Integrand size = 22, antiderivative size = 115 \[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=-\frac {3 i d (c+d x)^2}{2 b^2}-\frac {3 d (c+d x)^2 \cot (a+b x)}{2 b^2}-\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}+\frac {3 d^2 (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^4} \]
-3/2*I*d*(d*x+c)^2/b^2-3/2*d*(d*x+c)^2*cot(b*x+a)/b^2-1/2*(d*x+c)^3*csc(b* x+a)^2/b+3*d^2*(d*x+c)*ln(1-exp(2*I*(b*x+a)))/b^3-3/2*I*d^3*polylog(2,exp( 2*I*(b*x+a)))/b^4
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(115)=230\).
Time = 6.46 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.41 \[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=-\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}+\frac {3 c d^2 \csc (a) (-b x \cos (a)+\log (\cos (b x) \sin (a)+\cos (a) \sin (b x)) \sin (a))}{b^3 \left (\cos ^2(a)+\sin ^2(a)\right )}+\frac {3 \csc (a) \csc (a+b x) \left (c^2 d \sin (b x)+2 c d^2 x \sin (b x)+d^3 x^2 \sin (b x)\right )}{2 b^2}-\frac {3 d^3 \csc (a) \sec (a) \left (b^2 e^{i \arctan (\tan (a))} x^2+\frac {\left (i b x (-\pi +2 \arctan (\tan (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x+\arctan (\tan (a))) \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )+\pi \log (\cos (b x))+2 \arctan (\tan (a)) \log (\sin (b x+\arctan (\tan (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x+\arctan (\tan (a)))}\right )\right ) \tan (a)}{\sqrt {1+\tan ^2(a)}}\right )}{2 b^4 \sqrt {\sec ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}} \]
-1/2*((c + d*x)^3*Csc[a + b*x]^2)/b + (3*c*d^2*Csc[a]*(-(b*x*Cos[a]) + Log [Cos[b*x]*Sin[a] + Cos[a]*Sin[b*x]]*Sin[a]))/(b^3*(Cos[a]^2 + Sin[a]^2)) + (3*Csc[a]*Csc[a + b*x]*(c^2*d*Sin[b*x] + 2*c*d^2*x*Sin[b*x] + d^3*x^2*Sin [b*x]))/(2*b^2) - (3*d^3*Csc[a]*Sec[a]*(b^2*E^(I*ArcTan[Tan[a]])*x^2 + ((I *b*x*(-Pi + 2*ArcTan[Tan[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x + ArcT an[Tan[a]])*Log[1 - E^((2*I)*(b*x + ArcTan[Tan[a]]))] + Pi*Log[Cos[b*x]] + 2*ArcTan[Tan[a]]*Log[Sin[b*x + ArcTan[Tan[a]]]] + I*PolyLog[2, E^((2*I)*( b*x + ArcTan[Tan[a]]))])*Tan[a])/Sqrt[1 + Tan[a]^2]))/(2*b^4*Sqrt[Sec[a]^2 *(Cos[a]^2 + Sin[a]^2)])
Time = 0.60 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {4910, 3042, 4672, 3042, 25, 4202, 2620, 2715, 2838}
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.
| \(\displaystyle \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx\) |
| \(\Big \downarrow \) 4910 |
| \(\displaystyle \frac {3 d \int (c+d x)^2 \csc ^2(a+b x)dx}{2 b}-\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 d \int (c+d x)^2 \csc (a+b x)^2dx}{2 b}-\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 4672 |
| \(\displaystyle \frac {3 d \left (\frac {2 d \int (c+d x) \cot (a+b x)dx}{b}-\frac {(c+d x)^2 \cot (a+b x)}{b}\right )}{2 b}-\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3 d \left (\frac {2 d \int -\left ((c+d x) \tan \left (a+b x+\frac {\pi }{2}\right )\right )dx}{b}-\frac {(c+d x)^2 \cot (a+b x)}{b}\right )}{2 b}-\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {3 d \left (-\frac {2 d \int (c+d x) \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )dx}{b}-\frac {(c+d x)^2 \cot (a+b x)}{b}\right )}{2 b}-\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}\) |
| \(\Big \downarrow \) 4202 |
| \(\displaystyle -\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}+\frac {3 d \left (-\frac {(c+d x)^2 \cot (a+b x)}{b}-\frac {2 d \left (\frac {i (c+d x)^2}{2 d}-2 i \int \frac {e^{i (2 a+2 b x+\pi )} (c+d x)}{1+e^{i (2 a+2 b x+\pi )}}dx\right )}{b}\right )}{2 b}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle -\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}+\frac {3 d \left (-\frac {(c+d x)^2 \cot (a+b x)}{b}-\frac {2 d \left (\frac {i (c+d x)^2}{2 d}-2 i \left (\frac {i d \int \log \left (1+e^{i (2 a+2 b x+\pi )}\right )dx}{2 b}-\frac {i (c+d x) \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )\right )}{b}\right )}{2 b}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle -\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}+\frac {3 d \left (-\frac {(c+d x)^2 \cot (a+b x)}{b}-\frac {2 d \left (\frac {i (c+d x)^2}{2 d}-2 i \left (\frac {d \int e^{-i (2 a+2 b x+\pi )} \log \left (1+e^{i (2 a+2 b x+\pi )}\right )de^{i (2 a+2 b x+\pi )}}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )\right )}{b}\right )}{2 b}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle -\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}+\frac {3 d \left (-\frac {(c+d x)^2 \cot (a+b x)}{b}-\frac {2 d \left (\frac {i (c+d x)^2}{2 d}-2 i \left (-\frac {d \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{4 b^2}-\frac {i (c+d x) \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )\right )}{b}\right )}{2 b}\) |
-1/2*((c + d*x)^3*Csc[a + b*x]^2)/b + (3*d*(-(((c + d*x)^2*Cot[a + b*x])/b ) - (2*d*(((I/2)*(c + d*x)^2)/d - (2*I)*(((-1/2*I)*(c + d*x)*Log[1 + E^(I* (2*a + Pi + 2*b*x))])/b - (d*PolyLog[2, -E^(I*(2*a + Pi + 2*b*x))])/(4*b^2 ))))/b))/(2*b)
3.1.47.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I *((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^m*(E^(2*I*( e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt Q[m, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp [(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1) *Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[Cot[(a_.) + (b_.)*(x_)]^(p_.)*Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d _.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Csc[a + b*x]^n/(b*n)), x ] + Simp[d*(m/(b*n)) Int[(c + d*x)^(m - 1)*Csc[a + b*x]^n, x], x] /; Free Q[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (101 ) = 202\).
Time = 1.67 (sec) , antiderivative size = 409, normalized size of antiderivative = 3.56
| method | result | size |
| risch | \(\frac {2 b \,d^{3} x^{3} {\mathrm e}^{2 i \left (x b +a \right )}-3 i d^{3} x^{2} {\mathrm e}^{2 i \left (x b +a \right )}+6 b c \,d^{2} x^{2} {\mathrm e}^{2 i \left (x b +a \right )}-6 i c \,d^{2} x \,{\mathrm e}^{2 i \left (x b +a \right )}+6 b \,c^{2} d x \,{\mathrm e}^{2 i \left (x b +a \right )}-3 i c^{2} d \,{\mathrm e}^{2 i \left (x b +a \right )}+3 i d^{3} x^{2}+2 b \,c^{3} {\mathrm e}^{2 i \left (x b +a \right )}+6 i c \,d^{2} x +3 i c^{2} d}{b^{2} \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{2}}+\frac {3 d^{2} c \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{b^{3}}-\frac {6 d^{2} c \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{3}}+\frac {3 d^{2} c \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{3}}-\frac {3 i d^{3} x^{2}}{b^{2}}-\frac {6 i d^{3} x a}{b^{3}}-\frac {3 i d^{3} a^{2}}{b^{4}}+\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) x}{b^{3}}+\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (x b +a \right )}\right ) a}{b^{4}}-\frac {3 i d^{3} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (x b +a \right )}\right )}{b^{4}}+\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right ) x}{b^{3}}-\frac {3 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (x b +a \right )}\right )}{b^{4}}+\frac {6 d^{3} a \ln \left ({\mathrm e}^{i \left (x b +a \right )}\right )}{b^{4}}-\frac {3 d^{3} a \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{b^{4}}\) | \(409\) |
(2*b*d^3*x^3*exp(2*I*(b*x+a))-3*I*d^3*x^2*exp(2*I*(b*x+a))+6*b*c*d^2*x^2*e xp(2*I*(b*x+a))-6*I*c*d^2*x*exp(2*I*(b*x+a))+6*b*c^2*d*x*exp(2*I*(b*x+a))- 3*I*c^2*d*exp(2*I*(b*x+a))+3*I*d^3*x^2+2*b*c^3*exp(2*I*(b*x+a))+6*I*c*d^2* x+3*I*c^2*d)/b^2/(exp(2*I*(b*x+a))-1)^2+3*d^2/b^3*c*ln(exp(I*(b*x+a))+1)-6 *d^2/b^3*c*ln(exp(I*(b*x+a)))+3*d^2/b^3*c*ln(exp(I*(b*x+a))-1)-3*I*d^3/b^2 *x^2-6*I*d^3/b^3*x*a-3*I*d^3/b^4*a^2+3*d^3/b^3*ln(1-exp(I*(b*x+a)))*x+3*d^ 3/b^4*ln(1-exp(I*(b*x+a)))*a-3*I*d^3*polylog(2,exp(I*(b*x+a)))/b^4+3*d^3/b ^3*ln(exp(I*(b*x+a))+1)*x-3*I*d^3/b^4*polylog(2,-exp(I*(b*x+a)))+6*d^3/b^4 *a*ln(exp(I*(b*x+a)))-3*d^3/b^4*a*ln(exp(I*(b*x+a))-1)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 591 vs. \(2 (98) = 196\).
Time = 0.27 (sec) , antiderivative size = 591, normalized size of antiderivative = 5.14 \[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=\frac {b^{3} d^{3} x^{3} + 3 \, b^{3} c d^{2} x^{2} + 3 \, b^{3} c^{2} d x + b^{3} c^{3} + 3 \, {\left (b^{2} d^{3} x^{2} + 2 \, b^{2} c d^{2} x + b^{2} c^{2} d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 3 \, {\left (i \, d^{3} \cos \left (b x + a\right )^{2} - i \, d^{3}\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 3 \, {\left (-i \, d^{3} \cos \left (b x + a\right )^{2} + i \, d^{3}\right )} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 3 \, {\left (-i \, d^{3} \cos \left (b x + a\right )^{2} + i \, d^{3}\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - 3 \, {\left (i \, d^{3} \cos \left (b x + a\right )^{2} - i \, d^{3}\right )} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 3 \, {\left (b d^{3} x + b c d^{2} - {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{2}\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b d^{3} x + b c d^{2} - {\left (b d^{3} x + b c d^{2}\right )} \cos \left (b x + a\right )^{2}\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b c d^{2} - a d^{3} - {\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 3 \, {\left (b c d^{2} - a d^{3} - {\left (b c d^{2} - a d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) - 3 \, {\left (b d^{3} x + a d^{3} - {\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - 3 \, {\left (b d^{3} x + a d^{3} - {\left (b d^{3} x + a d^{3}\right )} \cos \left (b x + a\right )^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right )}{2 \, {\left (b^{4} \cos \left (b x + a\right )^{2} - b^{4}\right )}} \]
1/2*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + b^3*c^3 + 3*(b^2*d^3* x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*cos(b*x + a)*sin(b*x + a) - 3*(I*d^3*cos( b*x + a)^2 - I*d^3)*dilog(cos(b*x + a) + I*sin(b*x + a)) - 3*(-I*d^3*cos(b *x + a)^2 + I*d^3)*dilog(cos(b*x + a) - I*sin(b*x + a)) - 3*(-I*d^3*cos(b* x + a)^2 + I*d^3)*dilog(-cos(b*x + a) + I*sin(b*x + a)) - 3*(I*d^3*cos(b*x + a)^2 - I*d^3)*dilog(-cos(b*x + a) - I*sin(b*x + a)) - 3*(b*d^3*x + b*c* d^2 - (b*d^3*x + b*c*d^2)*cos(b*x + a)^2)*log(cos(b*x + a) + I*sin(b*x + a ) + 1) - 3*(b*d^3*x + b*c*d^2 - (b*d^3*x + b*c*d^2)*cos(b*x + a)^2)*log(co s(b*x + a) - I*sin(b*x + a) + 1) - 3*(b*c*d^2 - a*d^3 - (b*c*d^2 - a*d^3)* cos(b*x + a)^2)*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) - 3*(b*c *d^2 - a*d^3 - (b*c*d^2 - a*d^3)*cos(b*x + a)^2)*log(-1/2*cos(b*x + a) - 1 /2*I*sin(b*x + a) + 1/2) - 3*(b*d^3*x + a*d^3 - (b*d^3*x + a*d^3)*cos(b*x + a)^2)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) - 3*(b*d^3*x + a*d^3 - (b* d^3*x + a*d^3)*cos(b*x + a)^2)*log(-cos(b*x + a) - I*sin(b*x + a) + 1))/(b ^4*cos(b*x + a)^2 - b^4)
\[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=\int \left (c + d x\right )^{3} \cos {\left (a + b x \right )} \csc ^{3}{\left (a + b x \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1044 vs. \(2 (98) = 196\).
Time = 0.49 (sec) , antiderivative size = 1044, normalized size of antiderivative = 9.08 \[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=\text {Too large to display} \]
(6*b^2*c^2*d + 6*(b*d^3*x + b*c*d^2 + (b*d^3*x + b*c*d^2)*cos(4*b*x + 4*a) - 2*(b*d^3*x + b*c*d^2)*cos(2*b*x + 2*a) - (-I*b*d^3*x - I*b*c*d^2)*sin(4 *b*x + 4*a) - 2*(I*b*d^3*x + I*b*c*d^2)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) + 6*(b*c*d^2*cos(4*b*x + 4*a) - 2*b*c*d^2*cos(2*b* x + 2*a) + I*b*c*d^2*sin(4*b*x + 4*a) - 2*I*b*c*d^2*sin(2*b*x + 2*a) + b*c *d^2)*arctan2(sin(b*x + a), cos(b*x + a) - 1) - 6*(b*d^3*x*cos(4*b*x + 4*a ) - 2*b*d^3*x*cos(2*b*x + 2*a) + I*b*d^3*x*sin(4*b*x + 4*a) - 2*I*b*d^3*x* sin(2*b*x + 2*a) + b*d^3*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - 6*( b^2*d^3*x^2 + 2*b^2*c*d^2*x)*cos(4*b*x + 4*a) - 2*(2*I*b^3*d^3*x^3 + 2*I*b ^3*c^3 + 3*b^2*c^2*d + 3*(2*I*b^3*c*d^2 - b^2*d^3)*x^2 + 6*(I*b^3*c^2*d - b^2*c*d^2)*x)*cos(2*b*x + 2*a) - 6*(d^3*cos(4*b*x + 4*a) - 2*d^3*cos(2*b*x + 2*a) + I*d^3*sin(4*b*x + 4*a) - 2*I*d^3*sin(2*b*x + 2*a) + d^3)*dilog(- e^(I*b*x + I*a)) - 6*(d^3*cos(4*b*x + 4*a) - 2*d^3*cos(2*b*x + 2*a) + I*d^ 3*sin(4*b*x + 4*a) - 2*I*d^3*sin(2*b*x + 2*a) + d^3)*dilog(e^(I*b*x + I*a) ) - 3*(I*b*d^3*x + I*b*c*d^2 + (I*b*d^3*x + I*b*c*d^2)*cos(4*b*x + 4*a) + 2*(-I*b*d^3*x - I*b*c*d^2)*cos(2*b*x + 2*a) - (b*d^3*x + b*c*d^2)*sin(4*b* x + 4*a) + 2*(b*d^3*x + b*c*d^2)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + si n(b*x + a)^2 + 2*cos(b*x + a) + 1) - 3*(I*b*d^3*x + I*b*c*d^2 + (I*b*d^3*x + I*b*c*d^2)*cos(4*b*x + 4*a) + 2*(-I*b*d^3*x - I*b*c*d^2)*cos(2*b*x + 2* a) - (b*d^3*x + b*c*d^2)*sin(4*b*x + 4*a) + 2*(b*d^3*x + b*c*d^2)*sin(2...
\[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \cos \left (b x + a\right ) \csc \left (b x + a\right )^{3} \,d x } \]
Timed out. \[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=\int \frac {\cos \left (a+b\,x\right )\,{\left (c+d\,x\right )}^3}{{\sin \left (a+b\,x\right )}^3} \,d x \]