\(\int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx\) [72]

Optimal result

Integrand size = 31, antiderivative size = 175 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {(4 A-B) \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {8 (83 A-20 B) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-25 B) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(4 A-B) \tan (c+d x)}{a^4 d (1+\cos (c+d x))}-\frac {(A-B) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3} \]

[Out]

-(4*A-B)*arctanh(sin(d*x+c))/a^4/d+8/105*(83*A-20*B)*tan(d*x+c)/a^4/d-1/105*(88*A-25*B)*tan(d*x+c)/a^4/d/(1+co
s(d*x+c))^2-(4*A-B)*tan(d*x+c)/a^4/d/(1+cos(d*x+c))-1/7*(A-B)*tan(d*x+c)/d/(a+a*cos(d*x+c))^4-1/35*(12*A-5*B)*
tan(d*x+c)/a/d/(a+a*cos(d*x+c))^3

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3057, 2827, 3852, 8, 3855} \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {(4 A-B) \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {8 (83 A-20 B) \tan (c+d x)}{105 a^4 d}-\frac {(4 A-B) \tan (c+d x)}{a^4 d (\cos (c+d x)+1)}-\frac {(88 A-25 B) \tan (c+d x)}{105 a^4 d (\cos (c+d x)+1)^2}-\frac {(12 A-5 B) \tan (c+d x)}{35 a d (a \cos (c+d x)+a)^3}-\frac {(A-B) \tan (c+d x)}{7 d (a \cos (c+d x)+a)^4} \]

[In]

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^2)/(a + a*Cos[c + d*x])^4,x]

[Out]

-(((4*A - B)*ArcTanh[Sin[c + d*x]])/(a^4*d)) + (8*(83*A - 20*B)*Tan[c + d*x])/(105*a^4*d) - ((88*A - 25*B)*Tan
[c + d*x])/(105*a^4*d*(1 + Cos[c + d*x])^2) - ((4*A - B)*Tan[c + d*x])/(a^4*d*(1 + Cos[c + d*x])) - ((A - B)*T
an[c + d*x])/(7*d*(a + a*Cos[c + d*x])^4) - ((12*A - 5*B)*Tan[c + d*x])/(35*a*d*(a + a*Cos[c + d*x])^3)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3057

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[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] + Dist[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])

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}+\frac {\int \frac {(a (8 A-B)-4 a (A-B) \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx}{7 a^2} \\ & = -\frac {(A-B) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (2 a^2 (26 A-5 B)-3 a^2 (12 A-5 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{35 a^4} \\ & = -\frac {(88 A-25 B) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}+\frac {\int \frac {\left (a^3 (244 A-55 B)-2 a^3 (88 A-25 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{105 a^6} \\ & = -\frac {(88 A-25 B) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(4 A-B) \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {\int \left (8 a^4 (83 A-20 B)-105 a^4 (4 A-B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{105 a^8} \\ & = -\frac {(88 A-25 B) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(4 A-B) \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}+\frac {(8 (83 A-20 B)) \int \sec ^2(c+d x) \, dx}{105 a^4}-\frac {(4 A-B) \int \sec (c+d x) \, dx}{a^4} \\ & = -\frac {(4 A-B) \text {arctanh}(\sin (c+d x))}{a^4 d}-\frac {(88 A-25 B) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(4 A-B) \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )}-\frac {(8 (83 A-20 B)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{105 a^4 d} \\ & = -\frac {(4 A-B) \text {arctanh}(\sin (c+d x))}{a^4 d}+\frac {8 (83 A-20 B) \tan (c+d x)}{105 a^4 d}-\frac {(88 A-25 B) \tan (c+d x)}{105 a^4 d (1+\cos (c+d x))^2}-\frac {(A-B) \tan (c+d x)}{7 d (a+a \cos (c+d x))^4}-\frac {(12 A-5 B) \tan (c+d x)}{35 a d (a+a \cos (c+d x))^3}-\frac {(4 A-B) \tan (c+d x)}{d \left (a^4+a^4 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(595\) vs. \(2(175)=350\).

Time = 4.60 (sec) , antiderivative size = 595, normalized size of antiderivative = 3.40 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {26880 (4 A-B) \cos ^8\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 )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-245 (44 A-17 B) \sin \left (\frac {d x}{2}\right )+7 (2684 A-635 B) \sin \left (\frac {3 d x}{2}\right )-20524 A \sin \left (c-\frac {d x}{2}\right )+4795 B \sin \left (c-\frac {d x}{2}\right )+14644 A \sin \left (c+\frac {d x}{2}\right )-4795 B \sin \left (c+\frac {d x}{2}\right )-16660 A \sin \left (2 c+\frac {d x}{2}\right )+4165 B \sin \left (2 c+\frac {d x}{2}\right )-4690 A \sin \left (c+\frac {3 d x}{2}\right )+2275 B \sin \left (c+\frac {3 d x}{2}\right )+14378 A \sin \left (2 c+\frac {3 d x}{2}\right )-4445 B \sin \left (2 c+\frac {3 d x}{2}\right )-9100 A \sin \left (3 c+\frac {3 d x}{2}\right )+2275 B \sin \left (3 c+\frac {3 d x}{2}\right )+11668 A \sin \left (c+\frac {5 d x}{2}\right )-2785 B \sin \left (c+\frac {5 d x}{2}\right )-630 A \sin \left (2 c+\frac {5 d x}{2}\right )+735 B \sin \left (2 c+\frac {5 d x}{2}\right )+9358 A \sin \left (3 c+\frac {5 d x}{2}\right )-2785 B \sin \left (3 c+\frac {5 d x}{2}\right )-2940 A \sin \left (4 c+\frac {5 d x}{2}\right )+735 B \sin \left (4 c+\frac {5 d x}{2}\right )+4228 A \sin \left (2 c+\frac {7 d x}{2}\right )-1015 B \sin \left (2 c+\frac {7 d x}{2}\right )+315 A \sin \left (3 c+\frac {7 d x}{2}\right )+105 B \sin \left (3 c+\frac {7 d x}{2}\right )+3493 A \sin \left (4 c+\frac {7 d x}{2}\right )-1015 B \sin \left (4 c+\frac {7 d x}{2}\right )-420 A \sin \left (5 c+\frac {7 d x}{2}\right )+105 B \sin \left (5 c+\frac {7 d x}{2}\right )+664 A \sin \left (3 c+\frac {9 d x}{2}\right )-160 B \sin \left (3 c+\frac {9 d x}{2}\right )+105 A \sin \left (4 c+\frac {9 d x}{2}\right )+559 A \sin \left (5 c+\frac {9 d x}{2}\right )-160 B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{1680 a^4 d (1+\cos (c+d x))^4} \]

[In]

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^2)/(a + a*Cos[c + d*x])^4,x]

[Out]

(26880*(4*A - B)*Cos[(c + d*x)/2]^8*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]*(-245*(44*A - 17*B)*Sin[(d*x)/2] + 7*(2684*A - 635
*B)*Sin[(3*d*x)/2] - 20524*A*Sin[c - (d*x)/2] + 4795*B*Sin[c - (d*x)/2] + 14644*A*Sin[c + (d*x)/2] - 4795*B*Si
n[c + (d*x)/2] - 16660*A*Sin[2*c + (d*x)/2] + 4165*B*Sin[2*c + (d*x)/2] - 4690*A*Sin[c + (3*d*x)/2] + 2275*B*S
in[c + (3*d*x)/2] + 14378*A*Sin[2*c + (3*d*x)/2] - 4445*B*Sin[2*c + (3*d*x)/2] - 9100*A*Sin[3*c + (3*d*x)/2] +
2275*B*Sin[3*c + (3*d*x)/2] + 11668*A*Sin[c + (5*d*x)/2] - 2785*B*Sin[c + (5*d*x)/2] - 630*A*Sin[2*c + (5*d*x
)/2] + 735*B*Sin[2*c + (5*d*x)/2] + 9358*A*Sin[3*c + (5*d*x)/2] - 2785*B*Sin[3*c + (5*d*x)/2] - 2940*A*Sin[4*c
+ (5*d*x)/2] + 735*B*Sin[4*c + (5*d*x)/2] + 4228*A*Sin[2*c + (7*d*x)/2] - 1015*B*Sin[2*c + (7*d*x)/2] + 315*A
*Sin[3*c + (7*d*x)/2] + 105*B*Sin[3*c + (7*d*x)/2] + 3493*A*Sin[4*c + (7*d*x)/2] - 1015*B*Sin[4*c + (7*d*x)/2]
- 420*A*Sin[5*c + (7*d*x)/2] + 105*B*Sin[5*c + (7*d*x)/2] + 664*A*Sin[3*c + (9*d*x)/2] - 160*B*Sin[3*c + (9*d
*x)/2] + 105*A*Sin[4*c + (9*d*x)/2] + 559*A*Sin[5*c + (9*d*x)/2] - 160*B*Sin[5*c + (9*d*x)/2]))/(1680*a^4*d*(1
+ Cos[c + d*x])^4)

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.91

method	result	size
parallelrisch	\(\frac {13440 \cos \left (d x +c \right ) \left (A -\frac {B}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-13440 \cos \left (d x +c \right ) \left (A -\frac {B}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+332 \left (\sec ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (\frac {1650 A}{83}-\frac {390 B}{83}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {559 A}{83}-\frac {535 B}{332}\right ) \cos \left (3 d x +3 c \right )+\left (A -\frac {20 B}{83}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {2861 A}{83}-\frac {2645 B}{332}\right ) \cos \left (d x +c \right )+\frac {1672 A}{83}-\frac {370 B}{83}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3360 d \,a^{4} \cos \left (d x +c \right )}\)	\(160\)
derivativedivides	\(\frac {\left (-32 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}+\frac {7 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B +\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-15 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (32 A -8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{8 d \,a^{4}}\)	\(190\)
default	\(\frac {\left (-32 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{7}-\frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{7}+\frac {7 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B +\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+49 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-15 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (32 A -8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {8 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{8 d \,a^{4}}\)	\(190\)
norman	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 a d}+\frac {\left (7 A -5 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 a d}-\frac {5 \left (13 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {7 \left (17 A -5 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (71 A -11 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 a d}+\frac {\left (79 A -37 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{84 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{3}}+\frac {\left (4 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{4} d}-\frac {\left (4 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{4} d}\)	\(242\)
risch	\(\frac {2 i \left (420 A \,{\mathrm e}^{8 i \left (d x +c \right )}-105 B \,{\mathrm e}^{8 i \left (d x +c \right )}+2940 A \,{\mathrm e}^{7 i \left (d x +c \right )}-735 B \,{\mathrm e}^{7 i \left (d x +c \right )}+9100 A \,{\mathrm e}^{6 i \left (d x +c \right )}-2275 B \,{\mathrm e}^{6 i \left (d x +c \right )}+16660 A \,{\mathrm e}^{5 i \left (d x +c \right )}-4165 B \,{\mathrm e}^{5 i \left (d x +c \right )}+20524 A \,{\mathrm e}^{4 i \left (d x +c \right )}-4795 B \,{\mathrm e}^{4 i \left (d x +c \right )}+18788 A \,{\mathrm e}^{3 i \left (d x +c \right )}-4445 B \,{\mathrm e}^{3 i \left (d x +c \right )}+11668 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2785 B \,{\mathrm e}^{2 i \left (d x +c \right )}+4228 A \,{\mathrm e}^{i \left (d x +c \right )}-1015 B \,{\mathrm e}^{i \left (d x +c \right )}+664 A -160 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {4 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{4} d}+\frac {4 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{4} d}\)	\(323\)

[In]

int((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+cos(d*x+c)*a)^4,x,method=_RETURNVERBOSE)

[Out]

1/3360*(13440*cos(d*x+c)*(A-1/4*B)*ln(tan(1/2*d*x+1/2*c)-1)-13440*cos(d*x+c)*(A-1/4*B)*ln(tan(1/2*d*x+1/2*c)+1
)+332*sec(1/2*d*x+1/2*c)^6*((1650/83*A-390/83*B)*cos(2*d*x+2*c)+(559/83*A-535/332*B)*cos(3*d*x+3*c)+(A-20/83*B
)*cos(4*d*x+4*c)+(2861/83*A-2645/332*B)*cos(d*x+c)+1672/83*A-370/83*B)*tan(1/2*d*x+1/2*c))/d/a^4/cos(d*x+c)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (167) = 334\).

Time = 0.31 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.93 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {105 \, {\left ({\left (4 \, A - B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, {\left ({\left (4 \, A - B\right )} \cos \left (d x + c\right )^{5} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (4 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (8 \, {\left (83 \, A - 20 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (2236 \, A - 535 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (659 \, A - 155 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (296 \, A - 65 \, B\right )} \cos \left (d x + c\right ) + 105 \, A\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} + 4 \, a^{4} d \cos \left (d x + c\right )^{4} + 6 \, a^{4} d \cos \left (d x + c\right )^{3} + 4 \, a^{4} d \cos \left (d x + c\right )^{2} + a^{4} d \cos \left (d x + c\right )\right )}} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/210*(105*((4*A - B)*cos(d*x + c)^5 + 4*(4*A - B)*cos(d*x + c)^4 + 6*(4*A - B)*cos(d*x + c)^3 + 4*(4*A - B)*
cos(d*x + c)^2 + (4*A - B)*cos(d*x + c))*log(sin(d*x + c) + 1) - 105*((4*A - B)*cos(d*x + c)^5 + 4*(4*A - B)*c
os(d*x + c)^4 + 6*(4*A - B)*cos(d*x + c)^3 + 4*(4*A - B)*cos(d*x + c)^2 + (4*A - B)*cos(d*x + c))*log(-sin(d*x
+ c) + 1) - 2*(8*(83*A - 20*B)*cos(d*x + c)^4 + (2236*A - 535*B)*cos(d*x + c)^3 + 4*(659*A - 155*B)*cos(d*x +
c)^2 + 4*(296*A - 65*B)*cos(d*x + c) + 105*A)*sin(d*x + c))/(a^4*d*cos(d*x + c)^5 + 4*a^4*d*cos(d*x + c)^4 +
6*a^4*d*cos(d*x + c)^3 + 4*a^4*d*cos(d*x + c)^2 + a^4*d*cos(d*x + c))

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\cos ^{4}{\left (c + d x \right )} + 4 \cos ^{3}{\left (c + d x \right )} + 6 \cos ^{2}{\left (c + d x \right )} + 4 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\cos ^{4}{\left (c + d x \right )} + 4 \cos ^{3}{\left (c + d x \right )} + 6 \cos ^{2}{\left (c + d x \right )} + 4 \cos {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)**2/(a+a*cos(d*x+c))**4,x)

[Out]

(Integral(A*sec(c + d*x)**2/(cos(c + d*x)**4 + 4*cos(c + d*x)**3 + 6*cos(c + d*x)**2 + 4*cos(c + d*x) + 1), x)
+ Integral(B*cos(c + d*x)*sec(c + d*x)**2/(cos(c + d*x)**4 + 4*cos(c + d*x)**3 + 6*cos(c + d*x)**2 + 4*cos(c
+ d*x) + 1), x))/a**4

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.86 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} - \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {3360 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )} - 5 \, B {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{4}} + \frac {168 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{4}}\right )}}{840 \, d} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/840*(A*(1680*sin(d*x + c)/((a^4 - a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d
*x + c)/(cos(d*x + c) + 1) + 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5
+ 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 3360*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 3360*log(
sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4) - 5*B*((315*sin(d*x + c)/(cos(d*x + c) + 1) + 77*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 168*l
og(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^4 + 168*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^4))/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.28 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=-\frac {\frac {840 \, {\left (4 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac {840 \, {\left (4 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac {1680 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{4}} - \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

[In]

integrate((A+B*cos(d*x+c))*sec(d*x+c)^2/(a+a*cos(d*x+c))^4,x, algorithm="giac")

[Out]

-1/840*(840*(4*A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^4 - 840*(4*A - B)*log(abs(tan(1/2*d*x + 1/2*c) - 1)
)/a^4 + 1680*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^4) - (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 1
5*B*a^24*tan(1/2*d*x + 1/2*c)^7 + 147*A*a^24*tan(1/2*d*x + 1/2*c)^5 - 105*B*a^24*tan(1/2*d*x + 1/2*c)^5 + 805*
A*a^24*tan(1/2*d*x + 1/2*c)^3 - 385*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 5145*A*a^24*tan(1/2*d*x + 1/2*c) - 1575*B*
a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.35 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{8\,a^4}+\frac {5\,A-3\,B}{12\,a^4}+\frac {10\,A-2\,B}{24\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {A-B}{20\,a^4}+\frac {5\,A-3\,B}{40\,a^4}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-B}{2\,a^4}+\frac {3\,\left (5\,A-3\,B\right )}{8\,a^4}+\frac {10\,A-2\,B}{4\,a^4}+\frac {10\,A+2\,B}{8\,a^4}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4\,d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A-B\right )}{a^4\,d} \]

[In]

int((A + B*cos(c + d*x))/(cos(c + d*x)^2*(a + a*cos(c + d*x))^4),x)

[Out]

(tan(c/2 + (d*x)/2)^3*((A - B)/(8*a^4) + (5*A - 3*B)/(12*a^4) + (10*A - 2*B)/(24*a^4)))/d + (tan(c/2 + (d*x)/2
)^5*((A - B)/(20*a^4) + (5*A - 3*B)/(40*a^4)))/d + (tan(c/2 + (d*x)/2)*((A - B)/(2*a^4) + (3*(5*A - 3*B))/(8*a
^4) + (10*A - 2*B)/(4*a^4) + (10*A + 2*B)/(8*a^4)))/d + (tan(c/2 + (d*x)/2)^7*(A - B))/(56*a^4*d) - (2*A*tan(c
/2 + (d*x)/2))/(d*(a^4*tan(c/2 + (d*x)/2)^2 - a^4)) - (2*atanh(tan(c/2 + (d*x)/2))*(4*A - B))/(a^4*d)