\(\int \frac {(B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [276]

Optimal result

Integrand size = 40, antiderivative size = 196 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(13 B-6 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {8 (19 B-9 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 B-9 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

1/2*(13*B-6*C)*arctanh(sin(d*x+c))/a^3/d-8/15*(19*B-9*C)*tan(d*x+c)/a^3/d+1/2*(13*B-6*C)*sec(d*x+c)*tan(d*x+c)
/a^3/d-1/5*(B-C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15*(11*B-6*C)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos
(d*x+c))^2-4/15*(19*B-9*C)*sec(d*x+c)*tan(d*x+c)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {3108, 3057, 2827, 3853, 3855, 3852, 8} \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(13 B-6 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {8 (19 B-9 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \tan (c+d x) \sec (c+d x)}{2 a^3 d}-\frac {4 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(11 B-6 C) \tan (c+d x) \sec (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[In]

Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^3,x]

[Out]

((13*B - 6*C)*ArcTanh[Sin[c + d*x]])/(2*a^3*d) - (8*(19*B - 9*C)*Tan[c + d*x])/(15*a^3*d) + ((13*B - 6*C)*Sec[
c + d*x]*Tan[c + d*x])/(2*a^3*d) - ((B - C)*Sec[c + d*x]*Tan[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) - ((11*B -
6*C)*Sec[c + d*x]*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - (4*(19*B - 9*C)*Sec[c + d*x]*Tan[c + d*x])/
(15*d*(a^3 + a^3*Cos[c + d*x]))

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 3108

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*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 3853

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] + Dist[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]

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}& = \int \frac {(B+C \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx \\ & = -\frac {(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(a (7 B-2 C)-4 a (B-C) \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (a^2 (43 B-18 C)-3 a^2 (11 B-6 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 B-9 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \left (15 a^3 (13 B-6 C)-8 a^3 (19 B-9 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{15 a^6} \\ & = -\frac {(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 B-9 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(8 (19 B-9 C)) \int \sec ^2(c+d x) \, dx}{15 a^3}+\frac {(13 B-6 C) \int \sec ^3(c+d x) \, dx}{a^3} \\ & = \frac {(13 B-6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 B-9 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(13 B-6 C) \int \sec (c+d x) \, dx}{2 a^3}+\frac {(8 (19 B-9 C)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d} \\ & = \frac {(13 B-6 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {8 (19 B-9 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 B-9 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(610\) vs. \(2(196)=392\).

Time = 5.88 (sec) , antiderivative size = 610, normalized size of antiderivative = 3.11 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {1920 (13 B-6 C) \cos ^6\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 ^2(c+d x) \left ((-1235 B+870 C) \sin \left (\frac {d x}{2}\right )+5 (761 B-366 C) \sin \left (\frac {3 d x}{2}\right )-4329 B \sin \left (c-\frac {d x}{2}\right )+2094 C \sin \left (c-\frac {d x}{2}\right )+1989 B \sin \left (c+\frac {d x}{2}\right )-1314 C \sin \left (c+\frac {d x}{2}\right )-3575 B \sin \left (2 c+\frac {d x}{2}\right )+1650 C \sin \left (2 c+\frac {d x}{2}\right )-475 B \sin \left (c+\frac {3 d x}{2}\right )+450 C \sin \left (c+\frac {3 d x}{2}\right )+2005 B \sin \left (2 c+\frac {3 d x}{2}\right )-1230 C \sin \left (2 c+\frac {3 d x}{2}\right )-2275 B \sin \left (3 c+\frac {3 d x}{2}\right )+1050 C \sin \left (3 c+\frac {3 d x}{2}\right )+2673 B \sin \left (c+\frac {5 d x}{2}\right )-1278 C \sin \left (c+\frac {5 d x}{2}\right )+105 B \sin \left (2 c+\frac {5 d x}{2}\right )+90 C \sin \left (2 c+\frac {5 d x}{2}\right )+1593 B \sin \left (3 c+\frac {5 d x}{2}\right )-918 C \sin \left (3 c+\frac {5 d x}{2}\right )-975 B \sin \left (4 c+\frac {5 d x}{2}\right )+450 C \sin \left (4 c+\frac {5 d x}{2}\right )+1325 B \sin \left (2 c+\frac {7 d x}{2}\right )-630 C \sin \left (2 c+\frac {7 d x}{2}\right )+255 B \sin \left (3 c+\frac {7 d x}{2}\right )-60 C \sin \left (3 c+\frac {7 d x}{2}\right )+875 B \sin \left (4 c+\frac {7 d x}{2}\right )-480 C \sin \left (4 c+\frac {7 d x}{2}\right )-195 B \sin \left (5 c+\frac {7 d x}{2}\right )+90 C \sin \left (5 c+\frac {7 d x}{2}\right )+304 B \sin \left (3 c+\frac {9 d x}{2}\right )-144 C \sin \left (3 c+\frac {9 d x}{2}\right )+90 B \sin \left (4 c+\frac {9 d x}{2}\right )-30 C \sin \left (4 c+\frac {9 d x}{2}\right )+214 B \sin \left (5 c+\frac {9 d x}{2}\right )-114 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \]

[In]

Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d*x])^3,x]

[Out]

-1/480*(1920*(13*B - 6*C)*Cos[(c + d*x)/2]^6*(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]^2*((-1235*B + 870*C)*Sin[(d*x)/2] + 5*(76
1*B - 366*C)*Sin[(3*d*x)/2] - 4329*B*Sin[c - (d*x)/2] + 2094*C*Sin[c - (d*x)/2] + 1989*B*Sin[c + (d*x)/2] - 13
14*C*Sin[c + (d*x)/2] - 3575*B*Sin[2*c + (d*x)/2] + 1650*C*Sin[2*c + (d*x)/2] - 475*B*Sin[c + (3*d*x)/2] + 450
*C*Sin[c + (3*d*x)/2] + 2005*B*Sin[2*c + (3*d*x)/2] - 1230*C*Sin[2*c + (3*d*x)/2] - 2275*B*Sin[3*c + (3*d*x)/2
] + 1050*C*Sin[3*c + (3*d*x)/2] + 2673*B*Sin[c + (5*d*x)/2] - 1278*C*Sin[c + (5*d*x)/2] + 105*B*Sin[2*c + (5*d
*x)/2] + 90*C*Sin[2*c + (5*d*x)/2] + 1593*B*Sin[3*c + (5*d*x)/2] - 918*C*Sin[3*c + (5*d*x)/2] - 975*B*Sin[4*c
+ (5*d*x)/2] + 450*C*Sin[4*c + (5*d*x)/2] + 1325*B*Sin[2*c + (7*d*x)/2] - 630*C*Sin[2*c + (7*d*x)/2] + 255*B*S
in[3*c + (7*d*x)/2] - 60*C*Sin[3*c + (7*d*x)/2] + 875*B*Sin[4*c + (7*d*x)/2] - 480*C*Sin[4*c + (7*d*x)/2] - 19
5*B*Sin[5*c + (7*d*x)/2] + 90*C*Sin[5*c + (7*d*x)/2] + 304*B*Sin[3*c + (9*d*x)/2] - 144*C*Sin[3*c + (9*d*x)/2]
+ 90*B*Sin[4*c + (9*d*x)/2] - 30*C*Sin[4*c + (9*d*x)/2] + 214*B*Sin[5*c + (9*d*x)/2] - 114*C*Sin[5*c + (9*d*x
)/2]))/(a^3*d*(1 + Cos[c + d*x])^3)

Maple [A] (verified)

Time = 2.40 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(\frac {-1560 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (B -\frac {6 C}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+1560 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (B -\frac {6 C}{13}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-152 \left (\left (\frac {783 B}{76}-\frac {189 C}{38}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {717 B}{152}-\frac {9 C}{4}\right ) \cos \left (3 d x +3 c \right )+\left (B -\frac {9 C}{19}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {2331 B}{152}-\frac {573 C}{76}\right ) \cos \left (d x +c \right )+\frac {677 B}{76}-\frac {9 C}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d \,a^{3} \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(175\)
derivativedivides	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C -31 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 B -12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 B +12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\)	\(206\)
default	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C -31 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 B -12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 B +12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\)	\(206\)
norman	\(\frac {-\frac {\left (B -C \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (37 B -27 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (51 B -25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (109 B -45 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 a d}-\frac {7 \left (241 B -111 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}-\frac {\left (419 B -219 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (959 B -429 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}+\frac {\left (967 B -447 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{60 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} a^{2}}-\frac {\left (13 B -6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {\left (13 B -6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\)	\(295\)
risch	\(-\frac {i \left (195 B \,{\mathrm e}^{8 i \left (d x +c \right )}-90 C \,{\mathrm e}^{8 i \left (d x +c \right )}+975 B \,{\mathrm e}^{7 i \left (d x +c \right )}-450 C \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 B \,{\mathrm e}^{6 i \left (d x +c \right )}-1050 C \,{\mathrm e}^{6 i \left (d x +c \right )}+3575 B \,{\mathrm e}^{5 i \left (d x +c \right )}-1650 C \,{\mathrm e}^{5 i \left (d x +c \right )}+4329 B \,{\mathrm e}^{4 i \left (d x +c \right )}-2094 C \,{\mathrm e}^{4 i \left (d x +c \right )}+3805 B \,{\mathrm e}^{3 i \left (d x +c \right )}-1830 C \,{\mathrm e}^{3 i \left (d x +c \right )}+2673 B \,{\mathrm e}^{2 i \left (d x +c \right )}-1278 C \,{\mathrm e}^{2 i \left (d x +c \right )}+1325 B \,{\mathrm e}^{i \left (d x +c \right )}-630 C \,{\mathrm e}^{i \left (d x +c \right )}+304 B -144 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {13 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}+\frac {13 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}\)	\(324\)

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+cos(d*x+c)*a)^3,x,method=_RETURNVERBOSE)

[Out]

1/240*(-1560*(1+cos(2*d*x+2*c))*(B-6/13*C)*ln(tan(1/2*d*x+1/2*c)-1)+1560*(1+cos(2*d*x+2*c))*(B-6/13*C)*ln(tan(
1/2*d*x+1/2*c)+1)-152*((783/76*B-189/38*C)*cos(2*d*x+2*c)+(717/152*B-9/4*C)*cos(3*d*x+3*c)+(B-9/19*C)*cos(4*d*
x+4*c)+(2331/152*B-573/76*C)*cos(d*x+c)+677/76*B-9/2*C)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4)/d/a^3/(1+cos(
2*d*x+2*c))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.51 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (19 \, B - 9 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (239 \, B - 114 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (479 \, B - 234 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right ) - 15 \, B\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(15*((13*B - 6*C)*cos(d*x + c)^5 + 3*(13*B - 6*C)*cos(d*x + c)^4 + 3*(13*B - 6*C)*cos(d*x + c)^3 + (13*B
- 6*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((13*B - 6*C)*cos(d*x + c)^5 + 3*(13*B - 6*C)*cos(d*x + c)^4
+ 3*(13*B - 6*C)*cos(d*x + c)^3 + (13*B - 6*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(19*B - 9*C)*co
s(d*x + c)^4 + 3*(239*B - 114*C)*cos(d*x + c)^3 + (479*B - 234*C)*cos(d*x + c)^2 + 15*(3*B - 2*C)*cos(d*x + c)
- 15*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + a^3*d*cos(d*x
+ c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+a*cos(d*x+c))**3,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 377 vs. \(2 (184) = 368\).

Time = 0.23 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.92 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
+ 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(
cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*C*(40*sin(d*x + c)/((a^3 -
a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*
x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) +
1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.19 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (13 \, B - 6 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (13 \, B - 6 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(13*B - 6*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(13*B - 6*C)*log(abs(tan(1/2*d*x + 1/2*c) -
1))/a^3 + 60*(7*B*tan(1/2*d*x + 1/2*c)^3 - 2*C*tan(1/2*d*x + 1/2*c)^3 - 5*B*tan(1/2*d*x + 1/2*c) + 2*C*tan(1/2
*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^12*tan(1/2*d*x
+ 1/2*c)^5 + 40*B*a^12*tan(1/2*d*x + 1/2*c)^3 - 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*B*a^12*tan(1/2*d*x + 1/
2*c) - 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.10 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,B-2\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,B-2\,C\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^3}+\frac {3\,\left (5\,B-3\,C\right )}{4\,a^3}+\frac {10\,B-2\,C}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{4\,a^3}+\frac {5\,B-3\,C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (13\,B-6\,C\right )}{a^3\,d} \]

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a*cos(c + d*x))^3),x)

[Out]

(tan(c/2 + (d*x)/2)^3*(7*B - 2*C) - tan(c/2 + (d*x)/2)*(5*B - 2*C))/(d*(a^3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c
/2 + (d*x)/2)^2 + a^3)) - (tan(c/2 + (d*x)/2)*((3*(B - C))/(2*a^3) + (3*(5*B - 3*C))/(4*a^3) + (10*B - 2*C)/(4
*a^3)))/d - (tan(c/2 + (d*x)/2)^3*((B - C)/(4*a^3) + (5*B - 3*C)/(12*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(B - C))
/(20*a^3*d) + (atanh(tan(c/2 + (d*x)/2))*(13*B - 6*C))/(a^3*d)