Integrand size = 27, antiderivative size = 197 \[ \int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {9}{2} a^2 b x-\frac {15 b^3 x}{8}+\frac {a^3 \cos (c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {9 a^2 b \tan (c+d x)}{2 d}+\frac {15 b^3 \tan (c+d x)}{8 d}-\frac {3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d} \]
-9/2*a^2*b*x-15/8*b^3*x+a^3*cos(d*x+c)/d+6*a*b^2*cos(d*x+c)/d-a*b^2*cos(d* x+c)^3/d+a^3*sec(d*x+c)/d+3*a*b^2*sec(d*x+c)/d+9/2*a^2*b*tan(d*x+c)/d+15/8 *b^3*tan(d*x+c)/d-3/2*a^2*b*sin(d*x+c)^2*tan(d*x+c)/d-5/8*b^3*sin(d*x+c)^2 *tan(d*x+c)/d-1/4*b^3*sin(d*x+c)^4*tan(d*x+c)/d
Time = 0.92 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.75 \[ \int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {\sec (c+d x) \left (96 a^3+360 a b^2-24 b \left (12 a^2+5 b^2\right ) (c+d x) \cos (c+d x)+32 \left (a^3+5 a b^2\right ) \cos (2 (c+d x))-8 a b^2 \cos (4 (c+d x))+216 a^2 b \sin (c+d x)+80 b^3 \sin (c+d x)+24 a^2 b \sin (3 (c+d x))+15 b^3 \sin (3 (c+d x))-b^3 \sin (5 (c+d x))\right )}{64 d} \]
(Sec[c + d*x]*(96*a^3 + 360*a*b^2 - 24*b*(12*a^2 + 5*b^2)*(c + d*x)*Cos[c + d*x] + 32*(a^3 + 5*a*b^2)*Cos[2*(c + d*x)] - 8*a*b^2*Cos[4*(c + d*x)] + 216*a^2*b*Sin[c + d*x] + 80*b^3*Sin[c + d*x] + 24*a^2*b*Sin[3*(c + d*x)] + 15*b^3*Sin[3*(c + d*x)] - b^3*Sin[5*(c + d*x)]))/(64*d)
Time = 0.50 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 3391, 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.
| \(\displaystyle \int \sin (c+d x) \tan ^2(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 (a+b \sin (c+d x))^3}{\cos (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3391 |
| \(\displaystyle \int \left (a^3 \sin (c+d x) \tan ^2(c+d x)+3 a^2 b \sin ^2(c+d x) \tan ^2(c+d x)+3 a b^2 \sin ^3(c+d x) \tan ^2(c+d x)+b^3 \sin ^4(c+d x) \tan ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 \cos (c+d x)}{d}+\frac {a^3 \sec (c+d x)}{d}+\frac {9 a^2 b \tan (c+d x)}{2 d}-\frac {3 a^2 b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {9}{2} a^2 b x-\frac {a b^2 \cos ^3(c+d x)}{d}+\frac {6 a b^2 \cos (c+d x)}{d}+\frac {3 a b^2 \sec (c+d x)}{d}+\frac {15 b^3 \tan (c+d x)}{8 d}-\frac {b^3 \sin ^4(c+d x) \tan (c+d x)}{4 d}-\frac {5 b^3 \sin ^2(c+d x) \tan (c+d x)}{8 d}-\frac {15 b^3 x}{8}\) |
(-9*a^2*b*x)/2 - (15*b^3*x)/8 + (a^3*Cos[c + d*x])/d + (6*a*b^2*Cos[c + d* x])/d - (a*b^2*Cos[c + d*x]^3)/d + (a^3*Sec[c + d*x])/d + (3*a*b^2*Sec[c + d*x])/d + (9*a^2*b*Tan[c + d*x])/(2*d) + (15*b^3*Tan[c + d*x])/(8*d) - (3 *a^2*b*Sin[c + d*x]^2*Tan[c + d*x])/(2*d) - (5*b^3*Sin[c + d*x]^2*Tan[c + d*x])/(8*d) - (b^3*Sin[c + d*x]^4*Tan[c + d*x])/(4*d)
3.15.57.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (G tQ[m, 0] || IntegerQ[n])
Time = 1.03 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.77
| method | result | size |
| parallelrisch | \(\frac {\left (32 a^{3}+160 a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (24 a^{2} b +15 b^{3}\right ) \sin \left (3 d x +3 c \right )-8 a \,b^{2} \cos \left (4 d x +4 c \right )-b^{3} \sin \left (5 d x +5 c \right )+\left (-288 a^{2} b d x -120 b^{3} d x +128 a^{3}+512 a \,b^{2}\right ) \cos \left (d x +c \right )+\left (216 a^{2} b +80 b^{3}\right ) \sin \left (d x +c \right )+96 a^{3}+360 a \,b^{2}}{64 d \cos \left (d x +c \right )}\) | \(152\) |
| derivativedivides | \(\frac {a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a^{2} b \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a \,b^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) | \(214\) |
| default | \(\frac {a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a^{2} b \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a \,b^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+b^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) | \(214\) |
| risch | \(-\frac {9 a^{2} b x}{2}-\frac {15 b^{3} x}{8}-\frac {3 i b \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 d}-\frac {i b^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{4 d}+\frac {a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {21 \,{\mathrm e}^{i \left (d x +c \right )} a \,b^{2}}{8 d}+\frac {a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {21 \,{\mathrm e}^{-i \left (d x +c \right )} a \,b^{2}}{8 d}+\frac {3 i b \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 d}+\frac {i b^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}+\frac {2 i \left (-i a^{3} {\mathrm e}^{i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+3 a^{2} b +b^{3}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {b^{3} \sin \left (4 d x +4 c \right )}{32 d}-\frac {a \,b^{2} \cos \left (3 d x +3 c \right )}{4 d}\) | \(254\) |
| norman | \(\frac {-\frac {4 a^{3}+16 a \,b^{2}}{d}+\frac {3 b \left (12 a^{2}+5 b^{2}\right ) x}{8}-\frac {4 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 \left (4 a^{3}+16 a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (6 a^{3}+16 a \,b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 b \left (12 a^{2}+5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {2 b \left (12 a^{2}+5 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 b \left (12 a^{2}+5 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 b \left (12 a^{2}+5 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {9 b \left (12 a^{2}+5 b^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {3 b \left (12 a^{2}+5 b^{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {3 b \left (12 a^{2}+5 b^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {9 b \left (12 a^{2}+5 b^{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {3 b \left (12 a^{2}+5 b^{2}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {3 b \left (20 a^{2}+3 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(408\) |
1/64*((32*a^3+160*a*b^2)*cos(2*d*x+2*c)+(24*a^2*b+15*b^3)*sin(3*d*x+3*c)-8 *a*b^2*cos(4*d*x+4*c)-b^3*sin(5*d*x+5*c)+(-288*a^2*b*d*x-120*b^3*d*x+128*a ^3+512*a*b^2)*cos(d*x+c)+(216*a^2*b+80*b^3)*sin(d*x+c)+96*a^3+360*a*b^2)/d /cos(d*x+c)
Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.69 \[ \int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {8 \, a b^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (12 \, a^{2} b + 5 \, b^{3}\right )} d x \cos \left (d x + c\right ) - 8 \, a^{3} - 24 \, a b^{2} - 8 \, {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, b^{3} \cos \left (d x + c\right )^{4} - 24 \, a^{2} b - 8 \, b^{3} - 3 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{8 \, d \cos \left (d x + c\right )} \]
-1/8*(8*a*b^2*cos(d*x + c)^4 + 3*(12*a^2*b + 5*b^3)*d*x*cos(d*x + c) - 8*a ^3 - 24*a*b^2 - 8*(a^3 + 6*a*b^2)*cos(d*x + c)^2 + (2*b^3*cos(d*x + c)^4 - 24*a^2*b - 8*b^3 - 3*(4*a^2*b + 3*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*c os(d*x + c))
Timed out. \[ \int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.83 \[ \int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {12 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{2} b + 8 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a b^{2} + {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} b^{3} - 8 \, a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{8 \, d} \]
-1/8*(12*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c) )*a^2*b + 8*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a*b^2 + (15 *d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan( d*x + c)^2 + 1) - 8*tan(d*x + c))*b^3 - 8*a^3*(1/cos(d*x + c) + cos(d*x + c)))/d
Time = 0.45 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.71 \[ \int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=-\frac {3 \, {\left (12 \, a^{2} b + 5 \, b^{3}\right )} {\left (d x + c\right )} + \frac {16 \, {\left (3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} + 3 \, a b^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 24 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 136 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{3} - 40 \, a b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]
-1/8*(3*(12*a^2*b + 5*b^3)*(d*x + c) + 16*(3*a^2*b*tan(1/2*d*x + 1/2*c) + b^3*tan(1/2*d*x + 1/2*c) + a^3 + 3*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2 *(12*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 7*b^3*tan(1/2*d*x + 1/2*c)^7 - 8*a^3*t an(1/2*d*x + 1/2*c)^6 - 24*a*b^2*tan(1/2*d*x + 1/2*c)^6 + 12*a^2*b*tan(1/2 *d*x + 1/2*c)^5 + 15*b^3*tan(1/2*d*x + 1/2*c)^5 - 24*a^3*tan(1/2*d*x + 1/2 *c)^4 - 120*a*b^2*tan(1/2*d*x + 1/2*c)^4 - 12*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 15*b^3*tan(1/2*d*x + 1/2*c)^3 - 24*a^3*tan(1/2*d*x + 1/2*c)^2 - 136*a*b ^2*tan(1/2*d*x + 1/2*c)^2 - 12*a^2*b*tan(1/2*d*x + 1/2*c) - 7*b^3*tan(1/2* d*x + 1/2*c) - 8*a^3 - 40*a*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 15.33 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.64 \[ \int \sin (c+d x) (a+b \sin (c+d x))^3 \tan ^2(c+d x) \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a^2\,b+\frac {15\,b^3}{4}\right )+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a\,b^2+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (12\,a^3+32\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a^3+48\,a\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (24\,a^2\,b+10\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (9\,a^2\,b+\frac {15\,b^3}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (24\,a^2\,b+10\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (30\,a^2\,b+\frac {9\,b^3}{2}\right )+4\,a^3}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {3\,b\,\mathrm {atan}\left (\frac {3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (12\,a^2+5\,b^2\right )}{36\,a^2\,b+15\,b^3}\right )\,\left (12\,a^2+5\,b^2\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*(9*a^2*b + (15*b^3)/4) + 4*a^3*tan(c/2 + (d*x)/2)^6 + 16*a*b^2 + tan(c/2 + (d*x)/2)^4*(32*a*b^2 + 12*a^3) + tan(c/2 + (d*x)/2)^2 *(48*a*b^2 + 12*a^3) + tan(c/2 + (d*x)/2)^3*(24*a^2*b + 10*b^3) + tan(c/2 + (d*x)/2)^9*(9*a^2*b + (15*b^3)/4) + tan(c/2 + (d*x)/2)^7*(24*a^2*b + 10* b^3) + tan(c/2 + (d*x)/2)^5*(30*a^2*b + (9*b^3)/2) + 4*a^3)/(d*(3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 - tan(c/2 + (d*x)/2)^10 + 1)) - (3*b*atan((3*b*tan(c/2 + (d* x)/2)*(12*a^2 + 5*b^2))/(36*a^2*b + 15*b^3))*(12*a^2 + 5*b^2))/(4*d)