\(\int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx\) [321]

Optimal result

Integrand size = 28, antiderivative size = 157 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {15 a^3 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{2 f (c-c \sin (e+f x))^{9/2}}-\frac {5 a^3 \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {15 a^3 \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}} \] Output:

15/4*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^(5/2)/f+1/2*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(9/2)-5/4*a^3* 
cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(5/2)-15/4*a^3*cos(f*x+e)/c^2/f/(c-c*sin(f 
*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 8.49 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.19 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-5 \cos \left (\frac {1}{2} (e+f x)\right )-15 \cos \left (\frac {3}{2} (e+f x)\right )+2 \cos \left (\frac {5}{2} (e+f x)\right )-5 \sin \left (\frac {1}{2} (e+f x)\right )+(15+15 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) (-3+\cos (2 (e+f x))+4 \sin (e+f x))+15 \sin \left (\frac {3}{2} (e+f x)\right )+2 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{4 c^2 f (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(5/2),x]
 

Output:

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(-5*Cos[(e + f*x)/2] - 15*Cos[( 
3*(e + f*x))/2] + 2*Cos[(5*(e + f*x))/2] - 5*Sin[(e + f*x)/2] + (15 + 15*I 
)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(-3 + C 
os[2*(e + f*x)] + 4*Sin[e + f*x]) + 15*Sin[(3*(e + f*x))/2] + 2*Sin[(5*(e 
+ f*x))/2]))/(4*c^2*f*(-1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.04, 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, 3158, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3215

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3159

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{2 c f (c-c \sin (e+f x))^{9/2}}-\frac {5 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{7/2}}dx}{4 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{2 c f (c-c \sin (e+f x))^{9/2}}-\frac {5 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{7/2}}dx}{4 c^2}\right )\)

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3158

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{2 c f (c-c \sin (e+f x))^{9/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )}{4 c^2}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{2 c f (c-c \sin (e+f x))^{9/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )}{4 c^2}\right )\)

\(\Big \downarrow \) 3128

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{2 c f (c-c \sin (e+f x))^{9/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (-\frac {4 \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )}{4 c^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle a^3 c^3 \left (\frac {\cos ^5(e+f x)}{2 c f (c-c \sin (e+f x))^{9/2}}-\frac {5 \left (\frac {\cos ^3(e+f x)}{c f (c-c \sin (e+f x))^{5/2}}-\frac {3 \left (\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{2 c^2}\right )}{4 c^2}\right )\)

Input:

Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(5/2),x]
 

Output:

a^3*c^3*(Cos[e + f*x]^5/(2*c*f*(c - c*Sin[e + f*x])^(9/2)) - (5*(Cos[e + f 
*x]^3/(c*f*(c - c*Sin[e + f*x])^(5/2)) - (3*((2*Sqrt[2]*ArcTanh[(Sqrt[c]*C 
os[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(c^(3/2)*f) - (2*Cos[e + 
 f*x])/(c*f*Sqrt[c - c*Sin[e + f*x]])))/(2*c^2)))/(4*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 3158

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

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 = 2.10 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.52

method	result	size
default	\(-\frac {a^{3} \left (15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{2} c^{2}-30 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )-8 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}} \sin \left (f x +e \right )^{2}+15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+18 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c}+16 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}} \sin \left (f x +e \right )-36 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{4 c^{\frac {9}{2}} \left (-1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(239\)
parts	\(\text {Expression too large to display}\)	\(816\)

Input:

int((a+sin(f*x+e)*a)^3/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4/c^(9/2)*a^3*(15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/ 
c^(1/2))*sin(f*x+e)^2*c^2-30*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)* 
2^(1/2)/c^(1/2))*c^2*sin(f*x+e)-8*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)*sin(f*x 
+e)^2+15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^2 
+18*(c*(1+sin(f*x+e)))^(3/2)*c^(1/2)+16*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)*s 
in(f*x+e)-36*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2))*(c*(1+sin(f*x+e)))^(1/2)/(- 
1+sin(f*x+e))/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. 388 vs. \(2 (134) = 268\).

Time = 0.10 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.47 \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {15 \, \sqrt {2} {\left (a^{3} \cos \left (f x + e\right )^{3} + 3 \, a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) - 4 \, a^{3} - {\left (a^{3} \cos \left (f x + e\right )^{2} - 2 \, a^{3} \cos \left (f x + e\right ) - 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (4 \, a^{3} \cos \left (f x + e\right )^{3} - 13 \, a^{3} \cos \left (f x + e\right )^{2} - 13 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3} + {\left (4 \, a^{3} \cos \left (f x + e\right )^{2} + 17 \, a^{3} \cos \left (f x + e\right ) + 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{8 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")
 

Output:

1/8*(15*sqrt(2)*(a^3*cos(f*x + e)^3 + 3*a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x 
 + e) - 4*a^3 - (a^3*cos(f*x + e)^2 - 2*a^3*cos(f*x + e) - 4*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) + sin(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*(4*a^3*cos(f*x + e)^3 - 13*a^3*cos(f*x + 
 e)^2 - 13*a^3*cos(f*x + e) + 4*a^3 + (4*a^3*cos(f*x + e)^2 + 17*a^3*cos(f 
*x + e) + 4*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + 
 e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*c 
os(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")
 

Output:

integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),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. \[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(5/2),x)
 

Output:

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(5/2), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {c}\, a^{3} \left (-\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right )-\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right )-3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right )-3 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right )\right )}{c^{3}} \] Input:

int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x)
 

Output:

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