\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx\) [121]

Optimal result

Integrand size = 38, antiderivative size = 175 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {(5 A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{8 \sqrt {2} a^2 c^{3/2} f}+\frac {(5 A+B) \cos (e+f x)}{8 a^2 f (c-c \sin (e+f x))^{3/2}}-\frac {(5 A+B) \sec (e+f x)}{6 a^2 c f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 c^2 f} \] Output:

1/16*(5*A+B)*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e))^(1/2) 
)*2^(1/2)/a^2/c^(3/2)/f+1/8*(5*A+B)*cos(f*x+e)/a^2/f/(c-c*sin(f*x+e))^(3/2 
)-1/6*(5*A+B)*sec(f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^(1/2)-1/3*(A-B)*sec(f*x+ 
e)^3*(c-c*sin(f*x+e))^(1/2)/a^2/c^2/f
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.89 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.71 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-12 A \cos ^2(e+f x)+4 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+3 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-(3+3 i) \sqrt [4]{-1} (5 A+B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+6 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3\right )}{24 a^2 f (1+\sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \] Input:

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

Output:

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 
])*(-12*A*Cos[e + f*x]^2 + 4*(-A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2] 
)^2 + 3*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + 
Sin[(e + f*x)/2])^3 - (3 + 3*I)*(-1)^(1/4)*(5*A + B)*ArcTan[(1/2 + I/2)*(- 
1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*( 
Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 6*(A + B)*Sin[(e + f*x)/2]*(Cos[( 
e + f*x)/2] + Sin[(e + f*x)/2])^3))/(24*a^2*f*(1 + Sin[e + f*x])^2*(c - c* 
Sin[e + f*x])^(3/2))
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.94, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 3446, 3042, 3334, 3042, 3166, 3042, 3129, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

\(\displaystyle \frac {\int \sec ^4(e+f x) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}dx}{a^2 c^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

\(\displaystyle \frac {\frac {1}{6} c (5 A+B) \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} c (5 A+B) \int \frac {1}{\cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}}dx-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3166

\(\displaystyle \frac {\frac {1}{6} c (5 A+B) \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{6} c (5 A+B) \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3129

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3128

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

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{6} c (5 A+B) \left (\frac {3}{2} c \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}}{a^2 c^2}\)

Input:

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

Output:

(-1/3*((A - B)*Sec[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/f + ((5*A + B)*c*( 
-(Sec[e + f*x]/(f*Sqrt[c - c*Sin[e + f*x]])) + (3*c*(ArcTanh[(Sqrt[c]*Cos[ 
e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/(2*Sqrt[2]*c^(3/2)*f) + Cos[ 
e + f*x]/(2*f*(c - c*Sin[e + f*x])^(3/2))))/2))/6)/(a^2*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 3129

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

rule 3166

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. 
)*(x_)]], x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*S 
qrt[a + b*Sin[e + f*x]])), x] + Simp[a*((2*p + 1)/(2*g^2*(p + 1)))   Int[(g 
*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e 
, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*p]
 

rule 3334

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

rule 3446

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.60

method	result	size
default	\(\frac {-15 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) c -30 c^{\frac {5}{2}} \cos \left (f x +e \right )^{2} A -3 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) c -6 c^{\frac {5}{2}} \cos \left (f x +e \right )^{2} B +15 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c A +20 A \,c^{\frac {5}{2}} \sin \left (f x +e \right )+3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} c B +4 B \,c^{\frac {5}{2}} \sin \left (f x +e \right )+4 A \,c^{\frac {5}{2}}+20 B \,c^{\frac {5}{2}}}{48 c^{\frac {7}{2}} a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(280\)

Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(3/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/48*(-15*A*(c+c*sin(f*x+e))^(3/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1 
/2)/c^(1/2))*2^(1/2)*sin(f*x+e)*c-30*c^(5/2)*cos(f*x+e)^2*A-3*B*(c+c*sin(f 
*x+e))^(3/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*s 
in(f*x+e)*c-6*c^(5/2)*cos(f*x+e)^2*B+15*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e 
))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(3/2)*c*A+20*A*c^(5/2)*sin(f*x+ 
e)+3*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin( 
f*x+e))^(3/2)*c*B+4*B*c^(5/2)*sin(f*x+e)+4*A*c^(5/2)+20*B*c^(5/2))/c^(7/2) 
/a^2/(1+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.18 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {3 \, \sqrt {2} {\left (5 \, A + B\right )} \sqrt {c} \cos \left (f x + e\right )^{3} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (3 \, {\left (5 \, A + B\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (5 \, A + B\right )} \sin \left (f x + e\right ) - 2 \, A - 10 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{96 \, a^{2} c^{2} f \cos \left (f x + e\right )^{3}} \] Input:

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

Output:

1/96*(3*sqrt(2)*(5*A + B)*sqrt(c)*cos(f*x + e)^3*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*(3*(5 
*A + B)*cos(f*x + e)^2 - 2*(5*A + B)*sin(f*x + e) - 2*A - 10*B)*sqrt(-c*si 
n(f*x + e) + c))/(a^2*c^2*f*cos(f*x + e)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(c)*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**4 - 2*sin(e + f*x)* 
*2 + 1),x)*a + int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)* 
*4 - 2*sin(e + f*x)**2 + 1),x)*b))/(a**2*c**2)