\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx\) [319]

Optimal result

Integrand size = 37, antiderivative size = 292 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx=-\frac {(A c+3 B c-9 A d+5 B d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{2 \sqrt {2} a^{3/2} (c-d)^3 f}-\frac {\sqrt {d} \left (A d (5 c+3 d)-B \left (3 c^2+3 c d+2 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{3/2} (c-d)^3 (c+d)^{3/2} f}-\frac {(A-B) \cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}+\frac {d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{2 a (c-d)^2 (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

-1/4*(A*c-9*A*d+3*B*c+5*B*d)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*s 
in(f*x+e))^(1/2))*2^(1/2)/a^(3/2)/(c-d)^3/f-d^(1/2)*(A*d*(5*c+3*d)-B*(3*c^ 
2+3*c*d+2*d^2))*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)^3/(c+d)^(3/2)/f-1/2*(A-B)*cos(f*x+e)/(c-d)/f/(a 
+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))+1/2*d*(B*(3*c+d)-A*(c+3*d))*cos(f*x+ 
e)/a/(c-d)^2/(c+d)/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))
 

Mathematica [C] (warning: unable to verify)

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

Time = 15.95 (sec) , antiderivative size = 904, normalized size of antiderivative = 3.10 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(4*(A - B)*(c - d)*Sin[(e + f*x)/2] 
 + 2*(-A + B)*(c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + (2 + 2*I)*(- 
1)^(3/4)*(A*c + 3*B*c - 9*A*d + 5*B*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]*(- 
(A*d*(5*c + 3*d)) + B*(3*c^2 + 3*c*d + 2*d^2))*(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)/(c + d)^(3/2) + (Sqrt[d]*(A*d*(5*c + 3*d) - B*(3*c^2 + 3*c*d + 2*d^2 
))*(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*Sqr 
t[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*Lo 
g[-#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)/(c + d)^(3/2) + (4*(c - d)*d*(B*c - A* 
d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f...
 

Rubi [A] (verified)

Time = 1.73 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.09, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.351, Rules used = {3042, 3457, 27, 3042, 3463, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {\int -\frac {a (A c+3 B c-6 A d+2 B d)+3 a (A-B) d \sin (e+f x)}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^2}dx}{2 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (A c+3 B c-6 A d+2 B d)+3 a (A-B) d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^2}dx}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (A c+3 B c-6 A d+2 B d)+3 a (A-B) d \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))^2}dx}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {\frac {2 a d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac {\int -\frac {a^2 \left (A \left (c^2-7 d c-6 d^2\right )+B \left (3 c^2+5 d c+4 d^2\right )\right )-a^2 d (B (3 c+d)-A (c+3 d)) \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{a \left (c^2-d^2\right )}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {a^2 \left (A \left (c^2-7 d c-6 d^2\right )+B \left (3 c^2+5 d c+4 d^2\right )\right )-a^2 d (B (3 c+d)-A (c+3 d)) \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{a \left (c^2-d^2\right )}+\frac {2 a d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 \left (A \left (c^2-7 d c-6 d^2\right )+B \left (3 c^2+5 d c+4 d^2\right )\right )-a^2 d (B (3 c+d)-A (c+3 d)) \sin (e+f x)}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{a \left (c^2-d^2\right )}+\frac {2 a d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {\frac {a^2 (c+d) (A c-9 A d+3 B c+5 B d) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}+\frac {2 a d \left (A d (5 c+3 d)-B \left (3 c^2+3 c d+2 d^2\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{a \left (c^2-d^2\right )}+\frac {2 a d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^2 (c+d) (A c-9 A d+3 B c+5 B d) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}+\frac {2 a d \left (A d (5 c+3 d)-B \left (3 c^2+3 c d+2 d^2\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{a \left (c^2-d^2\right )}+\frac {2 a d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {\frac {2 a d \left (A d (5 c+3 d)-B \left (3 c^2+3 c d+2 d^2\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}-\frac {2 a^2 (c+d) (A c-9 A d+3 B c+5 B d) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}}{a \left (c^2-d^2\right )}+\frac {2 a d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {2 a d \left (A d (5 c+3 d)-B \left (3 c^2+3 c d+2 d^2\right )\right ) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}-\frac {\sqrt {2} a^{3/2} (c+d) (A c-9 A d+3 B c+5 B d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{a \left (c^2-d^2\right )}+\frac {2 a d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {-\frac {4 a^2 d \left (A d (5 c+3 d)-B \left (3 c^2+3 c d+2 d^2\right )\right ) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}-\frac {\sqrt {2} a^{3/2} (c+d) (A c-9 A d+3 B c+5 B d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{a \left (c^2-d^2\right )}+\frac {2 a d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {-\frac {4 a^{3/2} \sqrt {d} \left (A d (5 c+3 d)-B \left (3 c^2+3 c d+2 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d) \sqrt {c+d}}-\frac {\sqrt {2} a^{3/2} (c+d) (A c-9 A d+3 B c+5 B d) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{a \left (c^2-d^2\right )}+\frac {2 a d (B (3 c+d)-A (c+3 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))}\)

Input:

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

Output:

-1/2*((A - B)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])^(3/2)*(c + d*S 
in[e + f*x])) + ((-((Sqrt[2]*a^(3/2)*(c + d)*(A*c + 3*B*c - 9*A*d + 5*B*d) 
*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/((c - 
 d)*f)) - (4*a^(3/2)*Sqrt[d]*(A*d*(5*c + 3*d) - B*(3*c^2 + 3*c*d + 2*d^2)) 
*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))/(a*(c^2 - d^2)) + (2*a*d*(B*(3*c + d) - A* 
(c + 3*d))*Cos[e + f*x])/((c^2 - d^2)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Si 
n[e + f*x])))/(4*a^2*(c - d))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 3463

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
+ 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2))   Int[(a 
 + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 
1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], 
 x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq 
Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m 
 + 1/2, 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. \(2048\) vs. \(2(259)=518\).

Time = 0.52 (sec) , antiderivative size = 2049, normalized size of antiderivative = 7.02

method	result	size
default	\(\text {Expression too large to display}\)	\(2049\)

Input:

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

Output:

-1/4/a^(5/2)*(2*A*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*((c+d)*a*d)^(1/2)*c*d^ 
2-4*B*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*((c+d)*a*d)^(1/2)*c^2*d+6*B*(-a*(s 
in(f*x+e)-1))^(1/2)*a^(1/2)*((c+d)*a*d)^(1/2)*c*d^2+20*A*a^(3/2)*arctanh(( 
-a*(sin(f*x+e)-1))^(1/2)*d/((c+d)*a*d)^(1/2))*sin(f*x+e)^2*c*d^3-12*B*a^(3 
/2)*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/((c+d)*a*d)^(1/2))*sin(f*x+e)^2*c^ 
2*d^2-12*B*a^(3/2)*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/((c+d)*a*d)^(1/2))* 
sin(f*x+e)^2*c*d^3+A*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2) 
/a^(1/2))*((c+d)*a*d)^(1/2)*a*c^3-6*A*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*(( 
c+d)*a*d)^(1/2)*sin(f*x+e)*d^3+20*A*a^(3/2)*arctanh((-a*(sin(f*x+e)-1))^(1 
/2)*d/((c+d)*a*d)^(1/2))*sin(f*x+e)*c^2*d^2+32*A*a^(3/2)*arctanh((-a*(sin( 
f*x+e)-1))^(1/2)*d/((c+d)*a*d)^(1/2))*sin(f*x+e)*c*d^3-12*B*a^(3/2)*arctan 
h((-a*(sin(f*x+e)-1))^(1/2)*d/((c+d)*a*d)^(1/2))*sin(f*x+e)*c^3*d-24*B*a^( 
3/2)*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/((c+d)*a*d)^(1/2))*sin(f*x+e)*c^2 
*d^2-20*B*a^(3/2)*arctanh((-a*(sin(f*x+e)-1))^(1/2)*d/((c+d)*a*d)^(1/2))*s 
in(f*x+e)*c*d^3+3*B*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/ 
a^(1/2))*((c+d)*a*d)^(1/2)*a*c^3+2*B*(-a*(sin(f*x+e)-1))^(1/2)*a^(1/2)*((c 
+d)*a*d)^(1/2)*sin(f*x+e)*d^3+8*B*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^ 
(1/2)*2^(1/2)/a^(1/2))*((c+d)*a*d)^(1/2)*sin(f*x+e)^2*a*c*d^2-7*A*2^(1/2)* 
arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^(1/2)/a^(1/2))*((c+d)*a*d)^(1/2)*s 
in(f*x+e)*a*c^2*d+13*B*2^(1/2)*arctanh(1/2*(-a*(sin(f*x+e)-1))^(1/2)*2^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1559 vs. \(2 (259) = 518\).

Time = 4.01 (sec) , antiderivative size = 3403, normalized size of antiderivative = 11.65 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{4} d^{2}+2 \sin \left (f x +e \right )^{3} c d +2 \sin \left (f x +e \right )^{3} d^{2}+\sin \left (f x +e \right )^{2} c^{2}+4 \sin \left (f x +e \right )^{2} c d +\sin \left (f x +e \right )^{2} d^{2}+2 \sin \left (f x +e \right ) c^{2}+2 \sin \left (f x +e \right ) c d +c^{2}}d x \right ) a +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{4} d^{2}+2 \sin \left (f x +e \right )^{3} c d +2 \sin \left (f x +e \right )^{3} d^{2}+\sin \left (f x +e \right )^{2} c^{2}+4 \sin \left (f x +e \right )^{2} c d +\sin \left (f x +e \right )^{2} d^{2}+2 \sin \left (f x +e \right ) c^{2}+2 \sin \left (f x +e \right ) c d +c^{2}}d x \right ) b \right )}{a^{2}} \] Input:

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

Output:

(sqrt(a)*(int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)**4*d**2 + 2*sin(e + f*x 
)**3*c*d + 2*sin(e + f*x)**3*d**2 + sin(e + f*x)**2*c**2 + 4*sin(e + f*x)* 
*2*c*d + sin(e + f*x)**2*d**2 + 2*sin(e + f*x)*c**2 + 2*sin(e + f*x)*c*d + 
 c**2),x)*a + int((sqrt(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**4*d 
**2 + 2*sin(e + f*x)**3*c*d + 2*sin(e + f*x)**3*d**2 + sin(e + f*x)**2*c** 
2 + 4*sin(e + f*x)**2*c*d + sin(e + f*x)**2*d**2 + 2*sin(e + f*x)*c**2 + 2 
*sin(e + f*x)*c*d + c**2),x)*b))/a**2