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

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Rubi [A]  time = 0.05, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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])

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]

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*}

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Mathematica [A]  time = 0.03, size = 43, normalized size = 0.80 \[ \frac {((m+1) \sin (c+d x)-1) (a (\sin (c+d x)+1))^{m+1}}{a d (m+1) (m+2)} \]

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))

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fricas [A]  time = 0.45, size = 54, normalized size = 1.00 \[ -\frac {{\left ({\left (m + 1\right )} \cos \left (d x + c\right )^{2} - m \sin \left (d x + c\right ) - m\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{2} + 3 \, d m + 2 \, d} \]

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)

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giac [A]  time = 0.26, size = 92, normalized size = 1.70 \[ \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{m} m \sin \left (d x + c\right )^{2} + {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m \sin \left (d x + c\right ) + {\left (a \sin \left (d x + c\right ) + a\right )}^{m} \sin \left (d x + c\right )^{2} - {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{{\left (m^{2} + 3 \, m + 2\right )} d} \]

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

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maple [F]  time = 1.78, size = 0, normalized size = 0.00 \[ \int \cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]

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)

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maxima [A]  time = 0.91, size = 56, normalized size = 1.04 \[ \frac {{\left (a^{m} {\left (m + 1\right )} \sin \left (d x + c\right )^{2} + a^{m} m \sin \left (d x + c\right ) - a^{m}\right )} {\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{{\left (m^{2} + 3 \, m + 2\right )} d} \]

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)

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mupad [B]  time = 9.42, size = 62, normalized size = 1.15 \[ \frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (\frac {m}{2}+m\,\sin \left (c+d\,x\right )+\frac {m\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )}{2}+{\sin \left (c+d\,x\right )}^2-1\right )}{d\,\left (m^2+3\,m+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 4.80, size = 248, normalized size = 4.59 \[ \begin {cases} x \left (a \sin {\relax (c )} + a\right )^{m} \sin {\relax (c )} \cos {\relax (c )} & \text {for}\: d = 0 \\\frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {1}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} & \text {for}\: m = -2 \\- \frac {\log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a d} + \frac {\sin {\left (c + d x \right )}}{a d} & \text {for}\: m = -1 \\\frac {m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{2}{\left (c + d x \right )}}{d m^{2} + 3 d m + 2 d} + \frac {m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin {\left (c + d x \right )}}{d m^{2} + 3 d m + 2 d} + \frac {\left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{2}{\left (c + d x \right )}}{d m^{2} + 3 d m + 2 d} - \frac {\left (a \sin {\left (c + d x \right )} + a\right )^{m}}{d m^{2} + 3 d m + 2 d} & \text {otherwise} \end {cases} \]

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))

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