Integrand size = 29, antiderivative size = 106 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {a (3 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (A+B) \tan (c+d x)}{d}+\frac {a (3 A+4 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a (A+B) \tan ^3(c+d x)}{3 d} \] Output:
1/8*a*(3*A+4*B)*arctanh(sin(d*x+c))/d+a*(A+B)*tan(d*x+c)/d+1/8*a*(3*A+4*B) *sec(d*x+c)*tan(d*x+c)/d+1/4*a*A*sec(d*x+c)^3*tan(d*x+c)/d+1/3*a*(A+B)*tan (d*x+c)^3/d
Time = 0.47 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {a \left (3 (3 A+4 B) \text {arctanh}(\sin (c+d x))+\sec (c+d x) \left (9 A+12 B+8 (A+B) (2+\cos (2 (c+d x))) \sec (c+d x)+6 A \sec ^2(c+d x)\right ) \tan (c+d x)\right )}{24 d} \] Input:
Integrate[(a + a*Cos[c + d*x])*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]
Output:
(a*(3*(3*A + 4*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(9*A + 12*B + 8*(A + B)*(2 + Cos[2*(c + d*x)])*Sec[c + d*x] + 6*A*Sec[c + d*x]^2)*Tan[c + d*x ]))/(24*d)
Time = 0.66 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.96, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 2009, 4255, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+a) (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int \sec ^5(c+d x) \left ((a A+a B) \cos (c+d x)+a A+a B \cos ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a A+a B) \sin \left (c+d x+\frac {\pi }{2}\right )+a A+a B \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{4} \int (4 a (A+B)+a (3 A+4 B) \cos (c+d x)) \sec ^4(c+d x)dx+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {4 a (A+B)+a (3 A+4 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{4} \left (4 a (A+B) \int \sec ^4(c+d x)dx+a (3 A+4 B) \int \sec ^3(c+d x)dx\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+4 a (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a (A+B) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a (A+B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a (A+B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a (A+B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (a (3 A+4 B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a (A+B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
Input:
Int[(a + a*Cos[c + d*x])*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]
Output:
(a*A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*(3*A + 4*B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (4*a*(A + B)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/4
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
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]
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[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 10.74 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12
| method | result | size |
| parts | \(\frac {A a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (A a +B a \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(119\) |
| derivativedivides | \(\frac {-A a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(131\) |
| default | \(\frac {-A a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(131\) |
| parallelrisch | \(\frac {8 \left (-\frac {9 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {4 B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16}+\frac {9 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {4 B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16}+\left (A +B \right ) \sin \left (2 d x +2 c \right )+\frac {3 \left (\frac {3 A}{4}+B \right ) \sin \left (3 d x +3 c \right )}{8}+\frac {\left (A +B \right ) \sin \left (4 d x +4 c \right )}{4}+\frac {33 \left (A +\frac {4 B}{11}\right ) \sin \left (d x +c \right )}{32}\right ) a}{3 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(170\) |
| norman | \(\frac {-\frac {a \left (3 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {a \left (13 A -20 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}+\frac {a \left (13 A +12 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (29 A -4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {a \left (31 A +4 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}+\frac {a \left (47 A +20 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(226\) |
| risch | \(-\frac {i a \left (9 A \,{\mathrm e}^{7 i \left (d x +c \right )}+12 B \,{\mathrm e}^{7 i \left (d x +c \right )}+33 A \,{\mathrm e}^{5 i \left (d x +c \right )}+12 B \,{\mathrm e}^{5 i \left (d x +c \right )}-48 A \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B \,{\mathrm e}^{4 i \left (d x +c \right )}-33 A \,{\mathrm e}^{3 i \left (d x +c \right )}-12 B \,{\mathrm e}^{3 i \left (d x +c \right )}-64 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A \,{\mathrm e}^{i \left (d x +c \right )}-12 B \,{\mathrm e}^{i \left (d x +c \right )}-16 A -16 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 A a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}+\frac {3 A a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}\) | \(253\) |
Input:
int((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^5,x,method=_RETURNVERBOSE )
Output:
A*a/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+ta n(d*x+c)))-(A*a+B*a)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+B*a/d*(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 = 127, normalized size of antiderivative = 1.20 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 6 \, A a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="fri cas")
Output:
1/48*(3*(3*A + 4*B)*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*A + 4*B) *a*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*(A + B)*a*cos(d*x + c)^3 + 3*(3*A + 4*B)*a*cos(d*x + c)^2 + 8*(A + B)*a*cos(d*x + c) + 6*A*a)*sin(d *x + c))/(d*cos(d*x + c)^4)
\[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=a \left (\int A \sec ^{5}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int B \cos ^{2}{\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)**5,x)
Output:
a*(Integral(A*sec(c + d*x)**5, x) + Integral(A*cos(c + d*x)*sec(c + d*x)** 5, x) + Integral(B*cos(c + d*x)*sec(c + d*x)**5, x) + Integral(B*cos(c + d *x)**2*sec(c + d*x)**5, x))
Time = 0.04 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.54 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a - 3 \, A a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \] Input:
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="max ima")
Output:
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a + 16*(tan(d*x + c)^3 + 3*ta n(d*x + c))*B*a - 3*A*a*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c ) - 1)) - 12*B*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
Time = 0.34 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.77 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, A a + 4 \, B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, A a + 4 \, B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 49 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 28 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 31 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 52 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 39 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 36 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:
integrate((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="gia c")
Output:
1/24*(3*(3*A*a + 4*B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a + 4* B*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(9*A*a*tan(1/2*d*x + 1/2*c)^7 + 12*B*a*tan(1/2*d*x + 1/2*c)^7 - 49*A*a*tan(1/2*d*x + 1/2*c)^5 - 28*B*a*t an(1/2*d*x + 1/2*c)^5 + 31*A*a*tan(1/2*d*x + 1/2*c)^3 + 52*B*a*tan(1/2*d*x + 1/2*c)^3 - 39*A*a*tan(1/2*d*x + 1/2*c) - 36*B*a*tan(1/2*d*x + 1/2*c))/( tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 43.02 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.57 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {\left (-\frac {3\,A\,a}{4}-B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {49\,A\,a}{12}+\frac {7\,B\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {31\,A\,a}{12}-\frac {13\,B\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+3\,B\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A+4\,B\right )}{4\,d} \] Input:
int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x)))/cos(c + d*x)^5,x)
Output:
(tan(c/2 + (d*x)/2)*((13*A*a)/4 + 3*B*a) - tan(c/2 + (d*x)/2)^7*((3*A*a)/4 + B*a) - tan(c/2 + (d*x)/2)^3*((31*A*a)/12 + (13*B*a)/3) + tan(c/2 + (d*x )/2)^5*((49*A*a)/12 + (7*B*a)/3))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (a*ata nh(tan(c/2 + (d*x)/2))*(3*A + 4*B))/(4*d)
Time = 0.18 (sec) , antiderivative size = 479, normalized size of antiderivative = 4.52 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+a*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^5,x)
Output:
(a*( - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 12*cos (c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b + 18*cos(c + d*x)*lo g(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 24*cos(c + d*x)*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**2*b - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a - 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b - 18*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 24*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b + 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b - 9*cos(c + d*x)*sin(c + d*x)**3*a - 12*cos(c + d*x)*sin(c + d*x)**3*b + 15*cos(c + d*x)*sin(c + d*x)*a + 12*cos(c + d*x)*sin(c + d*x )*b + 16*sin(c + d*x)**5*a + 16*sin(c + d*x)**5*b - 40*sin(c + d*x)**3*a - 40*sin(c + d*x)**3*b + 24*sin(c + d*x)*a + 24*sin(c + d*x)*b))/(24*cos(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))