\(\int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx\) [261]

Optimal result

Integrand size = 25, antiderivative size = 41 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {a \sin ^{2+n}(c+d x)}{d (2+n)} \]

[Out]

a*sin(d*x+c)^(1+n)/d/(1+n)+a*sin(d*x+c)^(2+n)/d/(2+n)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2912, 45} \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^{n+1}(c+d x)}{d (n+1)}+\frac {a \sin ^{n+2}(c+d x)}{d (n+2)} \]

[In]

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

[Out]

(a*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (a*Sin[c + d*x]^(2 + n))/(d*(2 + n))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a \sin ^{1+n}(c+d x) (2+n+(1+n) \sin (c+d x))}{d (1+n) (2+n)} \]

[In]

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

[Out]

(a*Sin[c + d*x]^(1 + n)*(2 + n + (1 + n)*Sin[c + d*x]))/(d*(1 + n)*(2 + n))

Maple [A] (verified)

Time = 0.99 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.34

method	result	size
parallelrisch	\(-\frac {\left (\sin ^{n}\left (d x +c \right )\right ) \left (\cos \left (2 d x +2 c \right ) \left (1+n \right )+\left (-2 n -4\right ) \sin \left (d x +c \right )-n -1\right ) a}{2 d \left (1+n \right ) \left (2+n \right )}\)	\(55\)
derivativedivides	\(\frac {a \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {a \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}\)	\(56\)
default	\(\frac {a \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {a \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}\)	\(56\)

[In]

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

[Out]

-1/2*sin(d*x+c)^n*(cos(2*d*x+2*c)*(1+n)+(-2*n-4)*sin(d*x+c)-n-1)*a/d/(1+n)/(2+n)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.51 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {{\left ({\left (a n + a\right )} \cos \left (d x + c\right )^{2} - a n - {\left (a n + 2 \, a\right )} \sin \left (d x + c\right ) - a\right )} \sin \left (d x + c\right )^{n}}{d n^{2} + 3 \, d n + 2 \, d} \]

[In]

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

[Out]

-((a*n + a)*cos(d*x + c)^2 - a*n - (a*n + 2*a)*sin(d*x + c) - a)*sin(d*x + c)^n/(d*n^2 + 3*d*n + 2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (32) = 64\).

Time = 0.61 (sec) , antiderivative size = 190, normalized size of antiderivative = 4.63 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} x \left (a \sin {\left (c \right )} + a\right ) \sin ^{n}{\left (c \right )} \cos {\left (c \right )} & \text {for}\: d = 0 \\\frac {a \log {\left (\sin {\left (c + d x \right )} \right )}}{d} - \frac {a}{d \sin {\left (c + d x \right )}} & \text {for}\: n = -2 \\\frac {a \log {\left (\sin {\left (c + d x \right )} \right )}}{d} + \frac {a \sin {\left (c + d x \right )}}{d} & \text {for}\: n = -1 \\\frac {a n \sin ^{2}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{2} + 3 d n + 2 d} + \frac {a n \sin {\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{2} + 3 d n + 2 d} + \frac {a \sin ^{2}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{2} + 3 d n + 2 d} + \frac {2 a \sin {\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{2} + 3 d n + 2 d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*(a*sin(c) + a)*sin(c)**n*cos(c), Eq(d, 0)), (a*log(sin(c + d*x))/d - a/(d*sin(c + d*x)), Eq(n, -2
)), (a*log(sin(c + d*x))/d + a*sin(c + d*x)/d, Eq(n, -1)), (a*n*sin(c + d*x)**2*sin(c + d*x)**n/(d*n**2 + 3*d*
n + 2*d) + a*n*sin(c + d*x)*sin(c + d*x)**n/(d*n**2 + 3*d*n + 2*d) + a*sin(c + d*x)**2*sin(c + d*x)**n/(d*n**2
+ 3*d*n + 2*d) + 2*a*sin(c + d*x)*sin(c + d*x)**n/(d*n**2 + 3*d*n + 2*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.95 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {a \sin \left (d x + c\right )^{n + 2}}{n + 2} + \frac {a \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]

[In]

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

[Out]

(a*sin(d*x + c)^(n + 2)/(n + 2) + a*sin(d*x + c)^(n + 1)/(n + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.10 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {a \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2}}{n + 2} + \frac {a \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]

[In]

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

[Out]

(a*sin(d*x + c)^n*sin(d*x + c)^2/(n + 2) + a*sin(d*x + c)^(n + 1)/(n + 1))/d

Mupad [B] (verification not implemented)

Time = 10.36 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.63 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx=\frac {a\,{\sin \left (c+d\,x\right )}^n\,\left (n+4\,\sin \left (c+d\,x\right )+2\,n\,\sin \left (c+d\,x\right )+n\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )+2\,{\sin \left (c+d\,x\right )}^2\right )}{2\,d\,\left (n^2+3\,n+2\right )} \]

[In]

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

[Out]

(a*sin(c + d*x)^n*(n + 4*sin(c + d*x) + 2*n*sin(c + d*x) + n*(2*sin(c + d*x)^2 - 1) + 2*sin(c + d*x)^2))/(2*d*
(3*n + n^2 + 2))