\(\int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx\) [156]

Optimal result

Integrand size = 38, antiderivative size = 89 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=-\frac {2^{\frac {9}{4}+m} g \sqrt {g \cos (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-\frac {1}{4}-m,\frac {5}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {5}{4}-m} (a+a \sin (e+f x))^{1+m}}{a c f} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=-\frac {2^{\frac {9}{4}+m} g \sqrt {g \cos (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-\frac {1}{4}-m,\frac {5}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a (1+\sin (e+f x)))^m}{c f} \] Input:

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

Output:

-((2^(9/4 + m)*g*Sqrt[g*Cos[e + f*x]]*Hypergeometric2F1[1/4, -1/4 - m, 5/4 
, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/4 - m)*(a*(1 + Sin[e + f*x] 
))^m)/(c*f))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3319

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3168

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

\(\displaystyle -\frac {g 2^{m+\frac {9}{4}} \sqrt {g \cos (e+f x)} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-m-\frac {1}{4},\frac {5}{4},\frac {1}{2} (1-\sin (e+f x))\right )}{c f}\)

Input:

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

Output:

-((2^(9/4 + m)*g*Sqrt[g*Cos[e + f*x]]*Hypergeometric2F1[1/4, -1/4 - m, 5/4 
, (1 - Sin[e + f*x])/2]*(1 + Sin[e + f*x])^(-1/4 - m)*(a + a*Sin[e + f*x]) 
^m)/(c*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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{c-c \sin (e+f x)} \, dx=\frac {\sqrt {g}\, g \left (-2 \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {\cos \left (f x +e \right )}-4 \left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {\cos \left (f x +e \right )}\, \cos \left (f x +e \right )}{\sin \left (f x +e \right )^{2}-1}d x \right ) f m -\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {\cos \left (f x +e \right )}\, \sin \left (f x +e \right )^{3}}{\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}-\cos \left (f x +e \right )}d x \right ) f +\left (\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {\cos \left (f x +e \right )}\, \sin \left (f x +e \right )}{\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}-\cos \left (f x +e \right )}d x \right ) f \right )}{2 c f m} \] Input:

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