\(\int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\) [962]

Optimal result

Integrand size = 29, antiderivative size = 111 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {1}{16} a (6 A+B) x-\frac {a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac {a (6 A+B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a (6 A+B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d} \]

[Out]

1/16*a*(6*A+B)*x-1/30*a*(6*A+B)*cos(d*x+c)^5/d+1/16*a*(6*A+B)*cos(d*x+c)*sin(d*x+c)/d+1/24*a*(6*A+B)*cos(d*x+c
)^3*sin(d*x+c)/d-1/6*B*cos(d*x+c)^5*(a+a*sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2939, 2748, 2715, 8} \[ \int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac {a (6 A+B) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {a (6 A+B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a x (6 A+B)-\frac {B \cos ^5(c+d x) (a \sin (c+d x)+a)}{6 d} \]

[In]

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

[Out]

(a*(6*A + B)*x)/16 - (a*(6*A + B)*Cos[c + d*x]^5)/(30*d) + (a*(6*A + B)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a
*(6*A + B)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (B*Cos[c + d*x]^5*(a + a*Sin[c + d*x]))/(6*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x
] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2939

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x
] + Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; F
reeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[m + p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d}+\frac {1}{6} (6 A+B) \int \cos ^4(c+d x) (a+a \sin (c+d x)) \, dx \\ & = -\frac {a (6 A+B) \cos ^5(c+d x)}{30 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d}+\frac {1}{6} (a (6 A+B)) \int \cos ^4(c+d x) \, dx \\ & = -\frac {a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac {a (6 A+B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d}+\frac {1}{8} (a (6 A+B)) \int \cos ^2(c+d x) \, dx \\ & = -\frac {a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac {a (6 A+B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a (6 A+B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d}+\frac {1}{16} (a (6 A+B)) \int 1 \, dx \\ & = \frac {1}{16} a (6 A+B) x-\frac {a (6 A+B) \cos ^5(c+d x)}{30 d}+\frac {a (6 A+B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a (6 A+B) \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {B \cos ^5(c+d x) (a+a \sin (c+d x))}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.24 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a \cos (c+d x) \left (-36 A-36 B-\frac {60 (6 A+B) \arcsin \left (\frac {\sqrt {1-\sin (c+d x)}}{\sqrt {2}}\right )}{\sqrt {\cos ^2(c+d x)}}-48 (A+B) \cos (2 (c+d x))-12 (A+B) \cos (4 (c+d x))+210 A \sin (c+d x)+25 B \sin (c+d x)+30 A \sin (3 (c+d x))-10 B \sin (3 (c+d x))-5 B \sin (5 (c+d x))\right )}{480 d} \]

[In]

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

[Out]

(a*Cos[c + d*x]*(-36*A - 36*B - (60*(6*A + B)*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]])/Sqrt[Cos[c + d*x]^2] - 4
8*(A + B)*Cos[2*(c + d*x)] - 12*(A + B)*Cos[4*(c + d*x)] + 210*A*Sin[c + d*x] + 25*B*Sin[c + d*x] + 30*A*Sin[3
*(c + d*x)] - 10*B*Sin[3*(c + d*x)] - 5*B*Sin[5*(c + d*x)]))/(480*d)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {B a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {a A \left (\cos ^{5}\left (d x +c \right )\right )}{5}-\frac {B a \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(118\)
default	\(\frac {B a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{5}\left (d x +c \right )\right )}{6}+\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{24}+\frac {d x}{16}+\frac {c}{16}\right )-\frac {a A \left (\cos ^{5}\left (d x +c \right )\right )}{5}-\frac {B a \left (\cos ^{5}\left (d x +c \right )\right )}{5}+a A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(118\)
parallelrisch	\(\frac {\left (\frac {\left (-A -B \right ) \cos \left (3 d x +3 c \right )}{4}+\frac {\left (-A -B \right ) \cos \left (5 d x +5 c \right )}{20}+\left (A +\frac {B}{16}\right ) \sin \left (2 d x +2 c \right )+\frac {\left (A -\frac {B}{2}\right ) \sin \left (4 d x +4 c \right )}{8}-\frac {B \sin \left (6 d x +6 c \right )}{48}+\frac {\left (-A -B \right ) \cos \left (d x +c \right )}{2}+\frac {3 d x A}{2}+\frac {d x B}{4}-\frac {4 A}{5}-\frac {4 B}{5}\right ) a}{4 d}\)	\(118\)
risch	\(\frac {3 a x A}{8}+\frac {a B x}{16}-\frac {a A \cos \left (d x +c \right )}{8 d}-\frac {a \cos \left (d x +c \right ) B}{8 d}-\frac {\sin \left (6 d x +6 c \right ) B a}{192 d}-\frac {a \cos \left (5 d x +5 c \right ) A}{80 d}-\frac {a \cos \left (5 d x +5 c \right ) B}{80 d}+\frac {\sin \left (4 d x +4 c \right ) a A}{32 d}-\frac {\sin \left (4 d x +4 c \right ) B a}{64 d}-\frac {a \cos \left (3 d x +3 c \right ) A}{16 d}-\frac {a \cos \left (3 d x +3 c \right ) B}{16 d}+\frac {\sin \left (2 d x +2 c \right ) a A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B a}{64 d}\)	\(182\)
norman	\(\frac {\left (\frac {3}{8} a A +\frac {1}{16} B a \right ) x +\left (\frac {3}{8} a A +\frac {1}{16} B a \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {9}{4} a A +\frac {3}{8} B a \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {9}{4} a A +\frac {3}{8} B a \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {15}{2} a A +\frac {5}{4} B a \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {45}{8} a A +\frac {15}{16} B a \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {45}{8} a A +\frac {15}{16} B a \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 a A +2 B a}{5 d}-\frac {2 \left (a A +B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}-\frac {4 \left (a A +B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (a A +B a \right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 a A +2 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (4 a A +4 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a \left (2 A -13 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a \left (2 A -13 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (10 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {a \left (10 A -B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {a \left (42 A +47 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {a \left (42 A +47 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\)	\(436\)

[In]

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

[Out]

1/d*(B*a*(-1/6*sin(d*x+c)*cos(d*x+c)^5+1/24*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+1/16*d*x+1/16*c)-1/5*a*A*
cos(d*x+c)^5-1/5*B*a*cos(d*x+c)^5+a*A*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.73 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {48 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{5} - 15 \, {\left (6 \, A + B\right )} a d x + 5 \, {\left (8 \, B a \cos \left (d x + c\right )^{5} - 2 \, {\left (6 \, A + B\right )} a \cos \left (d x + c\right )^{3} - 3 \, {\left (6 \, A + B\right )} a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

[In]

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

[Out]

-1/240*(48*(A + B)*a*cos(d*x + c)^5 - 15*(6*A + B)*a*d*x + 5*(8*B*a*cos(d*x + c)^5 - 2*(6*A + B)*a*cos(d*x + c
)^3 - 3*(6*A + B)*a*cos(d*x + c))*sin(d*x + c))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 306 vs. \(2 (100) = 200\).

Time = 0.35 (sec) , antiderivative size = 306, normalized size of antiderivative = 2.76 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {3 A a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 A a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 A a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 A a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {A a \cos ^{5}{\left (c + d x \right )}}{5 d} + \frac {B a x \sin ^{6}{\left (c + d x \right )}}{16} + \frac {3 B a x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{16} + \frac {3 B a x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{16} + \frac {B a x \cos ^{6}{\left (c + d x \right )}}{16} + \frac {B a \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{16 d} + \frac {B a \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{6 d} - \frac {B a \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{16 d} - \frac {B a \cos ^{5}{\left (c + d x \right )}}{5 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right ) \cos ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**4*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

Piecewise((3*A*a*x*sin(c + d*x)**4/8 + 3*A*a*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*a*x*cos(c + d*x)**4/8 +
3*A*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*A*a*sin(c + d*x)*cos(c + d*x)**3/(8*d) - A*a*cos(c + d*x)**5/(5*
d) + B*a*x*sin(c + d*x)**6/16 + 3*B*a*x*sin(c + d*x)**4*cos(c + d*x)**2/16 + 3*B*a*x*sin(c + d*x)**2*cos(c + d
*x)**4/16 + B*a*x*cos(c + d*x)**6/16 + B*a*sin(c + d*x)**5*cos(c + d*x)/(16*d) + B*a*sin(c + d*x)**3*cos(c + d
*x)**3/(6*d) - B*a*sin(c + d*x)*cos(c + d*x)**5/(16*d) - B*a*cos(c + d*x)**5/(5*d), Ne(d, 0)), (x*(A + B*sin(c
))*(a*sin(c) + a)*cos(c)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=-\frac {192 \, A a \cos \left (d x + c\right )^{5} + 192 \, B a \cos \left (d x + c\right )^{5} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 12 \, d x + 12 \, c - 3 \, \sin \left (4 \, d x + 4 \, c\right )\right )} B a}{960 \, d} \]

[In]

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

[Out]

-1/960*(192*A*a*cos(d*x + c)^5 + 192*B*a*cos(d*x + c)^5 - 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x +
2*c))*A*a - 5*(4*sin(2*d*x + 2*c)^3 + 12*d*x + 12*c - 3*sin(4*d*x + 4*c))*B*a)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.20 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {1}{16} \, {\left (6 \, A a + B a\right )} x - \frac {B a \sin \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {{\left (A a + B a\right )} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {{\left (A a + B a\right )} \cos \left (3 \, d x + 3 \, c\right )}{16 \, d} - \frac {{\left (A a + B a\right )} \cos \left (d x + c\right )}{8 \, d} + \frac {{\left (2 \, A a - B a\right )} \sin \left (4 \, d x + 4 \, c\right )}{64 \, d} + \frac {{\left (16 \, A a + B a\right )} \sin \left (2 \, d x + 2 \, c\right )}{64 \, d} \]

[In]

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

[Out]

1/16*(6*A*a + B*a)*x - 1/192*B*a*sin(6*d*x + 6*c)/d - 1/80*(A*a + B*a)*cos(5*d*x + 5*c)/d - 1/16*(A*a + B*a)*c
os(3*d*x + 3*c)/d - 1/8*(A*a + B*a)*cos(d*x + c)/d + 1/64*(2*A*a - B*a)*sin(4*d*x + 4*c)/d + 1/64*(16*A*a + B*
a)*sin(2*d*x + 2*c)/d

Mupad [B] (verification not implemented)

Time = 12.67 (sec) , antiderivative size = 391, normalized size of antiderivative = 3.52 \[ \int \cos ^4(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx=\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,A+B\right )}{8\,\left (\frac {3\,A\,a}{4}+\frac {B\,a}{8}\right )}\right )\,\left (6\,A+B\right )}{8\,d}-\frac {a\,\left (6\,A+B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{8\,d}-\frac {\left (\frac {5\,A\,a}{4}-\frac {B\,a}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (2\,A\,a+2\,B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {7\,A\,a}{4}+\frac {47\,B\,a}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,a+2\,B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (\frac {A\,a}{2}-\frac {13\,B\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (4\,A\,a+4\,B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {13\,B\,a}{4}-\frac {A\,a}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (4\,A\,a+4\,B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (-\frac {7\,A\,a}{4}-\frac {47\,B\,a}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {2\,A\,a}{5}+\frac {2\,B\,a}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (\frac {B\,a}{8}-\frac {5\,A\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {2\,A\,a}{5}+\frac {2\,B\,a}{5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

(a*atan((a*tan(c/2 + (d*x)/2)*(6*A + B))/(8*((3*A*a)/4 + (B*a)/8)))*(6*A + B))/(8*d) - (a*(6*A + B)*(atan(tan(
c/2 + (d*x)/2)) - (d*x)/2))/(8*d) - ((2*A*a)/5 + (2*B*a)/5 - tan(c/2 + (d*x)/2)*((5*A*a)/4 - (B*a)/8) + tan(c/
2 + (d*x)/2)^4*(4*A*a + 4*B*a) + tan(c/2 + (d*x)/2)^8*(2*A*a + 2*B*a) + tan(c/2 + (d*x)/2)^6*(4*A*a + 4*B*a) +
tan(c/2 + (d*x)/2)^10*(2*A*a + 2*B*a) + tan(c/2 + (d*x)/2)^2*((2*A*a)/5 + (2*B*a)/5) - tan(c/2 + (d*x)/2)^5*(
(A*a)/2 - (13*B*a)/4) + tan(c/2 + (d*x)/2)^7*((A*a)/2 - (13*B*a)/4) + tan(c/2 + (d*x)/2)^11*((5*A*a)/4 - (B*a)
/8) - tan(c/2 + (d*x)/2)^3*((7*A*a)/4 + (47*B*a)/24) + tan(c/2 + (d*x)/2)^9*((7*A*a)/4 + (47*B*a)/24))/(d*(6*t
an(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 + 6*tan(c/2
+ (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))