\(\int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\) [298]

Optimal result

Integrand size = 37, antiderivative size = 191 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=-\frac {a^{3/2} \left (A d (c+3 d)-B \left (3 c^2+3 c d-2 d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{d^{5/2} (c+d)^{3/2} f}-\frac {a^2 (3 B c-A d+2 B d) \cos (e+f x)}{d^2 (c+d) f \sqrt {a+a \sin (e+f x)}}+\frac {a (B c-A d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{d (c+d) f (c+d \sin (e+f x))} \] Output:

-a^(3/2)*(A*d*(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))/d^(5/2)/(c+d)^(3/2)/f-a^2*(-A*d+ 
3*B*c+2*B*d)*cos(f*x+e)/d^2/(c+d)/f/(a+a*sin(f*x+e))^(1/2)+a*(-A*d+B*c)*co 
s(f*x+e)*(a+a*sin(f*x+e))^(1/2)/d/(c+d)/f/(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 = 13.43 (sec) , antiderivative size = 922, normalized size of antiderivative = 4.83 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

((a*(1 + Sin[e + f*x]))^(3/2)*(-8*B*Sqrt[d]*Cos[(e + f*x)/2] + ((A*d*(c + 
3*d) + B*(-3*c^2 - 3*c*d + 2*d^2))*((c + d)*(e + f*x - 2*Log[Sec[(e + f*x) 
/4]^2]) + Sqrt[c + d]*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & 
, (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f 
*x)/4]] - d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 
+ Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 - c*Sqrt 
[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[(e + f*x) 
/4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 + 3*d*Sqrt[c + d]*Log 
[-#1 + Tan[(e + f*x)/4]]*#1^2 - c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]* 
#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ]))/(c + d)^(5/2) + ((A*d*(c + 3*d 
) + B*(-3*c^2 - 3*c*d + 2*d^2))*(-((c + d)*(e + f*x - 2*Log[Sec[(e + f*x)/ 
4]^2])) + Sqrt[c + d]*RootSum[c + 4*d*#1 + 2*c*#1^2 - 4*d*#1^3 + c*#1^4 & 
, (-(c*Sqrt[d]*Log[-#1 + Tan[(e + f*x)/4]]) - d^(3/2)*Log[-#1 + Tan[(e + f 
*x)/4]] + d*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]] - 2*c*Sqrt[d]*Log[-#1 
+ Tan[(e + f*x)/4]]*#1 - 2*d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt 
[c + d]*Log[-#1 + Tan[(e + f*x)/4]]*#1 + c*Sqrt[d]*Log[-#1 + Tan[(e + f*x) 
/4]]*#1^2 + d^(3/2)*Log[-#1 + Tan[(e + f*x)/4]]*#1^2 - 3*d*Sqrt[c + d]*Log 
[-#1 + Tan[(e + f*x)/4]]*#1^2 + c*Sqrt[c + d]*Log[-#1 + Tan[(e + f*x)/4]]* 
#1^3)/(-d - c*#1 + 3*d*#1^2 - c*#1^3) & ]))/(c + d)^(5/2) + 8*B*Sqrt[d]*Si 
n[(e + f*x)/2] - (4*Sqrt[d]*(-c + d)*(-(B*c) + A*d)*(Cos[(e + f*x)/2] -...
 

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {3042, 3454, 27, 3042, 3460, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3460

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3252

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(a*(B*c - A*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(d*(c + d)*f*(c + d* 
Sin[e + f*x])) - ((2*a^(3/2)*(A*d*(c + 3*d) - B*(3*c^2 + 3*c*d - 2*d^2))*A 
rcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x] 
])])/(d^(3/2)*Sqrt[c + d]*f) + (2*a^2*(3*B*c - A*d + 2*B*d)*Cos[e + f*x])/ 
(d*f*Sqrt[a + a*Sin[e + f*x]]))/(2*d*(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 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 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 3454

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], 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] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3460

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(591\) vs. \(2(173)=346\).

Time = 0.54 (sec) , antiderivative size = 592, normalized size of antiderivative = 3.10

method	result	size
default	\(\frac {a \left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (-\sin \left (f x +e \right ) d \left (A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a c 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 \,d^{2}-3 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a \,c^{2}-3 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a c d +2 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a \,d^{2}+2 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, c +2 B \sqrt {\left (c +d \right ) a d}\, \sqrt {a -a \sin \left (f x +e \right )}\, d \right )-A \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a \,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 c \,d^{2}+3 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a \,c^{3}+3 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a \,c^{2} d -2 B \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, d}{\sqrt {c d a +d^{2} a}}\right ) a c \,d^{2}+A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, c d -A \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, d^{2}-3 B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, c^{2}-B \sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {\left (c +d \right ) a d}\, c d \right )}{d^{2} \left (c +d \right ) \left (c +d \sin \left (f x +e \right )\right ) \sqrt {\left (c +d \right ) a d}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(592\)

Input:

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

Output:

a*(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(1/2)*(-sin(f*x+e)*d*(A*arctanh((a-a* 
sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c*d+3*A*arctanh((a-a*sin(f*x+e) 
)^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*d^2-3*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/ 
(a*c*d+a*d^2)^(1/2))*a*c^2-3*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d 
^2)^(1/2))*a*c*d+2*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2)) 
*a*d^2+2*B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*c+2*B*((c+d)*a*d)^(1/2 
)*(a-a*sin(f*x+e))^(1/2)*d)-A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^ 
2)^(1/2))*a*c^2*d-3*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2) 
)*a*c*d^2+3*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c^3+ 
3*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c^2*d-2*B*arct 
anh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a*c*d^2+A*(a-a*sin(f*x+e 
))^(1/2)*((c+d)*a*d)^(1/2)*c*d-A*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)* 
d^2-3*B*(a-a*sin(f*x+e))^(1/2)*((c+d)*a*d)^(1/2)*c^2-B*(a-a*sin(f*x+e))^(1 
/2)*((c+d)*a*d)^(1/2)*c*d)/d^2/(c+d)/(c+d*sin(f*x+e))/((c+d)*a*d)^(1/2)/co 
s(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. 556 vs. \(2 (173) = 346\).

Time = 0.81 (sec) , antiderivative size = 1428, normalized size of antiderivative = 7.48 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*((3*B*a*c^3 - (A - 6*B)*a*c^2*d - (4*A - B)*a*c*d^2 - (3*A + 2*B)*a*d 
^3 - (3*B*a*c^2*d - (A - 3*B)*a*c*d^2 - (3*A + 2*B)*a*d^3)*cos(f*x + e)^2 
+ (3*B*a*c^3 - (A - 3*B)*a*c^2*d - (3*A + 2*B)*a*c*d^2)*cos(f*x + e) + (3* 
B*a*c^3 - (A - 6*B)*a*c^2*d - (4*A - B)*a*c*d^2 - (3*A + 2*B)*a*d^3 + (3*B 
*a*c^2*d - (A - 3*B)*a*c*d^2 - (3*A + 2*B)*a*d^3)*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)*cos(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*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + 4*(3*B*a*c^2 
 - (A + B)*a*c*d + (A - 2*B)*a*d^2 + 2*(B*a*c*d + B*a*d^2)*cos(f*x + e)^2 
+ (3*B*a*c^2 - (A - B)*a*c*d + A*a*d^2)*cos(f*x + e) - (3*B*a*c^2 - (A + B 
)*a*c*d + (A - 2*B)*a*d^2 - 2*(B*a*c*d + B*a*d^2)*cos(f*x + e))*sin(f*x + 
e))*sqrt(a*sin(f*x + e) + a))/((c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^2*d^2 + 
 c*d^3)*f*cos(f*x + e) - (c^2*d^2 + 2*c*d^3 + d^4)*f - ((c*d^3 + d^4)*f*co 
s(f*x + e) + (c^2*d^2 + 2*c*d^3 + d^4)*f)*sin(f*x + e)), -1/2*((3*B*a*c...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 364 vs. \(2 (173) = 346\).

Time = 0.24 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.91 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\frac {\sqrt {2} {\left (\frac {4 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{d^{2}} + \frac {\sqrt {2} {\left (3 \, B a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - A a c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, A a d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 2 \, B a d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \arctan \left (\frac {\sqrt {2} d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c d - d^{2}}}\right )}{{\left (c d^{2} + d^{3}\right )} \sqrt {-c d - d^{2}}} - \frac {2 \, {\left (B a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - A a c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - B a c d \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A a d^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (c d^{2} + d^{3}\right )} {\left (2 \, d \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )}}\right )} \sqrt {a}}{2 \, f} \] Input:

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

Output:

1/2*sqrt(2)*(4*B*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f 
*x + 1/2*e)/d^2 + sqrt(2)*(3*B*a*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 
 A*a*c*d*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*a*c*d*sgn(cos(-1/4*pi + 
 1/2*f*x + 1/2*e)) - 3*A*a*d^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 2*B*a 
*d^2*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*d^2 + d^3)*sqrt(-c*d - d^2)) - 2*(B* 
a*c^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e) - 
 A*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 
) - B*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) + A*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))/((c*d^2 + d^3)*(2*d*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 - c - d)))*s 
qrt(a)/f
 

Mupad [F(-1)]

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

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

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

Output:

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