\(\int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx\) [322]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F(-2)]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 157 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx=-\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} c^{7/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 \cos ^3(e+f x)}{12 f (c-c \sin (e+f x))^{7/2}}+\frac {5 a^3 \cos (e+f x)}{8 c^2 f (c-c \sin (e+f x))^{3/2}} \] Output:

-5/16*a^3*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^(7/2)/f+1/3*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(11/2)-5/12*a 
^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(7/2)+5/8*a^3*cos(f*x+e)/c^2/f/(c-c*sin 
(f*x+e))^(3/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 9.86 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.96 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (32 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-52 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+33 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+(15+15 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6+64 \sin \left (\frac {1}{2} (e+f x)\right )-104 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+66 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3}{24 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (c-c \sin (e+f x))^{7/2}} \] Input:

Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(7/2),x]
 

Output:

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(32*(Cos[(e + f*x)/2] - Sin[(e 
+ f*x)/2]) - 52*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 + 33*(Cos[(e + f*x 
)/2] - Sin[(e + f*x)/2])^5 + (15 + 15*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1 
)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6 + 
64*Sin[(e + f*x)/2] - 104*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + 
 f*x)/2] + 66*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(1 
 + Sin[e + f*x])^3)/(24*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*S 
in[e + f*x])^(7/2))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 3215, 3042, 3159, 3042, 3159, 3042, 3159, 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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^3}{(c-c \sin (e+f x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^3}{(c-c \sin (e+f x))^{7/2}}dx\)

\(\Big \downarrow \) 3215

\(\displaystyle a^3 c^3 \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{13/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \int \frac {\cos (e+f x)^6}{(c-c \sin (e+f x))^{13/2}}dx\)

\(\Big \downarrow \) 3159

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}}dx}{6 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{9/2}}dx}{6 c^2}\right )\)

\(\Big \downarrow \) 3159

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}}dx}{4 c^2}\right )}{6 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{5/2}}dx}{4 c^2}\right )}{6 c^2}\right )\)

\(\Big \downarrow \) 3159

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{2 c^2}\right )}{4 c^2}\right )}{6 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{2 c^2}\right )}{4 c^2}\right )}{6 c^2}\right )\)

\(\Big \downarrow \) 3128

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c^2 f}+\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}\right )}{4 c^2}\right )}{6 c^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{3 c f (c-c \sin (e+f x))^{11/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} c^{5/2} f}\right )}{4 c^2}\right )}{6 c^2}\right )\)

Input:

Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(7/2),x]
 

Output:

a^3*c^3*(Cos[e + f*x]^5/(3*c*f*(c - c*Sin[e + f*x])^(11/2)) - (5*(Cos[e + 
f*x]^3/(2*c*f*(c - c*Sin[e + f*x])^(7/2)) - (3*(-(ArcTanh[(Sqrt[c]*Cos[e + 
 f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/(Sqrt[2]*c^(5/2)*f)) + Cos[e + 
f*x]/(c*f*(c - c*Sin[e + f*x])^(3/2))))/(4*c^2)))/(6*c^2))
 

Defintions of rubi rules used

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 3159
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
 

rule 3215
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + 
 d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((Lt 
Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 
Maple [A] (verified)

Time = 2.13 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.56

method result size
default \(\frac {a^{3} \left (15 \sin \left (f x +e \right )^{3} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}-45 \sin \left (f x +e \right )^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}+45 \sin \left (f x +e \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}+66 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {c}-15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}-160 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}+120 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {5}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{48 c^{\frac {13}{2}} \left (-1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) \(245\)
parts \(\text {Expression too large to display}\) \(992\)

Input:

int((a+sin(f*x+e)*a)^3/(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/48*a^3/c^(13/2)*(15*sin(f*x+e)^3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^ 
(1/2)*2^(1/2)/c^(1/2))*c^3-45*sin(f*x+e)^2*2^(1/2)*arctanh(1/2*(c*(1+sin(f 
*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^3+45*sin(f*x+e)*2^(1/2)*arctanh(1/2*(c*(1 
+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^3+66*(c*(1+sin(f*x+e)))^(5/2)*c^(1/ 
2)-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^3-16 
0*(c*(1+sin(f*x+e)))^(3/2)*c^(3/2)+120*(c*(1+sin(f*x+e)))^(1/2)*c^(5/2))*( 
c*(1+sin(f*x+e)))^(1/2)/(-1+sin(f*x+e))^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. 440 vs. \(2 (134) = 268\).

Time = 0.10 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.80 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {15 \, \sqrt {2} {\left (a^{3} \cos \left (f x + e\right )^{4} - 3 \, a^{3} \cos \left (f x + e\right )^{3} - 8 \, a^{3} \cos \left (f x + e\right )^{2} + 4 \, a^{3} \cos \left (f x + e\right ) + 8 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{3} + 4 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} \cos \left (f x + e\right ) - 8 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (33 \, a^{3} \cos \left (f x + e\right )^{3} + 19 \, a^{3} \cos \left (f x + e\right )^{2} - 46 \, a^{3} \cos \left (f x + e\right ) - 32 \, a^{3} + {\left (33 \, a^{3} \cos \left (f x + e\right )^{2} + 14 \, a^{3} \cos \left (f x + e\right ) - 32 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{96 \, {\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f + {\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(7/2),x, algorithm="fricas")
 

Output:

1/96*(15*sqrt(2)*(a^3*cos(f*x + e)^4 - 3*a^3*cos(f*x + e)^3 - 8*a^3*cos(f* 
x + e)^2 + 4*a^3*cos(f*x + e) + 8*a^3 + (a^3*cos(f*x + e)^3 + 4*a^3*cos(f* 
x + e)^2 - 4*a^3*cos(f*x + e) - 8*a^3)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f 
*x + e)^2 - 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + si 
n(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 
 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2 
)) - 4*(33*a^3*cos(f*x + e)^3 + 19*a^3*cos(f*x + e)^2 - 46*a^3*cos(f*x + e 
) - 32*a^3 + (33*a^3*cos(f*x + e)^2 + 14*a^3*cos(f*x + e) - 32*a^3)*sin(f* 
x + e))*sqrt(-c*sin(f*x + e) + c))/(c^4*f*cos(f*x + e)^4 - 3*c^4*f*cos(f*x 
 + e)^3 - 8*c^4*f*cos(f*x + e)^2 + 4*c^4*f*cos(f*x + e) + 8*c^4*f + (c^4*f 
*cos(f*x + e)^3 + 4*c^4*f*cos(f*x + e)^2 - 4*c^4*f*cos(f*x + e) - 8*c^4*f) 
*sin(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(7/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(7/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(7/2),x, algorithm="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. \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \] Input:

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(7/2),x)
 

Output:

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(7/2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{7/2}} \, dx=\frac {\sqrt {c}\, a^{3} \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{4}-4 \sin \left (f x +e \right )^{3}+6 \sin \left (f x +e \right )^{2}-4 \sin \left (f x +e \right )+1}d x +\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{4}-4 \sin \left (f x +e \right )^{3}+6 \sin \left (f x +e \right )^{2}-4 \sin \left (f x +e \right )+1}d x +3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{4}-4 \sin \left (f x +e \right )^{3}+6 \sin \left (f x +e \right )^{2}-4 \sin \left (f x +e \right )+1}d x \right )+3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{4}-4 \sin \left (f x +e \right )^{3}+6 \sin \left (f x +e \right )^{2}-4 \sin \left (f x +e \right )+1}d x \right )\right )}{c^{4}} \] Input:

int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(7/2),x)
 

Output:

(sqrt(c)*a**3*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**4 - 4*sin(e + 
f*x)**3 + 6*sin(e + f*x)**2 - 4*sin(e + f*x) + 1),x) + int((sqrt( - sin(e 
+ f*x) + 1)*sin(e + f*x)**3)/(sin(e + f*x)**4 - 4*sin(e + f*x)**3 + 6*sin( 
e + f*x)**2 - 4*sin(e + f*x) + 1),x) + 3*int((sqrt( - sin(e + f*x) + 1)*si 
n(e + f*x)**2)/(sin(e + f*x)**4 - 4*sin(e + f*x)**3 + 6*sin(e + f*x)**2 - 
4*sin(e + f*x) + 1),x) + 3*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(s 
in(e + f*x)**4 - 4*sin(e + f*x)**3 + 6*sin(e + f*x)**2 - 4*sin(e + f*x) + 
1),x)))/c**4