\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx\) [318]

Optimal result

Integrand size = 37, antiderivative size = 187 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx=-\frac {(A (c-5 d)+B (3 c+d)) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-d)^2 f}+\frac {2 \sqrt {d} (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{3/2} (c-d)^2 \sqrt {c+d} f}-\frac {(A-B) \cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2}} \] Output:

-1/4*(A*(c-5*d)+B*(3*c+d))*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin 
(f*x+e))^(1/2))*2^(1/2)/a^(3/2)/(c-d)^2/f+2*d^(1/2)*(-A*d+B*c)*arctanh(a^( 
1/2)*d^(1/2)*cos(f*x+e)/(c+d)^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(3/2)/(c-d)^ 
2/(c+d)^(1/2)/f-1/2*(A-B)*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^(3/2)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 6.85 (sec) , antiderivative size = 781, normalized size of antiderivative = 4.18 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + 
f*x])),x]
 

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(A - B)*(c - d)*Sin[(e + f*x)/2] 
 + (-A + B)*(c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (1 + I)*(-1)^( 
3/4)*(A*(c - 5*d) + B*(3*c + d))*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[ 
(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (Sqrt[d]*(B*c - A 
*d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + RootSum[c + 4*d*#1 + 2*c*#1^2 - 
 4*d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) + Sqrt[d]*Sqrt[c 
+ d]*Log[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 2*Sq 
rt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 3*d*Log[-#1 + Tan[(e + 
f*x)/4]]*#1^2 - Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*L 
og[-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ])*(Cos 
[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/Sqrt[c + d] + (Sqrt[d]*(-(B*c) + A*d) 
*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + RootSum[c + 4*d*#1 + 2*c*#1^2 - 4* 
d*#1^3 + c*#1^4 & , (-(d*Log[-#1 + Tan[(e + f*x)/4]]) - Sqrt[d]*Sqrt[c + d 
]*Log[-#1 + Tan[(e + f*x)/4]] - c*Log[-#1 + Tan[(e + f*x)/4]]*#1 - 2*Sqrt[ 
d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + 3*d*Log[-#1 + Tan[(e + f*x 
)/4]]*#1^2 + Sqrt[d]*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Log[ 
-#1 + Tan[(e + f*x)/4]]*#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ])*(Cos[(e 
 + f*x)/2] + Sin[(e + f*x)/2])^2)/Sqrt[c + d]))/(2*(c - d)^2*f*(a*(1 + Sin 
[e + f*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {3042, 3457, 27, 3042, 3464, 3042, 3128, 219, 3252, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {\int -\frac {a (3 B c+A (c-4 d))+a (A-B) d \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{2 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (3 B c+A (c-4 d))+a (A-B) d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (3 B c+A (c-4 d))+a (A-B) d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {a (A (c-5 d)+B (3 c+d)) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}-\frac {4 d (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a (A (c-5 d)+B (3 c+d)) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}-\frac {4 d (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {-\frac {2 a (A (c-5 d)+B (3 c+d)) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}-\frac {4 d (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {4 d (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}-\frac {\sqrt {2} \sqrt {a} (A (c-5 d)+B (3 c+d)) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {8 a d (B c-A d) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}-\frac {\sqrt {2} \sqrt {a} (A (c-5 d)+B (3 c+d)) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {8 \sqrt {a} \sqrt {d} (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d) \sqrt {c+d}}-\frac {\sqrt {2} \sqrt {a} (A (c-5 d)+B (3 c+d)) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}\)

Input:

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])) 
,x]
 

Output:

(-((Sqrt[2]*Sqrt[a]*(A*(c - 5*d) + B*(3*c + d))*ArcTanh[(Sqrt[a]*Cos[e + f 
*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/((c - d)*f)) + (8*Sqrt[a]*Sqrt[d 
]*(B*c - A*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + 
 a*Sin[e + f*x]])])/((c - d)*Sqrt[c + d]*f))/(4*a^2*(c - d)) - ((A - B)*Co 
s[e + f*x])/(2*(c - d)*f*(a + a*Sin[e + f*x])^(3/2))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 3128

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d 
Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], 
 x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
 

rule 3252

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

rule 3457

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

rule 3464

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(623\) vs. \(2(158)=316\).

Time = 0.45 (sec) , antiderivative size = 624, normalized size of antiderivative = 3.34

method	result	size
default	\(-\frac {\left (\sin \left (f x +e \right ) \left (8 A \,a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) d^{2}-8 B \,a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) c d +A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {\left (c +d \right ) a d}\, \sqrt {2}\, a c -5 A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {\left (c +d \right ) a d}\, \sqrt {2}\, a d +3 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {\left (c +d \right ) a d}\, \sqrt {2}\, a c +B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {\left (c +d \right ) a d}\, \sqrt {2}\, a d \right )+8 A \,a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) d^{2}-8 B \,a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) c d +A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {\left (c +d \right ) a d}\, \sqrt {2}\, a c -5 A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {\left (c +d \right ) a d}\, \sqrt {2}\, a d +3 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {\left (c +d \right ) a d}\, \sqrt {2}\, a c +B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {\left (c +d \right ) a d}\, \sqrt {2}\, a d +2 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, \sqrt {\left (c +d \right ) a d}\, c -2 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, \sqrt {\left (c +d \right ) a d}\, d -2 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, \sqrt {\left (c +d \right ) a d}\, c +2 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {a}\, \sqrt {\left (c +d \right ) a d}\, d \right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{4 a^{\frac {5}{2}} \sqrt {\left (c +d \right ) a d}\, \left (c -d \right )^{2} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(624\)

Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x,method=_RET 
URNVERBOSE)
 

Output:

-1/4/a^(5/2)*(sin(f*x+e)*(8*A*a^(3/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a* 
c*d+a*d^2)^(1/2))*d^2-8*B*a^(3/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+ 
a*d^2)^(1/2))*c*d+A*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(( 
c+d)*a*d)^(1/2)*2^(1/2)*a*c-5*A*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2) 
/a^(1/2))*((c+d)*a*d)^(1/2)*2^(1/2)*a*d+3*B*arctanh(1/2*(a-a*sin(f*x+e))^( 
1/2)*2^(1/2)/a^(1/2))*((c+d)*a*d)^(1/2)*2^(1/2)*a*c+B*arctanh(1/2*(a-a*sin 
(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*((c+d)*a*d)^(1/2)*2^(1/2)*a*d)+8*A*a^(3/2) 
*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*d^2-8*B*a^(3/2)*arc 
tanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*c*d+A*arctanh(1/2*(a-a* 
sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*((c+d)*a*d)^(1/2)*2^(1/2)*a*c-5*A*arcta 
nh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*((c+d)*a*d)^(1/2)*2^(1/2)*a 
*d+3*B*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*((c+d)*a*d)^(1/ 
2)*2^(1/2)*a*c+B*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*((c+d 
)*a*d)^(1/2)*2^(1/2)*a*d+2*A*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*((c+d)*a*d)^(1 
/2)*c-2*A*(a-a*sin(f*x+e))^(1/2)*a^(1/2)*((c+d)*a*d)^(1/2)*d-2*B*(a-a*sin( 
f*x+e))^(1/2)*a^(1/2)*((c+d)*a*d)^(1/2)*c+2*B*(a-a*sin(f*x+e))^(1/2)*a^(1/ 
2)*((c+d)*a*d)^(1/2)*d)*(-a*(sin(f*x+e)-1))^(1/2)/((c+d)*a*d)^(1/2)/(c-d)^ 
2/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 638 vs. \(2 (158) = 316\).

Time = 1.67 (sec) , antiderivative size = 1561, normalized size of antiderivative = 8.35 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algo 
rithm="fricas")
 

Output:

[1/8*(sqrt(2)*(((A + 3*B)*c - (5*A - B)*d)*cos(f*x + e)^2 - 2*(A + 3*B)*c 
+ 2*(5*A - B)*d - ((A + 3*B)*c - (5*A - B)*d)*cos(f*x + e) - (2*(A + 3*B)* 
c - 2*(5*A - B)*d + ((A + 3*B)*c - (5*A - B)*d)*cos(f*x + e))*sin(f*x + e) 
)*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt 
(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) 
 - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + 
 e) - cos(f*x + e) - 2)) + 4*(2*B*a*c - 2*A*a*d - (B*a*c - A*a*d)*cos(f*x 
+ e)^2 + (B*a*c - A*a*d)*cos(f*x + e) + (2*B*a*c - 2*A*a*d + (B*a*c - A*a* 
d)*cos(f*x + e))*sin(f*x + e))*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 
 - (6*c*d + 7*d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - 4*((c*d + d^2)*cos 
(f*x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)*cos(f*x + e) + ( 
c^2 + 4*c*d + 3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f 
*x + e) + a)*sqrt(d/(a*c + a*d)) - (c^2 + 8*c*d + 9*d^2)*cos(f*x + e) + (d 
^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*d^2)*cos(f*x + e))*si 
n(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c 
*d - d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x 
+ e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*((A - B)*c - (A - B)*d + ((A 
- B)*c - (A - B)*d)*cos(f*x + e) - ((A - B)*c - (A - B)*d)*sin(f*x + e))*s 
qrt(a*sin(f*x + e) + a))/((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e)^2 
 - (a^2*c^2 - 2*a^2*c*d + a^2*d^2)*f*cos(f*x + e) - 2*(a^2*c^2 - 2*a^2*...
 

Sympy [F(-1)]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**(3/2)/(c+d*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algo 
rithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)/((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) 
 + c)), x)
 

Giac [F(-2)]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x, algo 
rithm="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. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\left (c+d\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))) 
,x)
 

Output:

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))) 
, x)
 

Reduce [F]

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

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e)),x)
 

Output:

(sqrt(a)*(int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)**3*d + sin(e + f*x)**2* 
c + 2*sin(e + f*x)**2*d + 2*sin(e + f*x)*c + sin(e + f*x)*d + c),x)*a + in 
t((sqrt(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3*d + sin(e + f*x)* 
*2*c + 2*sin(e + f*x)**2*d + 2*sin(e + f*x)*c + sin(e + f*x)*d + c),x)*b)) 
/a**2