Integrand size = 29, antiderivative size = 153 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {5 a^2 x}{8}+\frac {5 a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {4 a^2 \cos (c+d x)}{d}-\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \cot (c+d x)}{d}-\frac {a^2 \cot ^3(c+d x)}{3 d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{d}-\frac {5 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos (c+d x) \sin ^3(c+d x)}{4 d} \]
5/8*a^2*x+5*a^2*arctanh(cos(d*x+c))/d-4*a^2*cos(d*x+c)/d-2/3*a^2*cos(d*x+c )^3/d+a^2*cot(d*x+c)/d-1/3*a^2*cot(d*x+c)^3/d-a^2*cot(d*x+c)*csc(d*x+c)/d- 5/8*a^2*cos(d*x+c)*sin(d*x+c)/d+1/4*a^2*cos(d*x+c)*sin(d*x+c)^3/d
Time = 8.41 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.37 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 (1+\sin (c+d x))^2 \left (60 (c+d x)-432 \cos (c+d x)-16 \cos (3 (c+d x))+64 \cot \left (\frac {1}{2} (c+d x)\right )-24 \csc ^2\left (\frac {1}{2} (c+d x)\right )+480 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-480 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+24 \sec ^2\left (\frac {1}{2} (c+d x)\right )+32 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-2 \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)-24 \sin (2 (c+d x))-3 \sin (4 (c+d x))-64 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{96 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4} \]
(a^2*(1 + Sin[c + d*x])^2*(60*(c + d*x) - 432*Cos[c + d*x] - 16*Cos[3*(c + d*x)] + 64*Cot[(c + d*x)/2] - 24*Csc[(c + d*x)/2]^2 + 480*Log[Cos[(c + d* x)/2]] - 480*Log[Sin[(c + d*x)/2]] + 24*Sec[(c + d*x)/2]^2 + 32*Csc[c + d* x]^3*Sin[(c + d*x)/2]^4 - 2*Csc[(c + d*x)/2]^4*Sin[c + d*x] - 24*Sin[2*(c + d*x)] - 3*Sin[4*(c + d*x)] - 64*Tan[(c + d*x)/2]))/(96*d*(Cos[(c + d*x)/ 2] + Sin[(c + d*x)/2])^4)
Time = 0.43 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3351, 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.
| \(\displaystyle \int \cos ^2(c+d x) \cot ^4(c+d x) (a \sin (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^6 (a \sin (c+d x)+a)^2}{\sin (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (\csc ^4(c+d x) a^8-\sin ^4(c+d x) a^8+2 \csc ^3(c+d x) a^8-2 \sin ^3(c+d x) a^8-2 \csc ^2(c+d x) a^8+2 \sin ^2(c+d x) a^8-6 \csc (c+d x) a^8+6 \sin (c+d x) a^8\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {5 a^8 \text {arctanh}(\cos (c+d x))}{d}-\frac {2 a^8 \cos ^3(c+d x)}{3 d}-\frac {4 a^8 \cos (c+d x)}{d}-\frac {a^8 \cot ^3(c+d x)}{3 d}+\frac {a^8 \cot (c+d x)}{d}+\frac {a^8 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {5 a^8 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {a^8 \cot (c+d x) \csc (c+d x)}{d}+\frac {5 a^8 x}{8}}{a^6}\) |
((5*a^8*x)/8 + (5*a^8*ArcTanh[Cos[c + d*x]])/d - (4*a^8*Cos[c + d*x])/d - (2*a^8*Cos[c + d*x]^3)/(3*d) + (a^8*Cot[c + d*x])/d - (a^8*Cot[c + d*x]^3) /(3*d) - (a^8*Cot[c + d*x]*Csc[c + d*x])/d - (5*a^8*Cos[c + d*x]*Sin[c + d *x])/(8*d) + (a^8*Cos[c + d*x]*Sin[c + d*x]^3)/(4*d))/a^6
3.6.95.3.1 Defintions of rubi rules used
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 0]))
Time = 0.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.16
| method | result | size |
| parallelrisch | \(\frac {a^{2} \left (\csc ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\sec ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (960 \left (-3 \sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 d x \sin \left (3 d x +3 c \right )+360 d x \sin \left (d x +c \right )-1200 \sin \left (2 d x +2 c \right )+1088 \sin \left (3 d x +3 c \right )+384 \sin \left (4 d x +4 c \right )+16 \sin \left (6 d x +6 c \right )-45 \cos \left (d x +c \right )-193 \cos \left (3 d x +3 c \right )-15 \cos \left (5 d x +5 c \right )-3 \cos \left (7 d x +7 c \right )-3264 \sin \left (d x +c \right )\right )}{6144 d}\) | \(178\) |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+2 a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{5}\left (d x +c \right )\right )}{2}-\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {5 \cos \left (d x +c \right )}{2}-\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {4 \left (\cos ^{7}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )}+\frac {4 \left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(224\) |
| default | \(\frac {a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+2 a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{2 \sin \left (d x +c \right )^{2}}-\frac {\left (\cos ^{5}\left (d x +c \right )\right )}{2}-\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{6}-\frac {5 \cos \left (d x +c \right )}{2}-\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{3 \sin \left (d x +c \right )^{3}}+\frac {4 \left (\cos ^{7}\left (d x +c \right )\right )}{3 \sin \left (d x +c \right )}+\frac {4 \left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(224\) |
| risch | \(\frac {5 a^{2} x}{8}+\frac {i a^{2} {\mathrm e}^{4 i \left (d x +c \right )}}{64 d}+\frac {i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {9 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{4 d}-\frac {9 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{4 d}-\frac {i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {i a^{2} {\mathrm e}^{-4 i \left (d x +c \right )}}{64 d}+\frac {2 a^{2} \left (6 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}-6 i {\mathrm e}^{2 i \left (d x +c \right )}+4 i-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {a^{2} \cos \left (3 d x +3 c \right )}{6 d}\) | \(241\) |
| norman | \(\frac {-\frac {a^{2}}{24 d}-\frac {a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {3 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {15 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {15 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {3 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {11 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {a^{2} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {5 a^{2} x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {5 a^{2} x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {15 a^{2} x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{2} x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{2} x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {34 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {34 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {59 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {367 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}-\frac {5 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(386\) |
1/6144*a^2*csc(1/2*d*x+1/2*c)^3*sec(1/2*d*x+1/2*c)^3*(960*(-3*sin(d*x+c)+s in(3*d*x+3*c))*ln(tan(1/2*d*x+1/2*c))-120*d*x*sin(3*d*x+3*c)+360*d*x*sin(d *x+c)-1200*sin(2*d*x+2*c)+1088*sin(3*d*x+3*c)+384*sin(4*d*x+4*c)+16*sin(6* d*x+6*c)-45*cos(d*x+c)-193*cos(3*d*x+3*c)-15*cos(5*d*x+5*c)-3*cos(7*d*x+7* c)-3264*sin(d*x+c))/d
Time = 0.29 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.43 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {6 \, a^{2} \cos \left (d x + c\right )^{7} - 3 \, a^{2} \cos \left (d x + c\right )^{5} + 20 \, a^{2} \cos \left (d x + c\right )^{3} - 15 \, a^{2} \cos \left (d x + c\right ) + 60 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 60 \, {\left (a^{2} \cos \left (d x + c\right )^{2} - a^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - {\left (16 \, a^{2} \cos \left (d x + c\right )^{5} - 15 \, a^{2} d x \cos \left (d x + c\right )^{2} + 80 \, a^{2} \cos \left (d x + c\right )^{3} + 15 \, a^{2} d x - 120 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]
1/24*(6*a^2*cos(d*x + c)^7 - 3*a^2*cos(d*x + c)^5 + 20*a^2*cos(d*x + c)^3 - 15*a^2*cos(d*x + c) + 60*(a^2*cos(d*x + c)^2 - a^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 60*(a^2*cos(d*x + c)^2 - a^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - (16*a^2*cos(d*x + c)^5 - 15*a^2*d*x*cos(d*x + c)^2 + 80*a^2*cos(d*x + c)^3 + 15*a^2*d*x - 120*a^2*cos(d*x + c))*sin(d*x + c)) /((d*cos(d*x + c)^2 - d)*sin(d*x + c))
Timed out. \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]
Time = 0.32 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.24 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {4 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} + 3 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2} - 4 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{2}}{24 \, d} \]
-1/24*(4*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos( d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1))*a^2 + 3*(1 5*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan(d*x + c)^5 + 2*tan(d*x + c)^3 + tan(d*x + c)))*a^2 - 4*(15*d*x + 15*c + (15*tan(d*x + c)^4 + 10*tan(d*x + c)^2 - 2)/(tan(d*x + c)^5 + tan(d*x + c)^3))*a^2)/d
Time = 0.40 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.79 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, {\left (d x + c\right )} a^{2} - 120 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {220 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} + \frac {2 \, {\left (15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 144 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 336 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 304 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 112 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]
1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^2*tan(1/2*d*x + 1/2*c)^2 + 15*(d*x + c)*a^2 - 120*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 15*a^2*tan(1/2*d*x + 1 /2*c) + (220*a^2*tan(1/2*d*x + 1/2*c)^3 + 15*a^2*tan(1/2*d*x + 1/2*c)^2 - 6*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^3 + 2*(15*a^2*tan(1 /2*d*x + 1/2*c)^7 - 144*a^2*tan(1/2*d*x + 1/2*c)^6 - 9*a^2*tan(1/2*d*x + 1 /2*c)^5 - 336*a^2*tan(1/2*d*x + 1/2*c)^4 + 9*a^2*tan(1/2*d*x + 1/2*c)^3 - 304*a^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*tan(1/2*d*x + 1/2*c) - 112*a^2)/(t an(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 10.39 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.51 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,d}+\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {5\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {5\,a^2\,\mathrm {atan}\left (\frac {25\,a^4}{16\,\left (\frac {25\,a^4}{2}+\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {25\,a^4}{2}+\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d}-\frac {5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {-15\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+98\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {41\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+232\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {104\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {644\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {248\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^2}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+48\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )} \]
(a^2*tan(c/2 + (d*x)/2)^2)/(4*d) + (a^2*tan(c/2 + (d*x)/2)^3)/(24*d) - (5* a^2*log(tan(c/2 + (d*x)/2)))/d - (5*a^2*atan((25*a^4)/(16*((25*a^4)/2 + (2 5*a^4*tan(c/2 + (d*x)/2))/16)) - (25*a^4*tan(c/2 + (d*x)/2))/(2*((25*a^4)/ 2 + (25*a^4*tan(c/2 + (d*x)/2))/16))))/(4*d) - (5*a^2*tan(c/2 + (d*x)/2))/ (8*d) - ((248*a^2*tan(c/2 + (d*x)/2)^3)/3 - (11*a^2*tan(c/2 + (d*x)/2)^2)/ 3 - 8*a^2*tan(c/2 + (d*x)/2)^4 + (644*a^2*tan(c/2 + (d*x)/2)^5)/3 - (104*a ^2*tan(c/2 + (d*x)/2)^6)/3 + 232*a^2*tan(c/2 + (d*x)/2)^7 - (41*a^2*tan(c/ 2 + (d*x)/2)^8)/3 + 98*a^2*tan(c/2 + (d*x)/2)^9 - 15*a^2*tan(c/2 + (d*x)/2 )^10 + a^2/3 + 2*a^2*tan(c/2 + (d*x)/2))/(d*(8*tan(c/2 + (d*x)/2)^3 + 32*t an(c/2 + (d*x)/2)^5 + 48*tan(c/2 + (d*x)/2)^7 + 32*tan(c/2 + (d*x)/2)^9 + 8*tan(c/2 + (d*x)/2)^11))