\(\int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx\) [260]

Optimal result

Integrand size = 27, antiderivative size = 68 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {2 a^2 \sin ^{2+n}(c+d x)}{d (2+n)}+\frac {a^2 \sin ^{3+n}(c+d x)}{d (3+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2912, 45} \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^{n+1}(c+d x)}{d (n+1)}+\frac {2 a^2 \sin ^{n+2}(c+d x)}{d (n+2)}+\frac {a^2 \sin ^{n+3}(c+d x)}{d (n+3)} \]

[In]

Int[Cos[c + d*x]*Sin[c + d*x]^n*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (2*a^2*Sin[c + d*x]^(2 + n))/(d*(2 + n)) + (a^2*Sin[c + d*x]^(3 + n))
/(d*(3 + 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)^2 \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \left (a^2 \left (\frac {x}{a}\right )^n+2 a^2 \left (\frac {x}{a}\right )^{1+n}+a^2 \left (\frac {x}{a}\right )^{2+n}\right ) \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^2 \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {2 a^2 \sin ^{2+n}(c+d x)}{d (2+n)}+\frac {a^2 \sin ^{3+n}(c+d x)}{d (3+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \sin ^{1+n}(c+d x) \left (\frac {1}{1+n}+\frac {2 \sin (c+d x)}{2+n}+\frac {\sin ^2(c+d x)}{3+n}\right )}{d} \]

[In]

Integrate[Cos[c + d*x]*Sin[c + d*x]^n*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*Sin[c + d*x]^(1 + n)*((1 + n)^(-1) + (2*Sin[c + d*x])/(2 + n) + Sin[c + d*x]^2/(3 + n)))/d

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {a^{2} \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {2 a^{2} \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}\)	\(91\)
default	\(\frac {a^{2} \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {2 a^{2} \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}\)	\(91\)
parallelrisch	\(-\frac {\left (\left (n^{2}+4 n +3\right ) \cos \left (2 d x +2 c \right )+\left (\frac {1}{4} n^{2}+\frac {3}{4} n +\frac {1}{2}\right ) \sin \left (3 d x +3 c \right )+\left (-\frac {7}{4} n^{2}-\frac {29}{4} n -\frac {15}{2}\right ) \sin \left (d x +c \right )-n^{2}-4 n -3\right ) \left (\sin ^{n}\left (d x +c \right )\right ) a^{2}}{\left (3+n \right ) \left (1+n \right ) d \left (2+n \right )}\)	\(97\)

[In]

int(cos(d*x+c)*sin(d*x+c)^n*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

a^2/d/(1+n)*sin(d*x+c)*exp(n*ln(sin(d*x+c)))+a^2/d/(3+n)*sin(d*x+c)^3*exp(n*ln(sin(d*x+c)))+2*a^2/d/(2+n)*sin(
d*x+c)^2*exp(n*ln(sin(d*x+c)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.99 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {{\left (2 \, a^{2} n^{2} + 8 \, a^{2} n - 2 \, {\left (a^{2} n^{2} + 4 \, a^{2} n + 3 \, a^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, a^{2} + {\left (2 \, a^{2} n^{2} + 8 \, a^{2} n - {\left (a^{2} n^{2} + 3 \, a^{2} n + 2 \, a^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{d n^{3} + 6 \, d n^{2} + 11 \, d n + 6 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^n*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

(2*a^2*n^2 + 8*a^2*n - 2*(a^2*n^2 + 4*a^2*n + 3*a^2)*cos(d*x + c)^2 + 6*a^2 + (2*a^2*n^2 + 8*a^2*n - (a^2*n^2
+ 3*a^2*n + 2*a^2)*cos(d*x + c)^2 + 8*a^2)*sin(d*x + c))*sin(d*x + c)^n/(d*n^3 + 6*d*n^2 + 11*d*n + 6*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (56) = 112\).

Time = 1.07 (sec) , antiderivative size = 530, normalized size of antiderivative = 7.79 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\begin {cases} x \left (a \sin {\left (c \right )} + a\right )^{2} \sin ^{n}{\left (c \right )} \cos {\left (c \right )} & \text {for}\: d = 0 \\\frac {a^{2} \log {\left (\sin {\left (c + d x \right )} \right )}}{d} - \frac {2 a^{2}}{d \sin {\left (c + d x \right )}} - \frac {a^{2}}{2 d \sin ^{2}{\left (c + d x \right )}} & \text {for}\: n = -3 \\\frac {2 a^{2} \log {\left (\sin {\left (c + d x \right )} \right )}}{d} + \frac {a^{2} \sin {\left (c + d x \right )}}{d} - \frac {a^{2}}{d \sin {\left (c + d x \right )}} & \text {for}\: n = -2 \\\frac {a^{2} \log {\left (\sin {\left (c + d x \right )} \right )}}{d} + \frac {a^{2} \sin ^{2}{\left (c + d x \right )}}{2 d} + \frac {2 a^{2} \sin {\left (c + d x \right )}}{d} & \text {for}\: n = -1 \\\frac {a^{2} n^{2} \sin ^{3}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{3} + 6 d n^{2} + 11 d n + 6 d} + \frac {2 a^{2} n^{2} \sin ^{2}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{3} + 6 d n^{2} + 11 d n + 6 d} + \frac {a^{2} n^{2} \sin {\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{3} + 6 d n^{2} + 11 d n + 6 d} + \frac {3 a^{2} n \sin ^{3}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{3} + 6 d n^{2} + 11 d n + 6 d} + \frac {8 a^{2} n \sin ^{2}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{3} + 6 d n^{2} + 11 d n + 6 d} + \frac {5 a^{2} n \sin {\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{3} + 6 d n^{2} + 11 d n + 6 d} + \frac {2 a^{2} \sin ^{3}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{3} + 6 d n^{2} + 11 d n + 6 d} + \frac {6 a^{2} \sin ^{2}{\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{3} + 6 d n^{2} + 11 d n + 6 d} + \frac {6 a^{2} \sin {\left (c + d x \right )} \sin ^{n}{\left (c + d x \right )}}{d n^{3} + 6 d n^{2} + 11 d n + 6 d} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)**n*(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((x*(a*sin(c) + a)**2*sin(c)**n*cos(c), Eq(d, 0)), (a**2*log(sin(c + d*x))/d - 2*a**2/(d*sin(c + d*x)
) - a**2/(2*d*sin(c + d*x)**2), Eq(n, -3)), (2*a**2*log(sin(c + d*x))/d + a**2*sin(c + d*x)/d - a**2/(d*sin(c
+ d*x)), Eq(n, -2)), (a**2*log(sin(c + d*x))/d + a**2*sin(c + d*x)**2/(2*d) + 2*a**2*sin(c + d*x)/d, Eq(n, -1)
), (a**2*n**2*sin(c + d*x)**3*sin(c + d*x)**n/(d*n**3 + 6*d*n**2 + 11*d*n + 6*d) + 2*a**2*n**2*sin(c + d*x)**2
*sin(c + d*x)**n/(d*n**3 + 6*d*n**2 + 11*d*n + 6*d) + a**2*n**2*sin(c + d*x)*sin(c + d*x)**n/(d*n**3 + 6*d*n**
2 + 11*d*n + 6*d) + 3*a**2*n*sin(c + d*x)**3*sin(c + d*x)**n/(d*n**3 + 6*d*n**2 + 11*d*n + 6*d) + 8*a**2*n*sin
(c + d*x)**2*sin(c + d*x)**n/(d*n**3 + 6*d*n**2 + 11*d*n + 6*d) + 5*a**2*n*sin(c + d*x)*sin(c + d*x)**n/(d*n**
3 + 6*d*n**2 + 11*d*n + 6*d) + 2*a**2*sin(c + d*x)**3*sin(c + d*x)**n/(d*n**3 + 6*d*n**2 + 11*d*n + 6*d) + 6*a
**2*sin(c + d*x)**2*sin(c + d*x)**n/(d*n**3 + 6*d*n**2 + 11*d*n + 6*d) + 6*a**2*sin(c + d*x)*sin(c + d*x)**n/(
d*n**3 + 6*d*n**2 + 11*d*n + 6*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.93 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\frac {a^{2} \sin \left (d x + c\right )^{n + 3}}{n + 3} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{n + 2}}{n + 2} + \frac {a^{2} \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^n*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.59 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.10 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {\frac {a^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3}}{n + 3} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2}}{n + 2} + \frac {a^{2} \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^n*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.26 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.16 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\sin \left (c+d\,x\right )}^n\,\left (16\,n+30\,\sin \left (c+d\,x\right )-2\,\sin \left (3\,c+3\,d\,x\right )+29\,n\,\sin \left (c+d\,x\right )+16\,n\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )-3\,n\,\sin \left (3\,c+3\,d\,x\right )+7\,n^2\,\sin \left (c+d\,x\right )+4\,n^2\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )+4\,n^2+24\,{\sin \left (c+d\,x\right )}^2-n^2\,\sin \left (3\,c+3\,d\,x\right )\right )}{4\,d\,\left (n^3+6\,n^2+11\,n+6\right )} \]

[In]

int(cos(c + d*x)*sin(c + d*x)^n*(a + a*sin(c + d*x))^2,x)

[Out]

(a^2*sin(c + d*x)^n*(16*n + 30*sin(c + d*x) - 2*sin(3*c + 3*d*x) + 29*n*sin(c + d*x) + 16*n*(2*sin(c + d*x)^2
- 1) - 3*n*sin(3*c + 3*d*x) + 7*n^2*sin(c + d*x) + 4*n^2*(2*sin(c + d*x)^2 - 1) + 4*n^2 + 24*sin(c + d*x)^2 -
n^2*sin(3*c + 3*d*x)))/(4*d*(11*n + 6*n^2 + n^3 + 6))