3.6.93 \(\int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx\) [593]

3.6.93.1 Optimal result
3.6.93.2 Mathematica [A] (verified)
3.6.93.3 Rubi [A] (verified)
3.6.93.4 Maple [A] (verified)
3.6.93.5 Fricas [A] (verification not implemented)
3.6.93.6 Sympy [F(-1)]
3.6.93.7 Maxima [A] (verification not implemented)
3.6.93.8 Giac [A] (verification not implemented)
3.6.93.9 Mupad [B] (verification not implemented)

3.6.93.1 Optimal result

Integrand size = 29, antiderivative size = 158 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {25 a^2 x}{16}-\frac {2 a^2 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {a^2 \cot (c+d x)}{d}-\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}-\frac {7 a^2 \cos (c+d x) \sin ^3(c+d x)}{24 d}+\frac {a^2 \cos (c+d x) \sin ^5(c+d x)}{6 d} \]

output
-25/16*a^2*x-2*a^2*arctanh(cos(d*x+c))/d+2*a^2*cos(d*x+c)/d+2/3*a^2*cos(d* 
x+c)^3/d+2/5*a^2*cos(d*x+c)^5/d-a^2*cot(d*x+c)/d-7/16*a^2*cos(d*x+c)*sin(d 
*x+c)/d-7/24*a^2*cos(d*x+c)*sin(d*x+c)^3/d+1/6*a^2*cos(d*x+c)*sin(d*x+c)^5 
/d
 
3.6.93.2 Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.70 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {a^2 \left (-1500 c-1500 d x+2640 \cos (c+d x)+280 \cos (3 (c+d x))+24 \cos (5 (c+d x))-960 \cot (c+d x)-1920 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+1920 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-255 \sin (2 (c+d x))+15 \sin (4 (c+d x))+5 \sin (6 (c+d x))\right )}{960 d} \]

input
Integrate[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
 
output
(a^2*(-1500*c - 1500*d*x + 2640*Cos[c + d*x] + 280*Cos[3*(c + d*x)] + 24*C 
os[5*(c + d*x)] - 960*Cot[c + d*x] - 1920*Log[Cos[(c + d*x)/2]] + 1920*Log 
[Sin[(c + d*x)/2]] - 255*Sin[2*(c + d*x)] + 15*Sin[4*(c + d*x)] + 5*Sin[6* 
(c + d*x)]))/(960*d)
 
3.6.93.3 Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 162, 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.

\(\displaystyle \int \cos ^4(c+d x) \cot ^2(c+d x) (a \sin (c+d x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (c+d x)^6 (a \sin (c+d x)+a)^2}{\sin (c+d x)^2}dx\)

\(\Big \downarrow \) 3351

\(\displaystyle \frac {\int \left (-\sin ^6(c+d x) a^8-2 \sin ^5(c+d x) a^8+2 \sin ^4(c+d x) a^8+6 \sin ^3(c+d x) a^8+\csc ^2(c+d x) a^8+2 \csc (c+d x) a^8-6 \sin (c+d x) a^8-2 a^8\right )dx}{a^6}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {2 a^8 \text {arctanh}(\cos (c+d x))}{d}+\frac {2 a^8 \cos ^5(c+d x)}{5 d}+\frac {2 a^8 \cos ^3(c+d x)}{3 d}+\frac {2 a^8 \cos (c+d x)}{d}-\frac {a^8 \cot (c+d x)}{d}+\frac {a^8 \sin ^5(c+d x) \cos (c+d x)}{6 d}-\frac {7 a^8 \sin ^3(c+d x) \cos (c+d x)}{24 d}-\frac {7 a^8 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {25 a^8 x}{16}}{a^6}\)

input
Int[Cos[c + d*x]^4*Cot[c + d*x]^2*(a + a*Sin[c + d*x])^2,x]
 
output
((-25*a^8*x)/16 - (2*a^8*ArcTanh[Cos[c + d*x]])/d + (2*a^8*Cos[c + d*x])/d 
 + (2*a^8*Cos[c + d*x]^3)/(3*d) + (2*a^8*Cos[c + d*x]^5)/(5*d) - (a^8*Cot[ 
c + d*x])/d - (7*a^8*Cos[c + d*x]*Sin[c + d*x])/(16*d) - (7*a^8*Cos[c + d* 
x]*Sin[c + d*x]^3)/(24*d) + (a^8*Cos[c + d*x]*Sin[c + d*x]^5)/(6*d))/a^6
 

3.6.93.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3351
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]))
 
3.6.93.4 Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.87

method result size
parallelrisch \(-\frac {a^{2} \left (3000 d x \sin \left (d x +c \right )-3840 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )+2175 \cos \left (d x +c \right )-5888 \sin \left (d x +c \right )-24 \sin \left (6 d x +6 c \right )+5 \cos \left (7 d x +7 c \right )+10 \cos \left (5 d x +5 c \right )-256 \sin \left (4 d x +4 c \right )-2360 \sin \left (2 d x +2 c \right )-270 \cos \left (3 d x +3 c \right )\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )}{3840 d}\) \(138\)
derivativedivides \(\frac {a^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+2 a^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) \(166\)
default \(\frac {a^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+2 a^{2} \left (\frac {\left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{3}\left (d x +c \right )\right )}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\cos ^{7}\left (d x +c \right )}{\sin \left (d x +c \right )}-\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}\) \(166\)
risch \(-\frac {25 a^{2} x}{16}+\frac {11 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {11 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {17 i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d}+\frac {17 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{128 d}-\frac {2 i a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {a^{2} \sin \left (6 d x +6 c \right )}{192 d}+\frac {a^{2} \cos \left (5 d x +5 c \right )}{40 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{64 d}+\frac {7 a^{2} \cos \left (3 d x +3 c \right )}{24 d}\) \(208\)
norman \(\frac {\frac {56 a^{2} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2}}{2 d}-\frac {27 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {227 a^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}-\frac {5 a^{2} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a^{2} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {227 a^{2} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {27 a^{2} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {a^{2} \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {25 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}-\frac {75 a^{2} x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {375 a^{2} x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {125 a^{2} x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {375 a^{2} x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}-\frac {75 a^{2} x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {25 a^{2} x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16}+\frac {12 a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {36 a^{2} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {184 a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {124 a^{2} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {92 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) \(418\)

input
int(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
 
output
-1/3840*a^2*(3000*d*x*sin(d*x+c)-3840*ln(tan(1/2*d*x+1/2*c))*sin(d*x+c)+21 
75*cos(d*x+c)-5888*sin(d*x+c)-24*sin(6*d*x+6*c)+5*cos(7*d*x+7*c)+10*cos(5* 
d*x+5*c)-256*sin(4*d*x+4*c)-2360*sin(2*d*x+2*c)-270*cos(3*d*x+3*c))*sec(1/ 
2*d*x+1/2*c)*csc(1/2*d*x+1/2*c)/d
 
3.6.93.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.02 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {40 \, a^{2} \cos \left (d x + c\right )^{7} - 50 \, a^{2} \cos \left (d x + c\right )^{5} - 125 \, a^{2} \cos \left (d x + c\right )^{3} + 240 \, a^{2} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 240 \, a^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 375 \, a^{2} \cos \left (d x + c\right ) - {\left (96 \, a^{2} \cos \left (d x + c\right )^{5} + 160 \, a^{2} \cos \left (d x + c\right )^{3} - 375 \, a^{2} d x + 480 \, a^{2} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \sin \left (d x + c\right )} \]

input
integrate(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="frica 
s")
 
output
-1/240*(40*a^2*cos(d*x + c)^7 - 50*a^2*cos(d*x + c)^5 - 125*a^2*cos(d*x + 
c)^3 + 240*a^2*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 240*a^2*log(-1/2 
*cos(d*x + c) + 1/2)*sin(d*x + c) + 375*a^2*cos(d*x + c) - (96*a^2*cos(d*x 
 + c)^5 + 160*a^2*cos(d*x + c)^3 - 375*a^2*d*x + 480*a^2*cos(d*x + c))*sin 
(d*x + c))/(d*sin(d*x + c))
 
3.6.93.6 Sympy [F(-1)]

Timed out. \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\text {Timed out} \]

input
integrate(cos(d*x+c)**6*csc(d*x+c)**2*(a+a*sin(d*x+c))**2,x)
 
output
Timed out
 
3.6.93.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.09 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {64 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 120 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{2}}{960 \, d} \]

input
integrate(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxim 
a")
 
output
1/960*(64*(6*cos(d*x + c)^5 + 10*cos(d*x + c)^3 + 30*cos(d*x + c) - 15*log 
(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1))*a^2 - 5*(4*sin(2*d*x + 2*c) 
^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^2 - 120*( 
15*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)/d
 
3.6.93.8 Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.73 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=-\frac {375 \, {\left (d x + c\right )} a^{2} - 480 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) - 120 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {120 \, {\left (4 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1440 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4320 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 7360 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 150 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6720 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2976 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 736 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

input
integrate(cos(d*x+c)^6*csc(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac" 
)
 
output
-1/240*(375*(d*x + c)*a^2 - 480*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 120*a 
^2*tan(1/2*d*x + 1/2*c) + 120*(4*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d 
*x + 1/2*c) - 2*(105*a^2*tan(1/2*d*x + 1/2*c)^11 + 1440*a^2*tan(1/2*d*x + 
1/2*c)^10 + 595*a^2*tan(1/2*d*x + 1/2*c)^9 + 4320*a^2*tan(1/2*d*x + 1/2*c) 
^8 - 150*a^2*tan(1/2*d*x + 1/2*c)^7 + 7360*a^2*tan(1/2*d*x + 1/2*c)^6 + 15 
0*a^2*tan(1/2*d*x + 1/2*c)^5 + 6720*a^2*tan(1/2*d*x + 1/2*c)^4 - 595*a^2*t 
an(1/2*d*x + 1/2*c)^3 + 2976*a^2*tan(1/2*d*x + 1/2*c)^2 - 105*a^2*tan(1/2* 
d*x + 1/2*c) + 736*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d
 
3.6.93.9 Mupad [B] (verification not implemented)

Time = 10.55 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.54 \[ \int \cos ^4(c+d x) \cot ^2(c+d x) (a+a \sin (c+d x))^2 \, dx=\frac {2\,a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {25\,a^2\,\mathrm {atan}\left (\frac {625\,a^4}{64\,\left (\frac {25\,a^4}{2}+\frac {625\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}-\frac {25\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,\left (\frac {25\,a^4}{2}+\frac {625\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64}\right )}\right )}{8\,d}+\frac {\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{4}+24\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\frac {47\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{12}+72\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-\frac {35\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+\frac {368\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {35\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+112\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {299\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{12}+\frac {248\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}-\frac {31\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4}+\frac {184\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}-a^2}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d} \]

input
int((cos(c + d*x)^6*(a + a*sin(c + d*x))^2)/sin(c + d*x)^2,x)
 
output
(2*a^2*log(tan(c/2 + (d*x)/2)))/d + (25*a^2*atan((625*a^4)/(64*((25*a^4)/2 
 + (625*a^4*tan(c/2 + (d*x)/2))/64)) - (25*a^4*tan(c/2 + (d*x)/2))/(2*((25 
*a^4)/2 + (625*a^4*tan(c/2 + (d*x)/2))/64))))/(8*d) + ((248*a^2*tan(c/2 + 
(d*x)/2)^3)/5 - (31*a^2*tan(c/2 + (d*x)/2)^2)/4 - (299*a^2*tan(c/2 + (d*x) 
/2)^4)/12 + 112*a^2*tan(c/2 + (d*x)/2)^5 - (35*a^2*tan(c/2 + (d*x)/2)^6)/2 
 + (368*a^2*tan(c/2 + (d*x)/2)^7)/3 - (35*a^2*tan(c/2 + (d*x)/2)^8)/2 + 72 
*a^2*tan(c/2 + (d*x)/2)^9 + (47*a^2*tan(c/2 + (d*x)/2)^10)/12 + 24*a^2*tan 
(c/2 + (d*x)/2)^11 + (3*a^2*tan(c/2 + (d*x)/2)^12)/4 - a^2 + (184*a^2*tan( 
c/2 + (d*x)/2))/15)/(d*(2*tan(c/2 + (d*x)/2) + 12*tan(c/2 + (d*x)/2)^3 + 3 
0*tan(c/2 + (d*x)/2)^5 + 40*tan(c/2 + (d*x)/2)^7 + 30*tan(c/2 + (d*x)/2)^9 
 + 12*tan(c/2 + (d*x)/2)^11 + 2*tan(c/2 + (d*x)/2)^13)) + (a^2*tan(c/2 + ( 
d*x)/2))/(2*d)