\(\int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx\) [229]

Optimal result

Integrand size = 25, antiderivative size = 131 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {2} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}-\frac {2 \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}} \] Output:

2^(1/2)*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1/2)/cos(d*x+c)^(1/2)/(a+a*cos(d* 
x+c))^(1/2))/a^(1/2)/d+2/3*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^ 
(1/2)-2/3*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 7.82 (sec) , antiderivative size = 473, normalized size of antiderivative = 3.61 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=-\frac {2 \cot \left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (12 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \, _3F_2\left (2,2,\frac {7}{2};1,\frac {9}{2};\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right )+12 \operatorname {Hypergeometric2F1}\left (2,\frac {7}{2},\frac {9}{2},\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}\right ) \sin ^8\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (4-7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+3 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )\right )+7 \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3 \sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (15-20 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )+8 \sin ^4\left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \left (\text {arctanh}\left (\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}}\right ) \left (3-6 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )+\sqrt {\frac {\sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{-1+2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}} \left (-3+7 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )\right )\right )}{63 d \sqrt {a (1+\cos (c+d x))} \left (1-2 \sin ^2\left (\frac {c}{2}+\frac {d x}{2}\right )\right )^{7/2}} \] Input:

Integrate[1/(Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]),x]
 

Output:

(-2*Cot[c/2 + (d*x)/2]*Csc[c/2 + (d*x)/2]^4*(12*Cos[(c + d*x)/2]^4*Hyperge 
ometricPFQ[{2, 2, 7/2}, {1, 9/2}, Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + ( 
d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^8 + 12*Hypergeometric2F1[2, 7/2, 9/2, Sin[c 
/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*Sin[c/2 + (d*x)/2]^8*(4 - 7 
*Sin[c/2 + (d*x)/2]^2 + 3*Sin[c/2 + (d*x)/2]^4) + 7*(1 - 2*Sin[c/2 + (d*x) 
/2]^2)^3*Sqrt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(15 - 20 
*Sin[c/2 + (d*x)/2]^2 + 8*Sin[c/2 + (d*x)/2]^4)*(ArcTanh[Sqrt[Sin[c/2 + (d 
*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]]*(3 - 6*Sin[c/2 + (d*x)/2]^2) + Sq 
rt[Sin[c/2 + (d*x)/2]^2/(-1 + 2*Sin[c/2 + (d*x)/2]^2)]*(-3 + 7*Sin[c/2 + ( 
d*x)/2]^2))))/(63*d*Sqrt[a*(1 + Cos[c + d*x])]*(1 - 2*Sin[c/2 + (d*x)/2]^2 
)^(7/2))
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3258, 3042, 3463, 27, 3042, 3261, 218}

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.

\(\displaystyle \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\)

\(\Big \downarrow \) 3258

\(\displaystyle \frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a-2 a \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {\cos (c+d x) a+a}}dx}{3 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\int \frac {a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}\)

\(\Big \downarrow \) 3463

\(\displaystyle \frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 \int -\frac {3 a^2}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{a}+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{3 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-3 a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}\)

\(\Big \downarrow \) 3261

\(\displaystyle \frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {6 a^2 \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}}{3 a}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}-\frac {\frac {2 a \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}-\frac {3 \sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}}{3 a}\)

Input:

Int[1/(Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]),x]
 

Output:

(2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Cos[c + d*x]]) - ((-3* 
Sqrt[2]*Sqrt[a]*ArcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]* 
Sqrt[a + a*Cos[c + d*x]])])/d + (2*a*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*S 
qrt[a + a*Cos[c + d*x]]))/(3*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 3042

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

rule 3258

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. 
) + (f_.)*(x_)]], x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x]) 
^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(2* 
b*(n + 1)*(c^2 - d^2))   Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c 
*(n + 1) + b*d*(2*n + 3)*Sin[e + f*x], x]/Sqrt[a + b*Sin[e + f*x]]), x], x] 
 /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] 
&& NeQ[c^2 - d^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

rule 3261

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

rule 3463

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
+ 1)/(f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*(n + 1)*(c^2 - d^2))   Int[(a 
 + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 
1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e + f*x], x], 
 x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && Eq 
Q[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m 
 + 1/2, 0])
 

Maple [A] (verified)

Time = 6.79 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {a \left (\cos \left (d x +c \right )+1\right )}\, \left (\sin \left (d x +c \right ) \left (\cos \left (d x +c \right )-1\right ) \sqrt {2}+\arcsin \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (3 \cos \left (d x +c \right )^{2}+3 \cos \left (d x +c \right )\right )\right )}{3 d \cos \left (d x +c \right )^{\frac {3}{2}} \left (\cos \left (d x +c \right )+1\right ) a}\)	\(116\)

Input:

int(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/3/d*2^(1/2)*(a*(cos(d*x+c)+1))^(1/2)*(sin(d*x+c)*(cos(d*x+c)-1)*2^(1/2) 
+arcsin(cot(d*x+c)-csc(d*x+c))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(3*cos(d* 
x+c)^2+3*cos(d*x+c)))/cos(d*x+c)^(3/2)/(cos(d*x+c)+1)/a
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.11 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=-\frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - 1\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - \frac {3 \, \sqrt {2} {\left (a \cos \left (d x + c\right )^{3} + a \cos \left (d x + c\right )^{2}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a}}\right )}{\sqrt {a}}}{3 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

-1/3*(2*sqrt(a*cos(d*x + c) + a)*(cos(d*x + c) - 1)*sqrt(cos(d*x + c))*sin 
(d*x + c) - 3*sqrt(2)*(a*cos(d*x + c)^3 + a*cos(d*x + c)^2)*arctan(1/2*sqr 
t(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/((cos(d*x + 
c)^2 + cos(d*x + c))*sqrt(a)))/sqrt(a))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x 
+ c)^2)
 

Sympy [F]

\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {1}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \cos ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:

integrate(1/cos(d*x+c)**(5/2)/(a+a*cos(d*x+c))**(1/2),x)
 

Output:

Integral(1/(sqrt(a*(cos(c + d*x) + 1))*cos(c + d*x)**(5/2)), x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.43 (sec) , antiderivative size = 818, normalized size of antiderivative = 6.24 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")
 

Output:

1/3*(3*(sqrt(2)*cos(2*d*x + 2*c)^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 2*sqrt(2 
)*cos(2*d*x + 2*c) + sqrt(2))*arctan2(((abs(e^(I*d*x + I*c) + 1)^4 + cos(d 
*x + c)^4 + sin(d*x + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d* 
x + c) + 1)*abs(e^(I*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c 
)^2 - 2*cos(d*x + c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + 
c) + 1)^(1/4)*sin(1/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d 
*x + I*c) + 1)^2, (abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + 
 c)^2 - 2*cos(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sin(d*x + c))/a 
bs(e^(I*d*x + I*c) + 1), ((abs(e^(I*d*x + I*c) + 1)^4 + cos(d*x + c)^4 + s 
in(d*x + c)^4 + 2*(cos(d*x + c)^2 - sin(d*x + c)^2 - 2*cos(d*x + c) + 1)*a 
bs(e^(I*d*x + I*c) + 1)^2 - 4*cos(d*x + c)^3 + 2*(cos(d*x + c)^2 - 2*cos(d 
*x + c) + 1)*sin(d*x + c)^2 + 6*cos(d*x + c)^2 - 4*cos(d*x + c) + 1)^(1/4) 
*sqrt(a)*cos(1/2*arctan2(2*(cos(d*x + c) - 1)*sin(d*x + c)/abs(e^(I*d*x + 
I*c) + 1)^2, (abs(e^(I*d*x + I*c) + 1)^2 + cos(d*x + c)^2 - sin(d*x + c)^2 
 - 2*cos(d*x + c) + 1)/abs(e^(I*d*x + I*c) + 1)^2)) + sqrt(a)*cos(d*x + c) 
 - sqrt(a))/(sqrt(a)*abs(e^(I*d*x + I*c) + 1))) - 2*(cos(2*d*x + 2*c)^2 + 
sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*(cos(1/2*arctan2(sin(2* 
d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - (cos(d*x + c) - 3)*sin(1 
/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 4*(cos(2*d*x + 2*c) 
^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(3/2*arctan...
 

Giac [F]

\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(a*cos(d*x + c) + a)*cos(d*x + c)^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \] Input:

int(1/(cos(c + d*x)^(5/2)*(a + a*cos(c + d*x))^(1/2)),x)
 

Output:

int(1/(cos(c + d*x)^(5/2)*(a + a*cos(c + d*x))^(1/2)), x)
                                                                                    
                                                                                    
 

Reduce [F]

\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\cos \left (d x +c \right )+1}\, \sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}+\cos \left (d x +c \right )^{3}}d x \right )}{a} \] Input:

int(1/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2),x)
 

Output:

(sqrt(a)*int((sqrt(cos(c + d*x) + 1)*sqrt(cos(c + d*x)))/(cos(c + d*x)**4 
+ cos(c + d*x)**3),x))/a