\(\int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx\) [52]

Optimal result

Integrand size = 38, antiderivative size = 145 \[ \int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {4 c^2 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}+\frac {\cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 7.76 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (-1+\sin (e+f x)) \left (\cos (2 (e+f x))-32 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+12 \sin (e+f x)\right ) \sqrt {c-c \sin (e+f x)}}{4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (a (1+\sin (e+f x)))^{3/2}} \] Input:

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

Output:

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-1 + Sin[e + f*x])*(Cos[2*(e + 
 f*x)] - 32*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + 12*Sin[e + f*x])*Sq 
rt[c - c*Sin[e + f*x]])/(4*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(a*(1 
 + Sin[e + f*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 3320, 3042, 3219, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

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

Input:

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

Output:

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

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 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])
 

rule 3320

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((c*cos(f*x + e)^2*sin(f*x + e) - c*cos(f*x + e)^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]

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

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

Output:

Integral((-c*(sin(e + f*x) - 1))**(3/2)*cos(e + f*x)**2/(a*(sin(e + f*x) + 
 1))**(3/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 844 vs. \(2 (131) = 262\).

Time = 0.18 (sec) , antiderivative size = 844, normalized size of antiderivative = 5.82 \[ \int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(16*c^(3/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^(3/2) - 8*c^(3 
/2)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/a^(3/2) - (10*c^(3/2) + 1 
3*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 25*c^(3/2)*sin(f*x + e)^2/(cos 
(f*x + e) + 1)^2 + 20*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*c^( 
3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 9*c^(3/2)*sin(f*x + e)^5/(cos(f 
*x + e) + 1)^5)/(a^(3/2) + 2*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a 
^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^(3/2)*sin(f*x + e)^3/(cos 
(f*x + e) + 1)^3 + 3*a^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 2*a^(3/ 
2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + a^(3/2)*sin(f*x + e)^6/(cos(f*x + 
 e) + 1)^6) + (10*c^(3/2) + 11*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 1 
5*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 20*c^(3/2)*sin(f*x + e)^3/ 
(cos(f*x + e) + 1)^3 + 5*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7*c 
^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^(3/2) + 2*a^(3/2)*sin(f*x + 
 e)/(cos(f*x + e) + 1) + 3*a^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4 
*a^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^(3/2)*sin(f*x + e)^4/(c 
os(f*x + e) + 1)^4 + 2*a^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + a^(3/ 
2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6) - 2*(5*c^(3/2)*sin(f*x + e)/(cos(f 
*x + e) + 1) + 5*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*c^(3/2)*s 
in(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e 
) + 1)^4 + 5*c^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(a^(3/2) + 2*...
 

Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {2 \, \sqrt {a} c^{\frac {3}{2}} {\left (\frac {2 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{4}}\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:

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

Output:

-2*sqrt(a)*c^(3/2)*(2*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(a^2*sgn( 
cos(-1/4*pi + 1/2*f*x + 1/2*e))) + (a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e) 
)*sin(-1/4*pi + 1/2*f*x + 1/2*e)^4 + 2*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2 
*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2)/a^4)*sgn(sin(-1/4*pi + 1/2*f*x + 1/ 
2*e))/f
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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