\(\int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx\) [54]

Optimal result

Integrand size = 31, antiderivative size = 152 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {(7 A-4 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}-\frac {2 (8 A-5 B) \tan (c+d x)}{3 a^2 d}+\frac {(7 A-4 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2} \]

[Out]

1/2*(7*A-4*B)*arctanh(sin(d*x+c))/a^2/d-2/3*(8*A-5*B)*tan(d*x+c)/a^2/d+1/2*(7*A-4*B)*sec(d*x+c)*tan(d*x+c)/a^2
/d-1/3*(8*A-5*B)*sec(d*x+c)*tan(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))
^2

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3057, 2827, 3853, 3855, 3852, 8} \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {(7 A-4 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}-\frac {2 (8 A-5 B) \tan (c+d x)}{3 a^2 d}+\frac {(7 A-4 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \tan (c+d x) \sec (c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

((7*A - 4*B)*ArcTanh[Sin[c + d*x]])/(2*a^2*d) - (2*(8*A - 5*B)*Tan[c + d*x])/(3*a^2*d) + ((7*A - 4*B)*Sec[c +
d*x]*Tan[c + d*x])/(2*a^2*d) - ((8*A - 5*B)*Sec[c + d*x]*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((A - B)
*Sec[c + d*x]*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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 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}& = -\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {(a (5 A-2 B)-3 a (A-B) \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2} \\ & = -\frac {(8 A-5 B) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \left (3 a^2 (7 A-4 B)-2 a^2 (8 A-5 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{3 a^4} \\ & = -\frac {(8 A-5 B) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(2 (8 A-5 B)) \int \sec ^2(c+d x) \, dx}{3 a^2}+\frac {(7 A-4 B) \int \sec ^3(c+d x) \, dx}{a^2} \\ & = \frac {(7 A-4 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {(7 A-4 B) \int \sec (c+d x) \, dx}{2 a^2}+\frac {(2 (8 A-5 B)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 a^2 d} \\ & = \frac {(7 A-4 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}-\frac {2 (8 A-5 B) \tan (c+d x)}{3 a^2 d}+\frac {(7 A-4 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(8 A-5 B) \sec (c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(496\) vs. \(2(152)=304\).

Time = 2.94 (sec) , antiderivative size = 496, normalized size of antiderivative = 3.26 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {96 (7 A-4 B) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left (-14 (A-B) \sin \left (\frac {d x}{2}\right )+(97 A-64 B) \sin \left (\frac {3 d x}{2}\right )-126 A \sin \left (c-\frac {d x}{2}\right )+84 B \sin \left (c-\frac {d x}{2}\right )+42 A \sin \left (c+\frac {d x}{2}\right )-42 B \sin \left (c+\frac {d x}{2}\right )-98 A \sin \left (2 c+\frac {d x}{2}\right )+56 B \sin \left (2 c+\frac {d x}{2}\right )-3 A \sin \left (c+\frac {3 d x}{2}\right )+6 B \sin \left (c+\frac {3 d x}{2}\right )+37 A \sin \left (2 c+\frac {3 d x}{2}\right )-34 B \sin \left (2 c+\frac {3 d x}{2}\right )-63 A \sin \left (3 c+\frac {3 d x}{2}\right )+36 B \sin \left (3 c+\frac {3 d x}{2}\right )+75 A \sin \left (c+\frac {5 d x}{2}\right )-48 B \sin \left (c+\frac {5 d x}{2}\right )+15 A \sin \left (2 c+\frac {5 d x}{2}\right )-6 B \sin \left (2 c+\frac {5 d x}{2}\right )+39 A \sin \left (3 c+\frac {5 d x}{2}\right )-30 B \sin \left (3 c+\frac {5 d x}{2}\right )-21 A \sin \left (4 c+\frac {5 d x}{2}\right )+12 B \sin \left (4 c+\frac {5 d x}{2}\right )+32 A \sin \left (2 c+\frac {7 d x}{2}\right )-20 B \sin \left (2 c+\frac {7 d x}{2}\right )+12 A \sin \left (3 c+\frac {7 d x}{2}\right )-6 B \sin \left (3 c+\frac {7 d x}{2}\right )+20 A \sin \left (4 c+\frac {7 d x}{2}\right )-14 B \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{48 a^2 d (1+\cos (c+d x))^2} \]

[In]

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

[Out]

-1/48*(96*(7*A - 4*B)*Cos[(c + d*x)/2]^4*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Si
n[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*(-14*(A - B)*Sin[(d*x)/2] + (97*A - 64*B)*S
in[(3*d*x)/2] - 126*A*Sin[c - (d*x)/2] + 84*B*Sin[c - (d*x)/2] + 42*A*Sin[c + (d*x)/2] - 42*B*Sin[c + (d*x)/2]
- 98*A*Sin[2*c + (d*x)/2] + 56*B*Sin[2*c + (d*x)/2] - 3*A*Sin[c + (3*d*x)/2] + 6*B*Sin[c + (3*d*x)/2] + 37*A*
Sin[2*c + (3*d*x)/2] - 34*B*Sin[2*c + (3*d*x)/2] - 63*A*Sin[3*c + (3*d*x)/2] + 36*B*Sin[3*c + (3*d*x)/2] + 75*
A*Sin[c + (5*d*x)/2] - 48*B*Sin[c + (5*d*x)/2] + 15*A*Sin[2*c + (5*d*x)/2] - 6*B*Sin[2*c + (5*d*x)/2] + 39*A*S
in[3*c + (5*d*x)/2] - 30*B*Sin[3*c + (5*d*x)/2] - 21*A*Sin[4*c + (5*d*x)/2] + 12*B*Sin[4*c + (5*d*x)/2] + 32*A
*Sin[2*c + (7*d*x)/2] - 20*B*Sin[2*c + (7*d*x)/2] + 12*A*Sin[3*c + (7*d*x)/2] - 6*B*Sin[3*c + (7*d*x)/2] + 20*
A*Sin[4*c + (7*d*x)/2] - 14*B*Sin[4*c + (7*d*x)/2]))/(a^2*d*(1 + Cos[c + d*x])^2)

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(\frac {-42 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {4 B}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+42 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {4 B}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\frac {43 A}{60}-\frac {7 B}{15}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {4 A}{15}-\frac {B}{6}\right ) \cos \left (3 d x +3 c \right )+\left (A -\frac {7 B}{10}\right ) \cos \left (d x +c \right )+\frac {37 A}{60}-\frac {7 B}{15}\right ) \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d \,a^{2} \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(158\)
derivativedivides	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-5 A +2 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (7 A -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-7 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-5 A +2 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{2 d \,a^{2}}\)	\(177\)
default	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}+\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-5 A +2 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (7 A -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-7 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-5 A +2 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{2 d \,a^{2}}\)	\(177\)
norman	\(\frac {-\frac {\left (A -B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (10 A -7 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (13 A -9 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {2 \left (13 A -7 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {\left (16 A -7 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a}-\frac {\left (7 A -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}+\frac {\left (7 A -4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\)	\(217\)
risch	\(-\frac {i \left (21 A \,{\mathrm e}^{6 i \left (d x +c \right )}-12 B \,{\mathrm e}^{6 i \left (d x +c \right )}+63 A \,{\mathrm e}^{5 i \left (d x +c \right )}-36 B \,{\mathrm e}^{5 i \left (d x +c \right )}+98 A \,{\mathrm e}^{4 i \left (d x +c \right )}-56 B \,{\mathrm e}^{4 i \left (d x +c \right )}+126 A \,{\mathrm e}^{3 i \left (d x +c \right )}-84 B \,{\mathrm e}^{3 i \left (d x +c \right )}+97 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}+75 A \,{\mathrm e}^{i \left (d x +c \right )}-48 B \,{\mathrm e}^{i \left (d x +c \right )}+32 A -20 B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {7 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{2} d}-\frac {7 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{2} d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{2} d}\)	\(276\)

[In]

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

[Out]

1/12*(-42*(1+cos(2*d*x+2*c))*(A-4/7*B)*ln(tan(1/2*d*x+1/2*c)-1)+42*(1+cos(2*d*x+2*c))*(A-4/7*B)*ln(tan(1/2*d*x
+1/2*c)+1)-60*tan(1/2*d*x+1/2*c)*((43/60*A-7/15*B)*cos(2*d*x+2*c)+(4/15*A-1/6*B)*cos(3*d*x+3*c)+(A-7/10*B)*cos
(d*x+c)+37/60*A-7/15*B)*sec(1/2*d*x+1/2*c)^2)/d/a^2/(1+cos(2*d*x+2*c))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.50 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left ({\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A - 4 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (8 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (43 \, A - 28 \, B\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (A - B\right )} \cos \left (d x + c\right ) - 3 \, A\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

1/12*(3*((7*A - 4*B)*cos(d*x + c)^4 + 2*(7*A - 4*B)*cos(d*x + c)^3 + (7*A - 4*B)*cos(d*x + c)^2)*log(sin(d*x +
c) + 1) - 3*((7*A - 4*B)*cos(d*x + c)^4 + 2*(7*A - 4*B)*cos(d*x + c)^3 + (7*A - 4*B)*cos(d*x + c)^2)*log(-sin
(d*x + c) + 1) - 2*(4*(8*A - 5*B)*cos(d*x + c)^3 + (43*A - 28*B)*cos(d*x + c)^2 + 6*(A - B)*cos(d*x + c) - 3*A
)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (142) = 284\).

Time = 0.26 (sec) , antiderivative size = 336, normalized size of antiderivative = 2.21 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {A {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} - \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {21 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

[In]

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

[Out]

-1/6*(A*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + s
in(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c
)/(cos(d*x + c) + 1) - 1)/a^2) - B*((15*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)
/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 + 12*
sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.30 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {3 \, {\left (7 \, A - 4 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (7 \, A - 4 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(7*A - 4*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(7*A - 4*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a
^2 + 6*(5*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 - 3*A*tan(1/2*d*x + 1/2*c) + 2*B*tan(1/2*d*x +
1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^2) - (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 +
21*A*a^4*tan(1/2*d*x + 1/2*c) - 15*B*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (5\,A-2\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A-2\,B\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^2}+\frac {4\,A-2\,B}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (7\,A-4\,B\right )}{a^2\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*(5*A - 2*B) - tan(c/2 + (d*x)/2)*(3*A - 2*B))/(d*(a^2*tan(c/2 + (d*x)/2)^4 - 2*a^2*tan(c
/2 + (d*x)/2)^2 + a^2)) - (tan(c/2 + (d*x)/2)*((3*(A - B))/(2*a^2) + (4*A - 2*B)/(2*a^2)))/d - (tan(c/2 + (d*x
)/2)^3*(A - B))/(6*a^2*d) + (atanh(tan(c/2 + (d*x)/2))*(7*A - 4*B))/(a^2*d)