\(\int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx\) [268]

Optimal result

Integrand size = 25, antiderivative size = 65 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a x}{8}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d} \] Output:

1/8*a*x-1/3*a*cos(d*x+c)^3/d+1/8*a*cos(d*x+c)*sin(d*x+c)/d-1/4*a*cos(d*x+c 
)^3*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a (24 \cos (c+d x)+8 \cos (3 (c+d x))+3 (-4 d x+\sin (4 (c+d x))))}{96 d} \] Input:

Integrate[Cos[c + d*x]^2*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]
 

Output:

-1/96*(a*(24*Cos[c + d*x] + 8*Cos[3*(c + d*x)] + 3*(-4*d*x + Sin[4*(c + d* 
x)])))/d
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3317, 3042, 3045, 15, 3048, 3042, 3115, 24}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sin (c+d x) \cos ^2(c+d x) (a \sin (c+d x)+a) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x) \cos (c+d x)^2 (a \sin (c+d x)+a)dx\)

\(\Big \downarrow \) 3317

\(\displaystyle a \int \cos ^2(c+d x) \sin ^2(c+d x)dx+a \int \cos ^2(c+d x) \sin (c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int \cos (c+d x)^2 \sin (c+d x)dx+a \int \cos (c+d x)^2 \sin (c+d x)^2dx\)

\(\Big \downarrow \) 3045

\(\displaystyle a \int \cos (c+d x)^2 \sin (c+d x)^2dx-\frac {a \int \cos ^2(c+d x)d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 15

\(\displaystyle a \int \cos (c+d x)^2 \sin (c+d x)^2dx-\frac {a \cos ^3(c+d x)}{3 d}\)

\(\Big \downarrow \) 3048

\(\displaystyle a \left (\frac {1}{4} \int \cos ^2(c+d x)dx-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^3(c+d x)}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {1}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^3(c+d x)}{3 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle a \left (\frac {1}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^3(c+d x)}{3 d}\)

\(\Big \downarrow \) 24

\(\displaystyle a \left (\frac {1}{4} \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\sin (c+d x) \cos ^3(c+d x)}{4 d}\right )-\frac {a \cos ^3(c+d x)}{3 d}\)

Input:

Int[Cos[c + d*x]^2*Sin[c + d*x]*(a + a*Sin[c + d*x]),x]
 

Output:

-1/3*(a*Cos[c + d*x]^3)/d + a*(-1/4*(Cos[c + d*x]^3*Sin[c + d*x])/d + (x/2 
 + (Cos[c + d*x]*Sin[c + d*x])/(2*d))/4)
 

Defintions of rubi rules used

rule 15

Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ 
{a, m}, x] && NeQ[m, -1]
 

rule 24

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3045

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_ 
Symbol] :> Simp[-(a*f)^(-1)   Subst[Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], 
x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] && 
 !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])
 

rule 3048

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> Simp[(-a)*(b*Cos[e + f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 
1)/(b*f*(m + n))), x] + Simp[a^2*((m - 1)/(m + n))   Int[(b*Cos[e + f*x])^n 
*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] 
 && NeQ[m + n, 0] && IntegersQ[2*m, 2*n]
 

rule 3115

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] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3317

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a   Int[(g*Co 
s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d   Int[(g*Cos[e + f*x])^ 
p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
 

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66

method	result	size
parallelrisch	\(-\frac {a \left (-12 d x +3 \sin \left (4 d x +4 c \right )+8 \cos \left (3 d x +3 c \right )+24 \cos \left (d x +c \right )+32\right )}{96 d}\)	\(43\)
risch	\(\frac {a x}{8}-\frac {a \cos \left (d x +c \right )}{4 d}-\frac {a \sin \left (4 d x +4 c \right )}{32 d}-\frac {a \cos \left (3 d x +3 c \right )}{12 d}\)	\(48\)
derivativedivides	\(\frac {a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a \cos \left (d x +c \right )^{3}}{3}}{d}\)	\(57\)
default	\(\frac {a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{3}}{4}+\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a \cos \left (d x +c \right )^{3}}{3}}{d}\)	\(57\)
norman	\(\frac {\frac {a x}{8}-\frac {2 a}{3 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {7 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}-\frac {7 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}+\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {3 a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {a x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(205\)
orering	\(x \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right )-\frac {169 \left (-2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right ) d +\cos \left (d x +c \right )^{3} d \left (a +a \sin \left (d x +c \right )\right )+\cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) a d \right )}{144 d^{2}}+\frac {169 x \left (2 d^{2} \sin \left (d x +c \right )^{3} \left (a +a \sin \left (d x +c \right )\right )-7 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right ) d^{2}-5 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2} a \,d^{2}+2 \cos \left (d x +c \right )^{4} d^{2} a \right )}{144 d^{2}}-\frac {13 \left (20 d^{3} \sin \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )+12 d^{3} \sin \left (d x +c \right )^{3} a \cos \left (d x +c \right )-7 \cos \left (d x +c \right )^{3} d^{3} \left (a +a \sin \left (d x +c \right )\right )-25 \cos \left (d x +c \right )^{3} \sin \left (d x +c \right ) a \,d^{3}\right )}{72 d^{4}}+\frac {13 x \left (61 d^{4} \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )^{2}+131 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2} a \,d^{4}-20 d^{4} \sin \left (d x +c \right )^{3} \left (a +a \sin \left (d x +c \right )\right )-12 \sin \left (d x +c \right )^{4} a \,d^{4}-32 \cos \left (d x +c \right )^{4} a \,d^{4}\right )}{72 d^{4}}-\frac {61 d^{5} \cos \left (d x +c \right )^{3} \left (a +a \sin \left (d x +c \right )\right )+451 d^{5} \sin \left (d x +c \right ) a \cos \left (d x +c \right )^{3}-182 d^{5} \sin \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right ) \cos \left (d x +c \right )-330 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a \,d^{5}}{144 d^{6}}+\frac {x \left (-547 d^{6} \cos \left (d x +c \right )^{2} \left (a +a \sin \left (d x +c \right )\right ) \sin \left (d x +c \right )+512 \cos \left (d x +c \right )^{4} a \,d^{6}-2525 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{2} a \,d^{6}+182 d^{6} \sin \left (d x +c \right )^{3} \left (a +a \sin \left (d x +c \right )\right )+330 \sin \left (d x +c \right )^{4} a \,d^{6}\right )}{144 d^{6}}\)	\(607\)

Input:

int(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

-1/96*a*(-12*d*x+3*sin(4*d*x+4*c)+8*cos(3*d*x+3*c)+24*cos(d*x+c)+32)/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.78 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {8 \, a \cos \left (d x + c\right )^{3} - 3 \, a d x + 3 \, {\left (2 \, a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="fricas")
 

Output:

-1/24*(8*a*cos(d*x + c)^3 - 3*a*d*x + 3*(2*a*cos(d*x + c)^3 - 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. 119 vs. \(2 (56) = 112\).

Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.83 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {a \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin {\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**2*sin(d*x+c)*(a+a*sin(d*x+c)),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.60 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=-\frac {32 \, a \cos \left (d x + c\right )^{3} - 3 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a}{96 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="maxima")
 

Output:

-1/96*(32*a*cos(d*x + c)^3 - 3*(4*d*x + 4*c - sin(4*d*x + 4*c))*a)/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\frac {1}{8} \, a x - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{12 \, d} - \frac {a \cos \left (d x + c\right )}{4 \, d} - \frac {a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \] Input:

integrate(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c)),x, algorithm="giac")
 

Output:

1/8*a*x - 1/12*a*cos(3*d*x + 3*c)/d - 1/4*a*cos(d*x + c)/d - 1/32*a*sin(4* 
d*x + 4*c)/d
 

Mupad [B] (verification not implemented)

Time = 20.54 (sec) , antiderivative size = 198, normalized size of antiderivative = 3.05 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,x}{8}+\frac {\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\left (\frac {a\,\left (12\,c+12\,d\,x-48\right )}{24}-\frac {a\,\left (c+d\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\left (\frac {a\,\left (18\,c+18\,d\,x-48\right )}{24}-\frac {3\,a\,\left (c+d\,x\right )}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\left (\frac {a\,\left (12\,c+12\,d\,x-16\right )}{24}-\frac {a\,\left (c+d\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {a\,\left (3\,c+3\,d\,x-16\right )}{24}-\frac {a\,\left (c+d\,x\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \] Input:

int(cos(c + d*x)^2*sin(c + d*x)*(a + a*sin(c + d*x)),x)
 

Output:

(a*x)/8 + ((a*(3*c + 3*d*x - 16))/24 - (a*tan(c/2 + (d*x)/2))/4 - (a*(c + 
d*x))/8 + tan(c/2 + (d*x)/2)^2*((a*(12*c + 12*d*x - 16))/24 - (a*(c + d*x) 
)/2) + tan(c/2 + (d*x)/2)^6*((a*(12*c + 12*d*x - 48))/24 - (a*(c + d*x))/2 
) + tan(c/2 + (d*x)/2)^4*((a*(18*c + 18*d*x - 48))/24 - (3*a*(c + d*x))/4) 
 + (7*a*tan(c/2 + (d*x)/2)^3)/4 - (7*a*tan(c/2 + (d*x)/2)^5)/4 + (a*tan(c/ 
2 + (d*x)/2)^7)/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \left (6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 \cos \left (d x +c \right )+3 d x +8\right )}{24 d} \] Input:

int(cos(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c)),x)
 

Output:

(a*(6*cos(c + d*x)*sin(c + d*x)**3 + 8*cos(c + d*x)*sin(c + d*x)**2 - 3*co 
s(c + d*x)*sin(c + d*x) - 8*cos(c + d*x) + 3*d*x + 8))/(24*d)