\(\int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\) [86]

Optimal result

Integrand size = 36, antiderivative size = 115 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {a (A+5 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} c^{3/2} f}+\frac {a (A+B) \cos (e+f x)}{f (c-c \sin (e+f x))^{3/2}}+\frac {2 a B \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 6.20 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.37 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {a \sec (e+f x) \left (\sqrt {2} (A+5 B) \arctan \left (\frac {\sqrt {-c (1+\sin (e+f x))}}{\sqrt {2} \sqrt {c}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sqrt {-c (1+\sin (e+f x))}+2 \sqrt {c} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 (A+3 B-2 B \sin (e+f x))\right )}{2 c^{3/2} f \sqrt {c-c \sin (e+f x)}} \] Input:

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

Output:

(a*Sec[e + f*x]*(Sqrt[2]*(A + 5*B)*ArcTan[Sqrt[-(c*(1 + Sin[e + f*x]))]/(S 
qrt[2]*Sqrt[c])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sqrt[-(c*(1 + Sin 
[e + f*x]))] + 2*Sqrt[c]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2*(A + 3*B 
- 2*B*Sin[e + f*x])))/(2*c^(3/2)*f*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3042, 3446, 3042, 3336, 25, 3042, 3230, 3042, 3128, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3336

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3230

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

a*c*(((A + B)*Cos[e + f*x])/(c*f*(c - c*Sin[e + f*x])^(3/2)) - ((Sqrt[2]*( 
A + 5*B)*Sqrt[c]*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e 
+ f*x]])])/f - (4*B*c*Cos[e + f*x])/(f*Sqrt[c - c*Sin[e + f*x]]))/(2*c^3))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 3230

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( 
f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1))   Int[(a + b*Sin[e 
 + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] 
&& EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]
 

rule 3336

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*( 
(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[2*(b*c - a*d)*Cos 
[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(2*m + 3))), x] + Simp[1/(b^ 
3*(2*m + 3))   Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d* 
(2*m + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 
- b^2, 0] && LtQ[m, -3/2]
 

rule 3446

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin 
[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* 
d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] 
&& GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(226\) vs. \(2(102)=204\).

Time = 0.57 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.97

method	result	size
default	\(\frac {a \left (A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c +5 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c -A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -5 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -4 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\, B \sin \left (f x +e \right )+2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\, A +6 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\, B \right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{2 c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(227\)
parts	\(-\frac {A a \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )-\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{4 c^{\frac {7}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {B a \left (-7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c +8 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\, \sin \left (f x +e \right )+7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c -10 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{4 c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {a \left (A +B \right ) \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c -3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c +2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {c}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{4 c^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(393\)

Input:

int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/2/c^(5/2)*a*(A*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1 
/2))*sin(f*x+e)*c+5*B*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2) 
/c^(1/2))*sin(f*x+e)*c-A*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1 
/2)/c^(1/2))*c-5*B*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^ 
(1/2))*c-4*(c*(1+sin(f*x+e)))^(1/2)*c^(1/2)*B*sin(f*x+e)+2*(c*(1+sin(f*x+e 
)))^(1/2)*c^(1/2)*A+6*(c*(1+sin(f*x+e)))^(1/2)*c^(1/2)*B)*(c*(1+sin(f*x+e) 
))^(1/2)/cos(f*x+e)/(c-c*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. 318 vs. \(2 (102) = 204\).

Time = 0.09 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.77 \[ \int \frac {(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\frac {\sqrt {2} {\left ({\left (A + 5 \, B\right )} a c \cos \left (f x + e\right )^{2} - {\left (A + 5 \, B\right )} a c \cos \left (f x + e\right ) - 2 \, {\left (A + 5 \, B\right )} a c + {\left ({\left (A + 5 \, B\right )} a c \cos \left (f x + e\right ) + 2 \, {\left (A + 5 \, B\right )} a c\right )} \sin \left (f x + e\right )\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} - 4 \, {\left (2 \, B a \cos \left (f x + e\right )^{2} + {\left (A + 3 \, B\right )} a \cos \left (f x + e\right ) + {\left (A + B\right )} a - {\left (2 \, B a \cos \left (f x + e\right ) - {\left (A + B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f + {\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, algo 
rithm="fricas")
 

Output:

1/4*(sqrt(2)*((A + 5*B)*a*c*cos(f*x + e)^2 - (A + 5*B)*a*c*cos(f*x + e) - 
2*(A + 5*B)*a*c + ((A + 5*B)*a*c*cos(f*x + e) + 2*(A + 5*B)*a*c)*sin(f*x + 
 e))*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sq 
rt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 3*cos( 
f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x 
+ e) - 2))/sqrt(c) - 4*(2*B*a*cos(f*x + e)^2 + (A + 3*B)*a*cos(f*x + e) + 
(A + B)*a - (2*B*a*cos(f*x + e) - (A + B)*a)*sin(f*x + e))*sqrt(-c*sin(f*x 
 + e) + c))/(c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e) - 2*c^2*f + (c^2*f* 
cos(f*x + e) + 2*c^2*f)*sin(f*x + e))
 

Sympy [F]

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

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

Output:

a*(Integral(A/(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c*sqrt(-c*sin(e 
 + f*x) + c)), x) + Integral(A*sin(e + f*x)/(-c*sqrt(-c*sin(e + f*x) + c)* 
sin(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), x) + Integral(B*sin(e + f*x)/ 
(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c*sqrt(-c*sin(e + f*x) + c)), 
 x) + Integral(B*sin(e + f*x)**2/(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x 
) + c*sqrt(-c*sin(e + f*x) + c)), x))
 

Maxima [F]

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

integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(3/2),x, algo 
rithm="maxima")
 

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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