\(\int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [63]

Optimal result

Integrand size = 31, antiderivative size = 145 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {(3 A-B) \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {2 (36 A-11 B) \tan (c+d x)}{15 a^3 d}-\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )} \]

[Out]

-(3*A-B)*arctanh(sin(d*x+c))/a^3/d+2/15*(36*A-11*B)*tan(d*x+c)/a^3/d-1/5*(A-B)*tan(d*x+c)/d/(a+a*cos(d*x+c))^3
-1/15*(9*A-4*B)*tan(d*x+c)/a/d/(a+a*cos(d*x+c))^2-(3*A-B)*tan(d*x+c)/d/(a^3+a^3*cos(d*x+c))

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3057, 2827, 3852, 8, 3855} \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {(3 A-B) \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {2 (36 A-11 B) \tan (c+d x)}{15 a^3 d}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3 \cos (c+d x)+a^3\right )}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a \cos (c+d x)+a)^2}-\frac {(A-B) \tan (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[In]

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

[Out]

-(((3*A - B)*ArcTanh[Sin[c + d*x]])/(a^3*d)) + (2*(36*A - 11*B)*Tan[c + d*x])/(15*a^3*d) - ((A - B)*Tan[c + d*
x])/(5*d*(a + a*Cos[c + d*x])^3) - ((9*A - 4*B)*Tan[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - ((3*A - B)*Tan
[c + d*x])/(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 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)}{5 d (a+a \cos (c+d x))^3}+\frac {\int \frac {(a (6 A-B)-3 a (A-B) \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {\left (a^2 (27 A-7 B)-2 a^2 (9 A-4 B) \cos (c+d x)\right ) \sec ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {\int \left (2 a^3 (36 A-11 B)-15 a^3 (3 A-B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{15 a^6} \\ & = -\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {(2 (36 A-11 B)) \int \sec ^2(c+d x) \, dx}{15 a^3}-\frac {(3 A-B) \int \sec (c+d x) \, dx}{a^3} \\ & = -\frac {(3 A-B) \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{d \left (a^3+a^3 \cos (c+d x)\right )}-\frac {(2 (36 A-11 B)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^3 d} \\ & = -\frac {(3 A-B) \text {arctanh}(\sin (c+d x))}{a^3 d}+\frac {2 (36 A-11 B) \tan (c+d x)}{15 a^3 d}-\frac {(A-B) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(9 A-4 B) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(3 A-B) \tan (c+d x)}{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. \(482\) vs. \(2(145)=290\).

Time = 2.78 (sec) , antiderivative size = 482, normalized size of antiderivative = 3.32 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {960 (3 A-B) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec (c+d x) \left (-5 (51 A-32 B) \sin \left (\frac {d x}{2}\right )+(567 A-167 B) \sin \left (\frac {3 d x}{2}\right )-600 A \sin \left (c-\frac {d x}{2}\right )+170 B \sin \left (c-\frac {d x}{2}\right )+375 A \sin \left (c+\frac {d x}{2}\right )-170 B \sin \left (c+\frac {d x}{2}\right )-480 A \sin \left (2 c+\frac {d x}{2}\right )+160 B \sin \left (2 c+\frac {d x}{2}\right )-60 A \sin \left (c+\frac {3 d x}{2}\right )+75 B \sin \left (c+\frac {3 d x}{2}\right )+402 A \sin \left (2 c+\frac {3 d x}{2}\right )-167 B \sin \left (2 c+\frac {3 d x}{2}\right )-225 A \sin \left (3 c+\frac {3 d x}{2}\right )+75 B \sin \left (3 c+\frac {3 d x}{2}\right )+315 A \sin \left (c+\frac {5 d x}{2}\right )-95 B \sin \left (c+\frac {5 d x}{2}\right )+30 A \sin \left (2 c+\frac {5 d x}{2}\right )+15 B \sin \left (2 c+\frac {5 d x}{2}\right )+240 A \sin \left (3 c+\frac {5 d x}{2}\right )-95 B \sin \left (3 c+\frac {5 d x}{2}\right )-45 A \sin \left (4 c+\frac {5 d x}{2}\right )+15 B \sin \left (4 c+\frac {5 d x}{2}\right )+72 A \sin \left (2 c+\frac {7 d x}{2}\right )-22 B \sin \left (2 c+\frac {7 d x}{2}\right )+15 A \sin \left (3 c+\frac {7 d x}{2}\right )+57 A \sin \left (4 c+\frac {7 d x}{2}\right )-22 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{120 a^3 d (1+\cos (c+d x))^3} \]

[In]

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

[Out]

(960*(3*A - B)*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]*(-5*(51*A - 32*B)*Sin[(d*x)/2] + (567*A - 167*B)*Sin
[(3*d*x)/2] - 600*A*Sin[c - (d*x)/2] + 170*B*Sin[c - (d*x)/2] + 375*A*Sin[c + (d*x)/2] - 170*B*Sin[c + (d*x)/2
] - 480*A*Sin[2*c + (d*x)/2] + 160*B*Sin[2*c + (d*x)/2] - 60*A*Sin[c + (3*d*x)/2] + 75*B*Sin[c + (3*d*x)/2] +
402*A*Sin[2*c + (3*d*x)/2] - 167*B*Sin[2*c + (3*d*x)/2] - 225*A*Sin[3*c + (3*d*x)/2] + 75*B*Sin[3*c + (3*d*x)/
2] + 315*A*Sin[c + (5*d*x)/2] - 95*B*Sin[c + (5*d*x)/2] + 30*A*Sin[2*c + (5*d*x)/2] + 15*B*Sin[2*c + (5*d*x)/2
] + 240*A*Sin[3*c + (5*d*x)/2] - 95*B*Sin[3*c + (5*d*x)/2] - 45*A*Sin[4*c + (5*d*x)/2] + 15*B*Sin[4*c + (5*d*x
)/2] + 72*A*Sin[2*c + (7*d*x)/2] - 22*B*Sin[2*c + (7*d*x)/2] + 15*A*Sin[3*c + (7*d*x)/2] + 57*A*Sin[4*c + (7*d
*x)/2] - 22*B*Sin[4*c + (7*d*x)/2]))/(120*a^3*d*(1 + Cos[c + d*x])^3)

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.99

method	result	size
parallelrisch	\(\frac {3 \cos \left (d x +c \right ) \left (A -\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3 \cos \left (d x +c \right ) \left (A -\frac {B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {57 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\frac {\left (A -\frac {17 B}{57}\right ) \cos \left (2 d x +2 c \right )}{2}+\frac {\left (2 A -\frac {11 B}{18}\right ) \cos \left (3 d x +3 c \right )}{19}+\left (A -\frac {97 B}{342}\right ) \cos \left (d x +c \right )+\frac {67 A}{114}-\frac {17 B}{114}\right ) \left (\sec ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20}}{a^{3} d \cos \left (d x +c \right )}\)	\(143\)
derivativedivides	\(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A -\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (12 A -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\)	\(162\)
default	\(\frac {\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{5}+2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A -\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+17 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (12 A -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-12 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {4 A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{4 d \,a^{3}}\)	\(162\)
norman	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\left (3 A -2 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (15 A -2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (25 A -7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (42 A -17 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{10 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}}+\frac {\left (3 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}-\frac {\left (3 A -B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}\)	\(216\)
risch	\(\frac {2 i \left (45 A \,{\mathrm e}^{6 i \left (d x +c \right )}-15 B \,{\mathrm e}^{6 i \left (d x +c \right )}+225 A \,{\mathrm e}^{5 i \left (d x +c \right )}-75 B \,{\mathrm e}^{5 i \left (d x +c \right )}+480 A \,{\mathrm e}^{4 i \left (d x +c \right )}-160 B \,{\mathrm e}^{4 i \left (d x +c \right )}+600 A \,{\mathrm e}^{3 i \left (d x +c \right )}-170 B \,{\mathrm e}^{3 i \left (d x +c \right )}+567 A \,{\mathrm e}^{2 i \left (d x +c \right )}-167 B \,{\mathrm e}^{2 i \left (d x +c \right )}+315 A \,{\mathrm e}^{i \left (d x +c \right )}-95 B \,{\mathrm e}^{i \left (d x +c \right )}+72 A -22 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{3} d}-\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{3} d}\)	\(275\)

[In]

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

[Out]

3/20*(20*cos(d*x+c)*(A-1/3*B)*ln(tan(1/2*d*x+1/2*c)-1)-20*cos(d*x+c)*(A-1/3*B)*ln(tan(1/2*d*x+1/2*c)+1)+19*tan
(1/2*d*x+1/2*c)*(1/2*(A-17/57*B)*cos(2*d*x+2*c)+1/19*(2*A-11/18*B)*cos(3*d*x+3*c)+(A-97/342*B)*cos(d*x+c)+67/1
14*A-17/114*B)*sec(1/2*d*x+1/2*c)^4)/d/a^3/cos(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.88 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {15 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (3 \, A - B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - B\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A - B\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (36 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (57 \, A - 17 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (117 \, A - 32 \, B\right )} \cos \left (d x + c\right ) + 15 \, A\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}} \]

[In]

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

[Out]

-1/30*(15*((3*A - B)*cos(d*x + c)^4 + 3*(3*A - B)*cos(d*x + c)^3 + 3*(3*A - B)*cos(d*x + c)^2 + (3*A - B)*cos(
d*x + c))*log(sin(d*x + c) + 1) - 15*((3*A - B)*cos(d*x + c)^4 + 3*(3*A - B)*cos(d*x + c)^3 + 3*(3*A - B)*cos(
d*x + c)^2 + (3*A - B)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(2*(36*A - 11*B)*cos(d*x + c)^3 + 3*(57*A - 17
*B)*cos(d*x + c)^2 + (117*A - 32*B)*cos(d*x + c) + 15*A)*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x
+ c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (139) = 278\).

Time = 0.22 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.97 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, A {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - B {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

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

[Out]

1/60*(3*A*(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) - B*((105*s
in(d*x + c)/(cos(d*x + c) + 1) + 20*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 - 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 = 0.32 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.31 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\frac {60 \, {\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, {\left (3 \, A - B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {120 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 105 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

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

[Out]

-1/60*(60*(3*A - B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*(3*A - B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a
^3 + 120*A*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^
12*tan(1/2*d*x + 1/2*c)^5 + 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 - 20*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 255*A*a^12*t
an(1/2*d*x + 1/2*c) - 105*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.16 \[ \int \frac {(A+B \cos (c+d x)) \sec ^2(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{6\,a^3}+\frac {4\,A-2\,B}{12\,a^3}\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A}{2\,a^3}+\frac {3\,\left (A-B\right )}{4\,a^3}+\frac {4\,A-2\,B}{2\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}-\frac {2\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A-B\right )}{a^3\,d} \]

[In]

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

[Out]

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