-2 A b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-b^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 A b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+b^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (2 (c+d x)) \left (4 a A b (c+d x)-\left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\left (2 A b^2+2 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\left (a^2 A+2 b^2 C\right ) \sin (c+d x)+4 a b C \sin (2 (c+d x))+a^2 A \sin (3 (c+d x))\right )}{4 d} \] Input:
Integrate[Cos[c + d*x]*(a + b*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]
Output:
(Sec[c + d*x]^2*(4*a*A*b*c + 4*a*A*b*d*x - 2*A*b^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2*a^2*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - b^2 *C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*A*b^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 2*a^2*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + b^2*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + Cos[2*(c + d*x)]*(4*a*A*b *(c + d*x) - (2*A*b^2 + (2*a^2 + b^2)*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d *x)/2]] + (2*A*b^2 + 2*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x) /2]]) + (a^2*A + 2*b^2*C)*Sin[c + d*x] + 4*a*b*C*Sin[2*(c + d*x)] + a^2*A* Sin[3*(c + d*x)]))/(4*d)
Time = 0.47 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 4583, 3042, 4536, 2009}
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.
| \(\displaystyle \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4583 |
| \(\displaystyle \int (a+b \sec (c+d x)) \left (-b (2 A-C) \sec ^2(c+d x)+a C \sec (c+d x)+2 A b\right )dx+\frac {A \sin (c+d x) (a+b \sec (c+d x))^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (2 A-C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a C \csc \left (c+d x+\frac {\pi }{2}\right )+2 A b\right )dx+\frac {A \sin (c+d x) (a+b \sec (c+d x))^2}{d}\) |
| \(\Big \downarrow \) 4536 |
| \(\displaystyle \frac {1}{2} \int \left (-4 a b (A-C) \sec ^2(c+d x)+\left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \sec (c+d x)+4 a A b\right )dx+\frac {A \sin (c+d x) (a+b \sec (c+d x))^2}{d}-\frac {b^2 (2 A-C) \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {\left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {4 a b (A-C) \tan (c+d x)}{d}+4 a A b x\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^2}{d}-\frac {b^2 (2 A-C) \tan (c+d x) \sec (c+d x)}{2 d}\) |
Input:
Int[Cos[c + d*x]*(a + b*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]
Output:
(A*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/d - (b^2*(2*A - C)*Sec[c + d*x]*Ta n[c + d*x])/(2*d) + (4*a*A*b*x + ((2*A*b^2 + (2*a^2 + b^2)*C)*ArcTanh[Sin[ c + d*x]])/d - (4*a*b*(A - C)*Tan[c + d*x])/d)/2
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Simp[1/2 Int[Simp[2*A*a + (2*B*a + b*(2* A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, e, f, A, B, C}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*(C*n + A*(n + 1))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^ 2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && Gt Q[m, 0] && LeQ[n, -1]
Time = 1.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.03
| method | result | size |
| derivativedivides | \(\frac {a^{2} A \sin \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a A b \left (d x +c \right )+2 C a b \tan \left (d x +c \right )+A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(112\) |
| default | \(\frac {a^{2} A \sin \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a A b \left (d x +c \right )+2 C a b \tan \left (d x +c \right )+A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(112\) |
| parallelrisch | \(\frac {-2 \left (\left (A +\frac {C}{2}\right ) b^{2}+C \,a^{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (\left (A +\frac {C}{2}\right ) b^{2}+C \,a^{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 a A b x d \cos \left (2 d x +2 c \right )+4 C a b \sin \left (2 d x +2 c \right )+a^{2} A \sin \left (3 d x +3 c \right )+\left (a^{2} A +2 C \,b^{2}\right ) \sin \left (d x +c \right )+4 a A b x d}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(170\) |
| risch | \(2 a A b x -\frac {i a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{2} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i C b \left (b \,{\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )} a -b \,{\mathrm e}^{i \left (d x +c \right )}-4 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{2}}{2 d}\) | \(236\) |
| norman | \(\frac {\frac {\left (2 a^{2} A -4 C a b +C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (6 a^{2} A +4 C a b -C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {\left (2 a^{2} A +4 C a b +C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (6 a^{2} A -4 C a b -C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-2 a A b x +4 a A b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-4 a A b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+2 a A b x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (2 A \,b^{2}+2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 A \,b^{2}+2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(291\) |
Input:
int(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBO SE)
Output:
1/d*(a^2*A*sin(d*x+c)+C*a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a*A*b*(d*x+c)+2*C* a*b*tan(d*x+c)+A*b^2*ln(sec(d*x+c)+tan(d*x+c))+C*b^2*(1/2*sec(d*x+c)*tan(d *x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.28 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, A a b d x \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{2} + {\left (2 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, C a^{2} + {\left (2 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 4 \, C a b \cos \left (d x + c\right ) + C b^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="f ricas")
Output:
1/4*(8*A*a*b*d*x*cos(d*x + c)^2 + (2*C*a^2 + (2*A + C)*b^2)*cos(d*x + c)^2 *log(sin(d*x + c) + 1) - (2*C*a^2 + (2*A + C)*b^2)*cos(d*x + c)^2*log(-sin (d*x + c) + 1) + 2*(2*A*a^2*cos(d*x + c)^2 + 4*C*a*b*cos(d*x + c) + C*b^2) *sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \cos {\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)*(a+b*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)
Output:
Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**2*cos(c + d*x), x)
Time = 0.04 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.28 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (d x + c\right )} A a b - C b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{2} \sin \left (d x + c\right ) + 8 \, C a b \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="m axima")
Output:
1/4*(8*(d*x + c)*A*a*b - C*b^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*C*a^2*(log(sin(d*x + c) + 1 ) - log(sin(d*x + c) - 1)) + 2*A*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*A*a^2*sin(d*x + c) + 8*C*a*b*tan(d*x + c))/d
Time = 0.29 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.75 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (d x + c\right )} A a b + \frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + {\left (2 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, C a^{2} + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="g iac")
Output:
1/2*(4*(d*x + c)*A*a*b + 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)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*C*a^2 + 2*A*b^2 + C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 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
Time = 12.44 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.72 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+2\,A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{4}+\frac {C\,b^2\,\sin \left (c+d\,x\right )}{2}+C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \] Input:
int(cos(c + d*x)*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^2,x)
Output:
(2*(A*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + C*a^2*atanh(sin(c /2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (C*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/ 2 + (d*x)/2)))/2 + 2*A*a*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/d + ((A*a^2*sin(3*c + 3*d*x))/4 + (A*a^2*sin(c + d*x))/4 + (C*b^2*sin(c + d *x))/2 + C*a*b*sin(2*c + 2*d*x))/(d*(cos(2*c + 2*d*x)/2 + 1/2))
Time = 0.16 (sec) , antiderivative size = 379, normalized size of antiderivative = 3.48 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {-4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2} c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2} c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} c +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a \,b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2} c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} c -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2} c +2 \sin \left (d x +c \right )^{3} a^{3}+4 \sin \left (d x +c \right )^{2} a^{2} b c +4 \sin \left (d x +c \right )^{2} a^{2} b d x -2 \sin \left (d x +c \right ) a^{3}-\sin \left (d x +c \right ) b^{2} c -4 a^{2} b c -4 a^{2} b d x}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)
Output:
( - 4*cos(c + d*x)*sin(c + d*x)*a*b*c - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*c - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2 - lo g(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2*c + 2*log(tan((c + d*x)/2) - 1)*a**2*c + 2*log(tan((c + d*x)/2) - 1)*a*b**2 + log(tan((c + d*x)/2) - 1) *b**2*c + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*c + 2*log(tan(( c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**2 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2*c - 2*log(tan((c + d*x)/2) + 1)*a**2*c - 2*log(tan((c + d*x )/2) + 1)*a*b**2 - log(tan((c + d*x)/2) + 1)*b**2*c + 2*sin(c + d*x)**3*a* *3 + 4*sin(c + d*x)**2*a**2*b*c + 4*sin(c + d*x)**2*a**2*b*d*x - 2*sin(c + d*x)*a**3 - sin(c + d*x)*b**2*c - 4*a**2*b*c - 4*a**2*b*d*x)/(2*d*(sin(c + d*x)**2 - 1))