\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx\) [325]

Optimal result

Integrand size = 37, antiderivative size = 261 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=-\frac {\left (B \left (5 c^2-34 c d-3 d^2\right )+A \left (3 c^2-14 c d+43 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} (c-d)^3 f}-\frac {2 d^{3/2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{5/2} (c-d)^3 \sqrt {c+d} f}-\frac {(A-B) \cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 A c+5 B c-11 A d+3 B d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}} \] Output:

-1/32*(B*(5*c^2-34*c*d-3*d^2)+A*(3*c^2-14*c*d+43*d^2))*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^(5/2)/(c-d)^3/f-2*d^ 
(3/2)*(-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^(5/2)/(c-d)^3/(c+d)^(1/2)/f-1/4*(A-B)*cos(f*x+e)/(c-d)/f/( 
a+a*sin(f*x+e))^(5/2)-1/16*(3*A*c-11*A*d+5*B*c+3*B*d)*cos(f*x+e)/a/(c-d)^2 
/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 = 11.45 (sec) , antiderivative size = 912, normalized size of antiderivative = 3.49 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + 
f*x])),x]
 

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(A - B)*(c - d)^2*Sin[(e + f*x)/ 
2] + 4*(-A + B)*(c - d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(c - d 
)*(3*A*c + 5*B*c - 11*A*d + 3*B*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Si 
n[(e + f*x)/2])^2 - (c - d)*(3*A*c + 5*B*c - 11*A*d + 3*B*d)*(Cos[(e + f*x 
)/2] + Sin[(e + f*x)/2])^3 + (1 + I)*(-1)^(3/4)*(B*(5*c^2 - 34*c*d - 3*d^2 
) + A*(3*c^2 - 14*c*d + 43*d^2))*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[ 
(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 + (8*d^(3/2)*(-(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]*Sq 
rt[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])^4)/Sqrt[c + d] + (8*d^(3/2)*(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*S 
qrt[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) & ])*...
 

Rubi [A] (verified)

Time = 1.64 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.10, 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, 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.

\(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^{5/2} (c+d \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {\int -\frac {a (3 A c+5 B c-8 A d)+3 a (A-B) d \sin (e+f x)}{2 (\sin (e+f x) a+a)^{3/2} (c+d \sin (e+f x))}dx}{4 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (3 A c+5 B c-8 A d)+3 a (A-B) d \sin (e+f x)}{(\sin (e+f x) a+a)^{3/2} (c+d \sin (e+f x))}dx}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (3 A c+5 B c-8 A d)+3 a (A-B) d \sin (e+f x)}{(\sin (e+f x) a+a)^{3/2} (c+d \sin (e+f x))}dx}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {-\frac {\int -\frac {\left (B c (5 c-29 d)+A \left (3 c^2-11 d c+32 d^2\right )\right ) a^2+d (3 A c+5 B c-11 A d+3 B d) \sin (e+f x) a^2}{2 \sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{2 a^2 (c-d)}-\frac {a (3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\left (B c (5 c-29 d)+A \left (3 c^2-11 d c+32 d^2\right )\right ) a^2+d (3 A c+5 B c-11 A d+3 B d) \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{4 a^2 (c-d)}-\frac {a (3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (B c (5 c-29 d)+A \left (3 c^2-11 d c+32 d^2\right )\right ) a^2+d (3 A c+5 B c-11 A d+3 B d) \sin (e+f x) a^2}{\sqrt {\sin (e+f x) a+a} (c+d \sin (e+f x))}dx}{4 a^2 (c-d)}-\frac {a (3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3464

\(\displaystyle \frac {\frac {\frac {a^2 \left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}+\frac {32 a d^2 (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{4 a^2 (c-d)}-\frac {a (3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {a^2 \left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \int \frac {1}{\sqrt {\sin (e+f x) a+a}}dx}{c-d}+\frac {32 a d^2 (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}}{4 a^2 (c-d)}-\frac {a (3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {\frac {\frac {32 a d^2 (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}-\frac {2 a^2 \left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \int \frac {1}{2 a-\frac {a^2 \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}}{4 a^2 (c-d)}-\frac {a (3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {32 a d^2 (B c-A d) \int \frac {\sqrt {\sin (e+f x) a+a}}{c+d \sin (e+f x)}dx}{c-d}-\frac {\sqrt {2} a^{3/2} \left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{4 a^2 (c-d)}-\frac {a (3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 3252

\(\displaystyle \frac {\frac {-\frac {64 a^2 d^2 (B c-A d) \int \frac {1}{a (c+d)-\frac {a^2 d \cos ^2(e+f x)}{\sin (e+f x) a+a}}d\frac {a \cos (e+f x)}{\sqrt {\sin (e+f x) a+a}}}{f (c-d)}-\frac {\sqrt {2} a^{3/2} \left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}}{4 a^2 (c-d)}-\frac {a (3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {-\frac {\sqrt {2} a^{3/2} \left (A \left (3 c^2-14 c d+43 d^2\right )+B \left (5 c^2-34 c d-3 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d)}-\frac {64 a^{3/2} d^{3/2} (B c-A d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{f (c-d) \sqrt {c+d}}}{4 a^2 (c-d)}-\frac {a (3 A c-11 A d+5 B c+3 B d) \cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2}}}{8 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}}\)

Input:

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x])) 
,x]
 

Output:

-1/4*((A - B)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])^(5/2)) + ((-(( 
Sqrt[2]*a^(3/2)*(B*(5*c^2 - 34*c*d - 3*d^2) + A*(3*c^2 - 14*c*d + 43*d^2)) 
*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/((c - 
 d)*f)) - (64*a^(3/2)*d^(3/2)*(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*(3*A*c + 5*B*c - 11*A*d + 3*B*d)*Cos[e + f*x])/(2*(c - 
 d)*f*(a + a*Sin[e + f*x])^(3/2)))/(8*a^2*(c - d))
 

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. \(1417\) vs. \(2(228)=456\).

Time = 0.46 (sec) , antiderivative size = 1418, normalized size of antiderivative = 5.43

method	result	size
default	\(\text {Expression too large to display}\)	\(1418\)

Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x,method=_RET 
URNVERBOSE)
 

Output:

1/32/a^(9/2)*((-64*A*a^(5/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2 
)^(1/2))*d^3+64*B*a^(5/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^( 
1/2))*c*d^2+3*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^2*c^2-14*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^2*c*d+43*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^2*d^2+5*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^2* 
c^2-34*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^2*c*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^2*d^2)*cos(f*x+e)^2+2*sin(f*x+e)*(64*A*a^(5 
/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*d^3-64*B*a^(5/2) 
*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*c*d^2-3*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^2*c 
^2+14*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^2*c*d-43*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^2*d^2-5*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^2*c^2+34*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^2*c*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^2*d^2)+128*A*a^(5/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1146 vs. \(2 (228) = 456\).

Time = 3.52 (sec) , antiderivative size = 2577, normalized size of antiderivative = 9.87 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algo 
rithm="fricas")
 

Output:

[1/64*(sqrt(2)*(((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)* 
cos(f*x + e)^3 - 4*(3*A + 5*B)*c^2 + 8*(7*A + 17*B)*c*d - 4*(43*A - 3*B)*d 
^2 + 3*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + 
 e)^2 - 2*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f* 
x + e) - (4*(3*A + 5*B)*c^2 - 8*(7*A + 17*B)*c*d + 4*(43*A - 3*B)*d^2 - (( 
3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e)^2 + 2 
*((3*A + 5*B)*c^2 - 2*(7*A + 17*B)*c*d + (43*A - 3*B)*d^2)*cos(f*x + e))*s 
in(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*c 
os(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)) - 32*(4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d 
 - A*a*d^2)*cos(f*x + e)^3 - 3*(B*a*c*d - A*a*d^2)*cos(f*x + e)^2 + 2*(B*a 
*c*d - A*a*d^2)*cos(f*x + e) + (4*B*a*c*d - 4*A*a*d^2 - (B*a*c*d - A*a*d^2 
)*cos(f*x + e)^2 + 2*(B*a*c*d - A*a*d^2)*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)*co 
s(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))*sin(f*x + e))/(d^2*cos(f*x + e)^3 +...
 

Sympy [F(-1)]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**(5/2)/(c+d*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algo 
rithm="maxima")
 

Output:

integrate((B*sin(f*x + e) + A)/((a*sin(f*x + e) + a)^(5/2)*(d*sin(f*x + e) 
 + c)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2)/(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. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\left (c+d\,\sin \left (e+f\,x\right )\right )} \,d x \] Input:

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(5/2)*(c + d*sin(e + f*x))) 
,x)
 

Output:

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(5/2)*(c + d*sin(e + f*x))) 
, x)
 

Reduce [F]

\[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{4} d +\sin \left (f x +e \right )^{3} c +3 \sin \left (f x +e \right )^{3} d +3 \sin \left (f x +e \right )^{2} c +3 \sin \left (f x +e \right )^{2} d +3 \sin \left (f x +e \right ) c +\sin \left (f x +e \right ) d +c}d x \right ) a +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{4} d +\sin \left (f x +e \right )^{3} c +3 \sin \left (f x +e \right )^{3} d +3 \sin \left (f x +e \right )^{2} c +3 \sin \left (f x +e \right )^{2} d +3 \sin \left (f x +e \right ) c +\sin \left (f x +e \right ) d +c}d x \right ) b \right )}{a^{3}} \] Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x)
 

Output:

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