\(\int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx\) [292]

Optimal result

Integrand size = 37, antiderivative size = 192 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=-\frac {\sqrt {a} (3 A d+B (c+4 d)) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{4 d^{3/2} (c+d)^{5/2} f}+\frac {a (B c-A d) \cos (e+f x)}{2 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^2}-\frac {a (3 A d+B (c+4 d)) \cos (e+f x)}{4 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \] Output:

-1/4*a^(1/2)*(3*A*d+B*(c+4*d))*arctanh(a^(1/2)*d^(1/2)*cos(f*x+e)/(c+d)^(1 
/2)/(a+a*sin(f*x+e))^(1/2))/d^(3/2)/(c+d)^(5/2)/f+1/2*a*(-A*d+B*c)*cos(f*x 
+e)/d/(c+d)/f/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^2-1/4*a*(3*A*d+B*(c+ 
4*d))*cos(f*x+e)/d/(c+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 = 15.17 (sec) , antiderivative size = 967, normalized size of antiderivative = 5.04 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((1/16 + I/16)*Sqrt[a*(1 + Sin[e + f*x])]*(((3*A*d + B*(c + 4*d))*(Cos[e/2 
] + I*Sin[e/2])*((-1 + I)*x*Cos[e] + (RootSum[-d + (2*I)*c*E^(I*e)*#1^2 + 
d*E^((2*I)*e)*#1^4 & , ((1 + I)*d*Sqrt[E^((-I)*e)]*f*x - (2 - 2*I)*d*Sqrt[ 
E^((-I)*e)]*Log[E^((I/2)*f*x) - #1] - I*Sqrt[d]*Sqrt[c + d]*f*x*#1 + 2*Sqr 
t[d]*Sqrt[c + d]*Log[E^((I/2)*f*x) - #1]*#1 + ((1 - I)*c*f*x*#1^2)/Sqrt[E^ 
((-I)*e)] + ((2 + 2*I)*c*Log[E^((I/2)*f*x) - #1]*#1^2)/Sqrt[E^((-I)*e)] - 
Sqrt[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 - (2*I)*Sqrt[d]*Sqrt[c + d]*E^(I*e)*L 
og[E^((I/2)*f*x) - #1]*#1^3)/(d - I*c*E^(I*e)*#1^2) & ]*(Cos[e] + I*(-1 + 
Sin[e]))*Sqrt[Cos[e] - I*Sin[e]])/(4*f) + (1 + I)*x*Sin[e]))/((c + d)^(5/2 
)*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) + ((3*A*d + B*(c + 4 
*d))*(Cos[e/2] + I*Sin[e/2])*((1 - I)*x*Cos[e] - (1 + I)*x*Sin[e] + (RootS 
um[-d + (2*I)*c*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ((1 - I)*d*Sqrt[E^(( 
-I)*e)]*f*x + (2 + 2*I)*d*Sqrt[E^((-I)*e)]*Log[E^((I/2)*f*x) - #1] + Sqrt[ 
d]*Sqrt[c + d]*f*x*#1 + (2*I)*Sqrt[d]*Sqrt[c + d]*Log[E^((I/2)*f*x) - #1]* 
#1 - ((1 + I)*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 - 2*I)*c*Log[E^((I/2)*f*x 
) - #1]*#1^2)/Sqrt[E^((-I)*e)] - I*Sqrt[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 + 
2*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Log[E^((I/2)*f*x) - #1]*#1^3)/(d - I*c*E^(I* 
e)*#1^2) & ]*Sqrt[Cos[e] - I*Sin[e]]*(-1 - I*Cos[e] + Sin[e]))/(4*f)))/((c 
 + d)^(5/2)*(Cos[e] + I*(-1 + Sin[e]))*Sqrt[Cos[e] - I*Sin[e]]) - ((4 - 4* 
I)*Sqrt[d]*(-(B*c) + A*d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + ...
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {3042, 3459, 3042, 3251, 3042, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3459

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3251

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 3251

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e 
+ f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim 
p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2)))   Int[Sqrt[a + b*Sin[e 
+ f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, 
-1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
 

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 3459

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( 
f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp 
[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) 
*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* 
c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d))   Int[Sqrt[a + b*Sin[e + f*x] 
]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, 
-1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(627\) vs. \(2(168)=336\).

Time = 0.51 (sec) , antiderivative size = 628, normalized size of antiderivative = 3.27

method	result	size
default	\(\frac {\left (\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} d^{2} \left (3 A d +B c +4 B d \right ) \cos \left (f x +e \right )^{2}-2 \sin \left (f x +e \right ) \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c d \left (3 A d +B c +4 B d \right )+3 A \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, d^{2}-3 A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c^{2} d -3 A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} d^{3}+B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, c d +4 B \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {\left (c +d \right ) a d}\, d^{2}-a^{2} \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) B \,c^{3}-4 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c^{2} d -B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} c \,d^{2}-4 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a^{2} d^{3}-5 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a c d -5 A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a \,d^{2}+B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a \,c^{2}-3 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a c d -4 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, a \,d^{2}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (1+\sin \left (f x +e \right )\right )}{4 a \sqrt {\left (c +d \right ) a d}\, \left (c +d \sin \left (f x +e \right )\right )^{2} \left (c +d \right )^{2} d \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(628\)

Input:

int((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x,method=_R 
ETURNVERBOSE)
 

Output:

1/4/a*(arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^2*(3*A* 
d+B*c+4*B*d)*cos(f*x+e)^2-2*sin(f*x+e)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a 
*c*d+a*d^2)^(1/2))*a^2*c*d*(3*A*d+B*c+4*B*d)+3*A*(a-a*sin(f*x+e))^(3/2)*(( 
c+d)*a*d)^(1/2)*d^2-3*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/ 
2))*a^2*c^2*d-3*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^ 
2*d^3+B*(a-a*sin(f*x+e))^(3/2)*((c+d)*a*d)^(1/2)*c*d+4*B*(a-a*sin(f*x+e))^ 
(3/2)*((c+d)*a*d)^(1/2)*d^2-a^2*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a* 
d^2)^(1/2))*B*c^3-4*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2) 
)*a^2*c^2*d-B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c* 
d^2-4*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^3-5*A* 
(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*c*d-5*A*(a-a*sin(f*x+e))^(1/2)* 
((c+d)*a*d)^(1/2)*a*d^2+B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*c^2-3 
*B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*a*c*d-4*B*(a-a*sin(f*x+e))^(1/ 
2)*((c+d)*a*d)^(1/2)*a*d^2)*(-a*(sin(f*x+e)-1))^(1/2)*(1+sin(f*x+e))/((c+d 
)*a*d)^(1/2)/(c+d*sin(f*x+e))^2/(c+d)^2/d/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. 717 vs. \(2 (168) = 336\).

Time = 1.14 (sec) , antiderivative size = 1750, normalized size of antiderivative = 9.11 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, al 
gorithm="fricas")
 

Output:

[-1/16*((B*c^3 + 3*(A + 2*B)*c^2*d + 3*(2*A + 3*B)*c*d^2 + (3*A + 4*B)*d^3 
 - (B*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x + e)^3 - (2*B*c^2*d + 3*(2*A + 3*B) 
*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x + e)^2 + (B*c^3 + (3*A + 4*B)*c^2*d + B* 
c*d^2 + (3*A + 4*B)*d^3)*cos(f*x + e) + (B*c^3 + 3*(A + 2*B)*c^2*d + 3*(2* 
A + 3*B)*c*d^2 + (3*A + 4*B)*d^3 - (B*c*d^2 + (3*A + 4*B)*d^3)*cos(f*x + e 
)^2 + 2*(B*c^2*d + (3*A + 4*B)*c*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a/( 
c*d + 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*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3)*co 
s(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x + e) - (c^2*d + 4*c*d^2 + 
 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a 
)*sqrt(a/(c*d + d^2)) - (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)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*c 
os(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(B*c^2 - (5*A + B)*c*d 
 + (A + 4*B)*d^2 - (B*c*d + (3*A + 4*B)*d^2)*cos(f*x + e)^2 + (B*c^2 - (5* 
A + 2*B)*c*d - 2*A*d^2)*cos(f*x + e) - (B*c^2 - (5*A + B)*c*d + (A + 4*B)* 
d^2 + (B*c*d + (3*A + 4*B)*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x 
 + e) + a))/((c^2*d^3 + 2*c*d^4 + d^5)*f*cos(f*x + e)^3 + (2*c^3*d^2 + 5*c 
^2*d^3 + 4*c*d^4 + d^5)*f*cos(f*x + e)^2 - (c^4*d + 2*c^3*d^2 + 2*c^2*d...
 

Sympy [F(-1)]

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

integrate((a+a*sin(f*x+e))**(1/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))**3,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,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)^3, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (168) = 336\).

Time = 0.27 (sec) , antiderivative size = 422, normalized size of antiderivative = 2.20 \[ \int \frac {\sqrt {a+a \sin (e+f x)} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, al 
gorithm="giac")
 

Output:

-1/8*sqrt(2)*sqrt(a)*(sqrt(2)*(B*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3 
*A*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*B*d*sgn(cos(-1/4*pi + 1/2*f*x 
 + 1/2*e)))*arctan(sqrt(2)*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)/sqrt(-c*d - d^ 
2))/((c^2*d + 2*c*d^2 + d^3)*sqrt(-c*d - d^2)) + 2*(2*B*c*d*sgn(cos(-1/4*p 
i + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 6*A*d^2*sgn(cos(- 
1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + 8*B*d^2*sgn( 
cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^3 + B*c^2*s 
gn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 5*A*c* 
d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 3*B 
*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 
5*A*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) 
 - 4*B*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2 
*e))/((c^2*d + 2*c*d^2 + d^3)*(2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c - 
d)^2))/f
 

Mupad [F(-1)]

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

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2))/(c + d*sin(e + f*x)) 
^3,x)
 

Output:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(1/2))/(c + d*sin(e + f*x)) 
^3, x)
 

Reduce [F]

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

int((a+a*sin(f*x+e))^(1/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x)
 

Output:

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