\(\int (c+d x)^3 \csc ^2(a+b x) \, dx\) [28]

Optimal result
Mathematica [B] (warning: unable to verify)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 16, antiderivative size = 113 \[ \int (c+d x)^3 \csc ^2(a+b x) \, dx=-\frac {i (c+d x)^3}{b}-\frac {(c+d x)^3 \cot (a+b x)}{b}+\frac {3 d (c+d x)^2 \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {3 i d^2 (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3}+\frac {3 d^3 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^4} \] Output:

-I*(d*x+c)^3/b-(d*x+c)^3*cot(b*x+a)/b+3*d*(d*x+c)^2*ln(1-exp(2*I*(b*x+a))) 
/b^2-3*I*d^2*(d*x+c)*polylog(2,exp(2*I*(b*x+a)))/b^3+3/2*d^3*polylog(3,exp 
(2*I*(b*x+a)))/b^4
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(486\) vs. \(2(113)=226\).

Time = 7.02 (sec) , antiderivative size = 486, normalized size of antiderivative = 4.30 \[ \int (c+d x)^3 \csc ^2(a+b x) \, dx=-\frac {d^3 e^{i a} \csc (a) \left (2 b^3 e^{-2 i a} x^3+3 i b^2 \left (1-e^{-2 i a}\right ) x^2 \log \left (1-e^{-i (a+b x)}\right )+3 i b^2 \left (1-e^{-2 i a}\right ) x^2 \log \left (1+e^{-i (a+b x)}\right )-6 b \left (1-e^{-2 i a}\right ) x \operatorname {PolyLog}\left (2,-e^{-i (a+b x)}\right )-6 b \left (1-e^{-2 i a}\right ) x \operatorname {PolyLog}\left (2,e^{-i (a+b x)}\right )+6 i \left (1-e^{-2 i a}\right ) \operatorname {PolyLog}\left (3,-e^{-i (a+b x)}\right )+6 i \left (1-e^{-2 i a}\right ) \operatorname {PolyLog}\left (3,e^{-i (a+b x)}\right )\right )}{2 b^4}+\frac {3 c^2 d \csc (a) (-b x \cos (a)+\log (\cos (b x) \sin (a)+\cos (a) \sin (b x)) \sin (a))}{b^2 \left (\cos ^2(a)+\sin ^2(a)\right )}+\frac {\csc (a) \csc (a+b x) \left (c^3 \sin (b x)+3 c^2 d x \sin (b x)+3 c d^2 x^2 \sin (b x)+d^3 x^3 \sin (b x)\right )}{b}-\frac {3 c d^2 \csc (a) \sec (a) \left (b^2 e^{i \arctan (\tan (a))} x^2+\frac {\left (i b x (-\pi +2 \arctan (\tan (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x+\arctan (\tan (a))) \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )+\pi \log (\cos (b x))+2 \arctan (\tan (a)) \log (\sin (b x+\arctan (\tan (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x+\arctan (\tan (a)))}\right )\right ) \tan (a)}{\sqrt {1+\tan ^2(a)}}\right )}{b^3 \sqrt {\sec ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}} \] Input:

Integrate[(c + d*x)^3*Csc[a + b*x]^2,x]
 

Output:

-1/2*(d^3*E^(I*a)*Csc[a]*((2*b^3*x^3)/E^((2*I)*a) + (3*I)*b^2*(1 - E^((-2* 
I)*a))*x^2*Log[1 - E^((-I)*(a + b*x))] + (3*I)*b^2*(1 - E^((-2*I)*a))*x^2* 
Log[1 + E^((-I)*(a + b*x))] - 6*b*(1 - E^((-2*I)*a))*x*PolyLog[2, -E^((-I) 
*(a + b*x))] - 6*b*(1 - E^((-2*I)*a))*x*PolyLog[2, E^((-I)*(a + b*x))] + ( 
6*I)*(1 - E^((-2*I)*a))*PolyLog[3, -E^((-I)*(a + b*x))] + (6*I)*(1 - E^((- 
2*I)*a))*PolyLog[3, E^((-I)*(a + b*x))]))/b^4 + (3*c^2*d*Csc[a]*(-(b*x*Cos 
[a]) + Log[Cos[b*x]*Sin[a] + Cos[a]*Sin[b*x]]*Sin[a]))/(b^2*(Cos[a]^2 + Si 
n[a]^2)) + (Csc[a]*Csc[a + b*x]*(c^3*Sin[b*x] + 3*c^2*d*x*Sin[b*x] + 3*c*d 
^2*x^2*Sin[b*x] + d^3*x^3*Sin[b*x]))/b - (3*c*d^2*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 + ArcTan[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]))/(b^3*Sqrt[Sec[a]^2*(Cos[a]^2 + Sin[a]^2)])
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.33, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {3042, 4672, 3042, 25, 4202, 2620, 3011, 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.

\(\displaystyle \int (c+d x)^3 \csc ^2(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^3 \csc (a+b x)^2dx\)

\(\Big \downarrow \) 4672

\(\displaystyle \frac {3 d \int (c+d x)^2 \cot (a+b x)dx}{b}-\frac {(c+d x)^3 \cot (a+b x)}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 d \int -(c+d x)^2 \tan \left (a+b x+\frac {\pi }{2}\right )dx}{b}-\frac {(c+d x)^3 \cot (a+b x)}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {3 d \int (c+d x)^2 \tan \left (\frac {1}{2} (2 a+\pi )+b x\right )dx}{b}-\frac {(c+d x)^3 \cot (a+b x)}{b}\)

\(\Big \downarrow \) 4202

\(\displaystyle -\frac {(c+d x)^3 \cot (a+b x)}{b}-\frac {3 d \left (\frac {i (c+d x)^3}{3 d}-2 i \int \frac {e^{i (2 a+2 b x+\pi )} (c+d x)^2}{1+e^{i (2 a+2 b x+\pi )}}dx\right )}{b}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {(c+d x)^3 \cot (a+b x)}{b}-\frac {3 d \left (\frac {i (c+d x)^3}{3 d}-2 i \left (\frac {i d \int (c+d x) \log \left (1+e^{i (2 a+2 b x+\pi )}\right )dx}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )\right )}{b}\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {(c+d x)^3 \cot (a+b x)}{b}-\frac {3 d \left (\frac {i (c+d x)^3}{3 d}-2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {i d \int \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )dx}{2 b}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )\right )}{b}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {(c+d x)^3 \cot (a+b x)}{b}-\frac {3 d \left (\frac {i (c+d x)^3}{3 d}-2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {d \int e^{-i (2 a+2 b x+\pi )} \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )de^{i (2 a+2 b x+\pi )}}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )\right )}{b}\)

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {(c+d x)^3 \cot (a+b x)}{b}-\frac {3 d \left (\frac {i (c+d x)^3}{3 d}-2 i \left (\frac {i d \left (\frac {i (c+d x) \operatorname {PolyLog}\left (2,-e^{i (2 a+2 b x+\pi )}\right )}{2 b}-\frac {d \operatorname {PolyLog}\left (3,-e^{i (2 a+2 b x+\pi )}\right )}{4 b^2}\right )}{b}-\frac {i (c+d x)^2 \log \left (1+e^{i (2 a+2 b x+\pi )}\right )}{2 b}\right )\right )}{b}\)

Input:

Int[(c + d*x)^3*Csc[a + b*x]^2,x]
 

Output:

-(((c + d*x)^3*Cot[a + b*x])/b) - (3*d*(((I/3)*(c + d*x)^3)/d - (2*I)*(((- 
1/2*I)*(c + d*x)^2*Log[1 + E^(I*(2*a + Pi + 2*b*x))])/b + (I*d*(((I/2)*(c 
+ d*x)*PolyLog[2, -E^(I*(2*a + Pi + 2*b*x))])/b - (d*PolyLog[3, -E^(I*(2*a 
 + Pi + 2*b*x))])/(4*b^2)))/b)))/b
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2620
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]
 

rule 2720
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]]]
 

rule 3011
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]
 

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

rule 4202
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]
 

rule 4672
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]
 

rule 7143
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]
 
Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 540 vs. \(2 (103 ) = 206\).

Time = 1.16 (sec) , antiderivative size = 541, normalized size of antiderivative = 4.79

method result size
risch \(-\frac {12 i d^{2} c a x}{b^{2}}-\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {3 d \,c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}-\frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {6 d^{3} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 d^{3} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{4}}-\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{4}}+\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b^{2}}-\frac {2 i d^{3} x^{3}}{b}+\frac {4 i d^{3} a^{3}}{b^{4}}+\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}+\frac {3 d \,c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}+\frac {12 d^{2} c a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 d^{2} c a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {6 d^{2} c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}+\frac {6 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}+\frac {6 d^{2} c \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {6 i d^{2} c \,x^{2}}{b}-\frac {6 i d^{2} c \,a^{2}}{b^{3}}-\frac {6 i d^{2} c \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 i d^{2} c \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 i d^{3} a^{2} x}{b^{3}}-\frac {6 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {6 i d^{3} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {6 d^{3} \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}\) \(541\)

Input:

int((d*x+c)^3*csc(b*x+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-2*I*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/b/(exp(2*I*(b*x+a))-1)-12*I/b^2*d 
^2*c*a*x-6/b^4*d^3*a^2*ln(exp(I*(b*x+a)))+3/b^4*d^3*a^2*ln(exp(I*(b*x+a))- 
1)-3/b^4*d^3*ln(1-exp(I*(b*x+a)))*a^2+3/b^2*d^3*ln(exp(I*(b*x+a))+1)*x^2-2 
*I/b*d^3*x^3+4*I/b^4*d^3*a^3+12/b^3*d^2*c*a*ln(exp(I*(b*x+a)))+3/b^2*d^3*l 
n(1-exp(I*(b*x+a)))*x^2+3/b^2*d*c^2*ln(exp(I*(b*x+a))+1)-6/b^2*d*c^2*ln(ex 
p(I*(b*x+a)))+3/b^2*d*c^2*ln(exp(I*(b*x+a))-1)-6/b^3*d^2*c*a*ln(exp(I*(b*x 
+a))-1)+6/b^3*d^2*c*ln(1-exp(I*(b*x+a)))*a+6/b^2*d^2*c*ln(exp(I*(b*x+a))+1 
)*x+6/b^2*d^2*c*ln(1-exp(I*(b*x+a)))*x-6*I/b*d^2*c*x^2-6*I/b^3*d^2*c*a^2-6 
*I/b^3*d^2*c*polylog(2,-exp(I*(b*x+a)))-6*I/b^3*d^2*c*polylog(2,exp(I*(b*x 
+a)))+6*I/b^3*d^3*a^2*x-6*I/b^3*d^3*polylog(2,-exp(I*(b*x+a)))*x-6*I/b^3*d 
^3*polylog(2,exp(I*(b*x+a)))*x+6/b^4*d^3*polylog(3,exp(I*(b*x+a)))+6/b^4*d 
^3*polylog(3,-exp(I*(b*x+a)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 676 vs. \(2 (100) = 200\).

Time = 0.11 (sec) , antiderivative size = 676, normalized size of antiderivative = 5.98 \[ \int (c+d x)^3 \csc ^2(a+b x) \, dx =\text {Too large to display} \] Input:

integrate((d*x+c)^3*csc(b*x+a)^2,x, algorithm="fricas")
 

Output:

1/2*(6*d^3*polylog(3, cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) + 6*d^3* 
polylog(3, cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + 6*d^3*polylog(3, 
-cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) + 6*d^3*polylog(3, -cos(b*x + 
 a) - I*sin(b*x + a))*sin(b*x + a) - 6*(I*b*d^3*x + I*b*c*d^2)*dilog(cos(b 
*x + a) + I*sin(b*x + a))*sin(b*x + a) - 6*(-I*b*d^3*x - I*b*c*d^2)*dilog( 
cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) - 6*(-I*b*d^3*x - I*b*c*d^2)*d 
ilog(-cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - 6*(I*b*d^3*x + I*b*c*d 
^2)*dilog(-cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + 3*(b^2*d^3*x^2 + 
2*b^2*c*d^2*x + b^2*c^2*d)*log(cos(b*x + a) + I*sin(b*x + a) + 1)*sin(b*x 
+ a) + 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + b^2*c^2*d)*log(cos(b*x + a) - I*si 
n(b*x + a) + 1)*sin(b*x + a) + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*log(- 
1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) + 3*(b^2*c^2*d - 
 2*a*b*c*d^2 + a^2*d^3)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) + 1/2)* 
sin(b*x + a) + 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*log 
(-cos(b*x + a) + I*sin(b*x + a) + 1)*sin(b*x + a) + 3*(b^2*d^3*x^2 + 2*b^2 
*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*log(-cos(b*x + a) - I*sin(b*x + a) + 1)* 
sin(b*x + a) - 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) 
*cos(b*x + a))/(b^4*sin(b*x + a))
 

Sympy [F]

\[ \int (c+d x)^3 \csc ^2(a+b x) \, dx=\int \left (c + d x\right )^{3} \csc ^{2}{\left (a + b x \right )}\, dx \] Input:

integrate((d*x+c)**3*csc(b*x+a)**2,x)
 

Output:

Integral((c + d*x)**3*csc(a + b*x)**2, x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1654 vs. \(2 (100) = 200\).

Time = 0.18 (sec) , antiderivative size = 1654, normalized size of antiderivative = 14.64 \[ \int (c+d x)^3 \csc ^2(a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^3*csc(b*x+a)^2,x, algorithm="maxima")
 

Output:

1/2*(3*((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1) 
*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (cos(2*b*x + 
2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + 
 sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - 4*(b*x + a)*sin(2*b*x + 2*a))*c^2* 
d/((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*b) - 
 6*((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log 
(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (cos(2*b*x + 2*a) 
^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin 
(b*x + a)^2 - 2*cos(b*x + a) + 1) - 4*(b*x + a)*sin(2*b*x + 2*a))*a*c*d^2/ 
((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*b^2) + 
 3*((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log 
(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) + (cos(2*b*x + 2*a) 
^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*log(cos(b*x + a)^2 + sin 
(b*x + a)^2 - 2*cos(b*x + a) + 1) - 4*(b*x + a)*sin(2*b*x + 2*a))*a^2*d^3/ 
((cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 - 2*cos(2*b*x + 2*a) + 1)*b^3) - 
 2*c^3/tan(b*x + a) + 6*a*c^2*d/(b*tan(b*x + a)) - 6*a^2*c*d^2/(b^2*tan(b* 
x + a)) + 2*a^3*d^3/(b^3*tan(b*x + a)) - 2*(6*((b*x + a)^2*d^3 + 2*(b*c*d^ 
2 - a*d^3)*(b*x + a) - ((b*x + a)^2*d^3 + 2*(b*c*d^2 - a*d^3)*(b*x + a))*c 
os(2*b*x + 2*a) + (-I*(b*x + a)^2*d^3 + 2*(-I*b*c*d^2 + I*a*d^3)*(b*x + a) 
)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - 6*((b*x +...
 

Giac [F]

\[ \int (c+d x)^3 \csc ^2(a+b x) \, dx=\int { {\left (d x + c\right )}^{3} \csc \left (b x + a\right )^{2} \,d x } \] Input:

integrate((d*x+c)^3*csc(b*x+a)^2,x, algorithm="giac")
 

Output:

integrate((d*x + c)^3*csc(b*x + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \csc ^2(a+b x) \, dx=\int \frac {{\left (c+d\,x\right )}^3}{{\sin \left (a+b\,x\right )}^2} \,d x \] Input:

int((c + d*x)^3/sin(a + b*x)^2,x)
 

Output:

int((c + d*x)^3/sin(a + b*x)^2, x)
 

Reduce [F]

\[ \int (c+d x)^3 \csc ^2(a+b x) \, dx=\frac {-2 \cos \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,c^{3}-6 \cos \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,c^{2} d x -3 \cos \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b c \,d^{2} x^{2}-\cos \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,d^{3} x^{3}+3 \left (\int \frac {x^{2}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}d x \right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,d^{3}+6 \left (\int \frac {x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}d x \right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b c \,d^{2}-3 \left (\int \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) x^{2}d x \right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,d^{3}-6 \left (\int \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) x d x \right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b c \,d^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) c^{2} d +6 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) c^{2} d -3 \sin \left (b x +a \right ) b c \,d^{2} x^{2}-\sin \left (b x +a \right ) b \,d^{3} x^{3}+3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b c \,d^{2} x^{2}+\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b \,d^{3} x^{3}}{2 \sin \left (b x +a \right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) b^{2}} \] Input:

int((d*x+c)^3*csc(b*x+a)^2,x)
 

Output:

( - 2*cos(a + b*x)*tan((a + b*x)/2)*b*c**3 - 6*cos(a + b*x)*tan((a + b*x)/ 
2)*b*c**2*d*x - 3*cos(a + b*x)*tan((a + b*x)/2)*b*c*d**2*x**2 - cos(a + b* 
x)*tan((a + b*x)/2)*b*d**3*x**3 + 3*int(x**2/tan((a + b*x)/2),x)*sin(a + b 
*x)*tan((a + b*x)/2)*b*d**3 + 6*int(x/tan((a + b*x)/2),x)*sin(a + b*x)*tan 
((a + b*x)/2)*b*c*d**2 - 3*int(tan((a + b*x)/2)*x**2,x)*sin(a + b*x)*tan(( 
a + b*x)/2)*b*d**3 - 6*int(tan((a + b*x)/2)*x,x)*sin(a + b*x)*tan((a + b*x 
)/2)*b*c*d**2 - 6*log(tan((a + b*x)/2)**2 + 1)*sin(a + b*x)*tan((a + b*x)/ 
2)*c**2*d + 6*log(tan((a + b*x)/2))*sin(a + b*x)*tan((a + b*x)/2)*c**2*d - 
 3*sin(a + b*x)*b*c*d**2*x**2 - sin(a + b*x)*b*d**3*x**3 + 3*tan((a + b*x) 
/2)*b*c*d**2*x**2 + tan((a + b*x)/2)*b*d**3*x**3)/(2*sin(a + b*x)*tan((a + 
 b*x)/2)*b**2)