3.931 \(\int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=54 \[ \frac{(a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}-\frac{(a \sin (c+d x)+a)^{m+1}}{a d (m+1)} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.0526407, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2833, 12, 43} \[ \frac{(a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}-\frac{(a \sin (c+d x)+a)^{m+1}}{a d (m+1)} \]

Antiderivative was successfully verified.

[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 2833

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*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 12

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

Rule 43

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])

Rubi steps

\begin{align*} \int \cos (c+d x) \sin (c+d x) (a+a \sin (c+d x))^m \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x (a+x)^m}{a} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac{\operatorname{Subst}\left (\int x (a+x)^m \, dx,x,a \sin (c+d x)\right )}{a^2 d}\\ &=\frac{\operatorname{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]  time = 0.0266722, size = 43, normalized size = 0.8 \[ \frac{((m+1) \sin (c+d x)-1) (a (\sin (c+d x)+1))^{m+1}}{a d (m+1) (m+2)} \]

Antiderivative was successfully verified.

[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 [F]  time = 0.974, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.03093, size = 76, normalized size = 1.41 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [A]  time = 1.90926, size = 126, normalized size = 2.33 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 [A]  time = 7.61462, size = 248, normalized size = 4.59 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

Giac [B]  time = 1.22055, size = 162, normalized size = 3. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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
)