\(\int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx\) [922]

Optimal result

Integrand size = 31, antiderivative size = 96 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {(c-d)^2 (a+a \sin (e+f x))^{1+m}}{a f (1+m)}+\frac {2 (c-d) d (a+a \sin (e+f x))^{2+m}}{a^2 f (2+m)}+\frac {d^2 (a+a \sin (e+f x))^{3+m}}{a^3 f (3+m)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2912, 45} \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {d^2 (a \sin (e+f x)+a)^{m+3}}{a^3 f (m+3)}+\frac {2 d (c-d) (a \sin (e+f x)+a)^{m+2}}{a^2 f (m+2)}+\frac {(c-d)^2 (a \sin (e+f x)+a)^{m+1}}{a f (m+1)} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.86 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {(a (1+\sin (e+f x)))^{1+m} \left (\frac {a^2 (c-d)^2}{1+m}+\frac {2 a^2 (c-d) d (1+\sin (e+f x))}{2+m}+\frac {d^2 (a+a \sin (e+f x))^2}{3+m}\right )}{a^3 f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.78 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.60

method	result	size
parallelrisch	\(-\frac {\left (4 \left (1+m \right ) \left (\frac {d m}{2}+c \left (3+m \right )\right ) d \cos \left (2 f x +2 e \right )+d^{2} \left (2+m \right ) \left (1+m \right ) \sin \left (3 f x +3 e \right )+\left (\left (-3 m^{2}-m -6\right ) d^{2}-8 c m \left (3+m \right ) d -4 c^{2} \left (3+m \right ) \left (2+m \right )\right ) \sin \left (f x +e \right )+\left (-2 m^{2}-2 m -8\right ) d^{2}-4 c \left (3+m \right ) \left (-1+m \right ) d -4 c^{2} \left (3+m \right ) \left (2+m \right )\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{m}}{4 \left (m^{3}+6 m^{2}+11 m +6\right ) f}\)	\(154\)
derivativedivides	\(\frac {\left (c^{2} m^{2}+5 c^{2} m -2 c d m +6 c^{2}-6 c d +2 d^{2}\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {d^{2} \left (\sin ^{3}\left (f x +e \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (3+m \right )}+\frac {\left (c^{2} m^{2}+2 c d \,m^{2}+5 c^{2} m +6 c d m -2 d^{2} m +6 c^{2}\right ) \sin \left (f x +e \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {\left (2 c m +d m +6 c \right ) d \left (\sin ^{2}\left (f x +e \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{2}+5 m +6\right )}\)	\(226\)
default	\(\frac {\left (c^{2} m^{2}+5 c^{2} m -2 c d m +6 c^{2}-6 c d +2 d^{2}\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {d^{2} \left (\sin ^{3}\left (f x +e \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (3+m \right )}+\frac {\left (c^{2} m^{2}+2 c d \,m^{2}+5 c^{2} m +6 c d m -2 d^{2} m +6 c^{2}\right ) \sin \left (f x +e \right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{3}+6 m^{2}+11 m +6\right )}+\frac {\left (2 c m +d m +6 c \right ) d \left (\sin ^{2}\left (f x +e \right )\right ) {\mathrm e}^{m \ln \left (a +a \sin \left (f x +e \right )\right )}}{f \left (m^{2}+5 m +6\right )}\)	\(226\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.97 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {{\left ({\left (c^{2} + 2 \, c d + d^{2}\right )} m^{2} - {\left ({\left (2 \, c d + d^{2}\right )} m^{2} + 6 \, c d + {\left (8 \, c d + d^{2}\right )} m\right )} \cos \left (f x + e\right )^{2} + 6 \, c^{2} + 2 \, d^{2} + {\left (5 \, c^{2} + 6 \, c d + d^{2}\right )} m + {\left ({\left (c^{2} + 2 \, c d + d^{2}\right )} m^{2} - {\left (d^{2} m^{2} + 3 \, d^{2} m + 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, c^{2} + 2 \, d^{2} + {\left (5 \, c^{2} + 6 \, c d + d^{2}\right )} m\right )} \sin \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{f m^{3} + 6 \, f m^{2} + 11 \, f m + 6 \, f} \]

[In]

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

[Out]

((c^2 + 2*c*d + d^2)*m^2 - ((2*c*d + d^2)*m^2 + 6*c*d + (8*c*d + d^2)*m)*cos(f*x + e)^2 + 6*c^2 + 2*d^2 + (5*c
^2 + 6*c*d + d^2)*m + ((c^2 + 2*c*d + d^2)*m^2 - (d^2*m^2 + 3*d^2*m + 2*d^2)*cos(f*x + e)^2 + 6*c^2 + 2*d^2 +
(5*c^2 + 6*c*d + d^2)*m)*sin(f*x + e))*(a*sin(f*x + e) + a)^m/(f*m^3 + 6*f*m^2 + 11*f*m + 6*f)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1622 vs. \(2 (80) = 160\).

Time = 2.55 (sec) , antiderivative size = 1622, normalized size of antiderivative = 16.90 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\text {Too large to display} \]

[In]

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

[Out]

Piecewise((x*(c + d*sin(e))**2*(a*sin(e) + a)**m*cos(e), Eq(f, 0)), (-c**2/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*
f*sin(e + f*x) + 2*a**3*f) - 4*c*d*sin(e + f*x)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f)
- 2*c*d/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) + 2*d**2*log(sin(e + f*x) + 1)*sin(e + f
*x)**2/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) + 4*d**2*log(sin(e + f*x) + 1)*sin(e + f*
x)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) + 2*d**2*log(sin(e + f*x) + 1)/(2*a**3*f*sin(
e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f) + 4*d**2*sin(e + f*x)/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin
(e + f*x) + 2*a**3*f) + 3*d**2/(2*a**3*f*sin(e + f*x)**2 + 4*a**3*f*sin(e + f*x) + 2*a**3*f), Eq(m, -3)), (-c*
*2/(a**2*f*sin(e + f*x) + a**2*f) + 2*c*d*log(sin(e + f*x) + 1)*sin(e + f*x)/(a**2*f*sin(e + f*x) + a**2*f) +
2*c*d*log(sin(e + f*x) + 1)/(a**2*f*sin(e + f*x) + a**2*f) + 2*c*d/(a**2*f*sin(e + f*x) + a**2*f) - 2*d**2*log
(sin(e + f*x) + 1)*sin(e + f*x)/(a**2*f*sin(e + f*x) + a**2*f) - 2*d**2*log(sin(e + f*x) + 1)/(a**2*f*sin(e +
f*x) + a**2*f) + d**2*sin(e + f*x)**2/(a**2*f*sin(e + f*x) + a**2*f) - 2*d**2/(a**2*f*sin(e + f*x) + a**2*f),
Eq(m, -2)), (c**2*log(sin(e + f*x) + 1)/(a*f) - 2*c*d*log(sin(e + f*x) + 1)/(a*f) + 2*c*d*sin(e + f*x)/(a*f) +
d**2*log(sin(e + f*x) + 1)/(a*f) + d**2*sin(e + f*x)**2/(2*a*f) - d**2*sin(e + f*x)/(a*f), Eq(m, -1)), (c**2*
m**2*(a*sin(e + f*x) + a)**m*sin(e + f*x)/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + c**2*m**2*(a*sin(e + f*x) + a)*
*m/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + 5*c**2*m*(a*sin(e + f*x) + a)**m*sin(e + f*x)/(f*m**3 + 6*f*m**2 + 11*
f*m + 6*f) + 5*c**2*m*(a*sin(e + f*x) + a)**m/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + 6*c**2*(a*sin(e + f*x) + a)
**m*sin(e + f*x)/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + 6*c**2*(a*sin(e + f*x) + a)**m/(f*m**3 + 6*f*m**2 + 11*f
*m + 6*f) + 2*c*d*m**2*(a*sin(e + f*x) + a)**m*sin(e + f*x)**2/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + 2*c*d*m**2
*(a*sin(e + f*x) + a)**m*sin(e + f*x)/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + 8*c*d*m*(a*sin(e + f*x) + a)**m*sin
(e + f*x)**2/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + 6*c*d*m*(a*sin(e + f*x) + a)**m*sin(e + f*x)/(f*m**3 + 6*f*m
**2 + 11*f*m + 6*f) - 2*c*d*m*(a*sin(e + f*x) + a)**m/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + 6*c*d*(a*sin(e + f*
x) + a)**m*sin(e + f*x)**2/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) - 6*c*d*(a*sin(e + f*x) + a)**m/(f*m**3 + 6*f*m*
*2 + 11*f*m + 6*f) + d**2*m**2*(a*sin(e + f*x) + a)**m*sin(e + f*x)**3/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + d*
*2*m**2*(a*sin(e + f*x) + a)**m*sin(e + f*x)**2/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + 3*d**2*m*(a*sin(e + f*x)
+ a)**m*sin(e + f*x)**3/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + d**2*m*(a*sin(e + f*x) + a)**m*sin(e + f*x)**2/(f
*m**3 + 6*f*m**2 + 11*f*m + 6*f) - 2*d**2*m*(a*sin(e + f*x) + a)**m*sin(e + f*x)/(f*m**3 + 6*f*m**2 + 11*f*m +
6*f) + 2*d**2*(a*sin(e + f*x) + a)**m*sin(e + f*x)**3/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f) + 2*d**2*(a*sin(e +
f*x) + a)**m/(f*m**3 + 6*f*m**2 + 11*f*m + 6*f), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.78 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {\frac {2 \, {\left (a^{m} {\left (m + 1\right )} \sin \left (f x + e\right )^{2} + a^{m} m \sin \left (f x + e\right ) - a^{m}\right )} c d {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{2} + 3 \, m + 2} + \frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} a^{m} \sin \left (f x + e\right )^{3} + {\left (m^{2} + m\right )} a^{m} \sin \left (f x + e\right )^{2} - 2 \, a^{m} m \sin \left (f x + e\right ) + 2 \, a^{m}\right )} d^{2} {\left (\sin \left (f x + e\right ) + 1\right )}^{m}}{m^{3} + 6 \, m^{2} + 11 \, m + 6} + \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m + 1} c^{2}}{a {\left (m + 1\right )}}}{f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 435 vs. \(2 (96) = 192\).

Time = 0.34 (sec) , antiderivative size = 435, normalized size of antiderivative = 4.53 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {\frac {{\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m^{2} - 2 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m^{2} + {\left (a \sin \left (f x + e\right ) + a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a^{2} m^{2} + 3 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m - 8 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m + 5 \, {\left (a \sin \left (f x + e\right ) + a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a^{2} m + 2 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{3} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} - 6 \, {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a + 6 \, {\left (a \sin \left (f x + e\right ) + a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a^{2}\right )} d^{2}}{a^{2} m^{3} + 6 \, a^{2} m^{2} + 11 \, a^{2} m + 6 \, a^{2}} + \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m + 1} c^{2}}{m + 1} + \frac {2 \, {\left ({\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} m - {\left (a \sin \left (f x + e\right ) + a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a m + {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} - 2 \, {\left (a \sin \left (f x + e\right ) + a\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} a\right )} c d}{{\left (m^{2} + 3 \, m + 2\right )} a}}{a f} \]

[In]

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

[Out]

(((a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*m^2 - 2*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a*m^2 +
(a*sin(f*x + e) + a)*(a*sin(f*x + e) + a)^m*a^2*m^2 + 3*(a*sin(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m*m - 8*(a
*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a*m + 5*(a*sin(f*x + e) + a)*(a*sin(f*x + e) + a)^m*a^2*m + 2*(a*s
in(f*x + e) + a)^3*(a*sin(f*x + e) + a)^m - 6*(a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*a + 6*(a*sin(f*x +
e) + a)*(a*sin(f*x + e) + a)^m*a^2)*d^2/(a^2*m^3 + 6*a^2*m^2 + 11*a^2*m + 6*a^2) + (a*sin(f*x + e) + a)^(m +
1)*c^2/(m + 1) + 2*((a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m*m - (a*sin(f*x + e) + a)*(a*sin(f*x + e) + a
)^m*a*m + (a*sin(f*x + e) + a)^2*(a*sin(f*x + e) + a)^m - 2*(a*sin(f*x + e) + a)*(a*sin(f*x + e) + a)^m*a)*c*d
/((m^2 + 3*m + 2)*a))/(a*f)

Mupad [B] (verification not implemented)

Time = 11.40 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.18 \[ \int \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx=\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (20\,c^2\,m-12\,c\,d+2\,d^2\,m+24\,c^2\,\sin \left (e+f\,x\right )+6\,d^2\,\sin \left (e+f\,x\right )+24\,c^2+8\,d^2+4\,c^2\,m^2+2\,d^2\,m^2-2\,d^2\,\sin \left (3\,e+3\,f\,x\right )+20\,c^2\,m\,\sin \left (e+f\,x\right )+d^2\,m\,\sin \left (e+f\,x\right )-2\,d^2\,m\,\cos \left (2\,e+2\,f\,x\right )+4\,c^2\,m^2\,\sin \left (e+f\,x\right )-3\,d^2\,m\,\sin \left (3\,e+3\,f\,x\right )+3\,d^2\,m^2\,\sin \left (e+f\,x\right )+8\,c\,d\,m-2\,d^2\,m^2\,\cos \left (2\,e+2\,f\,x\right )-d^2\,m^2\,\sin \left (3\,e+3\,f\,x\right )-12\,c\,d\,\cos \left (2\,e+2\,f\,x\right )+4\,c\,d\,m^2+24\,c\,d\,m\,\sin \left (e+f\,x\right )-16\,c\,d\,m\,\cos \left (2\,e+2\,f\,x\right )+8\,c\,d\,m^2\,\sin \left (e+f\,x\right )-4\,c\,d\,m^2\,\cos \left (2\,e+2\,f\,x\right )\right )}{4\,f\,\left (m^3+6\,m^2+11\,m+6\right )} \]

[In]

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

[Out]

((a*(sin(e + f*x) + 1))^m*(20*c^2*m - 12*c*d + 2*d^2*m + 24*c^2*sin(e + f*x) + 6*d^2*sin(e + f*x) + 24*c^2 + 8
*d^2 + 4*c^2*m^2 + 2*d^2*m^2 - 2*d^2*sin(3*e + 3*f*x) + 20*c^2*m*sin(e + f*x) + d^2*m*sin(e + f*x) - 2*d^2*m*c
os(2*e + 2*f*x) + 4*c^2*m^2*sin(e + f*x) - 3*d^2*m*sin(3*e + 3*f*x) + 3*d^2*m^2*sin(e + f*x) + 8*c*d*m - 2*d^2
*m^2*cos(2*e + 2*f*x) - d^2*m^2*sin(3*e + 3*f*x) - 12*c*d*cos(2*e + 2*f*x) + 4*c*d*m^2 + 24*c*d*m*sin(e + f*x)
- 16*c*d*m*cos(2*e + 2*f*x) + 8*c*d*m^2*sin(e + f*x) - 4*c*d*m^2*cos(2*e + 2*f*x)))/(4*f*(11*m + 6*m^2 + m^3
+ 6))