\(\int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx\) [55]

Optimal result

Integrand size = 31, antiderivative size = 179 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {(10 A-7 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}+\frac {4 (3 A-2 B) \tan (c+d x)}{a^2 d}-\frac {(10 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {4 (3 A-2 B) \tan ^3(c+d x)}{3 a^2 d} \]

[Out]

-1/2*(10*A-7*B)*arctanh(sin(d*x+c))/a^2/d+4*(3*A-2*B)*tan(d*x+c)/a^2/d-1/2*(10*A-7*B)*sec(d*x+c)*tan(d*x+c)/a^
2/d-1/3*(10*A-7*B)*sec(d*x+c)^2*tan(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*cos(d
*x+c))^2+4/3*(3*A-2*B)*tan(d*x+c)^3/a^2/d

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 179, 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, 3853, 3855} \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {(10 A-7 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}+\frac {4 (3 A-2 B) \tan ^3(c+d x)}{3 a^2 d}+\frac {4 (3 A-2 B) \tan (c+d x)}{a^2 d}-\frac {(10 A-7 B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}-\frac {(10 A-7 B) \tan (c+d x) \sec ^2(c+d x)}{3 a^2 d (\cos (c+d x)+1)}-\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[In]

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

[Out]

-1/2*((10*A - 7*B)*ArcTanh[Sin[c + d*x]])/(a^2*d) + (4*(3*A - 2*B)*Tan[c + d*x])/(a^2*d) - ((10*A - 7*B)*Sec[c
+ d*x]*Tan[c + d*x])/(2*a^2*d) - ((10*A - 7*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - ((
A - B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (4*(3*A - 2*B)*Tan[c + d*x]^3)/(3*a^2*d)

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 ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \frac {(3 a (2 A-B)-4 a (A-B) \cos (c+d x)) \sec ^4(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2} \\ & = -\frac {(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {\int \left (12 a^2 (3 A-2 B)-3 a^2 (10 A-7 B) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx}{3 a^4} \\ & = -\frac {(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(10 A-7 B) \int \sec ^3(c+d x) \, dx}{a^2}+\frac {(4 (3 A-2 B)) \int \sec ^4(c+d x) \, dx}{a^2} \\ & = -\frac {(10 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {(10 A-7 B) \int \sec (c+d x) \, dx}{2 a^2}-\frac {(4 (3 A-2 B)) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{a^2 d} \\ & = -\frac {(10 A-7 B) \text {arctanh}(\sin (c+d x))}{2 a^2 d}+\frac {4 (3 A-2 B) \tan (c+d x)}{a^2 d}-\frac {(10 A-7 B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(10 A-7 B) \sec ^2(c+d x) \tan (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {4 (3 A-2 B) \tan ^3(c+d x)}{3 a^2 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(609\) vs. \(2(179)=358\).

Time = 4.02 (sec) , antiderivative size = 609, normalized size of antiderivative = 3.40 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {192 (10 A-7 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 ^3(c+d x) \left ((-6 A+45 B) \sin \left (\frac {d x}{2}\right )+(310 A-201 B) \sin \left (\frac {3 d x}{2}\right )-306 A \sin \left (c-\frac {d x}{2}\right )+195 B \sin \left (c-\frac {d x}{2}\right )+42 A \sin \left (c+\frac {d x}{2}\right )-51 B \sin \left (c+\frac {d x}{2}\right )-270 A \sin \left (2 c+\frac {d x}{2}\right )+189 B \sin \left (2 c+\frac {d x}{2}\right )+50 A \sin \left (c+\frac {3 d x}{2}\right )-B \sin \left (c+\frac {3 d x}{2}\right )+90 A \sin \left (2 c+\frac {3 d x}{2}\right )-81 B \sin \left (2 c+\frac {3 d x}{2}\right )-170 A \sin \left (3 c+\frac {3 d x}{2}\right )+119 B \sin \left (3 c+\frac {3 d x}{2}\right )+198 A \sin \left (c+\frac {5 d x}{2}\right )-129 B \sin \left (c+\frac {5 d x}{2}\right )+42 A \sin \left (2 c+\frac {5 d x}{2}\right )-9 B \sin \left (2 c+\frac {5 d x}{2}\right )+66 A \sin \left (3 c+\frac {5 d x}{2}\right )-57 B \sin \left (3 c+\frac {5 d x}{2}\right )-90 A \sin \left (4 c+\frac {5 d x}{2}\right )+63 B \sin \left (4 c+\frac {5 d x}{2}\right )+114 A \sin \left (2 c+\frac {7 d x}{2}\right )-75 B \sin \left (2 c+\frac {7 d x}{2}\right )+36 A \sin \left (3 c+\frac {7 d x}{2}\right )-15 B \sin \left (3 c+\frac {7 d x}{2}\right )+48 A \sin \left (4 c+\frac {7 d x}{2}\right )-39 B \sin \left (4 c+\frac {7 d x}{2}\right )-30 A \sin \left (5 c+\frac {7 d x}{2}\right )+21 B \sin \left (5 c+\frac {7 d x}{2}\right )+48 A \sin \left (3 c+\frac {9 d x}{2}\right )-32 B \sin \left (3 c+\frac {9 d x}{2}\right )+22 A \sin \left (4 c+\frac {9 d x}{2}\right )-12 B \sin \left (4 c+\frac {9 d x}{2}\right )+26 A \sin \left (5 c+\frac {9 d x}{2}\right )-20 B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{96 a^2 d (1+\cos (c+d x))^2} \]

[In]

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

[Out]

(192*(10*A - 7*B)*Cos[(c + d*x)/2]^4*(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]^3*((-6*A + 45*B)*Sin[(d*x)/2] + (310*A - 201*B)*S
in[(3*d*x)/2] - 306*A*Sin[c - (d*x)/2] + 195*B*Sin[c - (d*x)/2] + 42*A*Sin[c + (d*x)/2] - 51*B*Sin[c + (d*x)/2
] - 270*A*Sin[2*c + (d*x)/2] + 189*B*Sin[2*c + (d*x)/2] + 50*A*Sin[c + (3*d*x)/2] - B*Sin[c + (3*d*x)/2] + 90*
A*Sin[2*c + (3*d*x)/2] - 81*B*Sin[2*c + (3*d*x)/2] - 170*A*Sin[3*c + (3*d*x)/2] + 119*B*Sin[3*c + (3*d*x)/2] +
198*A*Sin[c + (5*d*x)/2] - 129*B*Sin[c + (5*d*x)/2] + 42*A*Sin[2*c + (5*d*x)/2] - 9*B*Sin[2*c + (5*d*x)/2] +
66*A*Sin[3*c + (5*d*x)/2] - 57*B*Sin[3*c + (5*d*x)/2] - 90*A*Sin[4*c + (5*d*x)/2] + 63*B*Sin[4*c + (5*d*x)/2]
+ 114*A*Sin[2*c + (7*d*x)/2] - 75*B*Sin[2*c + (7*d*x)/2] + 36*A*Sin[3*c + (7*d*x)/2] - 15*B*Sin[3*c + (7*d*x)/
2] + 48*A*Sin[4*c + (7*d*x)/2] - 39*B*Sin[4*c + (7*d*x)/2] - 30*A*Sin[5*c + (7*d*x)/2] + 21*B*Sin[5*c + (7*d*x
)/2] + 48*A*Sin[3*c + (9*d*x)/2] - 32*B*Sin[3*c + (9*d*x)/2] + 22*A*Sin[4*c + (9*d*x)/2] - 12*B*Sin[4*c + (9*d
*x)/2] + 26*A*Sin[5*c + (9*d*x)/2] - 20*B*Sin[5*c + (9*d*x)/2]))/(96*a^2*d*(1 + Cos[c + d*x])^2)

Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.09

method	result	size
parallelrisch	\(\frac {180 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A -\frac {7 B}{10}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-180 \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A -\frac {7 B}{10}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+24 \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\left (\frac {11 A}{4}-\frac {43 B}{24}\right ) \cos \left (3 d x +3 c \right )+\left (5 A -\frac {19 B}{6}\right ) \cos \left (2 d x +2 c \right )+\left (A -\frac {2 B}{3}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {95 A}{12}-\frac {39 B}{8}\right ) \cos \left (d x +c \right )+\frac {13 A}{3}-\frac {5 B}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{12 d \,a^{2} \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(196\)
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+9 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-10 A +7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {-6 A +2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {10 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {2 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {6 A -2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (10 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {10 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {2 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}}{2 d \,a^{2}}\)	\(222\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3}+9 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-7 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-10 A +7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {-6 A +2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {10 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {2 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {6 A -2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (10 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {10 A -5 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {2 A}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}}{2 d \,a^{2}}\)	\(222\)
norman	\(\frac {\frac {\left (A -B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (11 A -10 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\left (19 A -12 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (21 A -13 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {\left (25 A -19 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\left (97 A -71 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} a}+\frac {\left (10 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}-\frac {\left (10 A -7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\)	\(243\)
risch	\(\frac {i \left (30 A \,{\mathrm e}^{8 i \left (d x +c \right )}-21 B \,{\mathrm e}^{8 i \left (d x +c \right )}+90 A \,{\mathrm e}^{7 i \left (d x +c \right )}-63 B \,{\mathrm e}^{7 i \left (d x +c \right )}+170 A \,{\mathrm e}^{6 i \left (d x +c \right )}-119 B \,{\mathrm e}^{6 i \left (d x +c \right )}+270 A \,{\mathrm e}^{5 i \left (d x +c \right )}-189 B \,{\mathrm e}^{5 i \left (d x +c \right )}+306 A \,{\mathrm e}^{4 i \left (d x +c \right )}-195 B \,{\mathrm e}^{4 i \left (d x +c \right )}+310 A \,{\mathrm e}^{3 i \left (d x +c \right )}-201 B \,{\mathrm e}^{3 i \left (d x +c \right )}+198 A \,{\mathrm e}^{2 i \left (d x +c \right )}-129 B \,{\mathrm e}^{2 i \left (d x +c \right )}+114 A \,{\mathrm e}^{i \left (d x +c \right )}-75 B \,{\mathrm e}^{i \left (d x +c \right )}+48 A -32 B \right )}{3 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}-\frac {5 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a^{2} d}+\frac {5 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a^{2} d}\)	\(324\)

[In]

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

[Out]

1/12*(180*(1/3*cos(3*d*x+3*c)+cos(d*x+c))*(A-7/10*B)*ln(tan(1/2*d*x+1/2*c)-1)-180*(1/3*cos(3*d*x+3*c)+cos(d*x+
c))*(A-7/10*B)*ln(tan(1/2*d*x+1/2*c)+1)+24*sec(1/2*d*x+1/2*c)^2*((11/4*A-43/24*B)*cos(3*d*x+3*c)+(5*A-19/6*B)*
cos(2*d*x+2*c)+(A-2/3*B)*cos(4*d*x+4*c)+(95/12*A-39/8*B)*cos(d*x+c)+13/3*A-5/2*B)*tan(1/2*d*x+1/2*c))/d/a^2/(c
os(3*d*x+3*c)+3*cos(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.38 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {3 \, {\left ({\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (10 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{4} + {\left (66 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )^{2} - {\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]

[In]

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

[Out]

-1/12*(3*((10*A - 7*B)*cos(d*x + c)^5 + 2*(10*A - 7*B)*cos(d*x + c)^4 + (10*A - 7*B)*cos(d*x + c)^3)*log(sin(d
*x + c) + 1) - 3*((10*A - 7*B)*cos(d*x + c)^5 + 2*(10*A - 7*B)*cos(d*x + c)^4 + (10*A - 7*B)*cos(d*x + c)^3)*l
og(-sin(d*x + c) + 1) - 2*(16*(3*A - 2*B)*cos(d*x + c)^4 + (66*A - 43*B)*cos(d*x + c)^3 + 6*(2*A - B)*cos(d*x
+ c)^2 - (2*A - 3*B)*cos(d*x + c) + 2*A)*sin(d*x + c))/(a^2*d*cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*
cos(d*x + c)^3)

Sympy [F]

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

[In]

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

[Out]

(Integral(A*sec(c + d*x)**4/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sec(c + d*x)*
*4/(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. 425 vs. \(2 (169) = 338\).

Time = 0.23 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.37 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {A {\left (\frac {4 \, {\left (\frac {9 \, \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 {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \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 {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {30 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - B {\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 )}}{6 \, d} \]

[In]

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

[Out]

1/6*(A*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
- a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 30*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)
/a^2) - B*(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) +
sin(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))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.26 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (10 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (10 \, A - 7 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (30 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

[In]

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

[Out]

-1/6*(3*(10*A - 7*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(10*A - 7*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1)
)/a^2 + 2*(30*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 - 40*A*tan(1/2*d*x + 1/2*c)^3 + 24*B*tan(
1/2*d*x + 1/2*c)^3 + 18*A*tan(1/2*d*x + 1/2*c) - 9*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*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 + 27*A*a^4*tan(1/2*d*x + 1/2*c) - 21*B*a^4*ta
n(1/2*d*x + 1/2*c))/a^6)/d

Mupad [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.13 \[ \int \frac {(A+B \cos (c+d x)) \sec ^4(c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (A-B\right )}{a^2}+\frac {5\,A-3\,B}{2\,a^2}\right )}{d}-\frac {\left (10\,A-5\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (8\,B-\frac {40\,A}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A-3\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (10\,A-7\,B\right )}{a^2\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((2*(A - B))/a^2 + (5*A - 3*B)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^5*(10*A - 5*B) - tan(c/2
+ (d*x)/2)^3*((40*A)/3 - 8*B) + tan(c/2 + (d*x)/2)*(6*A - 3*B))/(d*(3*a^2*tan(c/2 + (d*x)/2)^2 - 3*a^2*tan(c/2
+ (d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 - a^2)) + (tan(c/2 + (d*x)/2)^3*(A - B))/(6*a^2*d) - (atanh(tan(c/2 +
(d*x)/2))*(10*A - 7*B))/(a^2*d)