Integrand size = 29, antiderivative size = 176 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {25 a^3 x}{8}+\frac {13 a^3 \text {arctanh}(\cos (c+d x))}{2 d}-\frac {5 a^3 \cos (c+d x)}{d}-\frac {2 a^3 \cos ^3(c+d x)}{3 d}+\frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \cot (c+d x)}{d}-\frac {a^3 \cot ^3(c+d x)}{3 d}-\frac {3 a^3 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {23 a^3 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {3 a^3 \cos (c+d x) \sin ^3(c+d x)}{4 d} \] Output:
-25/8*a^3*x+13/2*a^3*arctanh(cos(d*x+c))/d-5*a^3*cos(d*x+c)/d-2/3*a^3*cos( d*x+c)^3/d+1/5*a^3*cos(d*x+c)^5/d-a^3*cot(d*x+c)/d-1/3*a^3*cot(d*x+c)^3/d- 3/2*a^3*cot(d*x+c)*csc(d*x+c)/d-23/8*a^3*cos(d*x+c)*sin(d*x+c)/d+3/4*a^3*c os(d*x+c)*sin(d*x+c)^3/d
Time = 7.63 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.24 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3 (1+\sin (c+d x))^3 \left (-1500 (c+d x)-2580 \cos (c+d x)-50 \cos (3 (c+d x))+6 \cos (5 (c+d x))-160 \cot \left (\frac {1}{2} (c+d x)\right )-180 \csc ^2\left (\frac {1}{2} (c+d x)\right )+3120 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-3120 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+180 \sec ^2\left (\frac {1}{2} (c+d x)\right )+160 \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )-10 \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)-600 \sin (2 (c+d x))-45 \sin (4 (c+d x))+160 \tan \left (\frac {1}{2} (c+d x)\right )\right )}{480 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6} \] Input:
Integrate[Cos[c + d*x]^2*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]
Output:
(a^3*(1 + Sin[c + d*x])^3*(-1500*(c + d*x) - 2580*Cos[c + d*x] - 50*Cos[3* (c + d*x)] + 6*Cos[5*(c + d*x)] - 160*Cot[(c + d*x)/2] - 180*Csc[(c + d*x) /2]^2 + 3120*Log[Cos[(c + d*x)/2]] - 3120*Log[Sin[(c + d*x)/2]] + 180*Sec[ (c + d*x)/2]^2 + 160*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 - 10*Csc[(c + d*x)/ 2]^4*Sin[c + d*x] - 600*Sin[2*(c + d*x)] - 45*Sin[4*(c + d*x)] + 160*Tan[( c + d*x)/2]))/(480*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)
Time = 0.49 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.02, 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.
\(\displaystyle \int \cos ^2(c+d x) \cot ^4(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^6 (a \sin (c+d x)+a)^3}{\sin (c+d x)^4}dx\) |
\(\Big \downarrow \) 3351 |
\(\displaystyle \frac {\int \left (-\sin ^5(c+d x) a^9+\csc ^4(c+d x) a^9-3 \sin ^4(c+d x) a^9+3 \csc ^3(c+d x) a^9+8 \sin ^2(c+d x) a^9-8 \csc (c+d x) a^9+6 \sin (c+d x) a^9-6 a^9\right )dx}{a^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {13 a^9 \text {arctanh}(\cos (c+d x))}{2 d}+\frac {a^9 \cos ^5(c+d x)}{5 d}-\frac {2 a^9 \cos ^3(c+d x)}{3 d}-\frac {5 a^9 \cos (c+d x)}{d}-\frac {a^9 \cot ^3(c+d x)}{3 d}-\frac {a^9 \cot (c+d x)}{d}+\frac {3 a^9 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {23 a^9 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {3 a^9 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {25 a^9 x}{8}}{a^6}\) |
Input:
Int[Cos[c + d*x]^2*Cot[c + d*x]^4*(a + a*Sin[c + d*x])^3,x]
Output:
((-25*a^9*x)/8 + (13*a^9*ArcTanh[Cos[c + d*x]])/(2*d) - (5*a^9*Cos[c + d*x ])/d - (2*a^9*Cos[c + d*x]^3)/(3*d) + (a^9*Cos[c + d*x]^5)/(5*d) - (a^9*Co t[c + d*x])/d - (a^9*Cot[c + d*x]^3)/(3*d) - (3*a^9*Cot[c + d*x]*Csc[c + d *x])/(2*d) - (23*a^9*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (3*a^9*Cos[c + d*x ]*Sin[c + d*x]^3)/(4*d))/a^6
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]))
Result contains complex when optimal does not.
Time = 10.28 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.39
method | result | size |
risch | \(-\frac {25 a^{3} x}{8}+\frac {a^{3} {\mathrm e}^{5 i \left (d x +c \right )}}{160 d}+\frac {5 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {43 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{16 d}-\frac {43 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{16 d}-\frac {5 i a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {a^{3} {\mathrm e}^{-5 i \left (d x +c \right )}}{160 d}+\frac {a^{3} \left (9 \,{\mathrm e}^{5 i \left (d x +c \right )}+12 i {\mathrm e}^{2 i \left (d x +c \right )}-4 i-9 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}-\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {13 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {3 a^{3} \sin \left (4 d x +4 c \right )}{32 d}-\frac {5 a^{3} \cos \left (3 d x +3 c \right )}{48 d}\) | \(244\) |
derivativedivides | \(\frac {a^{3} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{7}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{5}}{2}-\frac {5 \cos \left (d x +c \right )^{3}}{6}-\frac {5 \cos \left (d x +c \right )}{2}-\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+a^{3} \left (-\frac {\cos \left (d x +c \right )^{7}}{3 \sin \left (d x +c \right )^{3}}+\frac {4 \cos \left (d x +c \right )^{7}}{3 \sin \left (d x +c \right )}+\frac {4 \left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(272\) |
default | \(\frac {a^{3} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+3 a^{3} \left (-\frac {\cos \left (d x +c \right )^{7}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{5}}{2}-\frac {5 \cos \left (d x +c \right )^{3}}{6}-\frac {5 \cos \left (d x +c \right )}{2}-\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+a^{3} \left (-\frac {\cos \left (d x +c \right )^{7}}{3 \sin \left (d x +c \right )^{3}}+\frac {4 \cos \left (d x +c \right )^{7}}{3 \sin \left (d x +c \right )}+\frac {4 \left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(272\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)^4*(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-25/8*a^3*x+1/160*a^3/d*exp(5*I*(d*x+c))+5/8*I*a^3/d*exp(2*I*(d*x+c))-43/1 6*a^3/d*exp(I*(d*x+c))-43/16*a^3/d*exp(-I*(d*x+c))-5/8*I*a^3/d*exp(-2*I*(d *x+c))+1/160*a^3/d*exp(-5*I*(d*x+c))+1/3*a^3*(9*exp(5*I*(d*x+c))+12*I*exp( 2*I*(d*x+c))-4*I-9*exp(I*(d*x+c)))/d/(exp(2*I*(d*x+c))-1)^3-13/2*a^3/d*ln( exp(I*(d*x+c))-1)+13/2*a^3/d*ln(exp(I*(d*x+c))+1)-3/32*a^3/d*sin(4*d*x+4*c )-5/48*a^3/d*cos(3*d*x+3*c)
Time = 0.11 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.31 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {90 \, a^{3} \cos \left (d x + c\right )^{7} + 75 \, a^{3} \cos \left (d x + c\right )^{5} - 500 \, a^{3} \cos \left (d x + c\right )^{3} + 375 \, a^{3} \cos \left (d x + c\right ) + 390 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 390 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + {\left (24 \, a^{3} \cos \left (d x + c\right )^{7} - 104 \, a^{3} \cos \left (d x + c\right )^{5} - 375 \, a^{3} d x \cos \left (d x + c\right )^{2} - 520 \, a^{3} \cos \left (d x + c\right )^{3} + 375 \, a^{3} d x + 780 \, a^{3} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
1/120*(90*a^3*cos(d*x + c)^7 + 75*a^3*cos(d*x + c)^5 - 500*a^3*cos(d*x + c )^3 + 375*a^3*cos(d*x + c) + 390*(a^3*cos(d*x + c)^2 - a^3)*log(1/2*cos(d* x + c) + 1/2)*sin(d*x + c) - 390*(a^3*cos(d*x + c)^2 - a^3)*log(-1/2*cos(d *x + c) + 1/2)*sin(d*x + c) + (24*a^3*cos(d*x + c)^7 - 104*a^3*cos(d*x + c )^5 - 375*a^3*d*x*cos(d*x + c)^2 - 520*a^3*cos(d*x + c)^3 + 375*a^3*d*x + 780*a^3*cos(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2 - d)*sin(d*x + c))
\[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=a^{3} \left (\int \cos ^{2}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx + \int 3 \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)**4*(a+a*sin(d*x+c))**3,x)
Output:
a**3*(Integral(cos(c + d*x)**2*cot(c + d*x)**4, x) + Integral(3*sin(c + d* x)*cos(c + d*x)**2*cot(c + d*x)**4, x) + Integral(3*sin(c + d*x)**2*cos(c + d*x)**2*cot(c + d*x)**4, x) + Integral(sin(c + d*x)**3*cos(c + d*x)**2*c ot(c + d*x)**4, x))
Time = 0.12 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.40 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {4 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} - 30 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{3} - 45 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a^{3} + 20 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 10 \, \tan \left (d x + c\right )^{2} - 2}{\tan \left (d x + c\right )^{5} + \tan \left (d x + c\right )^{3}}\right )} a^{3}}{120 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
1/120*(4*(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^3 - 30*(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^3 - 45*(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^3 + 20*(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^3)/d
Time = 0.24 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.66 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 45 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 375 \, {\left (d x + c\right )} a^{3} - 780 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + 45 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {5 \, {\left (286 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}} + \frac {2 \, {\left (345 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 720 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 330 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2880 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 3680 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 330 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2560 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 345 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 656 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^4*(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
1/120*(5*a^3*tan(1/2*d*x + 1/2*c)^3 + 45*a^3*tan(1/2*d*x + 1/2*c)^2 - 375* (d*x + c)*a^3 - 780*a^3*log(abs(tan(1/2*d*x + 1/2*c))) + 45*a^3*tan(1/2*d* x + 1/2*c) + 5*(286*a^3*tan(1/2*d*x + 1/2*c)^3 - 9*a^3*tan(1/2*d*x + 1/2*c )^2 - 9*a^3*tan(1/2*d*x + 1/2*c) - a^3)/tan(1/2*d*x + 1/2*c)^3 + 2*(345*a^ 3*tan(1/2*d*x + 1/2*c)^9 - 720*a^3*tan(1/2*d*x + 1/2*c)^8 + 330*a^3*tan(1/ 2*d*x + 1/2*c)^7 - 2880*a^3*tan(1/2*d*x + 1/2*c)^6 - 3680*a^3*tan(1/2*d*x + 1/2*c)^4 - 330*a^3*tan(1/2*d*x + 1/2*c)^3 - 2560*a^3*tan(1/2*d*x + 1/2*c )^2 - 345*a^3*tan(1/2*d*x + 1/2*c) - 656*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1) ^5)/d
Time = 33.90 (sec) , antiderivative size = 429, normalized size of antiderivative = 2.44 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}+\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {13\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {-43\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+99\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-\frac {86\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{3}+399\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\frac {95\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {1562\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {232\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+\frac {1114\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}+\frac {193\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {1537\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{15}+\frac {14\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3}{3}}{d\,\left (8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+80\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+80\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+40\,{\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 )}-\frac {25\,a^3\,\mathrm {atan}\left (\frac {625\,a^6}{16\,\left (\frac {325\,a^6}{4}-\frac {625\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}+\frac {325\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,\left (\frac {325\,a^6}{4}-\frac {625\,a^6\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16}\right )}\right )}{4\,d}+\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d} \] Input:
int(cos(c + d*x)^2*cot(c + d*x)^4*(a + a*sin(c + d*x))^3,x)
Output:
(3*a^3*tan(c/2 + (d*x)/2)^2)/(8*d) + (a^3*tan(c/2 + (d*x)/2)^3)/(24*d) - ( 13*a^3*log(tan(c/2 + (d*x)/2)))/(2*d) - ((14*a^3*tan(c/2 + (d*x)/2)^2)/3 + (1537*a^3*tan(c/2 + (d*x)/2)^3)/15 + (193*a^3*tan(c/2 + (d*x)/2)^4)/3 + ( 1114*a^3*tan(c/2 + (d*x)/2)^5)/3 + (232*a^3*tan(c/2 + (d*x)/2)^6)/3 + (156 2*a^3*tan(c/2 + (d*x)/2)^7)/3 + (95*a^3*tan(c/2 + (d*x)/2)^8)/3 + 399*a^3* tan(c/2 + (d*x)/2)^9 - (86*a^3*tan(c/2 + (d*x)/2)^10)/3 + 99*a^3*tan(c/2 + (d*x)/2)^11 - 43*a^3*tan(c/2 + (d*x)/2)^12 + a^3/3 + 3*a^3*tan(c/2 + (d*x )/2))/(d*(8*tan(c/2 + (d*x)/2)^3 + 40*tan(c/2 + (d*x)/2)^5 + 80*tan(c/2 + (d*x)/2)^7 + 80*tan(c/2 + (d*x)/2)^9 + 40*tan(c/2 + (d*x)/2)^11 + 8*tan(c/ 2 + (d*x)/2)^13)) - (25*a^3*atan((625*a^6)/(16*((325*a^6)/4 - (625*a^6*tan (c/2 + (d*x)/2))/16)) + (325*a^6*tan(c/2 + (d*x)/2))/(4*((325*a^6)/4 - (62 5*a^6*tan(c/2 + (d*x)/2))/16))))/(4*d) + (3*a^3*tan(c/2 + (d*x)/2))/(8*d)
Time = 2.29 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.01 \[ \int \cos ^2(c+d x) \cot ^4(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \left (24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}+90 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-345 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-656 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-80 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-180 \cos \left (d x +c \right ) \sin \left (d x +c \right )-40 \cos \left (d x +c \right )-780 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3}-375 \sin \left (d x +c \right )^{3} d x +881 \sin \left (d x +c \right )^{3}\right )}{120 \sin \left (d x +c \right )^{3} d} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)^4*(a+a*sin(d*x+c))^3,x)
Output:
(a**3*(24*cos(c + d*x)*sin(c + d*x)**7 + 90*cos(c + d*x)*sin(c + d*x)**6 + 32*cos(c + d*x)*sin(c + d*x)**5 - 345*cos(c + d*x)*sin(c + d*x)**4 - 656* cos(c + d*x)*sin(c + d*x)**3 - 80*cos(c + d*x)*sin(c + d*x)**2 - 180*cos(c + d*x)*sin(c + d*x) - 40*cos(c + d*x) - 780*log(tan((c + d*x)/2))*sin(c + d*x)**3 - 375*sin(c + d*x)**3*d*x + 881*sin(c + d*x)**3))/(120*sin(c + d* x)**3*d)