Integrand size = 30, antiderivative size = 192 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {a \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{f (c-c \sin (e+f x))^{3/2}}+\frac {12 a^4 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {6 a^3 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}+\frac {3 a^2 \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{2 c f \sqrt {c-c \sin (e+f x)}} \] Output:
a*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(3/2)+12*a^4*cos(f* x+e)*ln(1-sin(f*x+e))/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)+6* a^3*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin(f*x+e))^(1/2)+3/2*a^2*c os(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(1/2)
Time = 10.24 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.93 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (44+18 \cos (2 (e+f x))+192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (39-192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)+\sin (3 (e+f x))\right )}{8 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[(a + a*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(3/2),x]
Output:
-1/8*(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])] *(44 + 18*Cos[2*(e + f*x)] + 192*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + (39 - 192*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x] + Sin[3 *(e + f*x)]))/(c*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x ])*Sqrt[c - c*Sin[e + f*x]])
Time = 0.99 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {3042, 3218, 3042, 3219, 3042, 3219, 3042, 3216, 3042, 3146, 16}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{7/2}}{(c-c \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3218 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \int \frac {(\sin (e+f x) a+a)^{5/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \int \frac {(\sin (e+f x) a+a)^{5/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \left (2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \left (2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \left (2 a \left (2 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \left (2 a \left (2 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \left (2 a \left (\frac {2 a^2 c \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \left (2 a \left (\frac {2 a^2 c \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \left (2 a \left (-\frac {2 a^2 \cos (e+f x) \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {3 a \left (2 a \left (-\frac {2 a^2 \cos (e+f x) \log (c-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
Input:
Int[(a + a*Sin[e + f*x])^(7/2)/(c - c*Sin[e + f*x])^(3/2),x]
Output:
(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(f*(c - c*Sin[e + f*x])^(3/2)) - (3*a*(-1/2*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(f*Sqrt[c - c*Si n[e + f*x]]) + 2*a*((-2*a^2*Cos[e + f*x]*Log[c - c*Sin[e + f*x]])/(f*Sqrt[ a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a + a *Sin[e + f*x]])/(f*Sqrt[c - c*Sin[e + f*x]]))))/c
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x ]]*Sqrt[c + d*Sin[e + f*x]])) Int[Cos[e + f*x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 ]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^ (m - 1)*((c + d*Sin[e + f*x])^n/(f*(2*n + 1))), x] - Simp[b*((2*m - 1)/(d*( 2*n + 1))) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b ^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] && !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
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])
Time = 1.93 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.29
| method | result | size |
| default | \(\frac {\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}-6 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-12 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+12 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )+12 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )+3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+2\right ) a^{3} \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {4}}{f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c}\) | \(247\) |
Input:
int((a+sin(f*x+e)*a)^(7/2)/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOSE)
Output:
1/f*sec(1/4*Pi+1/2*f*x+1/2*e)*csc(1/4*Pi+1/2*f*x+1/2*e)*(cos(1/4*Pi+1/2*f* x+1/2*e)^6-6*cos(1/4*Pi+1/2*f*x+1/2*e)^4-12*cos(1/4*Pi+1/2*f*x+1/2*e)^2*ln (2/(cos(1/4*Pi+1/2*f*x+1/2*e)+1))+12*cos(1/4*Pi+1/2*f*x+1/2*e)^2*ln(-cot(1 /4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)+1)+12*cos(1/4*Pi+1/2*f*x+1/ 2*e)^2*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)-1)+3*cos(1/ 4*Pi+1/2*f*x+1/2*e)^2+2)*a^3*(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/(c*cos( 1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/c*4^(1/2)
\[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fric as")
Output:
integral((3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin( f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2), x)
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(7/2)/(c-c*sin(f*x+e))**(3/2),x)
Output:
Timed out
\[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="maxi ma")
Output:
integrate((a*sin(f*x + e) + a)^(7/2)/(-c*sin(f*x + e) + c)^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.89 \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 \, a^{\frac {7}{2}} \sqrt {c} {\left (\frac {6 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 4 \, c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{c^{4}} - \frac {2}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:
integrate((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="giac ")
Output:
-2*a^(7/2)*sqrt(c)*(6*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c^2*sgn( sin(-1/4*pi + 1/2*f*x + 1/2*e))) + (c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4*s gn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 4*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^ 2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))/c^4 - 2/((cos(-1/4*pi + 1/2*f*x + 1 /2*e)^2 - 1)*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))*sgn(cos(-1/4*pi + 1 /2*f*x + 1/2*e))/f
Timed out. \[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:
int((a + a*sin(e + f*x))^(7/2)/(c - c*sin(e + f*x))^(3/2),x)
Output:
int((a + a*sin(e + f*x))^(7/2)/(c - c*sin(e + f*x))^(3/2), x)
\[ \int \frac {(a+a \sin (e+f x))^{7/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a^{3} \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x +3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right )+3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right )+\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}-2 \sin \left (f x +e \right )+1}d x \right )}{c^{2}} \] Input:
int((a+a*sin(f*x+e))^(7/2)/(c-c*sin(f*x+e))^(3/2),x)
Output:
(sqrt(c)*sqrt(a)*a**3*(int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3)/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x) + 3*int((sqr t(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f* x)**2 - 2*sin(e + f*x) + 1),x) + 3*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin (e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**2 - 2*sin(e + f*x) + 1),x) + i nt((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x)**2 - 2 *sin(e + f*x) + 1),x)))/c**2