\(\int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx\) [365]

Optimal result

Integrand size = 27, antiderivative size = 94 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=-\frac {2 a^2 (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-2+m}}{d e \left (6-5 m+m^2\right )}-\frac {a (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (3-m)} \] Output:

-2*a^2*(e*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^(-2+m)/d/e/(m^2-5*m+6)-a*(e 
*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^(-1+m)/d/e/(3-m)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.77 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {e^3 \cos ^4(c+d x) (e \cos (c+d x))^{-2 m} (a (1+\sin (c+d x)))^m (-4+m+(-2+m) \sin (c+d x))}{d (-3+m) (-2+m) (1+\sin (c+d x))^2} \] Input:

Integrate[(e*Cos[c + d*x])^(3 - 2*m)*(a + a*Sin[c + d*x])^m,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 3153, 3042, 3152}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (a \sin (c+d x)+a)^m (e \cos (c+d x))^{3-2 m} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (c+d x)+a)^m (e \cos (c+d x))^{3-2 m}dx\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {2 a \int (e \cos (c+d x))^{3-2 m} (\sin (c+d x) a+a)^{m-1}dx}{3-m}-\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \int (e \cos (c+d x))^{3-2 m} (\sin (c+d x) a+a)^{m-1}dx}{3-m}-\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)}\)

\(\Big \downarrow \) 3152

\(\displaystyle -\frac {2 a^2 (a \sin (c+d x)+a)^{m-2} (e \cos (c+d x))^{4-2 m}}{d e (2-m) (3-m)}-\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)}\)

Input:

Int[(e*Cos[c + d*x])^(3 - 2*m)*(a + a*Sin[c + d*x])^m,x]
 

Output:

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

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3152

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x 
])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 
 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
 

rule 3153

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   Int[(g* 
Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && 
NeQ[m + p, 0]
 

Maple [F]

\[\int \left (e \cos \left (d x +c \right )\right )^{3-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}d x\]

Input:

int((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x)
 

Output:

int((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.81 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {{\left ({\left (m - 2\right )} \cos \left (d x + c\right )^{2} + {\left (m - 4\right )} \cos \left (d x + c\right ) + {\left ({\left (m - 2\right )} \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - 2\right )} \left (e \cos \left (d x + c\right )\right )^{-2 \, m + 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{2 \, d m^{2} - {\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right )^{2} - 10 \, d m + {\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right ) + {\left (2 \, d m^{2} - 10 \, d m + {\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right ) + 12 \, d\right )} \sin \left (d x + c\right ) + 12 \, d} \] Input:

integrate((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")
 

Output:

((m - 2)*cos(d*x + c)^2 + (m - 4)*cos(d*x + c) + ((m - 2)*cos(d*x + c) + 2 
)*sin(d*x + c) - 2)*(e*cos(d*x + c))^(-2*m + 3)*(a*sin(d*x + c) + a)^m/(2* 
d*m^2 - (d*m^2 - 5*d*m + 6*d)*cos(d*x + c)^2 - 10*d*m + (d*m^2 - 5*d*m + 6 
*d)*cos(d*x + c) + (2*d*m^2 - 10*d*m + (d*m^2 - 5*d*m + 6*d)*cos(d*x + c) 
+ 12*d)*sin(d*x + c) + 12*d)
 

Sympy [F]

\[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{3 - 2 m}\, dx \] Input:

integrate((e*cos(d*x+c))**(3-2*m)*(a+a*sin(d*x+c))**m,x)
 

Output:

Integral((a*(sin(c + d*x) + 1))**m*(e*cos(c + d*x))**(3 - 2*m), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (91) = 182\).

Time = 0.14 (sec) , antiderivative size = 351, normalized size of antiderivative = 3.73 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {{\left (a^{m} e^{3} {\left (m - 4\right )} - \frac {2 \, a^{m} e^{3} {\left (m - 6\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {a^{m} e^{3} {\left (m + 12\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{m} e^{3} {\left (m + 2\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {a^{m} e^{3} {\left (m + 12\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2 \, a^{m} e^{3} {\left (m - 6\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{m} e^{3} {\left (m - 4\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} e^{\left (-2 \, m \log \left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + m \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left ({\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} + \frac {3 \, {\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, {\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d} \] Input:

integrate((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")
 

Output:

(a^m*e^3*(m - 4) - 2*a^m*e^3*(m - 6)*sin(d*x + c)/(cos(d*x + c) + 1) - a^m 
*e^3*(m + 12)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 4*a^m*e^3*(m + 2)*sin( 
d*x + c)^3/(cos(d*x + c) + 1)^3 - a^m*e^3*(m + 12)*sin(d*x + c)^4/(cos(d*x 
 + c) + 1)^4 - 2*a^m*e^3*(m - 6)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a^m 
*e^3*(m - 4)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*e^(-2*m*log(-sin(d*x + c 
)/(cos(d*x + c) + 1) + 1) + m*log(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1) 
)/(((m^2 - 5*m + 6)*e^(2*m) + 3*(m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^2/(co 
s(d*x + c) + 1)^2 + 3*(m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^4/(cos(d*x + c) 
 + 1)^4 + (m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 16008 vs. \(2 (91) = 182\).

Time = 10.11 (sec) , antiderivative size = 16008, normalized size of antiderivative = 170.30 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\text {Too large to display} \] Input:

integrate((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")
 

Output:

2*(m*e^(m*log(2) - 2*m*log(4*abs(e)*abs(tan(1/8*pi - 1/4*d*x - 1/4*c))/(ta 
n(1/8*pi - 1/4*d*x - 1/4*c)^2 + 1)) + m*log(abs(a)) - 3*log(2) + 3*log(4*a 
bs(e)*abs(tan(1/8*pi - 1/4*d*x - 1/4*c))/(tan(1/8*pi - 1/4*d*x - 1/4*c)^2 
+ 1)))*tan(-3/8*pi + 1/2*pi*m*sgn(e*tan(1/2*d*x + 1/2*c)^2 - 2*e*tan(1/2*d 
*x + 1/2*c) + e)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(e) - 3/4*pi*sgn(e*tan 
(1/2*d*x + 1/2*c)^2 - 2*e*tan(1/2*d*x + 1/2*c) + e)*sgn(tan(1/2*d*x + 1/2* 
c)^2 - 1)*sgn(e) + pi*m*floor(1/2*d*x/pi + 1/2*c/pi - floor(1/2*d*x/pi + 1 
/2*c/pi + 1/2) + 1/4) + pi*m*floor(1/2*d*x/pi + 1/2*c/pi + 1/2) - pi*m*flo 
or(-1/4*sgn(a) + 1/2) + 1/2*pi*m*sgn(e*tan(1/2*d*x + 1/2*c)^2 - 2*e*tan(1/ 
2*d*x + 1/2*c) + e) - 1/4*pi*m*sgn(a) + 1/4*pi*m - 3/4*d*x - 3/2*pi*floor( 
1/2*d*x/pi + 1/2*c/pi - floor(1/2*d*x/pi + 1/2*c/pi + 1/2) + 1/4) - 3/2*pi 
*floor(1/2*d*x/pi + 1/2*c/pi + 1/2) - 3/4*pi*sgn(e*tan(1/2*d*x + 1/2*c)^2 
- 2*e*tan(1/2*d*x + 1/2*c) + e) - 3/4*c)^2*tan(1/2*d*x + 1/2*c)^6 + 4*m*e^ 
(m*log(2) - 2*m*log(4*abs(e)*abs(tan(1/8*pi - 1/4*d*x - 1/4*c))/(tan(1/8*p 
i - 1/4*d*x - 1/4*c)^2 + 1)) + m*log(abs(a)) - 3*log(2) + 3*log(4*abs(e)*a 
bs(tan(1/8*pi - 1/4*d*x - 1/4*c))/(tan(1/8*pi - 1/4*d*x - 1/4*c)^2 + 1)))* 
tan(-3/8*pi + 1/2*pi*m*sgn(e*tan(1/2*d*x + 1/2*c)^2 - 2*e*tan(1/2*d*x + 1/ 
2*c) + e)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(e) - 3/4*pi*sgn(e*tan(1/2*d* 
x + 1/2*c)^2 - 2*e*tan(1/2*d*x + 1/2*c) + e)*sgn(tan(1/2*d*x + 1/2*c)^2 - 
1)*sgn(e) + pi*m*floor(1/2*d*x/pi + 1/2*c/pi - floor(1/2*d*x/pi + 1/2*c...
 

Mupad [B] (verification not implemented)

Time = 18.34 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.56 \[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {e^3\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (14\,m-24\,\sin \left (c+d\,x\right )-36\,\sin \left (3\,c+3\,d\,x\right )-12\,\sin \left (5\,c+5\,d\,x\right )+24\,{\sin \left (2\,c+2\,d\,x\right )}^2-4\,{\sin \left (3\,c+3\,d\,x\right )}^2+8\,m\,\sin \left (c+d\,x\right )-17\,m\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )+12\,m\,\sin \left (3\,c+3\,d\,x\right )+4\,m\,\sin \left (5\,c+5\,d\,x\right )-2\,m\,\left (2\,{\sin \left (2\,c+2\,d\,x\right )}^2-1\right )+m\,\left (2\,{\sin \left (3\,c+3\,d\,x\right )}^2-1\right )+132\,{\sin \left (c+d\,x\right )}^2-128\right )}{8\,d\,{\left (-e\,\left (2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\right )}^{2\,m}\,\left (m^2-5\,m+6\right )\,\left (12\,{\sin \left (c+d\,x\right )}^2+15\,\sin \left (c+d\,x\right )-\sin \left (3\,c+3\,d\,x\right )+4\right )} \] Input:

int((e*cos(c + d*x))^(3 - 2*m)*(a + a*sin(c + d*x))^m,x)
 

Output:

(e^3*(a*(sin(c + d*x) + 1))^m*(14*m - 24*sin(c + d*x) - 36*sin(3*c + 3*d*x 
) - 12*sin(5*c + 5*d*x) + 24*sin(2*c + 2*d*x)^2 - 4*sin(3*c + 3*d*x)^2 + 8 
*m*sin(c + d*x) - 17*m*(2*sin(c + d*x)^2 - 1) + 12*m*sin(3*c + 3*d*x) + 4* 
m*sin(5*c + 5*d*x) - 2*m*(2*sin(2*c + 2*d*x)^2 - 1) + m*(2*sin(3*c + 3*d*x 
)^2 - 1) + 132*sin(c + d*x)^2 - 128))/(8*d*(-e*(2*sin(c/2 + (d*x)/2)^2 - 1 
))^(2*m)*(m^2 - 5*m + 6)*(15*sin(c + d*x) - sin(3*c + 3*d*x) + 12*sin(c + 
d*x)^2 + 4))
 

Reduce [F]

\[ \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx=\frac {\left (\int \frac {\left (\sin \left (d x +c \right ) a +a \right )^{m} \cos \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{2 m}}d x \right ) e^{3}}{e^{2 m}} \] Input:

int((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x)
 

Output:

(int(((sin(c + d*x)*a + a)**m*cos(c + d*x)**3)/cos(c + d*x)**(2*m),x)*e**3 
)/e**(2*m)