Integrand size = 33, antiderivative size = 115 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {A x}{a^3}-\frac {(A-B+C) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(7 A-2 B-3 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(22 A-2 B-3 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
A*x/a^3-1/5*(A-B+C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(7*A-2*B-3*C)*tan (d*x+c)/a/d/(a+a*sec(d*x+c))^2-1/15*(22*A-2*B-3*C)*tan(d*x+c)/d/(a^3+a^3*s ec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(115)=230\).
Time = 2.02 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.51 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (150 A d x \cos \left (\frac {d x}{2}\right )+150 A d x \cos \left (c+\frac {d x}{2}\right )+75 A d x \cos \left (c+\frac {3 d x}{2}\right )+75 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+15 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+15 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-370 A \sin \left (\frac {d x}{2}\right )+80 B \sin \left (\frac {d x}{2}\right )+30 C \sin \left (\frac {d x}{2}\right )+270 A \sin \left (c+\frac {d x}{2}\right )-60 B \sin \left (c+\frac {d x}{2}\right )-30 C \sin \left (c+\frac {d x}{2}\right )-230 A \sin \left (c+\frac {3 d x}{2}\right )+40 B \sin \left (c+\frac {3 d x}{2}\right )+30 C \sin \left (c+\frac {3 d x}{2}\right )+90 A \sin \left (2 c+\frac {3 d x}{2}\right )-30 B \sin \left (2 c+\frac {3 d x}{2}\right )-64 A \sin \left (2 c+\frac {5 d x}{2}\right )+14 B \sin \left (2 c+\frac {5 d x}{2}\right )+6 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{480 a^3 d} \]
(Sec[c/2]*Sec[(c + d*x)/2]^5*(150*A*d*x*Cos[(d*x)/2] + 150*A*d*x*Cos[c + ( d*x)/2] + 75*A*d*x*Cos[c + (3*d*x)/2] + 75*A*d*x*Cos[2*c + (3*d*x)/2] + 15 *A*d*x*Cos[2*c + (5*d*x)/2] + 15*A*d*x*Cos[3*c + (5*d*x)/2] - 370*A*Sin[(d *x)/2] + 80*B*Sin[(d*x)/2] + 30*C*Sin[(d*x)/2] + 270*A*Sin[c + (d*x)/2] - 60*B*Sin[c + (d*x)/2] - 30*C*Sin[c + (d*x)/2] - 230*A*Sin[c + (3*d*x)/2] + 40*B*Sin[c + (3*d*x)/2] + 30*C*Sin[c + (3*d*x)/2] + 90*A*Sin[2*c + (3*d*x )/2] - 30*B*Sin[2*c + (3*d*x)/2] - 64*A*Sin[2*c + (5*d*x)/2] + 14*B*Sin[2* c + (5*d*x)/2] + 6*C*Sin[2*c + (5*d*x)/2]))/(480*a^3*d)
Time = 0.70 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4540, 25, 3042, 4410, 25, 3042, 4407, 3042, 4281}
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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4540 |
| \(\displaystyle -\frac {\int -\frac {5 a A-a (2 A-2 B-3 C) \sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {5 a A-a (2 A-2 B-3 C) \sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {5 a A-a (2 A-2 B-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4410 |
| \(\displaystyle \frac {-\frac {\int -\frac {15 a^2 A-a^2 (7 A-2 B-3 C) \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (7 A-2 B-3 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {15 a^2 A-a^2 (7 A-2 B-3 C) \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (7 A-2 B-3 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {15 a^2 A-a^2 (7 A-2 B-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (7 A-2 B-3 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {\frac {15 a A x-a^2 (22 A-2 B-3 C) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (7 A-2 B-3 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {15 a A x-a^2 (22 A-2 B-3 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (7 A-2 B-3 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle \frac {\frac {15 a A x-\frac {a^2 (22 A-2 B-3 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {a (7 A-2 B-3 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A-B+C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
-1/5*((A - B + C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + (-1/3*(a*(7*A - 2*B - 3*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (15*a*A*x - (a^2* (22*A - 2*B - 3*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(5*a^2 )
3.5.70.3.1 Defintions of rubi rules used
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[(-(b*c - a*d))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Sim p[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*(2*m + 1) + (b*B*(m + 1) - a*(A*(m + 1) - C*m))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.63
| method | result | size |
| parallelrisch | \(\frac {3 \left (-A +B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+10 \left (2 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+15 \left (-7 A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+60 d x A}{60 a^{3} d}\) | \(73\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(127\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +8 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(127\) |
| norman | \(\frac {\frac {A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {A x}{a}-\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 a d}+\frac {\left (7 A -B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (23 A -13 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}-\frac {\left (25 A -5 B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{2}}\) | \(155\) |
| risch | \(\frac {A x}{a^{3}}-\frac {2 i \left (45 A \,{\mathrm e}^{4 i \left (d x +c \right )}-15 B \,{\mathrm e}^{4 i \left (d x +c \right )}+135 A \,{\mathrm e}^{3 i \left (d x +c \right )}-30 B \,{\mathrm e}^{3 i \left (d x +c \right )}-15 C \,{\mathrm e}^{3 i \left (d x +c \right )}+185 A \,{\mathrm e}^{2 i \left (d x +c \right )}-40 B \,{\mathrm e}^{2 i \left (d x +c \right )}-15 C \,{\mathrm e}^{2 i \left (d x +c \right )}+115 A \,{\mathrm e}^{i \left (d x +c \right )}-20 B \,{\mathrm e}^{i \left (d x +c \right )}-15 C \,{\mathrm e}^{i \left (d x +c \right )}+32 A -7 B -3 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(172\) |
1/60*(3*(-A+B-C)*tan(1/2*d*x+1/2*c)^5+10*(2*A-B)*tan(1/2*d*x+1/2*c)^3+15*( -7*A+B+C)*tan(1/2*d*x+1/2*c)+60*d*x*A)/a^3/d
Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.28 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, A d x \cos \left (d x + c\right )^{3} + 45 \, A d x \cos \left (d x + c\right )^{2} + 45 \, A d x \cos \left (d x + c\right ) + 15 \, A d x - {\left ({\left (32 \, A - 7 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, A - 2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 22 \, A - 2 \, B - 3 \, C\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
1/15*(15*A*d*x*cos(d*x + c)^3 + 45*A*d*x*cos(d*x + c)^2 + 45*A*d*x*cos(d*x + c) + 15*A*d*x - ((32*A - 7*B - 3*C)*cos(d*x + c)^2 + 3*(17*A - 2*B - 3* C)*cos(d*x + c) + 22*A - 2*B - 3*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
(Integral(A/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3
Time = 0.32 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.78 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - \frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]
-1/60*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d* x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 120*arctan(si n(d*x + c)/(cos(d*x + c) + 1))/a^3) - B*(15*sin(d*x + c)/(cos(d*x + c) + 1 ) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c ) + 1)^5)/a^3 - 3*C*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(c os(d*x + c) + 1)^5)/a^3)/d
Time = 0.32 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.33 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, {\left (d x + c\right )} A}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 10 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
1/60*(60*(d*x + c)*A/a^3 - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan (1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*A*a^12*tan(1/2* d*x + 1/2*c)^3 + 10*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d*x + 1/2*c) - 15*B*a^12*tan(1/2*d*x + 1/2*c) - 15*C*a^12*tan(1/2*d*x + 1/2*c ))/a^15)/d
Time = 16.34 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.37 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {A\,x}{a^3}+\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}-\frac {7\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}\right )+{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}\right )-\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}-\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]
(A*x)/a^3 + (cos(c/2 + (d*x)/2)^4*((B*sin(c/2 + (d*x)/2))/4 - (7*A*sin(c/2 + (d*x)/2))/4 + (C*sin(c/2 + (d*x)/2))/4) + cos(c/2 + (d*x)/2)^2*((A*sin( c/2 + (d*x)/2)^3)/3 - (B*sin(c/2 + (d*x)/2)^3)/6) - (A*sin(c/2 + (d*x)/2)^ 5)/20 + (B*sin(c/2 + (d*x)/2)^5)/20 - (C*sin(c/2 + (d*x)/2)^5)/20)/(a^3*d* cos(c/2 + (d*x)/2)^5)