\(\int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx\) [930]

Optimal result

Integrand size = 27, antiderivative size = 80 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a+a \sin (c+d x))^{1+m}}{a d (1+m)}-\frac {2 (a+a \sin (c+d x))^{2+m}}{a^2 d (2+m)}+\frac {(a+a \sin (c+d x))^{3+m}}{a^3 d (3+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 80, 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} \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {(a \sin (c+d x)+a)^{m+3}}{a^3 d (m+3)}-\frac {2 (a \sin (c+d x)+a)^{m+2}}{a^2 d (m+2)}+\frac {(a \sin (c+d x)+a)^{m+1}}{a d (m+1)} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx=-\frac {(a (1+\sin (c+d x)))^{1+m} \left (-6-3 m-m^2+\left (2+3 m+m^2\right ) \cos (2 (c+d x))+4 (1+m) \sin (c+d x)\right )}{2 a d (1+m) (2+m) (3+m)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.19

method	result	size
parallelrisch	\(-\frac {\left (\left (\frac {1}{2} m^{2}+\frac {3}{2} m +1\right ) \sin \left (3 d x +3 c \right )+\left (m^{2}+m \right ) \cos \left (2 d x +2 c \right )+\left (-\frac {1}{2} m -\frac {3}{2} m^{2}-3\right ) \sin \left (d x +c \right )-m^{2}-m -4\right ) \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{m}}{2 \left (3+m \right ) \left (2+m \right ) \left (1+m \right ) d}\)	\(95\)
derivativedivides	\(\frac {\left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (3+m \right )}+\frac {m \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{\left (m^{2}+5 m +6\right ) d}+\frac {2 \,{\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{3}+6 m^{2}+11 m +6\right )}-\frac {2 m \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{3}+6 m^{2}+11 m +6\right )}\)	\(145\)
default	\(\frac {\left (\sin ^{3}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (3+m \right )}+\frac {m \left (\sin ^{2}\left (d x +c \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{\left (m^{2}+5 m +6\right ) d}+\frac {2 \,{\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{3}+6 m^{2}+11 m +6\right )}-\frac {2 m \sin \left (d x +c \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (d x +c \right )\right )}}{d \left (m^{3}+6 m^{2}+11 m +6\right )}\)	\(145\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.16 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx=-\frac {{\left ({\left (m^{2} + m\right )} \cos \left (d x + c\right )^{2} - m^{2} + {\left ({\left (m^{2} + 3 \, m + 2\right )} \cos \left (d x + c\right )^{2} - m^{2} - m - 2\right )} \sin \left (d x + c\right ) - m - 2\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{d m^{3} + 6 \, d m^{2} + 11 \, d m + 6 \, d} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 697 vs. \(2 (65) = 130\).

Time = 2.13 (sec) , antiderivative size = 697, normalized size of antiderivative = 8.71 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx=\begin {cases} x \left (a \sin {\left (c \right )} + a\right )^{m} \sin ^{2}{\left (c \right )} \cos {\left (c \right )} & \text {for}\: d = 0 \\\frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {4 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {4 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {3}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: m = -3 \\- \frac {2 \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 {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {\sin ^{2}{\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} - \frac {2}{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 ^{2}{\left (c + d x \right )}}{2 a d} - \frac {\sin {\left (c + d x \right )}}{a d} & \text {for}\: m = -1 \\\frac {m^{2} \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{3}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {m^{2} \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {3 m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{3}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{2}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} - \frac {2 m \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin {\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {2 \left (a \sin {\left (c + d x \right )} + a\right )^{m} \sin ^{3}{\left (c + d x \right )}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} + \frac {2 \left (a \sin {\left (c + d x \right )} + a\right )^{m}}{d m^{3} + 6 d m^{2} + 11 d m + 6 d} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*(a*sin(c) + a)**m*sin(c)**2*cos(c), Eq(d, 0)), (2*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) + 4*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) + 2*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) + 4*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) + 3/(2*a*
*3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Eq(m, -3)), (-2*log(sin(c + d*x) + 1)*sin(c + d*x)/(
a**2*d*sin(c + d*x) + a**2*d) - 2*log(sin(c + d*x) + 1)/(a**2*d*sin(c + d*x) + a**2*d) + sin(c + d*x)**2/(a**2
*d*sin(c + d*x) + a**2*d) - 2/(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)**2/(2*a*d) - sin(c + d*x)/(a*d), Eq(m, -1)), (m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**3 + 6
*d*m**2 + 11*d*m + 6*d) + m**2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**2/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) + 3*
m*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) + m*(a*sin(c + d*x) + a)**m*sin(c
+ d*x)**2/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) - 2*m*(a*sin(c + d*x) + a)**m*sin(c + d*x)/(d*m**3 + 6*d*m**2 +
11*d*m + 6*d) + 2*(a*sin(c + d*x) + a)**m*sin(c + d*x)**3/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d) + 2*(a*sin(c + d*
x) + a)**m/(d*m**3 + 6*d*m**2 + 11*d*m + 6*d), True))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} a^{m} \sin \left (d x + c\right )^{3} + {\left (m^{2} + m\right )} a^{m} \sin \left (d x + c\right )^{2} - 2 \, a^{m} m \sin \left (d x + c\right ) + 2 \, a^{m}\right )} {\left (\sin \left (d x + c\right ) + 1\right )}^{m}}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (80) = 160\).

Time = 0.37 (sec) , antiderivative size = 287, normalized size of antiderivative = 3.59 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m^{2} - 2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m^{2} + {\left (a \sin \left (d x + c\right ) + a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m^{2} + 3 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} m - 8 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a m + 5 \, {\left (a \sin \left (d x + c\right ) + a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2} m + 2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} - 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a + 6 \, {\left (a \sin \left (d x + c\right ) + a\right )} {\left (a \sin \left (d x + c\right ) + a\right )}^{m} a^{2}}{{\left (a^{2} m^{3} + 6 \, a^{2} m^{2} + 11 \, a^{2} m + 6 \, a^{2}\right )} a d} \]

[In]

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

[Out]

((a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*m^2 - 2*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a*m^2 + (
a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^2*m^2 + 3*(a*sin(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m*m - 8*(a*
sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a*m + 5*(a*sin(d*x + c) + a)*(a*sin(d*x + c) + a)^m*a^2*m + 2*(a*si
n(d*x + c) + a)^3*(a*sin(d*x + c) + a)^m - 6*(a*sin(d*x + c) + a)^2*(a*sin(d*x + c) + a)^m*a + 6*(a*sin(d*x +
c) + a)*(a*sin(d*x + c) + a)^m*a^2)/((a^2*m^3 + 6*a^2*m^2 + 11*a^2*m + 6*a^2)*a*d)

Mupad [B] (verification not implemented)

Time = 10.71 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.72 \[ \int \cos (c+d x) \sin ^2(c+d x) (a+a \sin (c+d x))^m \, dx=\frac {{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (2\,m+6\,\sin \left (c+d\,x\right )-2\,\sin \left (3\,c+3\,d\,x\right )+m\,\sin \left (c+d\,x\right )+2\,m\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )-3\,m\,\sin \left (3\,c+3\,d\,x\right )+3\,m^2\,\sin \left (c+d\,x\right )+2\,m^2\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )+2\,m^2-m^2\,\sin \left (3\,c+3\,d\,x\right )+8\right )}{4\,d\,\left (m^3+6\,m^2+11\,m+6\right )} \]

[In]

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

[Out]

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