\(\int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx\) [414]

Optimal result

Integrand size = 26, antiderivative size = 86 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=-\frac {2^{\frac {1}{2}+m} a^2 c^2 \cos ^5(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^{-2+m}}{5 f} \] Output:

-1/5*2^(1/2+m)*a^2*c^2*cos(f*x+e)^5*hypergeom([5/2, 1/2-m],[7/2],1/2-1/2*s 
in(f*x+e))*(1+sin(f*x+e))^(-1/2-m)*(a+a*sin(f*x+e))^(-2+m)/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 8.79 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.90 \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\frac {(1+i) c^2 (a (1+\sin (e+f x)))^m \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left ((2-2 i) (5+2 m) \sec ^2\left (\frac {1}{2} (e+f x)\right ) (-3-m+m \sin (e+f x)) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \left (-i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^3+3 \operatorname {Hypergeometric2F1}\left (3+m,5+2 m,2 (3+m),\frac {(1-i) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{-i+\tan \left (\frac {1}{2} (e+f x)\right )}\right ) (-i \cos (e+f x)-\sin (e+f x))^m \left (i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )^4\right )}{f (1+2 m) (3+2 m) (5+2 m) \left (-i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (i+\tan \left (\frac {1}{2} (e+f x)\right )\right )^2} \] Input:

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

Output:

((1 + I)*c^2*(a*(1 + Sin[e + f*x]))^m*(1 + Tan[(e + f*x)/2])*((2 - 2*I)*(5 
 + 2*m)*Sec[(e + f*x)/2]^2*(-3 - m + m*Sin[e + f*x])*(-1 + Tan[(e + f*x)/2 
])*(-I + Tan[(e + f*x)/2])^3 + 3*Hypergeometric2F1[3 + m, 5 + 2*m, 2*(3 + 
m), ((1 - I)*(1 + Tan[(e + f*x)/2]))/(-I + Tan[(e + f*x)/2])]*((-I)*Cos[e 
+ f*x] - Sin[e + f*x])^m*(I + Tan[(e + f*x)/2])^2*(1 + Tan[(e + f*x)/2])^4 
))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(-I + Tan[(e + f*x)/2])^5*(I + Tan[(e 
+ f*x)/2])^2)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3215, 3042, 3168, 80, 79}

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.

\(\displaystyle \int (c-c \sin (e+f x))^2 (a \sin (e+f x)+a)^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c-c \sin (e+f x))^2 (a \sin (e+f x)+a)^mdx\)

\(\Big \downarrow \) 3215

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

\(\displaystyle \frac {a^4 c^2 \cos ^5(e+f x) \int (a-a \sin (e+f x))^{3/2} (\sin (e+f x) a+a)^{m-\frac {1}{2}}d\sin (e+f x)}{f (a-a \sin (e+f x))^{5/2} (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {a^4 c^2 2^{m-\frac {1}{2}} \cos ^5(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-3} \int \left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m-\frac {1}{2}} (a-a \sin (e+f x))^{3/2}d\sin (e+f x)}{f (a-a \sin (e+f x))^{5/2}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {a^3 c^2 2^{m+\frac {1}{2}} \cos ^5(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-3} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1}{2}-m,\frac {7}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{5 f}\)

Input:

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

Output:

-1/5*(2^(1/2 + m)*a^3*c^2*Cos[e + f*x]^5*Hypergeometric2F1[5/2, 1/2 - m, 7 
/2, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(1/2 - m)*(a + a*Sin[e + f*x] 
)^(-3 + m))/f
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 3042

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

rule 3168

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

rule 3215

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + 
 d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((Lt 
Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-(c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2)*(a*sin(f*x + e 
) + a)^m, x)
 

Sympy [F]

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

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

Output:

c**2*(Integral(-2*(a*sin(e + f*x) + a)**m*sin(e + f*x), x) + Integral((a*s 
in(e + f*x) + a)**m*sin(e + f*x)**2, x) + Integral((a*sin(e + f*x) + a)**m 
, x))
 

Maxima [F]

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

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

Output:

integrate((c*sin(f*x + e) - c)^2*(a*sin(f*x + e) + a)^m, x)
 

Giac [F]

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

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

Output:

integrate((c*sin(f*x + e) - c)^2*(a*sin(f*x + e) + a)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

c**2*(int((sin(e + f*x)*a + a)**m,x) + int((sin(e + f*x)*a + a)**m*sin(e + 
 f*x)**2,x) - 2*int((sin(e + f*x)*a + a)**m*sin(e + f*x),x))