\(\int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx\) [14]

Optimal result

Integrand size = 38, antiderivative size = 147 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {4 a^2 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 7.77 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{3/2} \left (-\cos (2 (e+f x))+32 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+12 \sin (e+f x)\right )}{4 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 151, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3320

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3216

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3146

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

\(\Big \downarrow \) 16

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral((a*(sin(e + f*x) + 1))**(3/2)*cos(e + f*x)**2/(-c*(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. \(2 (133) = 266\).

Time = 0.21 (sec) , antiderivative size = 844, normalized size of antiderivative = 5.74 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(16*a^(3/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(3/2) - 8*a^(3/ 
2)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/c^(3/2) + (10*a^(3/2) - 11 
*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 15*a^(3/2)*sin(f*x + e)^2/(cos( 
f*x + e) + 1)^2 - 20*a^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^(3/ 
2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 7*a^(3/2)*sin(f*x + e)^5/(cos(f*x 
 + e) + 1)^5)/(c^(3/2) - 2*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^( 
3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4*c^(3/2)*sin(f*x + e)^3/(cos(f 
*x + e) + 1)^3 + 3*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2*c^(3/2) 
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + c^(3/2)*sin(f*x + e)^6/(cos(f*x + e 
) + 1)^6) - (10*a^(3/2) - 13*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 25* 
a^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 20*a^(3/2)*sin(f*x + e)^3/(c 
os(f*x + e) + 1)^3 + 15*a^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 9*a^ 
(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(c^(3/2) - 2*c^(3/2)*sin(f*x + 
e)/(cos(f*x + e) + 1) + 3*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4* 
c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^(3/2)*sin(f*x + e)^4/(co 
s(f*x + e) + 1)^4 - 2*c^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + c^(3/2 
)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6) + 2*(5*a^(3/2)*sin(f*x + e)/(cos(f* 
x + e) + 1) - 5*a^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*a^(3/2)*si 
n(f*x + e)^3/(cos(f*x + e) + 1)^3 - 5*a^(3/2)*sin(f*x + e)^4/(cos(f*x + e) 
 + 1)^4 + 5*a^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/(c^(3/2) - 2*c...
 

Giac [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{3/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {2 \, a^{\frac {3}{2}} \sqrt {c} {\left (\frac {2 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{c^{4}}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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