3.261 \(\int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=41 \[ \frac {a \sin ^{n+1}(c+d x)}{d (n+1)}+\frac {a \sin ^{n+2}(c+d x)}{d (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2833, 43} \[ \frac {a \sin ^{n+1}(c+d x)}{d (n+1)}+\frac {a \sin ^{n+2}(c+d x)}{d (n+2)} \]

Antiderivative was successfully verified.

[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 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 ^n(c+d x) (a+a \sin (c+d x)) \, dx &=\frac {\operatorname {Subst}\left (\int \left (\frac {x}{a}\right )^n (a+x) \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {\operatorname {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*}

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Mathematica [A]  time = 0.30, size = 38, normalized size = 0.93 \[ \frac {a \sin ^{n+1}(c+d x) ((n+1) \sin (c+d x)+n+2)}{d (n+1) (n+2)} \]

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.50, size = 62, normalized size = 1.51 \[ -\frac {{\left ({\left (a n + a\right )} \cos \left (d x + c\right )^{2} - a n - {\left (a n + 2 \, a\right )} \sin \left (d x + c\right ) - a\right )} \sin \left (d x + c\right )^{n}}{d n^{2} + 3 \, d n + 2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.22, size = 86, normalized size = 2.10 \[ \frac {a n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2} + a n \sin \left (d x + c\right )^{n} \sin \left (d x + c\right ) + a \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2} + 2 \, a \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )}{{\left (n^{2} + 3 \, n + 2\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*n*sin(d*x + c)^n*sin(d*x + c)^2 + a*n*sin(d*x + c)^n*sin(d*x + c) + a*sin(d*x + c)^n*sin(d*x + c)^2 + 2*a*s
in(d*x + c)^n*sin(d*x + c))/((n^2 + 3*n + 2)*d)

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maple [F]  time = 3.79, size = 0, normalized size = 0.00 \[ \int \cos \left (d x +c \right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +a \sin \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 39, normalized size = 0.95 \[ \frac {\frac {a \sin \left (d x + c\right )^{n + 2}}{n + 2} + \frac {a \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 9.18, size = 67, normalized size = 1.63 \[ \frac {a\,{\sin \left (c+d\,x\right )}^n\,\left (n+4\,\sin \left (c+d\,x\right )+2\,n\,\sin \left (c+d\,x\right )+n\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )+2\,{\sin \left (c+d\,x\right )}^2\right )}{2\,d\,\left (n^2+3\,n+2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 5.08, size = 190, normalized size = 4.63 \[ \begin {cases} x \left (a \sin {\relax (c )} + a\right ) \sin ^{n}{\relax (c )} \cos {\relax (c )} & \text {for}\: d = 0 \\\frac {a \log {\left (\sin {\left (c + d x \right )} \right )}}{d} - \frac {a}{d \sin {\left (c + d x \right )}} & \text {for}\: n = -2 \\\frac {a \log {\left (\sin {\left (c + d x \right )} \right )}}{d} + \frac {a \sin {\left (c + d x \right )}}{d} & \text {for}\: n = -1 \\\frac {a n \sin ^{2}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{2} + 3 d n + 2 d} + \frac {a n \sin {\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{2} + 3 d n + 2 d} + \frac {a \sin ^{2}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{2} + 3 d n + 2 d} + \frac {2 a \sin {\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{2} + 3 d n + 2 d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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