\(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx\) [22]

Optimal result

Integrand size = 29, antiderivative size = 111 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {1}{2} a^3 (7 A+5 B) x+\frac {a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d} \]

[Out]

1/2*a^3*(7*A+5*B)*x+a^3*A*arctanh(sin(d*x+c))/d+5/2*a^3*(A+B)*sin(d*x+c)/d+1/3*a*B*(a+a*cos(d*x+c))^2*sin(d*x+
c)/d+1/6*(3*A+5*B)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3055, 3047, 3102, 2814, 3855} \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {(3 A+5 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac {1}{2} a^3 x (7 A+5 B)+\frac {a B \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[In]

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

[Out]

(a^3*(7*A + 5*B)*x)/2 + (a^3*A*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(A + B)*Sin[c + d*x])/(2*d) + (a*B*(a + a*Cos
[c + d*x])^2*Sin[c + d*x])/(3*d) + ((3*A + 5*B)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(6*d)

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3047

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

Rule 3055

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*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] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{3} \int (a+a \cos (c+d x))^2 (3 a A+a (3 A+5 B) \cos (c+d x)) \sec (c+d x) \, dx \\ & = \frac {a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{6} \int (a+a \cos (c+d x)) \left (6 a^2 A+15 a^2 (A+B) \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{6} \int \left (6 a^3 A+\left (6 a^3 A+15 a^3 (A+B)\right ) \cos (c+d x)+15 a^3 (A+B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {1}{6} \int \left (6 a^3 A+3 a^3 (7 A+5 B) \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {1}{2} a^3 (7 A+5 B) x+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^3 A\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^3 (7 A+5 B) x+\frac {a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (A+B) \sin (c+d x)}{2 d}+\frac {a B (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.18 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {a^3 \left (42 A d x+30 B d x-12 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 (4 A+5 B) \sin (c+d x)+3 (A+3 B) \sin (2 (c+d x))+B \sin (3 (c+d x))\right )}{12 d} \]

[In]

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

[Out]

(a^3*(42*A*d*x + 30*B*d*x - 12*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*A*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] + 9*(4*A + 5*B)*Sin[c + d*x] + 3*(A + 3*B)*Sin[2*(c + d*x)] + B*Sin[3*(c + d*x)]))/(12*d)

Maple [A] (verified)

Time = 2.55 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.84

method	result	size
parallelrisch	\(-\frac {a^{3} \left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {\left (-A -3 B \right ) \sin \left (2 d x +2 c \right )}{4}-\frac {B \sin \left (3 d x +3 c \right )}{12}+3 \left (-A -\frac {5 B}{4}\right ) \sin \left (d x +c \right )-\frac {7 \left (A +\frac {5 B}{7}\right ) x d}{2}\right )}{d}\)	\(93\)
parts	\(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (A \,a^{3}+3 B \,a^{3}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}\right ) \sin \left (d x +c \right )}{d}+\frac {B \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\)	\(131\)
derivativedivides	\(\frac {A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 A \,a^{3} \sin \left (d x +c \right )+3 B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{3} \left (d x +c \right )+3 B \,a^{3} \sin \left (d x +c \right )+A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \left (d x +c \right )}{d}\)	\(147\)
default	\(\frac {A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 A \,a^{3} \sin \left (d x +c \right )+3 B \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \,a^{3} \left (d x +c \right )+3 B \,a^{3} \sin \left (d x +c \right )+A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \left (d x +c \right )}{d}\)	\(147\)
risch	\(\frac {7 a^{3} A x}{2}+\frac {5 a^{3} B x}{2}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{3}}{2 d}-\frac {15 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{3}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{3}}{2 d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{3}}{8 d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{3}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}\)	\(189\)
norman	\(\frac {\left (\frac {7}{2} A \,a^{3}+\frac {5}{2} B \,a^{3}\right ) x +\left (\frac {7}{2} A \,a^{3}+\frac {5}{2} B \,a^{3}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (14 A \,a^{3}+10 B \,a^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (14 A \,a^{3}+10 B \,a^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (21 A \,a^{3}+15 B \,a^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{3} \left (7 A +11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {5 a^{3} \left (A +B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{3} \left (51 A +55 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{3} \left (57 A +73 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(276\)

[In]

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

[Out]

-a^3*(A*ln(tan(1/2*d*x+1/2*c)-1)-A*ln(tan(1/2*d*x+1/2*c)+1)+1/4*(-A-3*B)*sin(2*d*x+2*c)-1/12*B*sin(3*d*x+3*c)+
3*(-A-5/4*B)*sin(d*x+c)-7/2*(A+5/7*B)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.92 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {3 \, {\left (7 \, A + 5 \, B\right )} a^{3} d x + 3 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 2 \, {\left (9 \, A + 11 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

1/6*(3*(7*A + 5*B)*a^3*d*x + 3*A*a^3*log(sin(d*x + c) + 1) - 3*A*a^3*log(-sin(d*x + c) + 1) + (2*B*a^3*cos(d*x
+ c)^2 + 3*(A + 3*B)*a^3*cos(d*x + c) + 2*(9*A + 11*B)*a^3)*sin(d*x + c))/d

Sympy [F]

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

[In]

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

[Out]

a**3*(Integral(A*sec(c + d*x), x) + Integral(3*A*cos(c + d*x)*sec(c + d*x), x) + Integral(3*A*cos(c + d*x)**2*
sec(c + d*x), x) + Integral(A*cos(c + d*x)**3*sec(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + In
tegral(3*B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(3*B*cos(c + d*x)**3*sec(c + d*x), x) + Integral(B*cos(c
+ d*x)**4*sec(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 36 \, {\left (d x + c\right )} A a^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 12 \, {\left (d x + c\right )} B a^{3} + 12 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, A a^{3} \sin \left (d x + c\right ) + 36 \, B a^{3} \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 + 36*(d*x + c)*A*a^3 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^
3 + 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 12*(d*x + c)*B*a^3 + 12*A*a^3*log(sec(d*x + c) + tan(d*x + c))
+ 36*A*a^3*sin(d*x + c) + 36*B*a^3*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.62 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {6 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (7 \, A a^{3} + 5 \, B a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(6*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(7*A*a^3 + 5*
B*a^3)*(d*x + c) + 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 36*A*a^3*tan(1/2*d*x
+ 1/2*c)^3 + 40*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 21*A*a^3*tan(1/2*d*x + 1/2*c) + 33*B*a^3*tan(1/2*d*x + 1/2*c))
/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.60 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {15\,B\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {7\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

[In]

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

[Out]

(3*A*a^3*sin(c + d*x))/d + (15*B*a^3*sin(c + d*x))/(4*d) + (7*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (2*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (5*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (
d*x)/2)))/d + (A*a^3*sin(2*c + 2*d*x))/(4*d) + (3*B*a^3*sin(2*c + 2*d*x))/(4*d) + (B*a^3*sin(3*c + 3*d*x))/(12
*d)