Integrand size = 31, antiderivative size = 131 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {3 (A-B) \text {arctanh}(\sin (c+d x))}{2 a d}+\frac {(4 A-3 B) \tan (c+d x)}{a d}-\frac {3 (A-B) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {(4 A-3 B) \tan ^3(c+d x)}{3 a d} \]
-3/2*(A-B)*arctanh(sin(d*x+c))/a/d+(4*A-3*B)*tan(d*x+c)/a/d-3/2*(A-B)*sec( d*x+c)*tan(d*x+c)/a/d-(A-B)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d*x+c))+1/3 *(4*A-3*B)*tan(d*x+c)^3/a/d
Leaf count is larger than twice the leaf count of optimal. \(490\) vs. \(2(131)=262\).
Time = 3.70 (sec) , antiderivative size = 490, normalized size of antiderivative = 3.74 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (144 (A-B) \cos \left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sec (c) \sec ^3(c+d x) \left (6 (A+B) \sin \left (\frac {d x}{2}\right )+3 (13 A-9 B) \sin \left (\frac {3 d x}{2}\right )-24 A \sin \left (c-\frac {d x}{2}\right )+12 B \sin \left (c-\frac {d x}{2}\right )-6 A \sin \left (c+\frac {d x}{2}\right )+6 B \sin \left (c+\frac {d x}{2}\right )-24 A \sin \left (2 c+\frac {d x}{2}\right )+24 B \sin \left (2 c+\frac {d x}{2}\right )+21 A \sin \left (c+\frac {3 d x}{2}\right )-9 B \sin \left (c+\frac {3 d x}{2}\right )+9 A \sin \left (2 c+\frac {3 d x}{2}\right )-9 B \sin \left (2 c+\frac {3 d x}{2}\right )-9 A \sin \left (3 c+\frac {3 d x}{2}\right )+9 B \sin \left (3 c+\frac {3 d x}{2}\right )+7 A \sin \left (c+\frac {5 d x}{2}\right )-3 B \sin \left (c+\frac {5 d x}{2}\right )+A \sin \left (2 c+\frac {5 d x}{2}\right )+3 B \sin \left (2 c+\frac {5 d x}{2}\right )-3 A \sin \left (3 c+\frac {5 d x}{2}\right )+3 B \sin \left (3 c+\frac {5 d x}{2}\right )-9 A \sin \left (4 c+\frac {5 d x}{2}\right )+9 B \sin \left (4 c+\frac {5 d x}{2}\right )+16 A \sin \left (2 c+\frac {7 d x}{2}\right )-12 B \sin \left (2 c+\frac {7 d x}{2}\right )+10 A \sin \left (3 c+\frac {7 d x}{2}\right )-6 B \sin \left (3 c+\frac {7 d x}{2}\right )+6 A \sin \left (4 c+\frac {7 d x}{2}\right )-6 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )\right )}{48 a d (1+\cos (c+d x))} \]
(Cos[(c + d*x)/2]*(144*(A - B)*Cos[(c + d*x)/2]*(Log[Cos[(c + d*x)/2] - Si n[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec [c]*Sec[c + d*x]^3*(6*(A + B)*Sin[(d*x)/2] + 3*(13*A - 9*B)*Sin[(3*d*x)/2] - 24*A*Sin[c - (d*x)/2] + 12*B*Sin[c - (d*x)/2] - 6*A*Sin[c + (d*x)/2] + 6*B*Sin[c + (d*x)/2] - 24*A*Sin[2*c + (d*x)/2] + 24*B*Sin[2*c + (d*x)/2] + 21*A*Sin[c + (3*d*x)/2] - 9*B*Sin[c + (3*d*x)/2] + 9*A*Sin[2*c + (3*d*x)/ 2] - 9*B*Sin[2*c + (3*d*x)/2] - 9*A*Sin[3*c + (3*d*x)/2] + 9*B*Sin[3*c + ( 3*d*x)/2] + 7*A*Sin[c + (5*d*x)/2] - 3*B*Sin[c + (5*d*x)/2] + A*Sin[2*c + (5*d*x)/2] + 3*B*Sin[2*c + (5*d*x)/2] - 3*A*Sin[3*c + (5*d*x)/2] + 3*B*Sin [3*c + (5*d*x)/2] - 9*A*Sin[4*c + (5*d*x)/2] + 9*B*Sin[4*c + (5*d*x)/2] + 16*A*Sin[2*c + (7*d*x)/2] - 12*B*Sin[2*c + (7*d*x)/2] + 10*A*Sin[3*c + (7* d*x)/2] - 6*B*Sin[3*c + (7*d*x)/2] + 6*A*Sin[4*c + (7*d*x)/2] - 6*B*Sin[4* c + (7*d*x)/2])))/(48*a*d*(1 + Cos[c + d*x]))
Time = 0.70 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.90, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3457, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) (A+B \cos (c+d x))}{a \cos (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int (a (4 A-3 B)-3 a (A-B) \cos (c+d x)) \sec ^4(c+d x)dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (4 A-3 B)-3 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {a (4 A-3 B) \int \sec ^4(c+d x)dx-3 a (A-B) \int \sec ^3(c+d x)dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (4 A-3 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx-3 a (A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {-\frac {a (4 A-3 B) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}-3 a (A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-3 a (A-B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {a (4 A-3 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {-3 a (A-B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a (4 A-3 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-3 a (A-B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a (4 A-3 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {-3 a (A-B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a (4 A-3 B) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)}\) |
-(((A - B)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Cos[c + d*x]))) + (-3*a* (A - B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (a*(4*A - 3*B)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/a^2
3.1.46.3.1 Defintions of rubi rules used
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_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 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 = 1.40 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.30
| method | result | size |
| parallelrisch | \(\frac {27 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-27 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+44 \left (\frac {\left (4 A -3 B \right ) \cos \left (3 d x +3 c \right )}{11}+\frac {\left (7 A -3 B \right ) \cos \left (2 d x +2 c \right )}{22}+\left (A -\frac {6 B}{11}\right ) \cos \left (d x +c \right )+\frac {A}{2}-\frac {3 B}{22}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(170\) |
| derivativedivides | \(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {2 A -B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (\frac {3 A}{2}-\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 A}{2}-\frac {3 B}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {B -2 A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {\frac {5 A}{2}-\frac {3 B}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-\frac {3 A}{2}+\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\) | \(190\) |
| default | \(\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {2 A -B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (\frac {3 A}{2}-\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 A}{2}-\frac {3 B}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {B -2 A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {\frac {5 A}{2}-\frac {3 B}{2}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-\frac {3 A}{2}+\frac {3 B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}\) | \(190\) |
| norman | \(\frac {\frac {\left (A -B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (A -3 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {2 \left (2 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {\left (7 A -5 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (13 A -15 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {3 \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}-\frac {3 \left (A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\) | \(207\) |
| risch | \(\frac {i \left (9 A \,{\mathrm e}^{6 i \left (d x +c \right )}-9 B \,{\mathrm e}^{6 i \left (d x +c \right )}+9 A \,{\mathrm e}^{5 i \left (d x +c \right )}-9 B \,{\mathrm e}^{5 i \left (d x +c \right )}+24 A \,{\mathrm e}^{4 i \left (d x +c \right )}-24 B \,{\mathrm e}^{4 i \left (d x +c \right )}+24 A \,{\mathrm e}^{3 i \left (d x +c \right )}-12 B \,{\mathrm e}^{3 i \left (d x +c \right )}+39 A \,{\mathrm e}^{2 i \left (d x +c \right )}-27 B \,{\mathrm e}^{2 i \left (d x +c \right )}+7 A \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}+16 A -12 B \right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a d}+\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a d}\) | \(276\) |
1/6*(27*(1/3*cos(3*d*x+3*c)+cos(d*x+c))*(A-B)*ln(tan(1/2*d*x+1/2*c)-1)-27* (1/3*cos(3*d*x+3*c)+cos(d*x+c))*(A-B)*ln(tan(1/2*d*x+1/2*c)+1)+44*(1/11*(4 *A-3*B)*cos(3*d*x+3*c)+1/22*(7*A-3*B)*cos(2*d*x+2*c)+(A-6/11*B)*cos(d*x+c) +1/2*A-3/22*B)*tan(1/2*d*x+1/2*c))/a/d/(cos(3*d*x+3*c)+3*cos(d*x+c))
Time = 0.30 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.28 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {9 \, {\left ({\left (A - B\right )} \cos \left (d x + c\right )^{4} + {\left (A - B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, {\left ({\left (A - B\right )} \cos \left (d x + c\right )^{4} + {\left (A - B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (4 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \]
-1/12*(9*((A - B)*cos(d*x + c)^4 + (A - B)*cos(d*x + c)^3)*log(sin(d*x + c ) + 1) - 9*((A - B)*cos(d*x + c)^4 + (A - B)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(4*A - 3*B)*cos(d*x + c)^3 + (7*A - 3*B)*cos(d*x + c)^2 - (A - 3*B)*cos(d*x + c) + 2*A)*sin(d*x + c))/(a*d*cos(d*x + c)^4 + a*d*cos (d*x + c)^3)
\[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \]
(Integral(A*sec(c + d*x)**4/(cos(c + d*x) + 1), x) + Integral(B*cos(c + d* x)*sec(c + d*x)**4/(cos(c + d*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 368 vs. \(2 (125) = 250\).
Time = 0.22 (sec) , antiderivative size = 368, normalized size of antiderivative = 2.81 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {A {\left (\frac {2 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, B {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]
1/6*(A*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c )^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*sin (d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a* (cos(d*x + c) + 1))) - 3*B*(2*(sin(d*x + c)/(cos(d*x + c) + 1) - 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 3*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 2*sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d
Time = 0.32 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.39 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\frac {9 \, {\left (A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {9 \, {\left (A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \]
-1/6*(9*(A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 9*(A - B)*log(abs(t an(1/2*d*x + 1/2*c) - 1))/a - 6*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c))/a + 2*(15*A*tan(1/2*d*x + 1/2*c)^5 - 9*B*tan(1/2*d*x + 1/2*c)^5 - 16*A*tan(1/2*d*x + 1/2*c)^3 + 12*B*tan(1/2*d*x + 1/2*c)^3 + 9*A*tan(1/2*d* x + 1/2*c) - 3*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a)) /d
Time = 0.71 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.16 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\left (5\,A-3\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,B-\frac {16\,A}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A-B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A-B\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \]