\(\int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx\) [206]

Optimal result

Integrand size = 38, antiderivative size = 166 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {4 (A-B) c^2 \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}-\frac {2 (A-3 B) c^2 \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) \sqrt {c-c \sin (e+f x)}}-\frac {2 B c^2 \cos (e+f x) (a+a \sin (e+f x))^{2+m}}{a^2 f (5+2 m) \sqrt {c-c \sin (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 6.78 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.05 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^m \sqrt {c-c \sin (e+f x)} \left (50 A-39 B+40 A m-16 B m+8 A m^2-4 B m^2+B \left (3+8 m+4 m^2\right ) \cos (2 (e+f x))-2 (1+2 m) (5 A-9 B+2 A m-2 B m) \sin (e+f x)\right )}{f (1+2 m) (3+2 m) (5+2 m) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \] Input:

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

Output:

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^m*Sqrt[c - 
 c*Sin[e + f*x]]*(50*A - 39*B + 40*A*m - 16*B*m + 8*A*m^2 - 4*B*m^2 + B*(3 
 + 8*m + 4*m^2)*Cos[2*(e + f*x)] - 2*(1 + 2*m)*(5*A - 9*B + 2*A*m - 2*B*m) 
*Sin[e + f*x]))/(f*(1 + 2*m)*(3 + 2*m)*(5 + 2*m)*(Cos[(e + f*x)/2] - Sin[( 
e + f*x)/2]))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3452, 3042, 3219, 3042, 3217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3452

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3219

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3217

\(\displaystyle -\frac {(B (3-2 m)-A (2 m+5)) \left (\frac {8 c^2 \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) (2 m+3) \sqrt {c-c \sin (e+f x)}}+\frac {2 c \cos (e+f x) \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^m}{f (2 m+3)}\right )}{2 m+5}-\frac {2 B \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^m}{f (2 m+5)}\)

Input:

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

Output:

(-2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2))/(f*( 
5 + 2*m)) - ((B*(3 - 2*m) - A*(5 + 2*m))*((8*c^2*Cos[e + f*x]*(a + a*Sin[e 
 + f*x])^m)/(f*(1 + 2*m)*(3 + 2*m)*Sqrt[c - c*Sin[e + f*x]]) + (2*c*Cos[e 
+ f*x]*(a + a*Sin[e + f*x])^m*Sqrt[c - c*Sin[e + f*x]])/(f*(3 + 2*m))))/(5 
 + 2*m)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3217

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

rule 3219

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

rule 3452

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)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + 
n + 1))), x] - Simp[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1))   Int[( 
a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, 
e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, 
 -2^(-1)] && NeQ[m + n + 1, 0]
 

Maple [F]

\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{\frac {3}{2}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.89 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {2 \, {\left ({\left (4 \, B c m^{2} + 8 \, B c m + 3 \, B c\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (A + B\right )} c m + {\left (4 \, A c m^{2} + 12 \, {\left (A - B\right )} c m + {\left (5 \, A - 6 \, B\right )} c\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (5 \, A - 3 \, B\right )} c + {\left (4 \, {\left (A - B\right )} c m^{2} + 4 \, {\left (5 \, A - 3 \, B\right )} c m + {\left (25 \, A - 21 \, B\right )} c\right )} \cos \left (f x + e\right ) + {\left (8 \, {\left (A + B\right )} c m + {\left (4 \, B c m^{2} + 8 \, B c m + 3 \, B c\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left (5 \, A - 3 \, B\right )} c - {\left (4 \, {\left (A - B\right )} c m^{2} + 4 \, {\left (3 \, A - 5 \, B\right )} c m + {\left (5 \, A - 9 \, B\right )} c\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right ) - {\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \sin \left (f x + e\right ) + 15 \, f} \] Input:

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

Output:

2*((4*B*c*m^2 + 8*B*c*m + 3*B*c)*cos(f*x + e)^3 + 8*(A + B)*c*m + (4*A*c*m 
^2 + 12*(A - B)*c*m + (5*A - 6*B)*c)*cos(f*x + e)^2 + 4*(5*A - 3*B)*c + (4 
*(A - B)*c*m^2 + 4*(5*A - 3*B)*c*m + (25*A - 21*B)*c)*cos(f*x + e) + (8*(A 
 + B)*c*m + (4*B*c*m^2 + 8*B*c*m + 3*B*c)*cos(f*x + e)^2 + 4*(5*A - 3*B)*c 
 - (4*(A - B)*c*m^2 + 4*(3*A - 5*B)*c*m + (5*A - 9*B)*c)*cos(f*x + e))*sin 
(f*x + e))*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(8*f*m^3 + 36* 
f*m^2 + 46*f*m + (8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos(f*x + e) - (8*f* 
m^3 + 36*f*m^2 + 46*f*m + 15*f)*sin(f*x + e) + 15*f)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 498 vs. \(2 (160) = 320\).

Time = 0.15 (sec) , antiderivative size = 498, normalized size of antiderivative = 3.00 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx =\text {Too large to display} \] Input:

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

Output:

-2*((a^m*c^(3/2)*(2*m + 5) - a^m*c^(3/2)*(2*m - 3)*sin(f*x + e)/(cos(f*x + 
 e) + 1) - a^m*c^(3/2)*(2*m - 3)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^m 
*c^(3/2)*(2*m + 5)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*A*e^(2*m*log(sin(f 
*x + e)/(cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^ 
2 + 1))/((4*m^2 + 8*m + 3)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2) 
) - 2*(a^m*c^(3/2)*(2*m + 9) - 2*(2*m^2 + 9*m)*a^m*c^(3/2)*sin(f*x + e)/(c 
os(f*x + e) + 1) + (4*m^2 + 15)*a^m*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 
 1)^2 + (4*m^2 + 15)*a^m*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 2*( 
2*m^2 + 9*m)*a^m*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^m*c^(3/2) 
*(2*m + 9)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*B*e^(2*m*log(sin(f*x + e)/ 
(cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/ 
((8*m^3 + 36*m^2 + 46*m + (8*m^3 + 36*m^2 + 46*m + 15)*sin(f*x + e)^2/(cos 
(f*x + e) + 1)^2 + 15)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2)))/f
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 43.21 (sec) , antiderivative size = 480, normalized size of antiderivative = 2.89 \[ \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {c\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (45\,A-30\,B+28\,A\,m+4\,B\,m+4\,A\,m^2\right )}{f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {c\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (A\,45{}\mathrm {i}-B\,30{}\mathrm {i}+A\,m\,28{}\mathrm {i}+B\,m\,4{}\mathrm {i}+A\,m^2\,4{}\mathrm {i}\right )}{f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {B\,c\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (m^2\,4{}\mathrm {i}+m\,8{}\mathrm {i}+3{}\mathrm {i}\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {B\,c\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (4\,m^2+8\,m+3\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (2\,m+1\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (10\,A-15\,B+4\,A\,m-2\,B\,m\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}+\frac {c\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (2\,m+1\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (A\,10{}\mathrm {i}-B\,15{}\mathrm {i}+A\,m\,4{}\mathrm {i}-B\,m\,2{}\mathrm {i}\right )}{2\,f\,\left (m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}\right )}\right )}{{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (8\,m^3+36\,m^2+46\,m+15\right )}{m^3\,8{}\mathrm {i}+m^2\,36{}\mathrm {i}+m\,46{}\mathrm {i}+15{}\mathrm {i}}} \] Input:

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

Output:

((c - c*sin(e + f*x))^(1/2)*((c*exp(e*3i + f*x*3i)*(a + a*sin(e + f*x))^m* 
(45*A - 30*B + 28*A*m + 4*B*m + 4*A*m^2))/(f*(m*46i + m^2*36i + m^3*8i + 1 
5i)) + (c*exp(e*2i + f*x*2i)*(a + a*sin(e + f*x))^m*(A*45i - B*30i + A*m*2 
8i + B*m*4i + A*m^2*4i))/(f*(m*46i + m^2*36i + m^3*8i + 15i)) + (B*c*(a + 
a*sin(e + f*x))^m*(m*8i + m^2*4i + 3i))/(2*f*(m*46i + m^2*36i + m^3*8i + 1 
5i)) + (B*c*exp(e*5i + f*x*5i)*(a + a*sin(e + f*x))^m*(8*m + 4*m^2 + 3))/( 
2*f*(m*46i + m^2*36i + m^3*8i + 15i)) + (c*exp(e*1i + f*x*1i)*(2*m + 1)*(a 
 + a*sin(e + f*x))^m*(10*A - 15*B + 4*A*m - 2*B*m))/(2*f*(m*46i + m^2*36i 
+ m^3*8i + 15i)) + (c*exp(e*4i + f*x*4i)*(2*m + 1)*(a + a*sin(e + f*x))^m* 
(A*10i - B*15i + A*m*4i - B*m*2i))/(2*f*(m*46i + m^2*36i + m^3*8i + 15i))) 
)/(exp(e*3i + f*x*3i) + (exp(e*2i + f*x*2i)*(46*m + 36*m^2 + 8*m^3 + 15))/ 
(m*46i + m^2*36i + m^3*8i + 15i))
 

Reduce [F]

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

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

Output:

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