Integrand size = 31, antiderivative size = 108 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2 (3 a A+5 b B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (A b+a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 (A b+a B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 a A \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \] Output:
2/5*(3*A*a+5*B*b)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/3*(A*b+B*a)*In verseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d+2/3*(A*b+B*a)*cos(d*x+c)^(1/2)*sin( d*x+c)/d+2/5*a*A*cos(d*x+c)^(3/2)*sin(d*x+c)/d
Time = 1.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2 \left (3 (3 a A+5 b B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 (A b+a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (5 A b+5 a B+3 a A \cos (c+d x)) \sin (c+d x)\right )}{15 d} \] Input:
Integrate[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(2*(3*(3*a*A + 5*b*B)*EllipticE[(c + d*x)/2, 2] + 5*(A*b + a*B)*EllipticF[ (c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(5*A*b + 5*a*B + 3*a*A*Cos[c + d*x])* Sin[c + d*x]))/(15*d)
Time = 0.73 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.99, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 3433, 3042, 3447, 3042, 3502, 27, 3042, 3227, 3042, 3115, 3042, 3119, 3120}
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.
| \(\displaystyle \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \sqrt {\cos (c+d x)} (a \cos (c+d x)+b) (A \cos (c+d x)+B)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right ) \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )dx\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \int \sqrt {\cos (c+d x)} \left ((a B+A b) \cos (c+d x)+a A \cos ^2(c+d x)+b B\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left ((a B+A b) \sin \left (c+d x+\frac {\pi }{2}\right )+a A \sin \left (c+d x+\frac {\pi }{2}\right )^2+b B\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {\cos (c+d x)} (3 a A+5 b B+5 (A b+a B) \cos (c+d x))dx+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\cos (c+d x)} (3 a A+5 b B+5 (A b+a B) \cos (c+d x))dx+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 a A+5 b B+5 (A b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (5 (a B+A b) \int \cos ^{\frac {3}{2}}(c+d x)dx+(3 a A+5 b B) \int \sqrt {\cos (c+d x)}dx\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left ((3 a A+5 b B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 (a B+A b) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left ((3 a A+5 b B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 (a B+A b) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left ((3 a A+5 b B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+5 (a B+A b) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {1}{5} \left (5 (a B+A b) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+\frac {2 (3 a A+5 b B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {1}{5} \left (\frac {2 (3 a A+5 b B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+5 (a B+A b) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )\right )+\frac {2 a A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\) |
Input:
Int[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
Output:
(2*a*A*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d) + ((2*(3*a*A + 5*b*B)*Ellipt icE[(c + d*x)/2, 2])/d + 5*(A*b + a*B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d ) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
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]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(370\) vs. \(2(99)=198\).
Time = 25.63 (sec) , antiderivative size = 371, normalized size of antiderivative = 3.44
| method | result | size |
| default | \(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-24 A a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (24 A a +20 A b +20 B a \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-6 A a -10 A b -10 B a \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b -9 A \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a +5 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a -15 B \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b \right )}{15 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(371\) |
Input:
int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVER BOSE)
Output:
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*A*a*cos (1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(24*A*a+20*A*b+20*B*a)*sin(1/2*d*x+1/ 2*c)^4*cos(1/2*d*x+1/2*c)+(-6*A*a-10*A*b-10*B*a)*sin(1/2*d*x+1/2*c)^2*cos( 1/2*d*x+1/2*c)+5*A*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2) ^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b-9*A*(2*sin(1/2*d*x+1/2*c)^2 -1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 ))*a+5*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ell ipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-15*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2) *(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)/(-2 *sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*co s(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.62 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (3 \, A a \cos \left (d x + c\right ) + 5 \, B a + 5 \, A b\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, \sqrt {2} {\left (i \, B a + i \, A b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-i \, B a - i \, A b\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, \sqrt {2} {\left (-3 i \, A a - 5 i \, B b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, \sqrt {2} {\left (3 i \, A a + 5 i \, B b\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{15 \, d} \] Input:
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= "fricas")
Output:
1/15*(2*(3*A*a*cos(d*x + c) + 5*B*a + 5*A*b)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*sqrt(2)*(I*B*a + I*A*b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I *sin(d*x + c)) - 5*sqrt(2)*(-I*B*a - I*A*b)*weierstrassPInverse(-4, 0, cos (d*x + c) - I*sin(d*x + c)) - 3*sqrt(2)*(-3*I*A*a - 5*I*B*b)*weierstrassZe ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*s qrt(2)*(3*I*A*a + 5*I*B*b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
Timed out. \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**(5/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
Timed out
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= "maxima")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)*cos(d*x + c)^(5/2), x)
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm= "giac")
Output:
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)*cos(d*x + c)^(5/2), x)
Time = 0.63 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.19 \[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {2\,A\,b\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,a\,\left (\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )+\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )\right )}{3\,d}+\frac {2\,B\,b\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}-\frac {2\,A\,a\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:
int(cos(c + d*x)^(5/2)*(A + B/cos(c + d*x))*(a + b/cos(c + d*x)),x)
Output:
(2*A*b*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2)))/(3 *d) + (2*B*a*(cos(c + d*x)^(1/2)*sin(c + d*x) + ellipticF(c/2 + (d*x)/2, 2 )))/(3*d) + (2*B*b*ellipticE(c/2 + (d*x)/2, 2))/d - (2*A*a*cos(c + d*x)^(7 /2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))
\[ \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )^{2}d x \right ) b^{2}+2 \left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2} \sec \left (d x +c \right )d x \right ) a b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{2}d x \right ) a^{2} \] Input:
int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
Output:
int(sqrt(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x)**2,x)*b**2 + 2*int(sqr t(cos(c + d*x))*cos(c + d*x)**2*sec(c + d*x),x)*a*b + int(sqrt(cos(c + d*x ))*cos(c + d*x)**2,x)*a**2