3.421 \(\int \cos (c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=121 \[ -\frac {a^2 (2 A-2 B-3 C) \tan (c+d x)}{2 d}+\frac {a^2 (2 A+4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (2 A+B)-\frac {(2 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^2}{d} \]

[Out]

a^2*(2*A+B)*x+1/2*a^2*(2*A+4*B+3*C)*arctanh(sin(d*x+c))/d+A*(a+a*sec(d*x+c))^2*sin(d*x+c)/d-1/2*a^2*(2*A-2*B-3
*C)*tan(d*x+c)/d-1/2*(2*A-C)*(a^2+a^2*sec(d*x+c))*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4086, 3917, 3914, 3767, 8, 3770} \[ -\frac {a^2 (2 A-2 B-3 C) \tan (c+d x)}{2 d}+\frac {a^2 (2 A+4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+a^2 x (2 A+B)-\frac {(2 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

a^2*(2*A + B)*x + (a^2*(2*A + 4*B + 3*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (A*(a + a*Sec[c + d*x])^2*Sin[c + d*x]
)/d - (a^2*(2*A - 2*B - 3*C)*Tan[c + d*x])/(2*d) - ((2*A - C)*(a^2 + a^2*Sec[c + d*x])*Tan[c + d*x])/(2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3914

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

Rule 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4086

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}+\frac {\int (a+a \sec (c+d x))^2 (a (2 A+B)-a (2 A-C) \sec (c+d x)) \, dx}{a}\\ &=\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}+\frac {\int (a+a \sec (c+d x)) \left (2 a^2 (2 A+B)-a^2 (2 A-2 B-3 C) \sec (c+d x)\right ) \, dx}{2 a}\\ &=a^2 (2 A+B) x+\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}-\frac {1}{2} \left (a^2 (2 A-2 B-3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{2} \left (a^2 (2 A+4 B+3 C)\right ) \int \sec (c+d x) \, dx\\ &=a^2 (2 A+B) x+\frac {a^2 (2 A+4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}+\frac {\left (a^2 (2 A-2 B-3 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{2 d}\\ &=a^2 (2 A+B) x+\frac {a^2 (2 A+4 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {a^2 (2 A-2 B-3 C) \tan (c+d x)}{2 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 3.57, size = 365, normalized size = 3.02 \[ \frac {a^2 \cos ^4(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {2 (2 A+4 B+3 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (2 A+4 B+3 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+4 x (2 A+B)+\frac {4 A \sin (c) \cos (d x)}{d}+\frac {4 A \cos (c) \sin (d x)}{d}+\frac {4 (B+2 C) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 (B+2 C) \sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {C}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {C}{d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{8 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Cos[c + d*x]^4*Sec[(c + d*x)/2]^4*(1 + Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(4*(2*A +
B)*x - (2*(2*A + 4*B + 3*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(2*A + 4*B + 3*C)*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]])/d + (4*A*Cos[d*x]*Sin[c])/d + (4*A*Cos[c]*Sin[d*x])/d + C/(d*(Cos[(c + d*x)/2] - Si
n[(c + d*x)/2])^2) + (4*(B + 2*C)*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
) - C/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(B + 2*C)*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(
c + d*x)/2] + Sin[(c + d*x)/2]))))/(8*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 143, normalized size = 1.18 \[ \frac {4 \, {\left (2 \, A + B\right )} a^{2} d x \cos \left (d x + c\right )^{2} + {\left (2 \, A + 4 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + 4 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(2*A + B)*a^2*d*x*cos(d*x + c)^2 + (2*A + 4*B + 3*C)*a^2*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*A +
4*B + 3*C)*a^2*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^2*cos(d*x + c)^2 + 2*(B + 2*C)*a^2*cos(d*x + c
) + C*a^2)*sin(d*x + c))/(d*cos(d*x + c)^2)

________________________________________________________________________________________

giac [A]  time = 0.30, size = 204, normalized size = 1.69 \[ \frac {\frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (2 \, A a^{2} + B a^{2}\right )} {\left (d x + c\right )} + {\left (2 \, A a^{2} + 4 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a^{2} + 4 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*A*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(2*A*a^2 + B*a^2)*(d*x + c) + (2*A*a^2 + 4*
B*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*A*a^2 + 4*B*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x + 1/2*
c) - 1)) - 2*(2*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 3*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 2*B*a^2*tan(1/2*d*x + 1/2*c) -
5*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

________________________________________________________________________________________

maple [A]  time = 1.12, size = 166, normalized size = 1.37 \[ \frac {a^{2} A \sin \left (d x +c \right )}{d}+a^{2} B x +\frac {B \,a^{2} c}{d}+\frac {3 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+2 a^{2} A x +\frac {2 A \,a^{2} c}{d}+\frac {2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{2} C \tan \left (d x +c \right )}{d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} B \tan \left (d x +c \right )}{d}+\frac {a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*a^2*A*sin(d*x+c)+a^2*B*x+1/d*B*a^2*c+3/2/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+2*a^2*A*x+2/d*A*a^2*c+2/d*B*a^2
*ln(sec(d*x+c)+tan(d*x+c))+2/d*a^2*C*tan(d*x+c)+1/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+a^2*B*tan(d*x+c)/d+1/2/d*a
^2*C*sec(d*x+c)*tan(d*x+c)

________________________________________________________________________________________

maxima [A]  time = 0.40, size = 192, normalized size = 1.59 \[ \frac {8 \, {\left (d x + c\right )} A a^{2} + 4 \, {\left (d x + c\right )} B a^{2} - C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{2} \sin \left (d x + c\right ) + 4 \, B a^{2} \tan \left (d x + c\right ) + 8 \, C a^{2} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*(d*x + c)*A*a^2 + 4*(d*x + c)*B*a^2 - C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1
) + log(sin(d*x + c) - 1)) + 2*A*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*B*a^2*(log(sin(d*x +
c) + 1) - log(sin(d*x + c) - 1)) + 2*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*A*a^2*sin(d*x +
c) + 4*B*a^2*tan(d*x + c) + 8*C*a^2*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 3.83, size = 244, normalized size = 2.02 \[ \frac {\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+C\,a^2\,\sin \left (2\,c+2\,d\,x\right )+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{4}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{2}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {2\,\left (-2\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}-B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+\frac {C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((A*a^2*sin(3*c + 3*d*x))/4 + (B*a^2*sin(2*c + 2*d*x))/2 + C*a^2*sin(2*c + 2*d*x) + (A*a^2*sin(c + d*x))/4 + (
C*a^2*sin(c + d*x))/2)/(d*(cos(2*c + 2*d*x)/2 + 1/2)) - (2*(A*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)
/2))*1i - 2*A*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - B*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/
2)) + B*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i + (C*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2
+ (d*x)/2))*3i)/2))/d

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int A \cos {\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A*cos(c + d*x), x) + Integral(2*A*cos(c + d*x)*sec(c + d*x), x) + Integral(A*cos(c + d*x)*sec(c
+ d*x)**2, x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(2*B*cos(c + d*x)*sec(c + d*x)**2, x) + In
tegral(B*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(C*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(2*C*cos(c +
d*x)*sec(c + d*x)**3, x) + Integral(C*cos(c + d*x)*sec(c + d*x)**4, x))

________________________________________________________________________________________