\(\int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx\) [267]

Optimal result

Integrand size = 27, antiderivative size = 81 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a x}{8}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 d}+\frac {a \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d} \]

[Out]

1/8*a*x-1/3*a*cos(d*x+c)^3/d+1/5*a*cos(d*x+c)^5/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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2917, 2648, 2715, 8, 2645, 14} \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \cos ^5(c+d x)}{5 d}-\frac {a \cos ^3(c+d x)}{3 d}-\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {a x}{8} \]

[In]

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

[Out]

(a*x)/8 - (a*Cos[c + d*x]^3)/(3*d) + (a*Cos[c + d*x]^5)/(5*d) + (a*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*Cos[c
+ d*x]^3*Sin[c + d*x])/(4*d)

Rule 8

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(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 2648

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] + Dist[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 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 2917

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = a \int \cos ^2(c+d x) \sin ^2(c+d x) \, dx+a \int \cos ^2(c+d x) \sin ^3(c+d x) \, dx \\ & = -\frac {a \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} a \int \cos ^2(c+d x) \, dx-\frac {a \text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{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}+\frac {1}{8} a \int 1 \, dx-\frac {a \text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{d} \\ & = \frac {a x}{8}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \cos ^5(c+d x)}{5 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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.67 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a (60 c+60 d x-60 \cos (c+d x)-10 \cos (3 (c+d x))+6 \cos (5 (c+d x))-15 \sin (4 (c+d x)))}{480 d} \]

[In]

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

[Out]

(a*(60*c + 60*d*x - 60*Cos[c + d*x] - 10*Cos[3*(c + d*x)] + 6*Cos[5*(c + d*x)] - 15*Sin[4*(c + d*x)]))/(480*d)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.67

method	result	size
parallelrisch	\(\frac {a \left (60 d x +6 \cos \left (5 d x +5 c \right )-60 \cos \left (d x +c \right )-15 \sin \left (4 d x +4 c \right )-10 \cos \left (3 d x +3 c \right )-64\right )}{480 d}\)	\(54\)
risch	\(\frac {a x}{8}-\frac {a \cos \left (d x +c \right )}{8 d}+\frac {a \cos \left (5 d x +5 c \right )}{80 d}-\frac {a \sin \left (4 d x +4 c \right )}{32 d}-\frac {a \cos \left (3 d x +3 c \right )}{48 d}\)	\(63\)
derivativedivides	\(\frac {a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\)	\(77\)
default	\(\frac {a \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+a \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )}{d}\)	\(77\)
norman	\(\frac {\frac {a x}{8}-\frac {4 a}{15 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {3 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {3 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {5 a x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {a x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {4 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(220\)

[In]

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

[Out]

1/480*a*(60*d*x+6*cos(5*d*x+5*c)-60*cos(d*x+c)-15*sin(4*d*x+4*c)-10*cos(3*d*x+3*c)-64)/d

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {24 \, a \cos \left (d x + c\right )^{5} - 40 \, a \cos \left (d x + c\right )^{3} + 15 \, a d x - 15 \, {\left (2 \, a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

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

[Out]

1/120*(24*a*cos(d*x + c)^5 - 40*a*cos(d*x + c)^3 + 15*a*d*x - 15*(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. 144 vs. \(2 (70) = 140\).

Time = 0.23 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.78 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {a \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {a \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} - \frac {2 a \cos ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \sin ^{2}{\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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)**2*cos(c + d*x)**3/(3*d) - a*sin(c + d*x)*cos(c + d*x)**3/(8*d) - 2
*a*cos(c + d*x)**5/(15*d), Ne(d, 0)), (x*(a*sin(c) + a)*sin(c)**2*cos(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.64 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {32 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a + 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a}{480 \, d} \]

[In]

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

[Out]

1/480*(32*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a + 15*(4*d*x + 4*c - sin(4*d*x + 4*c))*a)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {1}{8} \, a x + \frac {a \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {a \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {a \cos \left (d x + c\right )}{8 \, d} - \frac {a \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} \]

[In]

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

[Out]

1/8*a*x + 1/80*a*cos(5*d*x + 5*c)/d - 1/48*a*cos(3*d*x + 3*c)/d - 1/8*a*cos(d*x + c)/d - 1/32*a*sin(4*d*x + 4*
c)/d

Mupad [B] (verification not implemented)

Time = 12.92 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.44 \[ \int \cos ^2(c+d x) \sin ^2(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,x}{8}+\frac {\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}-\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+\left (\frac {a\,\left (150\,c+150\,d\,x-480\right )}{120}-\frac {5\,a\,\left (c+d\,x\right )}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {a\,\left (150\,c+150\,d\,x+160\right )}{120}-\frac {5\,a\,\left (c+d\,x\right )}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\left (\frac {a\,\left (75\,c+75\,d\,x-160\right )}{120}-\frac {5\,a\,\left (c+d\,x\right )}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {a\,\left (15\,c+15\,d\,x-32\right )}{120}-\frac {a\,\left (c+d\,x\right )}{8}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

[In]

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

[Out]

(a*x)/8 + ((a*(15*c + 15*d*x - 32))/120 - (a*tan(c/2 + (d*x)/2))/4 - (a*(c + d*x))/8 + tan(c/2 + (d*x)/2)^2*((
a*(75*c + 75*d*x - 160))/120 - (5*a*(c + d*x))/8) + tan(c/2 + (d*x)/2)^4*((a*(150*c + 150*d*x + 160))/120 - (5
*a*(c + d*x))/4) + tan(c/2 + (d*x)/2)^6*((a*(150*c + 150*d*x - 480))/120 - (5*a*(c + d*x))/4) + (3*a*tan(c/2 +
(d*x)/2)^3)/2 - (3*a*tan(c/2 + (d*x)/2)^7)/2 + (a*tan(c/2 + (d*x)/2)^9)/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^5)