Integrand size = 21, antiderivative size = 123 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {3}{8} a \left (a^2+4 b^2\right ) x+\frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{4 d}+\frac {3 a \left (a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}-\frac {3 a^2 b \sin ^3(c+d x)}{4 d} \]
3/8*a*(a^2+4*b^2)*x+1/4*b*(11*a^2+4*b^2)*sin(d*x+c)/d+3/8*a*(a^2+4*b^2)*co s(d*x+c)*sin(d*x+c)/d+1/4*a^2*cos(d*x+c)^3*(a+b*sec(d*x+c))*sin(d*x+c)/d-3 /4*a^2*b*sin(d*x+c)^3/d
Time = 0.37 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {8 b \left (9 a^2+4 b^2\right ) \sin (c+d x)+a \left (12 a^2 c+48 b^2 c+12 a^2 d x+48 b^2 d x+8 \left (a^2+3 b^2\right ) \sin (2 (c+d x))+8 a b \sin (3 (c+d x))+a^2 \sin (4 (c+d x))\right )}{32 d} \]
(8*b*(9*a^2 + 4*b^2)*Sin[c + d*x] + a*(12*a^2*c + 48*b^2*c + 12*a^2*d*x + 48*b^2*d*x + 8*(a^2 + 3*b^2)*Sin[2*(c + d*x)] + 8*a*b*Sin[3*(c + d*x)] + a ^2*Sin[4*(c + d*x)]))/(32*d)
Time = 0.73 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.95, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4328, 3042, 4535, 3042, 3115, 24, 4532, 3042, 3492, 2009}
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.
| \(\displaystyle \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4328 |
| \(\displaystyle \frac {1}{4} \int \cos ^3(c+d x) \left (9 b a^2+3 \left (a^2+4 b^2\right ) \sec (c+d x) a+2 b \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right )dx+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {9 b a^2+3 \left (a^2+4 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 b \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{4} \left (3 a \left (a^2+4 b^2\right ) \int \cos ^2(c+d x)dx+\int \cos ^3(c+d x) \left (9 b a^2+2 b \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right )dx\right )+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (3 a \left (a^2+4 b^2\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\int \frac {9 b a^2+2 b \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\right )+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {9 b a^2+2 b \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+3 a \left (a^2+4 b^2\right ) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {9 b a^2+2 b \left (a^2+2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+3 a \left (a^2+4 b^2\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 4532 |
| \(\displaystyle \frac {1}{4} \left (\int \cos (c+d x) \left (9 a^2 b \cos ^2(c+d x)+2 b \left (a^2+2 b^2\right )\right )dx+3 a \left (a^2+4 b^2\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (9 a^2 b \sin \left (c+d x+\frac {\pi }{2}\right )^2+2 b \left (a^2+2 b^2\right )\right )dx+3 a \left (a^2+4 b^2\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3492 |
| \(\displaystyle \frac {1}{4} \left (3 a \left (a^2+4 b^2\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\int \left (b \left (11 a^2+4 b^2\right )-9 a^2 b \sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} \left (3 a \left (a^2+4 b^2\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {3 a^2 b \sin ^3(c+d x)-b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{d}\right )+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d}\) |
(a^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])*Sin[c + d*x])/(4*d) + (3*a*(a^2 + 4*b^2)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (-(b*(11*a^2 + 4*b^2)* Sin[c + d*x]) + 3*a^2*b*Sin[c + d*x]^3)/d)/4
3.5.73.3.1 Defintions of rubi rules used
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[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 ), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[a^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)* ((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2* (n + 1))*Csc[e + f*x] - b*(b^2*n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && ((Int egerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[ {e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Time = 0.85 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.73
| method | result | size |
| parallelrisch | \(\frac {8 \left (a^{3}+3 a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+8 a^{2} b \sin \left (3 d x +3 c \right )+a^{3} \sin \left (4 d x +4 c \right )+8 \left (9 a^{2} b +4 b^{3}\right ) \sin \left (d x +c \right )+12 a \left (a^{2}+4 b^{2}\right ) x d}{32 d}\) | \(90\) |
| derivativedivides | \(\frac {a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+3 a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{3} \sin \left (d x +c \right )}{d}\) | \(102\) |
| default | \(\frac {a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+3 a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{3} \sin \left (d x +c \right )}{d}\) | \(102\) |
| risch | \(\frac {3 a^{3} x}{8}+\frac {3 a \,b^{2} x}{2}+\frac {9 a^{2} b \sin \left (d x +c \right )}{4 d}+\frac {\sin \left (d x +c \right ) b^{3}}{d}+\frac {a^{3} \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{2} b \sin \left (3 d x +3 c \right )}{4 d}+\frac {a^{3} \sin \left (2 d x +2 c \right )}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) a \,b^{2}}{4 d}\) | \(113\) |
| norman | \(\frac {\left (\frac {3}{8} a^{3}+\frac {3}{2} a \,b^{2}\right ) x +\left (-\frac {3}{2} a^{3}-6 a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-\frac {3}{8} a^{3}-\frac {3}{2} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {3}{8} a^{3}-\frac {3}{2} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {3}{4} a^{3}+3 a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (\frac {3}{4} a^{3}+3 a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (\frac {3}{8} a^{3}+\frac {3}{2} a \,b^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-\frac {\left (5 a^{3}-24 a^{2} b +12 a \,b^{2}-8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {\left (5 a^{3}+24 a^{2} b +12 a \,b^{2}+8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (7 a^{3}-8 a^{2} b -12 a \,b^{2}-8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2 d}-\frac {\left (7 a^{3}+8 a^{2} b -12 a \,b^{2}+8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}+\frac {\left (13 a^{3}-8 a^{2} b +12 a \,b^{2}+8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (13 a^{3}+8 a^{2} b +12 a \,b^{2}-8 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\) | \(429\) |
1/32*(8*(a^3+3*a*b^2)*sin(2*d*x+2*c)+8*a^2*b*sin(3*d*x+3*c)+a^3*sin(4*d*x+ 4*c)+8*(9*a^2*b+4*b^3)*sin(d*x+c)+12*a*(a^2+4*b^2)*x*d)/d
Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} d x + {\left (2 \, a^{3} \cos \left (d x + c\right )^{3} + 8 \, a^{2} b \cos \left (d x + c\right )^{2} + 16 \, a^{2} b + 8 \, b^{3} + 3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]
1/8*(3*(a^3 + 4*a*b^2)*d*x + (2*a^3*cos(d*x + c)^3 + 8*a^2*b*cos(d*x + c)^ 2 + 16*a^2*b + 8*b^3 + 3*(a^3 + 4*a*b^2)*cos(d*x + c))*sin(d*x + c))/d
\[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{3} \cos ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.77 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2} + 32 \, b^{3} \sin \left (d x + c\right )}{32 \, d} \]
1/32*((12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3 - 32*(si n(d*x + c)^3 - 3*sin(d*x + c))*a^2*b + 24*(2*d*x + 2*c + sin(2*d*x + 2*c)) *a*b^2 + 32*b^3*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 297 vs. \(2 (113) = 226\).
Time = 0.32 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.41 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]
1/8*(3*(a^3 + 4*a*b^2)*(d*x + c) - 2*(5*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*a^ 2*b*tan(1/2*d*x + 1/2*c)^7 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 8*b^3*tan(1 /2*d*x + 1/2*c)^7 - 3*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 24*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*a*b ^2*tan(1/2*d*x + 1/2*c)^3 - 24*b^3*tan(1/2*d*x + 1/2*c)^3 - 5*a^3*tan(1/2* d*x + 1/2*c) - 24*a^2*b*tan(1/2*d*x + 1/2*c) - 12*a*b^2*tan(1/2*d*x + 1/2* c) - 8*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 16.57 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.03 \[ \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx=\frac {\left (-\frac {5\,a^3}{4}+6\,a^2\,b-3\,a\,b^2+2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,a^3}{4}+10\,a^2\,b-3\,a\,b^2+6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {3\,a^3}{4}+10\,a^2\,b+3\,a\,b^2+6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a^3}{4}+6\,a^2\,b+3\,a\,b^2+2\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atan}\left (\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b^2\right )}{4\,\left (\frac {3\,a^3}{4}+3\,a\,b^2\right )}\right )\,\left (a^2+4\,b^2\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)^3*(3*a*b^2 + 10*a^2*b - (3*a^3)/4 + 6*b^3) - tan(c/2 + (d*x)/2)^7*(3*a*b^2 - 6*a^2*b + (5*a^3)/4 - 2*b^3) + tan(c/2 + (d*x)/2)^5 *(10*a^2*b - 3*a*b^2 + (3*a^3)/4 + 6*b^3) + tan(c/2 + (d*x)/2)*(3*a*b^2 + 6*a^2*b + (5*a^3)/4 + 2*b^3))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d* x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (3*a*atan( (3*a*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2))/(4*(3*a*b^2 + (3*a^3)/4)))*(a^2 + 4 *b^2))/(4*d)