\(\int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx\) [163]

Optimal result

Integrand size = 40, antiderivative size = 111 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {2^{\frac {9}{4}+m} g^2 \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},-\frac {1}{4}-m,\frac {3}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {5}{4}-m} (a+a \sin (e+f x))^{1+m}}{a c f \sqrt {g \cos (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 3332, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3332

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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 3332

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^ 
IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f* 
x])^FracPart[m]/(g^(2*IntPart[m])*(g*Cos[e + f*x])^(2*FracPart[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, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] 
&& (FractionQ[m] ||  !FractionQ[n])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(g)*sqrt(c)*g*(2*(sin(e + f*x)*a + a)**m*sqrt( - sin(e + f*x) + 1)*sq 
rt(cos(e + f*x)) + 16*int(((sin(e + f*x)*a + a)**m*sqrt( - sin(e + f*x) + 
1)*sqrt(cos(e + f*x))*cos(e + f*x))/(4*sin(e + f*x)**3*m**2 - sin(e + f*x) 
**3 - 4*sin(e + f*x)**2*m**2 + sin(e + f*x)**2 - 4*sin(e + f*x)*m**2 + sin 
(e + f*x) + 4*m**2 - 1),x)*sin(e + f*x)*f*m**3 - 4*int(((sin(e + f*x)*a + 
a)**m*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*cos(e + f*x))/(4*sin(e 
+ f*x)**3*m**2 - sin(e + f*x)**3 - 4*sin(e + f*x)**2*m**2 + sin(e + f*x)** 
2 - 4*sin(e + f*x)*m**2 + sin(e + f*x) + 4*m**2 - 1),x)*sin(e + f*x)*f*m - 
 16*int(((sin(e + f*x)*a + a)**m*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f* 
x))*cos(e + f*x))/(4*sin(e + f*x)**3*m**2 - sin(e + f*x)**3 - 4*sin(e + f* 
x)**2*m**2 + sin(e + f*x)**2 - 4*sin(e + f*x)*m**2 + sin(e + f*x) + 4*m**2 
 - 1),x)*f*m**3 + 4*int(((sin(e + f*x)*a + a)**m*sqrt( - sin(e + f*x) + 1) 
*sqrt(cos(e + f*x))*cos(e + f*x))/(4*sin(e + f*x)**3*m**2 - sin(e + f*x)** 
3 - 4*sin(e + f*x)**2*m**2 + sin(e + f*x)**2 - 4*sin(e + f*x)*m**2 + sin(e 
 + f*x) + 4*m**2 - 1),x)*f*m + 8*int(((sin(e + f*x)*a + a)**m*sqrt( - sin( 
e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**3)/(4*cos(e + f*x)*sin(e + 
f*x)**3*m**2 - cos(e + f*x)*sin(e + f*x)**3 - 4*cos(e + f*x)*sin(e + f*x)* 
*2*m**2 + cos(e + f*x)*sin(e + f*x)**2 - 4*cos(e + f*x)*sin(e + f*x)*m**2 
+ cos(e + f*x)*sin(e + f*x) + 4*cos(e + f*x)*m**2 - cos(e + f*x)),x)*sin(e 
 + f*x)*f*m**2 - 4*int(((sin(e + f*x)*a + a)**m*sqrt( - sin(e + f*x) + ...