\(\int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx\) [157]

Optimal result

Integrand size = 38, antiderivative size = 93 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {2^{\frac {9}{4}+m} g^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4}-m,\frac {1}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {5}{4}-m} (a+a \sin (e+f x))^{2+m}}{3 a^2 c^2 f (g \cos (e+f x))^{3/2}} \] Output:

1/3*2^(9/4+m)*g^3*hypergeom([-3/4, -1/4-m],[1/4],1/2-1/2*sin(f*x+e))*(1+si 
n(f*x+e))^(-5/4-m)*(a+a*sin(f*x+e))^(2+m)/a^2/c^2/f/(g*cos(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=-\frac {2^{\frac {9}{4}+m} g \sqrt {g \cos (e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4}-m,\frac {1}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a (1+\sin (e+f x)))^m}{3 c^2 f (-1+\sin (e+f x))} \] Input:

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

Output:

-1/3*(2^(9/4 + m)*g*Sqrt[g*Cos[e + f*x]]*Hypergeometric2F1[-3/4, -1/4 - m, 
 1/4, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/4 - m)*(a*(1 + Sin[e + 
f*x]))^m)/(c^2*f*(-1 + Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3319, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3319

\(\displaystyle \frac {g^4 \int \frac {(\sin (e+f x) a+a)^{m+2}}{(g \cos (e+f x))^{5/2}}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {g^4 \int \frac {(\sin (e+f x) a+a)^{m+2}}{(g \cos (e+f x))^{5/2}}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3168

\(\displaystyle \frac {g^3 (a-a \sin (e+f x))^{3/4} (a \sin (e+f x)+a)^{3/4} \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{4}}}{(a-a \sin (e+f x))^{7/4}}d\sin (e+f x)}{c^2 f (g \cos (e+f x))^{3/2}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {g^3 2^{m+\frac {1}{4}} (a-a \sin (e+f x))^{3/4} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m+1} \int \frac {\left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m+\frac {1}{4}}}{(a-a \sin (e+f x))^{7/4}}d\sin (e+f x)}{c^2 f (g \cos (e+f x))^{3/2}}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {g^3 2^{m+\frac {9}{4}} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-m-\frac {1}{4},\frac {1}{4},\frac {1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f (g \cos (e+f x))^{3/2}}\)

Input:

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

Output:

(2^(9/4 + m)*g^3*Hypergeometric2F1[-3/4, -1/4 - m, 1/4, (1 - Sin[e + f*x]) 
/2]*(1 + Sin[e + f*x])^(-1/4 - m)*(a + a*Sin[e + f*x])^(1 + m))/(3*a*c^2*f 
*(g*Cos[e + f*x])^(3/2))
 

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 3319

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\text {too large to display} \] Input:

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

Output:

(sqrt(g)*g*( - 2*(sin(e + f*x)*a + a)**m*sqrt(cos(e + f*x)) + 4*int(((sin( 
e + f*x)*a + a)**m*sqrt(cos(e + f*x))*cos(e + f*x)*sin(e + f*x))/(sin(e + 
f*x)**3*m + sin(e + f*x)**3 - sin(e + f*x)**2*m - sin(e + f*x)**2 - sin(e 
+ f*x)*m - sin(e + f*x) + m + 1),x)*sin(e + f*x)*f*m**2 + 4*int(((sin(e + 
f*x)*a + a)**m*sqrt(cos(e + f*x))*cos(e + f*x)*sin(e + f*x))/(sin(e + f*x) 
**3*m + sin(e + f*x)**3 - sin(e + f*x)**2*m - sin(e + f*x)**2 - sin(e + f* 
x)*m - sin(e + f*x) + m + 1),x)*sin(e + f*x)*f*m - 4*int(((sin(e + f*x)*a 
+ a)**m*sqrt(cos(e + f*x))*cos(e + f*x)*sin(e + f*x))/(sin(e + f*x)**3*m + 
 sin(e + f*x)**3 - sin(e + f*x)**2*m - sin(e + f*x)**2 - sin(e + f*x)*m - 
sin(e + f*x) + m + 1),x)*f*m**2 - 4*int(((sin(e + f*x)*a + a)**m*sqrt(cos( 
e + f*x))*cos(e + f*x)*sin(e + f*x))/(sin(e + f*x)**3*m + sin(e + f*x)**3 
- sin(e + f*x)**2*m - sin(e + f*x)**2 - sin(e + f*x)*m - sin(e + f*x) + m 
+ 1),x)*f*m - int(((sin(e + f*x)*a + a)**m*sqrt(cos(e + f*x))*sin(e + f*x) 
**3)/(cos(e + f*x)*sin(e + f*x)**3*m + cos(e + f*x)*sin(e + f*x)**3 - cos( 
e + f*x)*sin(e + f*x)**2*m - cos(e + f*x)*sin(e + f*x)**2 - cos(e + f*x)*s 
in(e + f*x)*m - cos(e + f*x)*sin(e + f*x) + cos(e + f*x)*m + cos(e + f*x)) 
,x)*sin(e + f*x)*f*m - int(((sin(e + f*x)*a + a)**m*sqrt(cos(e + f*x))*sin 
(e + f*x)**3)/(cos(e + f*x)*sin(e + f*x)**3*m + cos(e + f*x)*sin(e + f*x)* 
*3 - cos(e + f*x)*sin(e + f*x)**2*m - cos(e + f*x)*sin(e + f*x)**2 - cos(e 
 + f*x)*sin(e + f*x)*m - cos(e + f*x)*sin(e + f*x) + cos(e + f*x)*m + c...