\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx\) [77]

Optimal result

Integrand size = 36, antiderivative size = 76 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{2}+m,\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^{1+m}}{a c f (3+2 m) \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 50.90 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\cos (e+f x) (a (1+\sin (e+f x)))^m \left (-6-4 m+(6+4 m) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right )+(1+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{2}+m,\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (1+\sin (e+f x))\right )}{2 c f (1+2 m) (3+2 m) \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

(Cos[e + f*x]*(a*(1 + Sin[e + f*x]))^m*(-6 - 4*m + (6 + 4*m)*Hypergeometri 
c2F1[1, 1/2 + m, 3/2 + m, (1 + Sin[e + f*x])/2] + (1 + 2*m)*Hypergeometric 
2F1[1, 3/2 + m, 5/2 + m, (1 + Sin[e + f*x])/2]*(1 + Sin[e + f*x])))/(2*c*f 
*(1 + 2*m)*(3 + 2*m)*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 3320, 3042, 3224, 3042, 3146, 78}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3224

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 78

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 78

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

rule 3042

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

rule 3146

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x, algori 
thm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x, algori 
thm="maxima")
 

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^m/(c-c*sin(f*x+e))^(3/2),x, algori 
thm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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