\(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx\) [84]

Optimal result

Integrand size = 38, antiderivative size = 117 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\frac {2^{-\frac {1}{2}-m} \cos ^3(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (3+2 m),\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2} (1+2 m)} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m}}{f (3+2 m)} \] Output:

2^(-1/2-m)*cos(f*x+e)^3*hypergeom([3/2+m, 3/2+m],[5/2+m],1/2+1/2*sin(f*x+e 
))*(1-sin(f*x+e))^(1/2+m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-2-m)/f/(3+ 
2*m)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 14.94 (sec) , antiderivative size = 453, normalized size of antiderivative = 3.87 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2-m} \, dx=\frac {2^{-1-m} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (a (1+\sin (e+f x)))^m (c-c \sin (e+f x))^{-2-m} \left (\frac {1-\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\frac {1}{1+\cos (e+f x)}}}\right )^{2 m} \left (\frac {1-\tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (e+f x)\right )}}\right )^{-2 m} \left (\frac {i \operatorname {Hypergeometric2F1}\left (1,-2 m,1-2 m,-\frac {i \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{1+\tan \left (\frac {1}{2} (e+f x)\right )}\right )}{m}-\frac {i \operatorname {Hypergeometric2F1}\left (1,-2 m,1-2 m,\frac {i \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{1+\tan \left (\frac {1}{2} (e+f x)\right )}\right )}{m}+2^{1+2 m} \left (\frac {2 \operatorname {Hypergeometric2F1}\left (-2 m,-2 (1+m),1-2 m,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{m}-\frac {4 \operatorname {Hypergeometric2F1}\left (-1-2 m,-2 (1+m),-2 m,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )}{(1+2 m) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}-\frac {2 \operatorname {Hypergeometric2F1}\left (-1-2 m,1-2 m,2-2 m,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{-1+2 m}-\frac {\operatorname {Hypergeometric2F1}\left (1-2 m,-2 m,2-2 m,\frac {1}{2} \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \left (-1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{-1+2 m}\right ) \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )^{-2 m}\right )}{f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.29, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3320, 3042, 3224, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-1}dx}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{m+1} (c-c \sin (e+f x))^{-m-1}dx}{a c}\)

\(\Big \downarrow \) 3224

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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 3224

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e 
+ f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPart[m]/Cos[e + f*x]^(2*FracP 
art[m]))   Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; F 
reeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] 
 && (FractionQ[m] ||  !FractionQ[n])
 

rule 3320

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(a^(p/ 
2)*c^(p/2))   Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(n + 
p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && 
EqQ[a^2 - b^2, 0] && IntegerQ[p/2]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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