Integrand size = 20, antiderivative size = 197 \[ \int (c+d x)^3 \csc (a+b x) \sec (a+b x) \, dx=-\frac {2 (c+d x)^3 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}+\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4} \]
-2*(d*x+c)^3*arctanh(exp(2*I*(b*x+a)))/b+3/2*I*d*(d*x+c)^2*polylog(2,-exp( 2*I*(b*x+a)))/b^2-3/2*I*d*(d*x+c)^2*polylog(2,exp(2*I*(b*x+a)))/b^2-3/2*d^ 2*(d*x+c)*polylog(3,-exp(2*I*(b*x+a)))/b^3+3/2*d^2*(d*x+c)*polylog(3,exp(2 *I*(b*x+a)))/b^3-3/4*I*d^3*polylog(4,-exp(2*I*(b*x+a)))/b^4+3/4*I*d^3*poly log(4,exp(2*I*(b*x+a)))/b^4
Time = 1.08 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.78 \[ \int (c+d x)^3 \csc (a+b x) \sec (a+b x) \, dx=\frac {-8 b^3 c^3 \text {arctanh}\left (e^{2 i (a+b x)}\right )+12 b^3 c^2 d x \log \left (1-e^{2 i (a+b x)}\right )+12 b^3 c d^2 x^2 \log \left (1-e^{2 i (a+b x)}\right )+4 b^3 d^3 x^3 \log \left (1-e^{2 i (a+b x)}\right )-12 b^3 c^2 d x \log \left (1+e^{2 i (a+b x)}\right )-12 b^3 c d^2 x^2 \log \left (1+e^{2 i (a+b x)}\right )-4 b^3 d^3 x^3 \log \left (1+e^{2 i (a+b x)}\right )+6 i b^2 d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )-6 i b^2 d (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )-6 b d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )+6 b c d^2 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )+6 b d^3 x \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )-3 i d^3 \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )+3 i d^3 \operatorname {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^4} \]
(-8*b^3*c^3*ArcTanh[E^((2*I)*(a + b*x))] + 12*b^3*c^2*d*x*Log[1 - E^((2*I) *(a + b*x))] + 12*b^3*c*d^2*x^2*Log[1 - E^((2*I)*(a + b*x))] + 4*b^3*d^3*x ^3*Log[1 - E^((2*I)*(a + b*x))] - 12*b^3*c^2*d*x*Log[1 + E^((2*I)*(a + b*x ))] - 12*b^3*c*d^2*x^2*Log[1 + E^((2*I)*(a + b*x))] - 4*b^3*d^3*x^3*Log[1 + E^((2*I)*(a + b*x))] + (6*I)*b^2*d*(c + d*x)^2*PolyLog[2, -E^((2*I)*(a + b*x))] - (6*I)*b^2*d*(c + d*x)^2*PolyLog[2, E^((2*I)*(a + b*x))] - 6*b*d^ 2*(c + d*x)*PolyLog[3, -E^((2*I)*(a + b*x))] + 6*b*c*d^2*PolyLog[3, E^((2* I)*(a + b*x))] + 6*b*d^3*x*PolyLog[3, E^((2*I)*(a + b*x))] - (3*I)*d^3*Pol yLog[4, -E^((2*I)*(a + b*x))] + (3*I)*d^3*PolyLog[4, E^((2*I)*(a + b*x))]) /(4*b^4)
Time = 0.71 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {4919, 3042, 4671, 3011, 7163, 2720, 7143}
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.
| \(\displaystyle \int (c+d x)^3 \csc (a+b x) \sec (a+b x) \, dx\) |
| \(\Big \downarrow \) 4919 |
| \(\displaystyle 2 \int (c+d x)^3 \csc (2 a+2 b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle 2 \int (c+d x)^3 \csc (2 a+2 b x)dx\) |
| \(\Big \downarrow \) 4671 |
| \(\displaystyle 2 \left (-\frac {3 d \int (c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )dx}{2 b}+\frac {3 d \int (c+d x)^2 \log \left (1+e^{2 i (a+b x)}\right )dx}{2 b}-\frac {(c+d x)^3 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}\right )\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle 2 \left (\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \int (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \int (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )dx}{b}\right )}{2 b}-\frac {(c+d x)^3 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}\right )\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle 2 \left (\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \left (\frac {i d \int \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \left (\frac {i d \int \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )dx}{2 b}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {(c+d x)^3 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}\right )\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle 2 \left (\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \left (\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \left (\frac {d \int e^{-2 i (a+b x)} \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )de^{2 i (a+b x)}}{4 b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {(c+d x)^3 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}\right )\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle 2 \left (-\frac {(c+d x)^3 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}+\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \left (\frac {d \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}-\frac {3 d \left (\frac {i (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b}-\frac {i d \left (\frac {d \operatorname {PolyLog}\left (4,e^{2 i (a+b x)}\right )}{4 b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b}\right )}{b}\right )}{2 b}\right )\) |
2*(-(((c + d*x)^3*ArcTanh[E^((2*I)*(a + b*x))])/b) + (3*d*(((I/2)*(c + d*x )^2*PolyLog[2, -E^((2*I)*(a + b*x))])/b - (I*d*(((-1/2*I)*(c + d*x)*PolyLo g[3, -E^((2*I)*(a + b*x))])/b + (d*PolyLog[4, -E^((2*I)*(a + b*x))])/(4*b^ 2)))/b))/(2*b) - (3*d*(((I/2)*(c + d*x)^2*PolyLog[2, E^((2*I)*(a + b*x))]) /b - (I*d*(((-1/2*I)*(c + d*x)*PolyLog[3, E^((2*I)*(a + b*x))])/b + (d*Pol yLog[4, E^((2*I)*(a + b*x))])/(4*b^2)))/b))/(2*b))
3.3.29.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f) Int[(c + d*x )^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG tQ[m, 0]
Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Simp[2^n Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n , x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. )*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a + b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F])) Int[(e + f*x) ^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c , d, e, f, n, p}, x] && GtQ[m, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 815 vs. \(2 (167 ) = 334\).
Time = 1.30 (sec) , antiderivative size = 816, normalized size of antiderivative = 4.14
-3/4*I*d^3*polylog(4,-exp(2*I*(b*x+a)))/b^4-6*I/b^2*c*d^2*polylog(2,exp(I* (b*x+a)))*x-6*I/b^2*c*d^2*polylog(2,-exp(I*(b*x+a)))*x-1/b*c^3*ln(exp(2*I* (b*x+a))+1)-3/b*d*c^2*ln(exp(2*I*(b*x+a))+1)*x-3/b*c*d^2*ln(exp(2*I*(b*x+a ))+1)*x^2+3/2*I/b^2*d^3*polylog(2,-exp(2*I*(b*x+a)))*x^2+3/2*I/b^2*d*c^2*p olylog(2,-exp(2*I*(b*x+a)))+3/b^3*c*d^2*a^2*ln(exp(I*(b*x+a))-1)-3/b^2*c^2 *d*a*ln(exp(I*(b*x+a))-1)+3/b^2*d*c^2*ln(1-exp(I*(b*x+a)))*a-3/b^3*c*d^2*l n(1-exp(I*(b*x+a)))*a^2-1/b*d^3*ln(exp(2*I*(b*x+a))+1)*x^3-3/2/b^3*d^3*pol ylog(3,-exp(2*I*(b*x+a)))*x-3/2/b^3*c*d^2*polylog(3,-exp(2*I*(b*x+a)))+1/b *c^3*ln(exp(I*(b*x+a))+1)+1/b*c^3*ln(exp(I*(b*x+a))-1)+3/b*d*c^2*ln(1-exp( I*(b*x+a)))*x+3/b*d*c^2*ln(exp(I*(b*x+a))+1)*x+3/b*c*d^2*ln(1-exp(I*(b*x+a )))*x^2+3/b*c*d^2*ln(exp(I*(b*x+a))+1)*x^2-3*I/b^2*d^3*polylog(2,-exp(I*(b *x+a)))*x^2-3*I/b^2*d*c^2*polylog(2,exp(I*(b*x+a)))-3*I/b^2*d*c^2*polylog( 2,-exp(I*(b*x+a)))-3*I/b^2*d^3*polylog(2,exp(I*(b*x+a)))*x^2-1/b^4*d^3*a^3 *ln(exp(I*(b*x+a))-1)+1/b^4*d^3*ln(1-exp(I*(b*x+a)))*a^3+6/b^3*d^3*polylog (3,exp(I*(b*x+a)))*x+6/b^3*d^3*polylog(3,-exp(I*(b*x+a)))*x+1/b*d^3*ln(1-e xp(I*(b*x+a)))*x^3+1/b*d^3*ln(exp(I*(b*x+a))+1)*x^3+6/b^3*c*d^2*polylog(3, exp(I*(b*x+a)))+6/b^3*c*d^2*polylog(3,-exp(I*(b*x+a)))+6*I/b^4*d^3*polylog (4,-exp(I*(b*x+a)))+6*I*d^3*polylog(4,exp(I*(b*x+a)))/b^4+3*I/b^2*c*d^2*po lylog(2,-exp(2*I*(b*x+a)))*x
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1786 vs. \(2 (161) = 322\).
Time = 0.38 (sec) , antiderivative size = 1786, normalized size of antiderivative = 9.07 \[ \int (c+d x)^3 \csc (a+b x) \sec (a+b x) \, dx=\text {Too large to display} \]
1/2*(6*I*d^3*polylog(4, cos(b*x + a) + I*sin(b*x + a)) - 6*I*d^3*polylog(4 , cos(b*x + a) - I*sin(b*x + a)) + 6*I*d^3*polylog(4, I*cos(b*x + a) + sin (b*x + a)) - 6*I*d^3*polylog(4, I*cos(b*x + a) - sin(b*x + a)) - 6*I*d^3*p olylog(4, -I*cos(b*x + a) + sin(b*x + a)) + 6*I*d^3*polylog(4, -I*cos(b*x + a) - sin(b*x + a)) - 6*I*d^3*polylog(4, -cos(b*x + a) + I*sin(b*x + a)) + 6*I*d^3*polylog(4, -cos(b*x + a) - I*sin(b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*dilog(cos(b*x + a) + I*sin(b*x + a)) - 3*(- I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2*d)*dilog(cos(b*x + a) - I*sin( b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*dilog(I*cos( b*x + a) + sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2 *d)*dilog(I*cos(b*x + a) - sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d ^2*x - I*b^2*c^2*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - 3*(I*b^2*d^3*x ^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*dilog(-I*cos(b*x + a) - sin(b*x + a)) - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2*d)*dilog(-cos(b*x + a) + I*sin(b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I*b^2*c^2*d)*dilog (-cos(b*x + a) - I*sin(b*x + a)) + (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)*log(cos(b*x + a) + I*sin(b*x + a) + 1) - (b^3*c^3 - 3*a *b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x + a) + I*sin(b*x + a) + I) + (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)*log(cos(b*x + a) - I*sin(b*x + a) + 1) - (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 ...
\[ \int (c+d x)^3 \csc (a+b x) \sec (a+b x) \, dx=\int \left (c + d x\right )^{3} \csc {\left (a + b x \right )} \sec {\left (a + b x \right )}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1078 vs. \(2 (161) = 322\).
Time = 0.51 (sec) , antiderivative size = 1078, normalized size of antiderivative = 5.47 \[ \int (c+d x)^3 \csc (a+b x) \sec (a+b x) \, dx=\text {Too large to display} \]
-1/6*(3*c^3*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2)) - 9*a*c^2*d*(l og(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b + 9*a^2*c*d^2*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b^2 - 3*a^3*d^3*(log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2))/b^3 + (6*I*d^3*polylog(4, -e^(2*I*b*x + 2*I*a)) - 36*I*d^3*polylog(4, -e^(I*b*x + I*a)) - 36*I*d^3*polylog(4, e^(I*b*x + I *a)) - 2*(-4*I*(b*x + a)^3*d^3 + 9*(-I*b*c*d^2 + I*a*d^3)*(b*x + a)^2 + 9* (-I*b^2*c^2*d + 2*I*a*b*c*d^2 - I*a^2*d^3)*(b*x + a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 6*(I*(b*x + a)^3*d^3 + 3*(I*b*c*d^2 - I*a*d^ 3)*(b*x + a)^2 + 3*(I*b^2*c^2*d - 2*I*a*b*c*d^2 + I*a^2*d^3)*(b*x + a))*ar ctan2(sin(b*x + a), cos(b*x + a) + 1) - 6*(-I*(b*x + a)^3*d^3 + 3*(-I*b*c* d^2 + I*a*d^3)*(b*x + a)^2 + 3*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 - I*a^2*d^3)* (b*x + a))*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - 3*(3*I*b^2*c^2*d - 6 *I*a*b*c*d^2 + 4*I*(b*x + a)^2*d^3 + 3*I*a^2*d^3 + 6*(I*b*c*d^2 - I*a*d^3) *(b*x + a))*dilog(-e^(2*I*b*x + 2*I*a)) - 18*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 - I*(b*x + a)^2*d^3 - I*a^2*d^3 + 2*(-I*b*c*d^2 + I*a*d^3)*(b*x + a))*dil og(-e^(I*b*x + I*a)) - 18*(-I*b^2*c^2*d + 2*I*a*b*c*d^2 - I*(b*x + a)^2*d^ 3 - I*a^2*d^3 + 2*(-I*b*c*d^2 + I*a*d^3)*(b*x + a))*dilog(e^(I*b*x + I*a)) + (4*(b*x + a)^3*d^3 + 9*(b*c*d^2 - a*d^3)*(b*x + a)^2 + 9*(b^2*c^2*d - 2 *a*b*c*d^2 + a^2*d^3)*(b*x + a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a) ^2 + 2*cos(2*b*x + 2*a) + 1) - 3*((b*x + a)^3*d^3 + 3*(b*c*d^2 - a*d^3)...
\[ \int (c+d x)^3 \csc (a+b x) \sec (a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \csc \left (b x + a\right ) \sec \left (b x + a\right ) \,d x } \]
Timed out. \[ \int (c+d x)^3 \csc (a+b x) \sec (a+b x) \, dx=\int \frac {{\left (c+d\,x\right )}^3}{\cos \left (a+b\,x\right )\,\sin \left (a+b\,x\right )} \,d x \]