\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)}}{a+b \cos (c+d x)} \, dx\) [1485]

Optimal result

Integrand size = 43, antiderivative size = 157 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{a+b \cos (c+d x)} \, dx=\frac {2 C \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b d}+\frac {2 (b B-a C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 d}+\frac {2 \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 (a+b) d} \] Output:

2*C*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2 
)/b/d+2*(B*b-C*a)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))* 
sec(d*x+c)^(1/2)/b^2/d+2*(A*b^2-a*(B*b-C*a))*cos(d*x+c)^(1/2)*EllipticPi(s 
in(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*sec(d*x+c)^(1/2)/b^2/(a+b)/d
 

Mathematica [A] (warning: unable to verify)

Time = 2.17 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.76 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{a+b \cos (c+d x)} \, dx=\frac {\cot (c+d x) \left (-a b C \sec ^{\frac {3}{2}}(c+d x)-a b C \cos (2 (c+d x)) \sec ^{\frac {3}{2}}(c+d x)+a b C \sec ^{\frac {7}{2}}(c+d x)+a b C \cos (2 (c+d x)) \sec ^{\frac {7}{2}}(c+d x)-2 a b C E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {-\tan ^2(c+d x)}+2 b (A b+a C) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-2 A b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}+2 a b B \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}-2 a^2 C \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {-\tan ^2(c+d x)}\right )}{a b^2 d} \] Input:

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]])/(a 
+ b*Cos[c + d*x]),x]
 

Output:

(Cot[c + d*x]*(-(a*b*C*Sec[c + d*x]^(3/2)) - a*b*C*Cos[2*(c + d*x)]*Sec[c 
+ d*x]^(3/2) + a*b*C*Sec[c + d*x]^(7/2) + a*b*C*Cos[2*(c + d*x)]*Sec[c + d 
*x]^(7/2) - 2*a*b*C*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c 
+ d*x]^2] + 2*b*(A*b + a*C)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt 
[-Tan[c + d*x]^2] - 2*A*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], 
 -1]*Sqrt[-Tan[c + d*x]^2] + 2*a*b*B*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c 
+ d*x]]], -1]*Sqrt[-Tan[c + d*x]^2] - 2*a^2*C*EllipticPi[-(a/b), ArcSin[Sq 
rt[Sec[c + d*x]]], -1]*Sqrt[-Tan[c + d*x]^2]))/(a*b^2*d)
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.78, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 4709, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

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.

\(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{a+b \cos (c+d x)}dx\)

\(\Big \downarrow \) 4709

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3538

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {C \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {A b+(b B-a C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {A b+(b B-a C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {C \int \sqrt {\cos (c+d x)}dx}{b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {A b+(b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {C \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {A b+(b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\right )\)

\(\Big \downarrow \) 3481

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {(b B-a C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {(b B-a C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\right )\)

\(\Big \downarrow \) 3284

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {2 \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}+\frac {2 (b B-a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\right )\)

Input:

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]])/(a + b*Co 
s[c + d*x]),x]
 

Output:

Sqrt[Cos[c + d*x]]*((2*C*EllipticE[(c + d*x)/2, 2])/(b*d) + ((2*(b*B - a*C 
)*EllipticF[(c + d*x)/2, 2])/(b*d) + (2*(A*b^2 - a*(b*B - a*C))*EllipticPi 
[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/b)*Sqrt[Sec[c + d*x]]
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b* 
Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, 
 B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

rule 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(322\) vs. \(2(150)=300\).

Time = 4.59 (sec) , antiderivative size = 323, normalized size of antiderivative = 2.06

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \left (A \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) b^{2}+B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}-B \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) a b -C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+C \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) a^{2}\right )}{b^{2} \left (a -b \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(323\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c)),x,me 
thod=_RETURNVERBOSE)
 

Output:

-2*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+1/ 
2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*(A*EllipticPi(cos(1/2*d*x+ 
1/2*c),-2*b/(a-b),2^(1/2))*b^2+B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b 
-B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-B*EllipticPi(cos(1/2*d*x+1/2* 
c),-2*b/(a-b),2^(1/2))*a*b-C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+C*E 
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-C*EllipticE(cos(1/2*d*x+1/2*c),2^ 
(1/2))*a*b+C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+C*EllipticPi(cos(1/ 
2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^2)/b^2/(a-b)/(-2*sin(1/2*d*x+1/2*c)^4+s 
in(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^( 
1/2)/d
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c) 
),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}{a + b \cos {\left (c + d x \right )}}\, dx \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(1/2)/(a+b*cos(d*x+ 
c)),x)
 

Output:

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sqrt(sec(c + d*x))/(a + 
b*cos(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c) 
),x, algorithm="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(sec(d*x + c))/(b*co 
s(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c) 
),x, algorithm="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sqrt(sec(d*x + c))/(b*co 
s(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{a+b \cos (c+d x)} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:

int(((1/cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + 
b*cos(c + d*x)),x)
                                                                                    
                                                                                    
 

Output:

int(((1/cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + 
b*cos(c + d*x)), x)
 

Reduce [F]

\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{a+b \cos (c+d x)} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\cos \left (d x +c \right ) b +a}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )}{\cos \left (d x +c \right ) b +a}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right ) b +a}d x \right ) c \] Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+b*cos(d*x+c)),x)
 

Output:

int(sqrt(sec(c + d*x))/(cos(c + d*x)*b + a),x)*a + int((sqrt(sec(c + d*x)) 
*cos(c + d*x))/(cos(c + d*x)*b + a),x)*b + int((sqrt(sec(c + d*x))*cos(c + 
 d*x)**2)/(cos(c + d*x)*b + a),x)*c