\(\int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx\) [1087]

Optimal result

Integrand size = 35, antiderivative size = 347 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}+\frac {4 b \cos (e+f x)}{3 a \left (a^2-b^2\right ) f \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}-\frac {4 b \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (1+\csc (e+f x))}{a-b}} E\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right )|-\frac {a+b}{a-b}\right ) \tan (e+f x)}{3 a^3 \sqrt {a+b} \sqrt {d} f}-\frac {4 \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (1+\csc (e+f x))}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right ),-\frac {a+b}{a-b}\right ) \tan (e+f x)}{3 a^2 \sqrt {a+b} \sqrt {d} f} \] Output:

2/3*cos(f*x+e)*(d*sin(f*x+e))^(1/2)/a/d/f/(a+b*sin(f*x+e))^(3/2)+4/3*b*cos 
(f*x+e)/a/(a^2-b^2)/f/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(1/2)-4/3*b*(a 
*(1-csc(f*x+e))/(a+b))^(1/2)*(a*(1+csc(f*x+e))/(a-b))^(1/2)*EllipticE(d^(1 
/2)*(a+b*sin(f*x+e))^(1/2)/(a+b)^(1/2)/(d*sin(f*x+e))^(1/2),(-(a+b)/(a-b)) 
^(1/2))*tan(f*x+e)/a^3/(a+b)^(1/2)/d^(1/2)/f-4/3*(a*(1-csc(f*x+e))/(a+b))^ 
(1/2)*(a*(1+csc(f*x+e))/(a-b))^(1/2)*EllipticF(d^(1/2)*(a+b*sin(f*x+e))^(1 
/2)/(a+b)^(1/2)/(d*sin(f*x+e))^(1/2),(-(a+b)/(a-b))^(1/2))*tan(f*x+e)/a^2/ 
(a+b)^(1/2)/d^(1/2)/f
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3348\) vs. \(2(347)=694\).

Time = 30.06 (sec) , antiderivative size = 3348, normalized size of antiderivative = 9.65 \[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\text {Result too large to show} \] Input:

Integrate[Cos[e + f*x]^2/(Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^(5/2)) 
,x]
 

Output:

(Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]]*((2*Cos[e + f*x])/(3*a*(a + b*Sin[e 
 + f*x])^2) - (4*b^2*Cos[e + f*x])/(3*a^2*(a^2 - b^2)*(a + b*Sin[e + f*x]) 
)))/(f*Sqrt[d*Sin[e + f*x]]) + (4*Sqrt[a + b*Sin[e + f*x]]*((2*Sqrt[a + b* 
Sin[e + f*x]])/(3*a*(a^2 - b^2)*Sqrt[Sin[e + f*x]]) - (4*b*Sqrt[Sin[e + f* 
x]]*Sqrt[a + b*Sin[e + f*x]])/(3*a^2*(a^2 - b^2)))*(-2*b*Sin[(e + f*x)/2]^ 
2 - (2*a*(-(b*EllipticE[ArcSin[Sqrt[(-b + Sqrt[-a^2 + b^2] - a*Tan[(e + f* 
x)/2])/Sqrt[-a^2 + b^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(-b + Sqrt[-a^2 + 
b^2])]*Tan[(e + f*x)/2]) + a*EllipticF[ArcSin[Sqrt[(b + Sqrt[-a^2 + b^2] + 
 a*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(b + 
 Sqrt[-a^2 + b^2])]*Sqrt[(a*Tan[(e + f*x)/2])/(-b + Sqrt[-a^2 + b^2])]*Sqr 
t[-((a*Tan[(e + f*x)/2])/(b + Sqrt[-a^2 + b^2]))]))/(Sqrt[-a^2 + b^2]*Sqrt 
[(a*Sec[(e + f*x)/2]^2*(a + b*Sin[e + f*x]))/(a^2 - b^2)]*Sqrt[(a*Tan[(e + 
 f*x)/2])/(-b + Sqrt[-a^2 + b^2])])))/(3*a^2*(a^2 - b^2)*f*Sqrt[d*Sin[e + 
f*x]]*((2*b*Cos[e + f*x]*(-2*b*Sin[(e + f*x)/2]^2 - (2*a*(-(b*EllipticE[Ar 
cSin[Sqrt[(-b + Sqrt[-a^2 + b^2] - a*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]]/S 
qrt[2]], (2*Sqrt[-a^2 + b^2])/(-b + Sqrt[-a^2 + b^2])]*Tan[(e + f*x)/2]) + 
 a*EllipticF[ArcSin[Sqrt[(b + Sqrt[-a^2 + b^2] + a*Tan[(e + f*x)/2])/Sqrt[ 
-a^2 + b^2]]/Sqrt[2]], (2*Sqrt[-a^2 + b^2])/(b + Sqrt[-a^2 + b^2])]*Sqrt[( 
a*Tan[(e + f*x)/2])/(-b + Sqrt[-a^2 + b^2])]*Sqrt[-((a*Tan[(e + f*x)/2])/( 
b + Sqrt[-a^2 + b^2]))]))/(Sqrt[-a^2 + b^2]*Sqrt[(a*Sec[(e + f*x)/2]^2*...
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3366, 3042, 3279, 3042, 3477, 3042, 3295, 3473}

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.

\(\displaystyle \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3366

\(\displaystyle \frac {2 \int \frac {1}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{3/2}}dx}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \frac {1}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{3/2}}dx}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3279

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3477

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

\(\displaystyle \frac {2 \left (\frac {d \left (b \int \frac {\sin (e+f x)+1}{(d \sin (e+f x))^{3/2} \sqrt {a+b \sin (e+f x)}}dx-\frac {2 (a-b) \sqrt {a+b} \tan (e+f x) \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (\csc (e+f x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right ),-\frac {a+b}{a-b}\right )}{a d^{3/2} f}\right )}{a^2-b^2}+\frac {2 b \cos (e+f x)}{f \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}\right )}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3473

\(\displaystyle \frac {2 \left (\frac {d \left (-\frac {2 b (a-b) \sqrt {a+b} \tan (e+f x) \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (\csc (e+f x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right )|-\frac {a+b}{a-b}\right )}{a^2 d^{3/2} f}-\frac {2 (a-b) \sqrt {a+b} \tan (e+f x) \sqrt {\frac {a (1-\csc (e+f x))}{a+b}} \sqrt {\frac {a (\csc (e+f x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {d} \sqrt {a+b \sin (e+f x)}}{\sqrt {a+b} \sqrt {d \sin (e+f x)}}\right ),-\frac {a+b}{a-b}\right )}{a d^{3/2} f}\right )}{a^2-b^2}+\frac {2 b \cos (e+f x)}{f \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)} \sqrt {a+b \sin (e+f x)}}\right )}{3 a}+\frac {2 \cos (e+f x) \sqrt {d \sin (e+f x)}}{3 a d f (a+b \sin (e+f x))^{3/2}}\)

Input:

Int[Cos[e + f*x]^2/(Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^(5/2)),x]
 

Output:

(2*Cos[e + f*x]*Sqrt[d*Sin[e + f*x]])/(3*a*d*f*(a + b*Sin[e + f*x])^(3/2)) 
 + (2*((2*b*Cos[e + f*x])/((a^2 - b^2)*f*Sqrt[d*Sin[e + f*x]]*Sqrt[a + b*S 
in[e + f*x]]) + (d*((-2*(a - b)*b*Sqrt[a + b]*Sqrt[(a*(1 - Csc[e + f*x]))/ 
(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*EllipticE[ArcSin[(Sqrt[d]*Sq 
rt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[d*Sin[e + f*x]])], -((a + b)/(a 
- b))]*Tan[e + f*x])/(a^2*d^(3/2)*f) - (2*(a - b)*Sqrt[a + b]*Sqrt[(a*(1 - 
 Csc[e + f*x]))/(a + b)]*Sqrt[(a*(1 + Csc[e + f*x]))/(a - b)]*EllipticF[Ar 
cSin[(Sqrt[d]*Sqrt[a + b*Sin[e + f*x]])/(Sqrt[a + b]*Sqrt[d*Sin[e + f*x]]) 
], -((a + b)/(a - b))]*Tan[e + f*x])/(a*d^(3/2)*f)))/(a^2 - b^2)))/(3*a)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3279

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(3/2)), x_Symbol] :> Simp[2*b*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + 
b*Sin[e + f*x]]*Sqrt[d*Sin[e + f*x]])), x] + Simp[d/(a^2 - b^2)   Int[(b + 
a*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*(d*Sin[e + f*x])^(3/2)), x], x] / 
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

rule 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

rule 3366

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*( 
x_)])^(m_))/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-g)*(g* 
Cos[e + f*x])^(p - 1)*Sqrt[d*Sin[e + f*x]]*((a + b*Sin[e + f*x])^(m + 1)/(a 
*d*f*(m + 1))), x] + Simp[g^2*((2*m + 3)/(2*a*(m + 1)))   Int[(g*Cos[e + f* 
x])^(p - 2)*((a + b*Sin[e + f*x])^(m + 1)/Sqrt[d*Sin[e + f*x]]), x], x] /; 
FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && EqQ[m + 
p + 1/2, 0]
 

rule 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[ArcSin[Sqrt[c + 
d*Sin[e + f*x]]/Sqrt[b*Sin[e + f*x]]/Rt[(c + d)/b, 2]], -(c + d)/(c - d)], 
x] /; FreeQ[{b, c, d, e, f, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e 
, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, B]
 

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(2093\) vs. \(2(301)=602\).

Time = 4.05 (sec) , antiderivative size = 2094, normalized size of antiderivative = 6.03

method	result	size
default	\(\text {Expression too large to display}\)	\(2094\)

Input:

int(cos(f*x+e)^2/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(5/2),x,method=_RET 
URNVERBOSE)
 

Output:

-1/3/f*2^(1/2)/(a^2-b^2)/a^3*(a+b*sin(f*x+e))^(1/2)*((-4*cos(f*x+e)-4)*b^2 
*EllipticE((-(a*cot(f*x+e)-(-a^2+b^2)^(1/2)-b-a*csc(f*x+e))/(b+(-a^2+b^2)^ 
(1/2)))^(1/2),1/2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*( 
-(a*cot(f*x+e)-(-a^2+b^2)^(1/2)-b-a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2 
)*(1/(-a^2+b^2)^(1/2)*(a*cot(f*x+e)+(-a^2+b^2)^(1/2)-b-a*csc(f*x+e)))^(1/2 
)*(1/(b+(-a^2+b^2)^(1/2))*a*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*(-a^2+b^2)^(1/ 
2)*a+(-4*cos(f*x+e)-4)*sin(f*x+e)*(-a^2+b^2)^(1/2)*(-(a*cot(f*x+e)-(-a^2+b 
^2)^(1/2)-b-a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2)*(1/(-a^2+b^2)^(1/2)* 
(a*cot(f*x+e)+(-a^2+b^2)^(1/2)-b-a*csc(f*x+e)))^(1/2)*(1/(b+(-a^2+b^2)^(1/ 
2))*a*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*EllipticE((-(a*cot(f*x+e)-(-a^2+b^2) 
^(1/2)-b-a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2),1/2*2^(1/2)*((b+(-a^2+b 
^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*b^3+(4*cos(f*x+e)+4)*b*EllipticE((-(a* 
cot(f*x+e)-(-a^2+b^2)^(1/2)-b-a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2),1/ 
2*2^(1/2)*((b+(-a^2+b^2)^(1/2))/(-a^2+b^2)^(1/2))^(1/2))*(-(a*cot(f*x+e)-( 
-a^2+b^2)^(1/2)-b-a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^(1/2)*(1/(-a^2+b^2)^ 
(1/2)*(a*cot(f*x+e)+(-a^2+b^2)^(1/2)-b-a*csc(f*x+e)))^(1/2)*(1/(b+(-a^2+b^ 
2)^(1/2))*a*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*a^3+(4*cos(f*x+e)+4)*sin(f*x+e 
)*(-(a*cot(f*x+e)-(-a^2+b^2)^(1/2)-b-a*csc(f*x+e))/(b+(-a^2+b^2)^(1/2)))^( 
1/2)*(1/(-a^2+b^2)^(1/2)*(a*cot(f*x+e)+(-a^2+b^2)^(1/2)-b-a*csc(f*x+e)))^( 
1/2)*(1/(b+(-a^2+b^2)^(1/2))*a*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*Elliptic...
 

Fricas [F]

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

integrate(cos(f*x+e)^2/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(5/2),x, algo 
rithm="fricas")
 

Output:

integral(sqrt(b*sin(f*x + e) + a)*sqrt(d*sin(f*x + e))*cos(f*x + e)^2/(b^3 
*d*cos(f*x + e)^4 - (3*a^2*b + 2*b^3)*d*cos(f*x + e)^2 + (3*a^2*b + b^3)*d 
 - (3*a*b^2*d*cos(f*x + e)^2 - (a^3 + 3*a*b^2)*d)*sin(f*x + e)), x)
 

Sympy [F(-1)]

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

integrate(cos(f*x+e)**2/(d*sin(f*x+e))**(1/2)/(a+b*sin(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(cos(f*x+e)^2/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(5/2),x, algo 
rithm="maxima")
 

Output:

integrate(cos(f*x + e)^2/((b*sin(f*x + e) + a)^(5/2)*sqrt(d*sin(f*x + e))) 
, x)
 

Giac [F(-1)]

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

integrate(cos(f*x+e)^2/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(5/2),x, algo 
rithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int(cos(e + f*x)^2/((d*sin(e + f*x))^(1/2)*(a + b*sin(e + f*x))^(5/2)),x)
 

Output:

int(cos(e + f*x)^2/((d*sin(e + f*x))^(1/2)*(a + b*sin(e + f*x))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\cos ^2(e+f x)}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )}\, \sqrt {\sin \left (f x +e \right ) b +a}\, \cos \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{4} b^{3}+3 \sin \left (f x +e \right )^{3} a \,b^{2}+3 \sin \left (f x +e \right )^{2} a^{2} b +\sin \left (f x +e \right ) a^{3}}d x \right )}{d} \] Input:

int(cos(f*x+e)^2/(d*sin(f*x+e))^(1/2)/(a+b*sin(f*x+e))^(5/2),x)
 

Output:

(sqrt(d)*int((sqrt(sin(e + f*x))*sqrt(sin(e + f*x)*b + a)*cos(e + f*x)**2) 
/(sin(e + f*x)**4*b**3 + 3*sin(e + f*x)**3*a*b**2 + 3*sin(e + f*x)**2*a**2 
*b + sin(e + f*x)*a**3),x))/d