\(\int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\) [398]

Optimal result

Integrand size = 30, antiderivative size = 143 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {4 c^3 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a+a \sin (e+f x))^{3/2}} \] Output:

-4*c^3*cos(f*x+e)*ln(1+sin(f*x+e))/a/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x 
+e))^(1/2)-2*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^(1 
/2)-c*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 8.49 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.07 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (7+\cos (2 (e+f x))+16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 \left (-1+8 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{3/2}} \] Input:

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

Output:

-1/2*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*( 
7 + Cos[2*(e + f*x)] + 16*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + 2*(-1 
 + 8*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sin[e + f*x]))/(f*(Cos[(e + 
 f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3218, 3042, 3219, 3042, 3216, 3042, 3146, 16}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3218

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

\(\displaystyle -\frac {2 c \left (\frac {2 c^2 \cos (e+f x) \log (a \sin (e+f x)+a)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}}\)

Input:

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

Output:

-((c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(f*(a + a*Sin[e + f*x])^(3/2 
))) - (2*c*((2*c^2*Cos[e + f*x]*Log[a + a*Sin[e + f*x]])/(f*Sqrt[a + a*Sin 
[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*Sqrt[c - c*Sin[e + 
f*x]])/(f*Sqrt[a + a*Sin[e + f*x]])))/a
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 3216

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) 
+ (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x 
]]*Sqrt[c + d*Sin[e + f*x]]))   Int[Cos[e + f*x]/(c + d*Sin[e + f*x]), x], 
x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 
]
 

rule 3218

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
(m - 1)*((c + d*Sin[e + f*x])^n/(f*(2*n + 1))), x] - Simp[b*((2*m - 1)/(d*( 
2*n + 1)))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), 
 x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b 
^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && GtQ[2*m + 
n + 1, 0])
 

rule 3219

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^ 
(m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n 
))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre 
eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I 
GtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m]) &&  !( 
ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
 

Maple [A] (verified)

Time = 1.94 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.31

method	result	size
default	\(-\frac {\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (-2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+8 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-8 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{2} \sqrt {4}}{2 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a}\)	\(188\)

Input:

int((c-c*sin(f*x+e))^(5/2)/(a+sin(f*x+e)*a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/f*sec(1/4*Pi+1/2*f*x+1/2*e)*csc(1/4*Pi+1/2*f*x+1/2*e)*(-2*cos(1/4*Pi+ 
1/2*f*x+1/2*e)^4+8*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e) 
)*sin(1/4*Pi+1/2*f*x+1/2*e)^2-8*ln(2/(cos(1/4*Pi+1/2*f*x+1/2*e)+1))*sin(1/ 
4*Pi+1/2*f*x+1/2*e)^2+5*cos(1/4*Pi+1/2*f*x+1/2*e)^2-1)*(c*cos(1/4*Pi+1/2*f 
*x+1/2*e)^2)^(1/2)*c^2/(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/a*4^(1/2)
 

Fricas [F]

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

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fric 
as")
 

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxi 
ma")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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