Integrand size = 30, antiderivative size = 191 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {12 c^4 \cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {6 c^3 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a f \sqrt {a+a \sin (e+f x)}}-\frac {3 c^2 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a+a \sin (e+f x))^{3/2}} \] Output:
-12*c^4*cos(f*x+e)*ln(1+sin(f*x+e))/a/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f* x+e))^(1/2)-6*c^3*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^( 1/2)-3/2*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a/f/(a+a*sin(f*x+e))^(1/2)- c*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/f/(a+a*sin(f*x+e))^(3/2)
Time = 10.03 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.85 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {c^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (-44-18 \cos (2 (e+f x))-192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (39-192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)+\sin (3 (e+f x))\right )}{8 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{3/2}} \] Input:
Integrate[(c - c*Sin[e + f*x])^(7/2)/(a + a*Sin[e + f*x])^(3/2),x]
Output:
(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*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)]))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^ (3/2))
Time = 0.97 (sec) , antiderivative size = 190, 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.
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^{7/2}}{(a \sin (e+f x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^{7/2}}{(a \sin (e+f x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3218 |
| \(\displaystyle -\frac {3 c \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {\sin (e+f x) a+a}}dx}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 c \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {\sin (e+f x) a+a}}dx}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle -\frac {3 c \left (2 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 c \left (2 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle -\frac {3 c \left (2 c \left (2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 c \left (2 c \left (2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle -\frac {3 c \left (2 c \left (\frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {3 c \left (2 c \left (\frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle -\frac {3 c \left (2 c \left (\frac {2 c^2 \cos (e+f x) \int \frac {1}{\sin (e+f x) a+a}d(a \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {3 c \left (2 c \left (\frac {2 c^2 \cos (e+f x) \log (a \sin (e+f x)+a)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )+\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{2 f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
Input:
Int[(c - c*Sin[e + f*x])^(7/2)/(a + a*Sin[e + f*x])^(3/2),x]
Output:
-((c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(f*(a + a*Sin[e + f*x])^(3/2 ))) - (3*c*((c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(2*f*Sqrt[a + a*Si n[e + f*x]]) + 2*c*((2*c^2*Cos[e + f*x]*Log[a + a*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*Sqrt[c - c* Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]))))/a
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 = 2.05 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.07
| method | result | size |
| default | \(-\frac {\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \left (-\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}-3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+12 \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-12 \ln \left (\frac {2}{\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+10 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-4\right ) \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{3} \sqrt {4}}{f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a}\) | \(204\) |
Input:
int((c-c*sin(f*x+e))^(7/2)/(a+sin(f*x+e)*a)^(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)^6-3*cos(1/4*Pi+1/2*f*x+1/2*e)^4+12*ln(-cot(1/4*Pi+1/2*f*x+1/2*e )+csc(1/4*Pi+1/2*f*x+1/2*e))*sin(1/4*Pi+1/2*f*x+1/2*e)^2-12*ln(2/(cos(1/4* Pi+1/2*f*x+1/2*e)+1))*sin(1/4*Pi+1/2*f*x+1/2*e)^2+10*cos(1/4*Pi+1/2*f*x+1/ 2*e)^2-4)*(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*c^3/(a*sin(1/4*Pi+1/2*f*x+ 1/2*e)^2)^(1/2)/a*4^(1/2)
\[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="fric as")
Output:
integral((3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin( f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)
Timed out. \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((c-c*sin(f*x+e))**(7/2)/(a+a*sin(f*x+e))**(3/2),x)
Output:
Timed out
\[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="maxi ma")
Output:
integrate((-c*sin(f*x + e) + c)^(7/2)/(a*sin(f*x + e) + a)^(3/2), x)
Time = 0.15 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.89 \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {2 \, \sqrt {a} c^{\frac {7}{2}} {\left (\frac {6 \, \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 4 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{4}} - \frac {2}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \] Input:
integrate((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x, algorithm="giac ")
Output:
2*sqrt(a)*c^(7/2)*(6*log(-sin(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(a^2*sgn(c os(-1/4*pi + 1/2*f*x + 1/2*e))) + (a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) *sin(-1/4*pi + 1/2*f*x + 1/2*e)^4 + 4*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2* e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^2)/a^4 - 2/((sin(-1/4*pi + 1/2*f*x + 1/ 2*e)^2 - 1)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))))*sgn(sin(-1/4*pi + 1/ 2*f*x + 1/2*e))/f
Timed out. \[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{7/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:
int((c - c*sin(e + f*x))^(7/2)/(a + a*sin(e + f*x))^(3/2),x)
Output:
int((c - c*sin(e + f*x))^(7/2)/(a + a*sin(e + f*x))^(3/2), x)
\[ \int \frac {(c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, c^{3} \left (-\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right )+3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right )-3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right )+\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right )}{a^{2}} \] Input:
int((c-c*sin(f*x+e))^(7/2)/(a+a*sin(f*x+e))^(3/2),x)
Output:
(sqrt(c)*sqrt(a)*c**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(( sqrt(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) + int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1),x)))/a**2