\(\int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx\) [312]

Optimal result

Integrand size = 37, antiderivative size = 207 \[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx=-\frac {\sqrt {2} (A-B) \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^2 f}+\frac {\left (A d (3 c+d)-B \left (c^2+c d+2 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d)^2 \sqrt {d} (c+d)^{3/2} f}-\frac {(B c-A d) \cos (e+f x)}{\left (c^2-d^2\right ) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

-2^(1/2)*(A-B)*arctanh(1/2*a^(1/2)*cos(f*x+e)*2^(1/2)/(a+a*sin(f*x+e))^(1/ 
2))/a^(1/2)/(c-d)^2/f+(A*d*(3*c+d)-B*(c^2+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^(1/2)/(c-d)^2/d^(1/2) 
/(c+d)^(3/2)/f-(-A*d+B*c)*cos(f*x+e)/(c^2-d^2)/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 = 10.61 (sec) , antiderivative size = 736, normalized size of antiderivative = 3.56 \[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((8 + 8*I)*(-1)^(3/4)*(A - B)*ArcTa 
nh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])] - ((-(A*d*(3*c + d)) + 
B*(c^2 + 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) & ]))/(Sqrt[d]*(c + d)^(3/2)) + ((-(A*d*(3*c + d)) + B*(c^2 + 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 + T 
an[(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) & 
 ]))/(Sqrt[d]*(c + d)^(3/2)) - (4*(c - d)*(B*c - A*d)*(Cos[(e + f*x)/2] - 
Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x]))))/(4*(c - d)^2*f*Sqrt[a* 
(1 + Sin[e + f*x])])
 

Rubi [A] (verified)

Time = 1.11 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {3042, 3463, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3463

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3464

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 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. \(898\) vs. \(2(182)=364\).

Time = 0.45 (sec) , antiderivative size = 899, normalized size of antiderivative = 4.34

method	result	size
default	\(\text {Expression too large to display}\)	\(899\)

Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^2,x,method=_R 
ETURNVERBOSE)
 

Output:

(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)/a^(5/2)*(sin(f*x+e)*d*(3*A*arctan 
h((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c*d+A*arctanh((a-a 
*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*d^2-B*arctanh((a-a*sin(f 
*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c^2-B*arctanh((a-a*sin(f*x+e)) 
^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*c*d-2*B*arctanh((a-a*sin(f*x+e))^(1/ 
2)*d/(a*c*d+a*d^2)^(1/2))*a^(5/2)*d^2-A*((c+d)*a*d)^(1/2)*arctanh(1/2*(a-a 
*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*c-A*((c+d)*a*d)^(1/2)*arct 
anh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*d+B*((c+d)*a*d 
)^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*c+ 
B*((c+d)*a*d)^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^ 
(1/2)*a^2*d)+3*A*a^(5/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1 
/2))*c^2*d+A*a^(5/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2)) 
*c*d^2-B*a^(5/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*c^3 
-B*a^(5/2)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*c^2*d-2*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+A*a^( 
3/2)*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*c*d-A*a^(3/2)*(a-a*sin(f*x+e 
))^(1/2)*((c+d)*a*d)^(1/2)*d^2-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-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-B*a^(3/2)*(a-a* 
sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*c^2+B*a^(3/2)*(a-a*sin(f*x+e))^(1/2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 952 vs. \(2 (182) = 364\).

Time = 1.96 (sec) , antiderivative size = 2159, normalized size of antiderivative = 10.43 \[ \int \frac {A+B \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^2,x, al 
gorithm="fricas")
 

Output:

[-1/4*((B*c^3 - (3*A - 2*B)*c^2*d - (4*A - 3*B)*c*d^2 - (A - 2*B)*d^3 - (B 
*c^2*d - (3*A - B)*c*d^2 - (A - 2*B)*d^3)*cos(f*x + e)^2 + (B*c^3 - (3*A - 
 B)*c^2*d - (A - 2*B)*c*d^2)*cos(f*x + e) + (B*c^3 - (3*A - 2*B)*c^2*d - ( 
4*A - 3*B)*c*d^2 - (A - 2*B)*d^3 + (B*c^2*d - (3*A - B)*c*d^2 - (A - 2*B)* 
d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*c*d + a*d^2)*log((a*d^2*cos(f*x + 
e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)^2 - 4*sq 
rt(a*c*d + a*d^2)*(d*cos(f*x + e)^2 - (c + 2*d)*cos(f*x + e) + (d*cos(f*x 
+ e) + c + 3*d)*sin(f*x + e) - c - 3*d)*sqrt(a*sin(f*x + e) + a) - (a*c^2 
+ 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c* 
d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e))*sin(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)*c 
os(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))) - 2*sqrt(2)*((A - B)*a*c^3*d + 3*(A - B)*a*c^2*d^2 + 3*(A 
 - B)*a*c*d^3 + (A - B)*a*d^4 - ((A - B)*a*c^2*d^2 + 2*(A - B)*a*c*d^3 + ( 
A - B)*a*d^4)*cos(f*x + e)^2 + ((A - B)*a*c^3*d + 2*(A - B)*a*c^2*d^2 + (A 
 - B)*a*c*d^3)*cos(f*x + e) + ((A - B)*a*c^3*d + 3*(A - B)*a*c^2*d^2 + 3*( 
A - B)*a*c*d^3 + (A - B)*a*d^4 + ((A - B)*a*c^2*d^2 + 2*(A - B)*a*c*d^3 + 
(A - B)*a*d^4)*cos(f*x + e))*sin(f*x + e))*log(-(cos(f*x + e)^2 - (cos(f*x 
 + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) 
 - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 - (c...
 

Sympy [F(-1)]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**(1/2)/(c+d*sin(f*x+e))**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^2,x, al 
gorithm="maxima")
 

Output:

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

Giac [F(-2)]

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

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

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^ 
2),x)
 

Output:

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^(1/2)*(c + d*sin(e + f*x))^ 
2), x)
 

Reduce [F]

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

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^2,x)
 

Output:

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