\(\int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx\) [259]

Optimal result

Integrand size = 27, antiderivative size = 91 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {3 a^3 \sin ^{2+n}(c+d x)}{d (2+n)}+\frac {3 a^3 \sin ^{3+n}(c+d x)}{d (3+n)}+\frac {a^3 \sin ^{4+n}(c+d x)}{d (4+n)} \]

[Out]

a^3*sin(d*x+c)^(1+n)/d/(1+n)+3*a^3*sin(d*x+c)^(2+n)/d/(2+n)+3*a^3*sin(d*x+c)^(3+n)/d/(3+n)+a^3*sin(d*x+c)^(4+n
)/d/(4+n)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 91, 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} \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^{n+1}(c+d x)}{d (n+1)}+\frac {3 a^3 \sin ^{n+2}(c+d x)}{d (n+2)}+\frac {3 a^3 \sin ^{n+3}(c+d x)}{d (n+3)}+\frac {a^3 \sin ^{n+4}(c+d x)}{d (n+4)} \]

[In]

Int[Cos[c + d*x]*Sin[c + d*x]^n*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*Sin[c + d*x]^(1 + n))/(d*(1 + n)) + (3*a^3*Sin[c + d*x]^(2 + n))/(d*(2 + n)) + (3*a^3*Sin[c + d*x]^(3 + n
))/(d*(3 + n)) + (a^3*Sin[c + d*x]^(4 + n))/(d*(4 + 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)^3 \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \left (a^3 \left (\frac {x}{a}\right )^n+3 a^3 \left (\frac {x}{a}\right )^{1+n}+3 a^3 \left (\frac {x}{a}\right )^{2+n}+a^3 \left (\frac {x}{a}\right )^{3+n}\right ) \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {a^3 \sin ^{1+n}(c+d x)}{d (1+n)}+\frac {3 a^3 \sin ^{2+n}(c+d x)}{d (2+n)}+\frac {3 a^3 \sin ^{3+n}(c+d x)}{d (3+n)}+\frac {a^3 \sin ^{4+n}(c+d x)}{d (4+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 \sin ^{1+n}(c+d x) \left (\frac {1}{1+n}+\frac {3 \sin (c+d x)}{2+n}+\frac {3 \sin ^2(c+d x)}{3+n}+\frac {\sin ^3(c+d x)}{4+n}\right )}{d} \]

[In]

Integrate[Cos[c + d*x]*Sin[c + d*x]^n*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*Sin[c + d*x]^(1 + n)*((1 + n)^(-1) + (3*Sin[c + d*x])/(2 + n) + (3*Sin[c + d*x]^2)/(3 + n) + Sin[c + d*x]
^3/(4 + n)))/d

Maple [A] (verified)

Time = 1.99 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.34

method	result	size
derivativedivides	\(\frac {a^{3} \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {a^{3} \left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (4+n \right )}+\frac {3 a^{3} \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}+\frac {3 a^{3} \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (3+n \right )}\)	\(122\)
default	\(\frac {a^{3} \sin \left (d x +c \right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (1+n \right )}+\frac {a^{3} \left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (4+n \right )}+\frac {3 a^{3} \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (2+n \right )}+\frac {3 a^{3} \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (\sin \left (d x +c \right )\right )}}{d \left (3+n \right )}\)	\(122\)
parallelrisch	\(-\frac {2 \left (\sin ^{n}\left (d x +c \right )\right ) \left (\left (n^{3}+\frac {15}{2} n^{2}+17 n +\frac {21}{2}\right ) \cos \left (2 d x +2 c \right )-\frac {\left (3+n \right ) \left (2+n \right ) \left (1+n \right ) \cos \left (4 d x +4 c \right )}{16}+\left (\frac {3}{8} n^{3}+\frac {21}{8} n^{2}+\frac {21}{4} n +3\right ) \sin \left (3 d x +3 c \right )+\left (-21-\frac {99}{8} n^{2}-\frac {115}{4} n -\frac {13}{8} n^{3}\right ) \sin \left (d x +c \right )-\frac {15 \left (1+n \right ) \left (n +\frac {18}{5}\right ) \left (3+n \right )}{16}\right ) a^{3}}{\left (n^{2}+4 n +3\right ) d \left (n^{2}+6 n +8\right )}\)	\(139\)

[In]

int(cos(d*x+c)*sin(d*x+c)^n*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

a^3/d/(1+n)*sin(d*x+c)*exp(n*ln(sin(d*x+c)))+a^3/d/(4+n)*sin(d*x+c)^4*exp(n*ln(sin(d*x+c)))+3*a^3/d/(2+n)*sin(
d*x+c)^2*exp(n*ln(sin(d*x+c)))+3*a^3/d/(3+n)*sin(d*x+c)^3*exp(n*ln(sin(d*x+c)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (91) = 182\).

Time = 0.28 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.31 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {{\left (4 \, a^{3} n^{3} + 30 \, a^{3} n^{2} + {\left (a^{3} n^{3} + 6 \, a^{3} n^{2} + 11 \, a^{3} n + 6 \, a^{3}\right )} \cos \left (d x + c\right )^{4} + 68 \, a^{3} n + 42 \, a^{3} - {\left (5 \, a^{3} n^{3} + 36 \, a^{3} n^{2} + 79 \, a^{3} n + 48 \, a^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (4 \, a^{3} n^{3} + 30 \, a^{3} n^{2} + 68 \, a^{3} n + 48 \, a^{3} - 3 \, {\left (a^{3} n^{3} + 7 \, a^{3} n^{2} + 14 \, a^{3} n + 8 \, a^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sin \left (d x + c\right )^{n}}{d n^{4} + 10 \, d n^{3} + 35 \, d n^{2} + 50 \, d n + 24 \, d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^n*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

(4*a^3*n^3 + 30*a^3*n^2 + (a^3*n^3 + 6*a^3*n^2 + 11*a^3*n + 6*a^3)*cos(d*x + c)^4 + 68*a^3*n + 42*a^3 - (5*a^3
*n^3 + 36*a^3*n^2 + 79*a^3*n + 48*a^3)*cos(d*x + c)^2 + (4*a^3*n^3 + 30*a^3*n^2 + 68*a^3*n + 48*a^3 - 3*(a^3*n
^3 + 7*a^3*n^2 + 14*a^3*n + 8*a^3)*cos(d*x + c)^2)*sin(d*x + c))*sin(d*x + c)^n/(d*n^4 + 10*d*n^3 + 35*d*n^2 +
50*d*n + 24*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1061 vs. \(2 (76) = 152\).

Time = 1.94 (sec) , antiderivative size = 1061, normalized size of antiderivative = 11.66 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)**n*(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((x*(a*sin(c) + a)**3*sin(c)**n*cos(c), Eq(d, 0)), (a**3*log(sin(c + d*x))/d - 3*a**3/(d*sin(c + d*x)
) - 3*a**3/(2*d*sin(c + d*x)**2) - a**3/(3*d*sin(c + d*x)**3), Eq(n, -4)), (3*a**3*log(sin(c + d*x))/d + a**3*
sin(c + d*x)/d - 3*a**3/(d*sin(c + d*x)) - a**3/(2*d*sin(c + d*x)**2), Eq(n, -3)), (3*a**3*log(sin(c + d*x))/d
+ a**3*sin(c + d*x)**2/(2*d) + 3*a**3*sin(c + d*x)/d - a**3/(d*sin(c + d*x)), Eq(n, -2)), (a**3*log(sin(c + d
*x))/d + a**3*sin(c + d*x)**3/(3*d) + 3*a**3*sin(c + d*x)**2/(2*d) + 3*a**3*sin(c + d*x)/d, Eq(n, -1)), (a**3*
n**3*sin(c + d*x)**4*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 3*a**3*n**3*sin(c + d*
x)**3*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 3*a**3*n**3*sin(c + d*x)**2*sin(c + d
*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + a**3*n**3*sin(c + d*x)*sin(c + d*x)**n/(d*n**4 + 10*
d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 6*a**3*n**2*sin(c + d*x)**4*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n
**2 + 50*d*n + 24*d) + 21*a**3*n**2*sin(c + d*x)**3*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n +
24*d) + 24*a**3*n**2*sin(c + d*x)**2*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 9*a**
3*n**2*sin(c + d*x)*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 11*a**3*n*sin(c + d*x)*
*4*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 42*a**3*n*sin(c + d*x)**3*sin(c + d*x)**
n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 57*a**3*n*sin(c + d*x)**2*sin(c + d*x)**n/(d*n**4 + 10*d*
n**3 + 35*d*n**2 + 50*d*n + 24*d) + 26*a**3*n*sin(c + d*x)*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 5
0*d*n + 24*d) + 6*a**3*sin(c + d*x)**4*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 24*a
**3*sin(c + d*x)**3*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 36*a**3*sin(c + d*x)**2
*sin(c + d*x)**n/(d*n**4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d) + 24*a**3*sin(c + d*x)*sin(c + d*x)**n/(d*n*
*4 + 10*d*n**3 + 35*d*n**2 + 50*d*n + 24*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\frac {a^{3} \sin \left (d x + c\right )^{n + 4}}{n + 4} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{n + 3}}{n + 3} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{n + 2}}{n + 2} + \frac {a^{3} \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^n*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

(a^3*sin(d*x + c)^(n + 4)/(n + 4) + 3*a^3*sin(d*x + c)^(n + 3)/(n + 3) + 3*a^3*sin(d*x + c)^(n + 2)/(n + 2) +
a^3*sin(d*x + c)^(n + 1)/(n + 1))/d

Giac [A] (verification not implemented)

none

Time = 0.49 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.11 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {\frac {a^{3} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{4}}{n + 4} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{3}}{n + 3} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{n} \sin \left (d x + c\right )^{2}}{n + 2} + \frac {a^{3} \sin \left (d x + c\right )^{n + 1}}{n + 1}}{d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^n*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

(a^3*sin(d*x + c)^n*sin(d*x + c)^4/(n + 4) + 3*a^3*sin(d*x + c)^n*sin(d*x + c)^3/(n + 3) + 3*a^3*sin(d*x + c)^
n*sin(d*x + c)^2/(n + 2) + a^3*sin(d*x + c)^(n + 1)/(n + 1))/d

Mupad [B] (verification not implemented)

Time = 12.21 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.66 \[ \int \cos (c+d x) \sin ^n(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3\,{\sin \left (c+d\,x\right )}^n\,\left (261\,n+336\,\sin \left (c+d\,x\right )-168\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )-48\,\sin \left (3\,c+3\,d\,x\right )+460\,n\,\sin \left (c+d\,x\right )-272\,n\,\cos \left (2\,c+2\,d\,x\right )+11\,n\,\cos \left (4\,c+4\,d\,x\right )-84\,n\,\sin \left (3\,c+3\,d\,x\right )+198\,n^2\,\sin \left (c+d\,x\right )+26\,n^3\,\sin \left (c+d\,x\right )+114\,n^2+15\,n^3-120\,n^2\,\cos \left (2\,c+2\,d\,x\right )-16\,n^3\,\cos \left (2\,c+2\,d\,x\right )+6\,n^2\,\cos \left (4\,c+4\,d\,x\right )+n^3\,\cos \left (4\,c+4\,d\,x\right )-42\,n^2\,\sin \left (3\,c+3\,d\,x\right )-6\,n^3\,\sin \left (3\,c+3\,d\,x\right )+162\right )}{8\,d\,\left (n^4+10\,n^3+35\,n^2+50\,n+24\right )} \]

[In]

int(cos(c + d*x)*sin(c + d*x)^n*(a + a*sin(c + d*x))^3,x)

[Out]

(a^3*sin(c + d*x)^n*(261*n + 336*sin(c + d*x) - 168*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) - 48*sin(3*c + 3*d*x
) + 460*n*sin(c + d*x) - 272*n*cos(2*c + 2*d*x) + 11*n*cos(4*c + 4*d*x) - 84*n*sin(3*c + 3*d*x) + 198*n^2*sin(
c + d*x) + 26*n^3*sin(c + d*x) + 114*n^2 + 15*n^3 - 120*n^2*cos(2*c + 2*d*x) - 16*n^3*cos(2*c + 2*d*x) + 6*n^2
*cos(4*c + 4*d*x) + n^3*cos(4*c + 4*d*x) - 42*n^2*sin(3*c + 3*d*x) - 6*n^3*sin(3*c + 3*d*x) + 162))/(8*d*(50*n
+ 35*n^2 + 10*n^3 + n^4 + 24))