\(\int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx\) [929]

Optimal result

Integrand size = 27, antiderivative size = 108 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx=-\frac {(a+a \sin (c+d x))^{1+m}}{a d (1+m)}+\frac {3 (a+a \sin (c+d x))^{2+m}}{a^2 d (2+m)}-\frac {3 (a+a \sin (c+d x))^{3+m}}{a^3 d (3+m)}+\frac {(a+a \sin (c+d x))^{4+m}}{a^4 d (4+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2912, 12, 45} \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a \sin (c+d x)+a)^{m+4}}{a^4 d (m+4)}-\frac {3 (a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)}+\frac {3 (a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}-\frac {(a \sin (c+d x)+a)^{m+1}}{a d (m+1)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.87 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a (1+\sin (c+d x)))^{1+m} \left (-6+6 (1+m) \sin (c+d x)-3 \left (2+3 m+m^2\right ) \sin ^2(c+d x)+\left (6+11 m+6 m^2+m^3\right ) \sin ^3(c+d x)\right )}{a d (1+m) (2+m) (3+m) (4+m)} \]

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^(1 + m)*(-6 + 6*(1 + m)*Sin[c + d*x] - 3*(2 + 3*m + m^2)*Sin[c + d*x]^2 + (6 + 11*m +
6*m^2 + m^3)*Sin[c + d*x]^3))/(a*d*(1 + m)*(2 + m)*(3 + m)*(4 + m))

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\(-\frac {\left (\left (m^{3}+3 m^{2}+8 m +6\right ) \cos \left (2 d x +2 c \right )-\frac {\left (3+m \right ) \left (2+m \right ) \left (1+m \right ) \cos \left (4 d x +4 c \right )}{4}+\left (\frac {1}{2} m^{3}+\frac {3}{2} m^{2}+m \right ) \sin \left (3 d x +3 c \right )-\frac {3 \left (m \sin \left (d x +c \right )+\frac {m}{2}-\frac {1}{2}\right ) \left (m^{2}+3 m +10\right )}{2}\right ) \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{m}}{2 \left (4+m \right ) \left (m^{3}+6 m^{2}+11 m +6\right ) d}\)	\(127\)
derivativedivides	\(\frac {\left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (4+m \right )}+\frac {m \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{2}+7 m +12\right )}-\frac {6 \,{\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{4}+10 m^{3}+35 m^{2}+50 m +24\right )}+\frac {6 m \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{4}+10 m^{3}+35 m^{2}+50 m +24\right )}-\frac {3 m \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{\left (m^{3}+9 m^{2}+26 m +24\right ) d}\)	\(198\)
default	\(\frac {\left (\sin ^{4}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (4+m \right )}+\frac {m \left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{2}+7 m +12\right )}-\frac {6 \,{\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{4}+10 m^{3}+35 m^{2}+50 m +24\right )}+\frac {6 m \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{4}+10 m^{3}+35 m^{2}+50 m +24\right )}-\frac {3 m \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{\left (m^{3}+9 m^{2}+26 m +24\right ) d}\)	\(198\)

[In]

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

[Out]

-1/2*((m^3+3*m^2+8*m+6)*cos(2*d*x+2*c)-1/4*(3+m)*(2+m)*(1+m)*cos(4*d*x+4*c)+(1/2*m^3+3/2*m^2+m)*sin(3*d*x+3*c)
-3/2*(m*sin(d*x+c)+1/2*m-1/2)*(m^2+3*m+10))*(a*(1+sin(d*x+c)))^m/(4+m)/(m^3+6*m^2+11*m+6)/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.30 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} \cos \left (d x + c\right )^{4} + m^{3} - {\left (2 \, m^{3} + 9 \, m^{2} + 19 \, m + 12\right )} \cos \left (d x + c\right )^{2} + 3 \, m^{2} + {\left (m^{3} - {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} \cos \left (d x + c\right )^{2} + 3 \, m^{2} + 8 \, m\right )} \sin \left (d x + c\right ) + 8 \, m\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{4} + 10 \, d m^{3} + 35 \, d m^{2} + 50 \, d m + 24 \, d} \]

[In]

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

[Out]

((m^3 + 6*m^2 + 11*m + 6)*cos(d*x + c)^4 + m^3 - (2*m^3 + 9*m^2 + 19*m + 12)*cos(d*x + c)^2 + 3*m^2 + (m^3 - (
m^3 + 3*m^2 + 2*m)*cos(d*x + c)^2 + 3*m^2 + 8*m)*sin(d*x + c) + 8*m)*(a*sin(d*x + c) + a)^m/(d*m^4 + 10*d*m^3
+ 35*d*m^2 + 50*d*m + 24*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1508 vs. \(2 (88) = 176\).

Time = 4.79 (sec) , antiderivative size = 1508, normalized size of antiderivative = 13.96 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((x*(a*sin(c) + a)**m*sin(c)**3*cos(c), Eq(d, 0)), (6*log(sin(c + d*x) + 1)*sin(c + d*x)**3/(6*a**4*d
*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 18*log(sin(c + d*x) + 1)*s
in(c + d*x)**2/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d) + 18
*log(sin(c + d*x) + 1)*sin(c + d*x)/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c +
d*x) + 6*a**4*d) + 6*log(sin(c + d*x) + 1)/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*s
in(c + d*x) + 6*a**4*d) + 18*sin(c + d*x)**2/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d
*sin(c + d*x) + 6*a**4*d) + 27*sin(c + d*x)/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*
sin(c + d*x) + 6*a**4*d) + 11/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) +
6*a**4*d), Eq(m, -4)), (-6*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c +
d*x) + 2*a**3*d) - 12*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) +
2*a**3*d) - 6*log(sin(c + d*x) + 1)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 2*sin(c +
d*x)**3/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 12*sin(c + d*x)/(2*a**3*d*sin(c + d*x)
**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) - 9/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Eq(
m, -3)), (6*log(sin(c + d*x) + 1)*sin(c + d*x)/(2*a**2*d*sin(c + d*x) + 2*a**2*d) + 6*log(sin(c + d*x) + 1)/(2
*a**2*d*sin(c + d*x) + 2*a**2*d) + sin(c + d*x)**3/(2*a**2*d*sin(c + d*x) + 2*a**2*d) - 3*sin(c + d*x)**2/(2*a
**2*d*sin(c + d*x) + 2*a**2*d) + 6/(2*a**2*d*sin(c + d*x) + 2*a**2*d), Eq(m, -2)), (-log(sin(c + d*x) + 1)/(a*
d) + sin(c + d*x)**3/(3*a*d) - sin(c + d*x)**2/(2*a*d) + sin(c + d*x)/(a*d), Eq(m, -1)), (m**3*(a*sin(c + d*x)
+ a)**m*sin(c + d*x)**4/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) + m**3*(a*sin(c + d*x) + a)**m*sin(c
+ d*x)**3/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) + 6*m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**4/(
d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) + 3*m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**4 + 10*
d*m**3 + 35*d*m**2 + 50*d*m + 24*d) - 3*m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**2/(d*m**4 + 10*d*m**3 + 35*
d*m**2 + 50*d*m + 24*d) + 11*m*(a*sin(c + d*x) + a)**m*sin(c + d*x)**4/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*
m + 24*d) + 2*m*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) - 3*m
*(a*sin(c + d*x) + a)**m*sin(c + d*x)**2/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) + 6*m*(a*sin(c + d*x
) + a)**m*sin(c + d*x)/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) + 6*(a*sin(c + d*x) + a)**m*sin(c + d*
x)**4/(d*m**4 + 10*d*m**3 + 35*d*m**2 + 50*d*m + 24*d) - 6*(a*sin(c + d*x) + a)**m/(d*m**4 + 10*d*m**3 + 35*d*
m**2 + 50*d*m + 24*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.10 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left ({\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} a^{m} \sin \left (d x + c\right )^{4} + {\left (m^{3} + 3 \, m^{2} + 2 \, m\right )} a^{m} \sin \left (d x + c\right )^{3} - 3 \, {\left (m^{2} + m\right )} a^{m} \sin \left (d x + c\right )^{2} + 6 \, a^{m} m \sin \left (d x + c\right ) - 6 \, a^{m}\right )} {\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{{\left (m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24\right )} d} \]

[In]

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

[Out]

((m^3 + 6*m^2 + 11*m + 6)*a^m*sin(d*x + c)^4 + (m^3 + 3*m^2 + 2*m)*a^m*sin(d*x + c)^3 - 3*(m^2 + m)*a^m*sin(d*
x + c)^2 + 6*a^m*m*sin(d*x + c) - 6*a^m)*(sin(d*x + c) + 1)^m/((m^4 + 10*m^3 + 35*m^2 + 50*m + 24)*d)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 508 vs. \(2 (108) = 216\).

Time = 0.37 (sec) , antiderivative size = 508, normalized size of antiderivative = 4.70 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{3} - 3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m^{3} + 3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m^{3} - {\left (a \sin \left (d x + c\right ) + a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{3} m^{3} + 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{2} - 21 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m^{2} + 24 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m^{2} - 9 \, {\left (a \sin \left (d x + c\right ) + a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{3} m^{2} + 11 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m - 42 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m + 57 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m - 26 \, {\left (a \sin \left (d x + c\right ) + a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{3} m + 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} - 24 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a + 36 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} - 24 \, {\left (a \sin \left (d x + c\right ) + a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{3}}{{\left (a^{3} m^{4} + 10 \, a^{3} m^{3} + 35 \, a^{3} m^{2} + 50 \, a^{3} m + 24 \, a^{3}\right )} a d} \]

[In]

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

[Out]

((a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*m^3 - 3*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a*m^3 + 3
*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^2*m^3 - (a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^3*m^3 +
6*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*m^2 - 21*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a*m^2
+ 24*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^2*m^2 - 9*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^3
*m^2 + 11*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m*m - 42*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a
*m + 57*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^2*m - 26*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a
^3*m + 6*(a*sin(d*x + c) + a)^4*(a*sin(d*x + c) + a)^m - 24*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*a +
36*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a^2 - 24*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^3)/((a
^3*m^4 + 10*a^3*m^3 + 35*a^3*m^2 + 50*a^3*m + 24*a^3)*a*d)

Mupad [B] (verification not implemented)

Time = 12.52 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.07 \[ \int \cos (c+d x) \sin ^3(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (21\,m-24\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+60\,m\,\sin \left (c+d\,x\right )-32\,m\,\cos \left (2\,c+2\,d\,x\right )+11\,m\,\cos \left (4\,c+4\,d\,x\right )-4\,m\,\sin \left (3\,c+3\,d\,x\right )+18\,m^2\,\sin \left (c+d\,x\right )+6\,m^3\,\sin \left (c+d\,x\right )+6\,m^2+3\,m^3-12\,m^2\,\cos \left (2\,c+2\,d\,x\right )-4\,m^3\,\cos \left (2\,c+2\,d\,x\right )+6\,m^2\,\cos \left (4\,c+4\,d\,x\right )+m^3\,\cos \left (4\,c+4\,d\,x\right )-6\,m^2\,\sin \left (3\,c+3\,d\,x\right )-2\,m^3\,\sin \left (3\,c+3\,d\,x\right )-30\right )}{8\,d\,\left (m^4+10\,m^3+35\,m^2+50\,m+24\right )} \]

[In]

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

[Out]

((a*(sin(c + d*x) + 1))^m*(21*m - 24*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + 60*m*sin(c + d*x) - 32*m*cos(2*c
+ 2*d*x) + 11*m*cos(4*c + 4*d*x) - 4*m*sin(3*c + 3*d*x) + 18*m^2*sin(c + d*x) + 6*m^3*sin(c + d*x) + 6*m^2 + 3
*m^3 - 12*m^2*cos(2*c + 2*d*x) - 4*m^3*cos(2*c + 2*d*x) + 6*m^2*cos(4*c + 4*d*x) + m^3*cos(4*c + 4*d*x) - 6*m^
2*sin(3*c + 3*d*x) - 2*m^3*sin(3*c + 3*d*x) - 30))/(8*d*(50*m + 35*m^2 + 10*m^3 + m^4 + 24))