Integrand size = 33, antiderivative size = 112 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=a b (A+2 C) x+\frac {b^2 C \text {arctanh}(\sin (c+d x))}{d}+\frac {\left (2 A b^2+a^2 (2 A+3 C)\right ) \sin (c+d x)}{3 d}+\frac {a A b \cos (c+d x) \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d} \] Output:
a*b*(A+2*C)*x+b^2*C*arctanh(sin(d*x+c))/d+1/3*(2*A*b^2+a^2*(2*A+3*C))*sin( d*x+c)/d+1/3*a*A*b*cos(d*x+c)*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+b*sec(d*x +c))^2*sin(d*x+c)/d
Time = 0.86 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.29 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {12 a A b c+24 a b c C+12 a A b d x+24 a b C d x-12 b^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 \left (4 A b^2+a^2 (3 A+4 C)\right ) \sin (c+d x)+6 a A b \sin (2 (c+d x))+a^2 A \sin (3 (c+d x))}{12 d} \] Input:
Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]
Output:
(12*a*A*b*c + 24*a*b*c*C + 12*a*A*b*d*x + 24*a*b*C*d*x - 12*b^2*C*Log[Cos[ (c + d*x)/2] - Sin[(c + d*x)/2]] + 12*b^2*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 3*(4*A*b^2 + a^2*(3*A + 4*C))*Sin[c + d*x] + 6*a*A*b*Sin[2*(c + d*x)] + a^2*A*Sin[3*(c + d*x)])/(12*d)
Time = 0.87 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4583, 3042, 4562, 27, 3042, 4535, 24, 3042, 4533, 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.
| \(\displaystyle \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 4583 |
| \(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (3 b C \sec ^2(c+d x)+a (2 A+3 C) \sec (c+d x)+2 A b\right )dx+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (3 b C \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (2 A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 A b\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {1}{3} \left (\frac {a A b \sin (c+d x) \cos (c+d x)}{d}-\frac {1}{2} \int -2 \cos (c+d x) \left ((2 A+3 C) a^2+3 b (A+2 C) \sec (c+d x) a+2 A b^2+3 b^2 C \sec ^2(c+d x)\right )dx\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\int \cos (c+d x) \left ((2 A+3 C) a^2+3 b (A+2 C) \sec (c+d x) a+2 A b^2+3 b^2 C \sec ^2(c+d x)\right )dx+\frac {a A b \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {(2 A+3 C) a^2+3 b (A+2 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 A b^2+3 b^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a A b \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{3} \left (\int \cos (c+d x) \left ((2 A+3 C) a^2+2 A b^2+3 b^2 C \sec ^2(c+d x)\right )dx+3 a b (A+2 C) \int 1dx+\frac {a A b \sin (c+d x) \cos (c+d x)}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} \left (\int \cos (c+d x) \left ((2 A+3 C) a^2+2 A b^2+3 b^2 C \sec ^2(c+d x)\right )dx+\frac {a A b \sin (c+d x) \cos (c+d x)}{d}+3 a b x (A+2 C)\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\int \frac {(2 A+3 C) a^2+2 A b^2+3 b^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a A b \sin (c+d x) \cos (c+d x)}{d}+3 a b x (A+2 C)\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {1}{3} \left (3 b^2 C \int \sec (c+d x)dx+\frac {\left (a^2 (2 A+3 C)+2 A b^2\right ) \sin (c+d x)}{d}+\frac {a A b \sin (c+d x) \cos (c+d x)}{d}+3 a b x (A+2 C)\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (3 b^2 C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\left (a^2 (2 A+3 C)+2 A b^2\right ) \sin (c+d x)}{d}+\frac {a A b \sin (c+d x) \cos (c+d x)}{d}+3 a b x (A+2 C)\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {\left (a^2 (2 A+3 C)+2 A b^2\right ) \sin (c+d x)}{d}+\frac {a A b \sin (c+d x) \cos (c+d x)}{d}+3 a b x (A+2 C)+\frac {3 b^2 C \text {arctanh}(\sin (c+d x))}{d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + (3*a*b*(A + 2*C)*x + (3*b^2*C*ArcTanh[Sin[c + d*x]])/d + ((2*A*b^2 + a^2*(2*A + 3*C)) *Sin[c + d*x])/d + (a*A*b*Cos[c + d*x]*Sin[c + d*x])/d)/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si mp[1/(d*n) Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
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 = 0.81 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{2} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \sin \left (d x +c \right )+2 a A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \left (d x +c \right )+A \,b^{2} \sin \left (d x +c \right )+C \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(106\) |
| default | \(\frac {\frac {a^{2} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \sin \left (d x +c \right )+2 a A b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \left (d x +c \right )+A \,b^{2} \sin \left (d x +c \right )+C \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(106\) |
| parallelrisch | \(\frac {-12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+12 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}+6 a A b \sin \left (2 d x +2 c \right )+a^{2} A \sin \left (3 d x +3 c \right )+\left (\left (9 A +12 C \right ) a^{2}+12 A \,b^{2}\right ) \sin \left (d x +c \right )+12 a b \left (A +2 C \right ) x d}{12 d}\) | \(107\) |
| risch | \(a A b x +2 a b x C -\frac {3 i a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C \,a^{2}}{2 d}+\frac {3 i a^{2} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{2}}{d}+\frac {a^{2} A \sin \left (3 d x +3 c \right )}{12 d}+\frac {a A b \sin \left (2 d x +2 c \right )}{2 d}\) | \(205\) |
| norman | \(\frac {\left (-a A b -2 C a b \right ) x +\left (-3 a A b -6 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (a A b +2 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (3 a A b +6 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {4 \left (a^{2} A -a A b -A \,b^{2}-C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 \left (a^{2} A -a A b +A \,b^{2}+C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {4 \left (a^{2} A +a A b -A \,b^{2}-C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {2 \left (a^{2} A +a A b +A \,b^{2}+C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (7 a^{2} A -9 a A b +3 A \,b^{2}+3 C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {2 \left (7 a^{2} A +9 a A b +3 A \,b^{2}+3 C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}+\frac {C \,b^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {C \,b^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(387\) |
Input:
int(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x,method=_RETURNVER BOSE)
Output:
1/d*(1/3*a^2*A*(2+cos(d*x+c)^2)*sin(d*x+c)+C*a^2*sin(d*x+c)+2*a*A*b*(1/2*c os(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+2*C*a*b*(d*x+c)+A*b^2*sin(d*x+c)+C*b^2 *ln(sec(d*x+c)+tan(d*x+c)))
Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {6 \, {\left (A + 2 \, C\right )} a b d x + 3 \, C b^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C b^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{2} \cos \left (d x + c\right )^{2} + 3 \, A a b \cos \left (d x + c\right ) + {\left (2 \, A + 3 \, C\right )} a^{2} + 3 \, A b^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm= "fricas")
Output:
1/6*(6*(A + 2*C)*a*b*d*x + 3*C*b^2*log(sin(d*x + c) + 1) - 3*C*b^2*log(-si n(d*x + c) + 1) + 2*(A*a^2*cos(d*x + c)^2 + 3*A*a*b*cos(d*x + c) + (2*A + 3*C)*a^2 + 3*A*b^2)*sin(d*x + c))/d
Timed out. \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*(a+b*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 12 \, {\left (d x + c\right )} C a b - 3 \, C b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, C a^{2} \sin \left (d x + c\right ) - 6 \, A b^{2} \sin \left (d x + c\right )}{6 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm= "maxima")
Output:
-1/6*(2*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 3*(2*d*x + 2*c + sin(2*d *x + 2*c))*A*a*b - 12*(d*x + c)*C*a*b - 3*C*b^2*(log(sin(d*x + c) + 1) - l og(sin(d*x + c) - 1)) - 6*C*a^2*sin(d*x + c) - 6*A*b^2*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 256 vs. \(2 (106) = 212\).
Time = 0.24 (sec) , antiderivative size = 256, normalized size of antiderivative = 2.29 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, C b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, C b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (A a b + 2 \, C a b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm= "giac")
Output:
1/3*(3*C*b^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*C*b^2*log(abs(tan(1/2* d*x + 1/2*c) - 1)) + 3*(A*a*b + 2*C*a*b)*(d*x + c) + 2*(3*A*a^2*tan(1/2*d* x + 1/2*c)^5 + 3*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*A*a*b*tan(1/2*d*x + 1/2* c)^5 + 3*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 2*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 6 *C*a^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*A*a^2*t an(1/2*d*x + 1/2*c) + 3*C*a^2*tan(1/2*d*x + 1/2*c) + 3*A*a*b*tan(1/2*d*x + 1/2*c) + 3*A*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 12.13 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.52 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,C\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {2\,A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \] Input:
int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^2,x)
Output:
(3*A*a^2*sin(c + d*x))/(4*d) + (A*b^2*sin(c + d*x))/d + (C*a^2*sin(c + d*x ))/d + (2*C*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^2*s in(3*c + 3*d*x))/(12*d) + (2*A*a*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x) /2)))/d + (4*C*a*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a*b *sin(2*c + 2*d*x))/(2*d)
Time = 0.16 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.22 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2} c +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2} c -\sin \left (d x +c \right )^{3} a^{3}+3 \sin \left (d x +c \right ) a^{3}+3 \sin \left (d x +c \right ) a^{2} c +3 \sin \left (d x +c \right ) a \,b^{2}+3 a^{2} b c +3 a^{2} b d x +6 a b \,c^{2}+6 a b c d x}{3 d} \] Input:
int(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)
Output:
(3*cos(c + d*x)*sin(c + d*x)*a**2*b - 3*log(tan((c + d*x)/2) - 1)*b**2*c + 3*log(tan((c + d*x)/2) + 1)*b**2*c - sin(c + d*x)**3*a**3 + 3*sin(c + d*x )*a**3 + 3*sin(c + d*x)*a**2*c + 3*sin(c + d*x)*a*b**2 + 3*a**2*b*c + 3*a* *2*b*d*x + 6*a*b*c**2 + 6*a*b*c*d*x)/(3*d)