\(\int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx\) [931]

Optimal result

Integrand size = 25, antiderivative size = 54 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx=-\frac {(a+a \sin (c+d x))^{1+m}}{a d (1+m)}+\frac {(a+a \sin (c+d x))^{2+m}}{a^2 d (2+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2912, 12, 45} \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}-\frac {(a \sin (c+d x)+a)^{m+1}}{a d (m+1)} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 \frac {x (a+x)^m}{a} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int x (a+x)^m \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = \frac {\text {Subst}\left (\int \left (-a (a+x)^m+(a+x)^{1+m}\right ) \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = -\frac {(a+a \sin (c+d x))^{1+m}}{a d (1+m)}+\frac {(a+a \sin (c+d x))^{2+m}}{a^2 d (2+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.80 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a (1+\sin (c+d x)))^{1+m} (-1+(1+m) \sin (c+d x))}{a d (1+m) (2+m)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02

method	result	size
parallelrisch	\(-\frac {\left (a \left (1+\sin \left (d x +c \right )\right )\right )^{m} \left (\left (1+m \right ) \cos \left (2 d x +2 c \right )-2 m \sin \left (d x +c \right )-m +1\right )}{2 d \left (m^{2}+3 m +2\right )}\)	\(55\)
derivativedivides	\(\frac {\left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (2+m \right )}+\frac {m \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{2}+3 m +2\right )}-\frac {{\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{2}+3 m +2\right )}\)	\(97\)
default	\(\frac {\left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (2+m \right )}+\frac {m \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{2}+3 m +2\right )}-\frac {{\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{2}+3 m +2\right )}\)	\(97\)

[In]

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

[Out]

-1/2*(a*(1+sin(d*x+c)))^m*((1+m)*cos(2*d*x+2*c)-2*m*sin(d*x+c)-m+1)/d/(m^2+3*m+2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx=-\frac {{\left ({\left (m + 1\right )} \cos \left (d x + c\right )^{2} - m \sin \left (d x + c\right ) - m\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{2} + 3 \, d m + 2 \, d} \]

[In]

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

[Out]

-((m + 1)*cos(d*x + c)^2 - m*sin(d*x + c) - m)*(a*sin(d*x + c) + a)^m/(d*m^2 + 3*d*m + 2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 248 vs. \(2 (41) = 82\).

Time = 0.96 (sec) , antiderivative size = 248, normalized size of antiderivative = 4.59 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx=\begin {cases} x \left (a \sin {\left (c \right )} + a\right )^{m} \sin {\left (c \right )} \cos {\left (c \right )} & \text {for}\: d = 0 \\\frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {1}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} & \text {for}\: m = -2 \\- \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a d} + \frac {\sin {\left (c + d x \right )}}{a d} & \text {for}\: m = -1 \\\frac {m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{2}{\left (c + d x \right )}}{d m^{2} + 3 d m + 2 d} + \frac {m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin {\left (c + d x \right )}}{d m^{2} + 3 d m + 2 d} + \frac {\left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{2}{\left (c + d x \right )}}{d m^{2} + 3 d m + 2 d} - \frac {\left (a \sin {\left (c + d x \right )} + a\right )^{m}}{d m^{2} + 3 d m + 2 d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.04 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a^{m} {\left (m + 1\right )} \sin \left (d x + c\right )^{2} + a^{m} m \sin \left (d x + c\right ) - a^{m}\right )} {\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{{\left (m^{2} + 3 \, m + 2\right )} d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (54) = 108\).

Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.22 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m - {\left (a \sin \left (d x + c\right ) + a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m + {\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} - 2 \, {\left (a \sin \left (d x + c\right ) + a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a}{{\left (m^{2} + 3 \, m + 2\right )} a^{2} d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.92 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.15 \[ \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (\frac {m}{2}+m\,\sin \left (c+d\,x\right )+\frac {m\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )}{2}+{\sin \left (c+d\,x\right )}^2-1\right )}{d\,\left (m^2+3\,m+2\right )} \]

[In]

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

[Out]

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