\(\int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx\) [277]

Optimal result

Integrand size = 27, antiderivative size = 91 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 x}{4}-\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{4 d}-\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{2 d} \]

[Out]

1/4*a^2*x-2/3*a^2*cos(d*x+c)^3/d+1/5*a^2*cos(d*x+c)^5/d+1/4*a^2*cos(d*x+c)*sin(d*x+c)/d-1/2*a^2*cos(d*x+c)^3*s
in(d*x+c)/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2939, 2757, 2748, 2715, 8} \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {a^2 \cos ^3(c+d x)}{6 d}-\frac {\cos ^3(c+d x) \left (a^2 \sin (c+d x)+a^2\right )}{10 d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{4 d}+\frac {a^2 x}{4}-\frac {\cos ^3(c+d x) (a \sin (c+d x)+a)^2}{5 d} \]

[In]

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

[Out]

(a^2*x)/4 - (a^2*Cos[c + d*x]^3)/(6*d) + (a^2*Cos[c + d*x]*Sin[c + d*x])/(4*d) - (Cos[c + d*x]^3*(a + a*Sin[c
+ d*x])^2)/(5*d) - (Cos[c + d*x]^3*(a^2 + a^2*Sin[c + d*x]))/(10*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 2757

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

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 {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}+\frac {2}{5} \int \cos ^2(c+d x) (a+a \sin (c+d x))^2 \, dx \\ & = -\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{10 d}+\frac {1}{2} a \int \cos ^2(c+d x) (a+a \sin (c+d x)) \, dx \\ & = -\frac {a^2 \cos ^3(c+d x)}{6 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{10 d}+\frac {1}{2} a^2 \int \cos ^2(c+d x) \, dx \\ & = -\frac {a^2 \cos ^3(c+d x)}{6 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{4 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{10 d}+\frac {1}{4} a^2 \int 1 \, dx \\ & = \frac {a^2 x}{4}-\frac {a^2 \cos ^3(c+d x)}{6 d}+\frac {a^2 \cos (c+d x) \sin (c+d x)}{4 d}-\frac {\cos ^3(c+d x) (a+a \sin (c+d x))^2}{5 d}-\frac {\cos ^3(c+d x) \left (a^2+a^2 \sin (c+d x)\right )}{10 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.63 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (-90 \cos (c+d x)-25 \cos (3 (c+d x))+3 (20 c+20 d x+\cos (5 (c+d x))-5 \sin (4 (c+d x))))}{240 d} \]

[In]

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

[Out]

(a^2*(-90*Cos[c + d*x] - 25*Cos[3*(c + d*x)] + 3*(20*c + 20*d*x + Cos[5*(c + d*x)] - 5*Sin[4*(c + d*x)])))/(24
0*d)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62

method	result	size
parallelrisch	\(-\frac {a^{2} \left (-60 d x +90 \cos \left (d x +c \right )+15 \sin \left (4 d x +4 c \right )+25 \cos \left (3 d x +3 c \right )-3 \cos \left (5 d x +5 c \right )+112\right )}{240 d}\)	\(56\)
risch	\(\frac {a^{2} x}{4}-\frac {3 a^{2} \cos \left (d x +c \right )}{8 d}+\frac {a^{2} \cos \left (5 d x +5 c \right )}{80 d}-\frac {a^{2} \sin \left (4 d x +4 c \right )}{16 d}-\frac {5 a^{2} \cos \left (3 d x +3 c \right )}{48 d}\)	\(73\)
derivativedivides	\(\frac {a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(95\)
default	\(\frac {a^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )-\frac {a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}}{d}\)	\(95\)
norman	\(\frac {\frac {a^{2} x}{4}-\frac {14 a^{2}}{15 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {3 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a^{2} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{2} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{2} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{2} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a^{2} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {4 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(267\)

[In]

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

[Out]

-1/240*a^2*(-60*d*x+90*cos(d*x+c)+15*sin(4*d*x+4*c)+25*cos(3*d*x+3*c)-3*cos(5*d*x+5*c)+112)/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {12 \, a^{2} \cos \left (d x + c\right )^{5} - 40 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} d x - 15 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{3} - a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, d} \]

[In]

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

[Out]

1/60*(12*a^2*cos(d*x + c)^5 - 40*a^2*cos(d*x + c)^3 + 15*a^2*d*x - 15*(2*a^2*cos(d*x + c)^3 - a^2*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. 172 vs. \(2 (80) = 160\).

Time = 0.23 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.89 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {2 a^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right )^{2} \sin {\left (c \right )} \cos ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*x*sin(c + d*x)**4/4 + a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + a**2*x*cos(c + d*x)**4/4 + a*
*2*sin(c + d*x)**3*cos(c + d*x)/(4*d) - a**2*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - a**2*sin(c + d*x)*cos(c +
d*x)**3/(4*d) - 2*a**2*cos(c + d*x)**5/(15*d) - a**2*cos(c + d*x)**3/(3*d), Ne(d, 0)), (x*(a*sin(c) + a)**2*s
in(c)*cos(c)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {80 \, a^{2} \cos \left (d x + c\right )^{3} - 16 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{2} - 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a^{2}}{240 \, d} \]

[In]

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

[Out]

-1/240*(80*a^2*cos(d*x + c)^3 - 16*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^2 - 15*(4*d*x + 4*c - sin(4*d*x + 4
*c))*a^2)/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.79 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {1}{4} \, a^{2} x + \frac {a^{2} \cos \left (5 \, d x + 5 \, c\right )}{80 \, d} - \frac {5 \, a^{2} \cos \left (3 \, d x + 3 \, c\right )}{48 \, d} - \frac {3 \, a^{2} \cos \left (d x + c\right )}{8 \, d} - \frac {a^{2} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} \]

[In]

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

[Out]

1/4*a^2*x + 1/80*a^2*cos(5*d*x + 5*c)/d - 5/48*a^2*cos(3*d*x + 3*c)/d - 3/8*a^2*cos(d*x + c)/d - 1/16*a^2*sin(
4*d*x + 4*c)/d

Mupad [B] (verification not implemented)

Time = 12.74 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.88 \[ \int \cos ^2(c+d x) \sin (c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,x}{4}-\frac {\frac {a^2\,\left (c+d\,x\right )}{4}-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}-\frac {a^2\,\left (15\,c+15\,d\,x-56\right )}{60}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{4}-\frac {a^2\,\left (75\,c+75\,d\,x-120\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{4}-\frac {a^2\,\left (75\,c+75\,d\,x-160\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (150\,c+150\,d\,x-80\right )}{60}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {5\,a^2\,\left (c+d\,x\right )}{2}-\frac {a^2\,\left (150\,c+150\,d\,x-480\right )}{60}\right )+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^5} \]

[In]

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

[Out]

(a^2*x)/4 - ((a^2*(c + d*x))/4 - 3*a^2*tan(c/2 + (d*x)/2)^3 + 3*a^2*tan(c/2 + (d*x)/2)^7 - (a^2*tan(c/2 + (d*x
)/2)^9)/2 - (a^2*(15*c + 15*d*x - 56))/60 + tan(c/2 + (d*x)/2)^8*((5*a^2*(c + d*x))/4 - (a^2*(75*c + 75*d*x -
120))/60) + tan(c/2 + (d*x)/2)^2*((5*a^2*(c + d*x))/4 - (a^2*(75*c + 75*d*x - 160))/60) + tan(c/2 + (d*x)/2)^4
*((5*a^2*(c + d*x))/2 - (a^2*(150*c + 150*d*x - 80))/60) + tan(c/2 + (d*x)/2)^6*((5*a^2*(c + d*x))/2 - (a^2*(1
50*c + 150*d*x - 480))/60) + (a^2*tan(c/2 + (d*x)/2))/2)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^5)