\(\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx\) [159]

Optimal result

Integrand size = 40, antiderivative size = 114 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2^{\frac {9}{4}+m} a^2 c^2 (g \cos (e+f x))^{15/2} \operatorname {Hypergeometric2F1}\left (\frac {15}{4},-\frac {1}{4}-m,\frac {19}{4},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac {5}{4}-m} (a+a \sin (e+f x))^{-2+m} \sqrt {c-c \sin (e+f x)}}{15 f g^6} \] Output:

-1/15*2^(9/4+m)*a^2*c^2*(g*cos(f*x+e))^(15/2)*hypergeom([15/4, -1/4-m],[19 
/4],1/2-1/2*sin(f*x+e))*sec(f*x+e)*(1+sin(f*x+e))^(-5/4-m)*(a+a*sin(f*x+e) 
)^(-2+m)*(c-c*sin(f*x+e))^(1/2)/f/g^6
 

Mathematica [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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.

\(\displaystyle \int (c-c \sin (e+f x))^{5/2} (g \cos (e+f x))^{3/2} (a \sin (e+f x)+a)^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c-c \sin (e+f x))^{5/2} (g \cos (e+f x))^{3/2} (a \sin (e+f x)+a)^mdx\)

\(\Big \downarrow \) 3332

\(\displaystyle \frac {a^2 c^2 \sec (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)} \int (g \cos (e+f x))^{13/2} (\sin (e+f x) a+a)^{m-\frac {5}{2}}dx}{g^5}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a^2 c^2 \sec (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)} \int (g \cos (e+f x))^{13/2} (\sin (e+f x) a+a)^{m-\frac {5}{2}}dx}{g^5}\)

\(\Big \downarrow \) 3168

\(\displaystyle \frac {a^4 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{15/2} \int (a-a \sin (e+f x))^{11/4} (\sin (e+f x) a+a)^{m+\frac {1}{4}}d\sin (e+f x)}{f g^6 (a-a \sin (e+f x))^{15/4} (a \sin (e+f x)+a)^{13/4}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {a^4 c^2 2^{m+\frac {1}{4}} \sec (e+f x) \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{15/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-3} \int \left (\frac {1}{2} \sin (e+f x)+\frac {1}{2}\right )^{m+\frac {1}{4}} (a-a \sin (e+f x))^{11/4}d\sin (e+f x)}{f g^6 (a-a \sin (e+f x))^{15/4}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {a^3 c^2 2^{m+\frac {9}{4}} \sec (e+f x) \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{15/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-3} \operatorname {Hypergeometric2F1}\left (\frac {15}{4},-m-\frac {1}{4},\frac {19}{4},\frac {1}{2} (1-\sin (e+f x))\right )}{15 f g^6}\)

Input:

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

Output:

-1/15*(2^(9/4 + m)*a^3*c^2*(g*Cos[e + f*x])^(15/2)*Hypergeometric2F1[15/4, 
 -1/4 - m, 19/4, (1 - Sin[e + f*x])/2]*Sec[e + f*x]*(1 + Sin[e + f*x])^(-1 
/4 - m)*(a + a*Sin[e + f*x])^(-3 + m)*Sqrt[c - c*Sin[e + f*x]])/(f*g^6)
 

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]

\[\int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}d x\]

Input:

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x)
 

Output:

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x)
 

Fricas [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x 
, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:

integrate((g*cos(f*x+e))**(3/2)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(5/2 
),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x 
, algorithm="maxima")
 

Output:

integrate((g*cos(f*x + e))^(3/2)*(-c*sin(f*x + e) + c)^(5/2)*(a*sin(f*x + 
e) + a)^m, x)
 

Giac [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x 
, algorithm="giac")
 

Output:

integrate((g*cos(f*x + e))^(3/2)*(-c*sin(f*x + e) + c)^(5/2)*(a*sin(f*x + 
e) + a)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \] Input:

int((g*cos(e + f*x))^(3/2)*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^(5/ 
2),x)
 

Output:

int((g*cos(e + f*x))^(3/2)*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^(5/ 
2), x)
 

Reduce [F]

\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\sqrt {g}\, \sqrt {c}\, c^{2} g \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {-\sin \left (f x +e \right )+1}\, \sqrt {\cos \left (f x +e \right )}\, \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}d x -2 \left (\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {-\sin \left (f x +e \right )+1}\, \sqrt {\cos \left (f x +e \right )}\, \cos \left (f x +e \right ) \sin \left (f x +e \right )d x \right )+\int \left (a +a \sin \left (f x +e \right )\right )^{m} \sqrt {-\sin \left (f x +e \right )+1}\, \sqrt {\cos \left (f x +e \right )}\, \cos \left (f x +e \right )d x \right ) \] Input:

int((g*cos(f*x+e))^(3/2)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(5/2),x)
 

Output:

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