\(\int (c+d x)^2 \csc ^2(a+b x) \sec (a+b x) \, dx\) [236]

Optimal result

Integrand size = 22, antiderivative size = 226 \[ \int (c+d x)^2 \csc ^2(a+b x) \sec (a+b x) \, dx=-\frac {2 i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b}-\frac {4 d (c+d x) \text {arctanh}\left (e^{i (a+b x)}\right )}{b^2}-\frac {(c+d x)^2 \csc (a+b x)}{b}+\frac {2 i d^2 \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^3}+\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^2}-\frac {2 i d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^2}-\frac {2 i d^2 \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^3}-\frac {2 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3} \] Output:

-2*I*(d*x+c)^2*arctan(exp(I*(b*x+a)))/b-4*d*(d*x+c)*arctanh(exp(I*(b*x+a)) 
)/b^2-(d*x+c)^2*csc(b*x+a)/b+2*I*d^2*polylog(2,-exp(I*(b*x+a)))/b^3+2*I*d* 
(d*x+c)*polylog(2,-I*exp(I*(b*x+a)))/b^2-2*I*d*(d*x+c)*polylog(2,I*exp(I*( 
b*x+a)))/b^2-2*I*d^2*polylog(2,exp(I*(b*x+a)))/b^3-2*d^2*polylog(3,-I*exp( 
I*(b*x+a)))/b^3+2*d^2*polylog(3,I*exp(I*(b*x+a)))/b^3
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(467\) vs. \(2(226)=452\).

Time = 5.36 (sec) , antiderivative size = 467, normalized size of antiderivative = 2.07 \[ \int (c+d x)^2 \csc ^2(a+b x) \sec (a+b x) \, dx=\frac {-4 i b^2 c^2 \arctan \left (e^{i (a+b x)}\right )-8 b c d \text {arctanh}\left (\cos (a)-\sin (a) \tan \left (\frac {b x}{2}\right )\right )+8 d^2 \arctan (\tan (a)) \text {arctanh}\left (\cos (a)-\sin (a) \tan \left (\frac {b x}{2}\right )\right )-2 b^2 (c+d x)^2 \csc (a)+4 b^2 c d x \log \left (1-i e^{i (a+b x)}\right )+2 b^2 d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )-4 b^2 c d x \log \left (1+i e^{i (a+b x)}\right )-2 b^2 d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+4 i b d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )-4 i b d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )-4 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )+4 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )+\frac {4 d^2 \left ((b x+\arctan (\tan (a))) \left (\log \left (1-e^{i (b x+\arctan (\tan (a)))}\right )-\log \left (1+e^{i (b x+\arctan (\tan (a)))}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i (b x+\arctan (\tan (a)))}\right )-i \operatorname {PolyLog}\left (2,e^{i (b x+\arctan (\tan (a)))}\right )\right ) \sec (a)}{\sqrt {\sec ^2(a)}}+b^2 (c+d x)^2 \csc \left (\frac {a}{2}\right ) \csc \left (\frac {1}{2} (a+b x)\right ) \sin \left (\frac {b x}{2}\right )-b^2 (c+d x)^2 \sec \left (\frac {a}{2}\right ) \sec \left (\frac {1}{2} (a+b x)\right ) \sin \left (\frac {b x}{2}\right )}{2 b^3} \] Input:

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

Output:

((-4*I)*b^2*c^2*ArcTan[E^(I*(a + b*x))] - 8*b*c*d*ArcTanh[Cos[a] - Sin[a]* 
Tan[(b*x)/2]] + 8*d^2*ArcTan[Tan[a]]*ArcTanh[Cos[a] - Sin[a]*Tan[(b*x)/2]] 
 - 2*b^2*(c + d*x)^2*Csc[a] + 4*b^2*c*d*x*Log[1 - I*E^(I*(a + b*x))] + 2*b 
^2*d^2*x^2*Log[1 - I*E^(I*(a + b*x))] - 4*b^2*c*d*x*Log[1 + I*E^(I*(a + b* 
x))] - 2*b^2*d^2*x^2*Log[1 + I*E^(I*(a + b*x))] + (4*I)*b*d*(c + d*x)*Poly 
Log[2, (-I)*E^(I*(a + b*x))] - (4*I)*b*d*(c + d*x)*PolyLog[2, I*E^(I*(a + 
b*x))] - 4*d^2*PolyLog[3, (-I)*E^(I*(a + b*x))] + 4*d^2*PolyLog[3, I*E^(I* 
(a + b*x))] + (4*d^2*((b*x + ArcTan[Tan[a]])*(Log[1 - E^(I*(b*x + ArcTan[T 
an[a]]))] - Log[1 + E^(I*(b*x + ArcTan[Tan[a]]))]) + I*PolyLog[2, -E^(I*(b 
*x + ArcTan[Tan[a]]))] - I*PolyLog[2, E^(I*(b*x + ArcTan[Tan[a]]))])*Sec[a 
])/Sqrt[Sec[a]^2] + b^2*(c + d*x)^2*Csc[a/2]*Csc[(a + b*x)/2]*Sin[(b*x)/2] 
 - b^2*(c + d*x)^2*Sec[a/2]*Sec[(a + b*x)/2]*Sin[(b*x)/2])/(2*b^3)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4920, 7292, 27, 7293, 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.

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

\(\Big \downarrow \) 4920

\(\displaystyle -2 d \int (c+d x) \left (\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\csc (a+b x)}{b}\right )dx+\frac {(c+d x)^2 \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

\(\Big \downarrow \) 7292

\(\displaystyle -2 d \int \frac {(c+d x) (\text {arctanh}(\sin (a+b x))-\csc (a+b x))}{b}dx+\frac {(c+d x)^2 \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 d \int (c+d x) (\text {arctanh}(\sin (a+b x))-\csc (a+b x))dx}{b}+\frac {(c+d x)^2 \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {2 d \int ((c+d x) \text {arctanh}(\sin (a+b x))-(c+d x) \csc (a+b x))dx}{b}+\frac {(c+d x)^2 \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 d \left (\frac {i (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{d}+\frac {2 (c+d x) \text {arctanh}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x)^2 \text {arctanh}(\sin (a+b x))}{2 d}-\frac {i d \operatorname {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}+\frac {i d \operatorname {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}+\frac {d \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^2}-\frac {d \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^2}-\frac {i (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b}+\frac {i (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b}\right )}{b}+\frac {(c+d x)^2 \text {arctanh}(\sin (a+b x))}{b}-\frac {(c+d x)^2 \csc (a+b x)}{b}\)

Input:

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

Output:

((c + d*x)^2*ArcTanh[Sin[a + b*x]])/b - ((c + d*x)^2*Csc[a + b*x])/b - (2* 
d*((I*(c + d*x)^2*ArcTan[E^(I*(a + b*x))])/d + (2*(c + d*x)*ArcTanh[E^(I*( 
a + b*x))])/b + ((c + d*x)^2*ArcTanh[Sin[a + b*x]])/(2*d) - (I*d*PolyLog[2 
, -E^(I*(a + b*x))])/b^2 - (I*(c + d*x)*PolyLog[2, (-I)*E^(I*(a + b*x))])/ 
b + (I*(c + d*x)*PolyLog[2, I*E^(I*(a + b*x))])/b + (I*d*PolyLog[2, E^(I*( 
a + b*x))])/b^2 + (d*PolyLog[3, (-I)*E^(I*(a + b*x))])/b^2 - (d*PolyLog[3, 
 I*E^(I*(a + b*x))])/b^2))/b
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 4920

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b 
_.)*(x_)]^(p_.), x_Symbol] :> Module[{u = IntHide[Csc[a + b*x]^n*Sec[a + b* 
x]^p, x]}, Simp[(c + d*x)^m   u, x] - Simp[d*m   Int[(c + d*x)^(m - 1)*u, x 
], x]] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p] && GtQ[m, 0] && NeQ[n, 
p]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (201 ) = 402\).

Time = 0.53 (sec) , antiderivative size = 556, normalized size of antiderivative = 2.46

method	result	size
risch	\(\frac {2 i c d \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 c d \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}+\frac {2 d^{2} \operatorname {polylog}\left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i c d \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 i \left (x^{2} d^{2}+2 c d x +c^{2}\right ) {\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {2 d^{2} \operatorname {polylog}\left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i d^{2} \operatorname {dilog}\left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i d^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {a^{2} d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {2 c d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}-\frac {2 c d \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\frac {a^{2} d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}-\frac {2 d^{2} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {4 i c d a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 i d^{2} \operatorname {dilog}\left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}-\frac {2 i c^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}\)	\(556\)

Input:

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

Output:

2*I/b^2*c*d*polylog(2,-I*exp(I*(b*x+a)))+2/b*c*d*ln(1-I*exp(I*(b*x+a)))*x+ 
2/b^2*c*d*ln(1-I*exp(I*(b*x+a)))*a+2*d^2*polylog(3,I*exp(I*(b*x+a)))/b^3-2 
*d^2/b^2*ln(exp(I*(b*x+a))+1)*x-2*I/b^2*d^2*polylog(2,I*exp(I*(b*x+a)))*x- 
2*I/b^2*c*d*polylog(2,I*exp(I*(b*x+a)))-2*I*(d^2*x^2+2*c*d*x+c^2)*exp(I*(b 
*x+a))/b/(exp(2*I*(b*x+a))-1)+1/b*d^2*ln(1-I*exp(I*(b*x+a)))*x^2-2*d^2*pol 
ylog(3,-I*exp(I*(b*x+a)))/b^3+2*I/b^3*d^2*dilog(exp(I*(b*x+a)))+2*I/b^2*d^ 
2*polylog(2,-I*exp(I*(b*x+a)))*x-1/b^3*a^2*d^2*ln(1-I*exp(I*(b*x+a)))-2*I/ 
b^3*d^2*a^2*arctan(exp(I*(b*x+a)))-2*d/b^2*c*ln(exp(I*(b*x+a))+1)-2/b*c*d* 
ln(I*exp(I*(b*x+a))+1)*x-1/b*d^2*ln(I*exp(I*(b*x+a))+1)*x^2-2/b^2*c*d*ln(I 
*exp(I*(b*x+a))+1)*a+2*d/b^2*c*ln(exp(I*(b*x+a))-1)+1/b^3*a^2*d^2*ln(I*exp 
(I*(b*x+a))+1)-2*d^2/b^3*a*ln(exp(I*(b*x+a))-1)+4*I/b^2*c*d*a*arctan(exp(I 
*(b*x+a)))+2*I/b^3*d^2*dilog(exp(I*(b*x+a))+1)-2*I/b*c^2*arctan(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. 1067 vs. \(2 (188) = 376\).

Time = 0.15 (sec) , antiderivative size = 1067, normalized size of antiderivative = 4.72 \[ \int (c+d x)^2 \csc ^2(a+b x) \sec (a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(2*b^2*d^2*x^2 + 4*b^2*c*d*x + 2*b^2*c^2 + 2*I*d^2*dilog(cos(b*x + a) 
 + I*sin(b*x + a))*sin(b*x + a) - 2*I*d^2*dilog(cos(b*x + a) - I*sin(b*x + 
 a))*sin(b*x + a) + 2*I*d^2*dilog(-cos(b*x + a) + I*sin(b*x + a))*sin(b*x 
+ a) - 2*I*d^2*dilog(-cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + 2*d^2* 
polylog(3, I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) - 2*d^2*polylog(3, 
I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a) + 2*d^2*polylog(3, -I*cos(b*x 
+ a) + sin(b*x + a))*sin(b*x + a) - 2*d^2*polylog(3, -I*cos(b*x + a) - sin 
(b*x + a))*sin(b*x + a) + 2*(I*b*d^2*x + I*b*c*d)*dilog(I*cos(b*x + a) + s 
in(b*x + a))*sin(b*x + a) + 2*(I*b*d^2*x + I*b*c*d)*dilog(I*cos(b*x + a) - 
 sin(b*x + a))*sin(b*x + a) + 2*(-I*b*d^2*x - I*b*c*d)*dilog(-I*cos(b*x + 
a) + sin(b*x + a))*sin(b*x + a) + 2*(-I*b*d^2*x - I*b*c*d)*dilog(-I*cos(b* 
x + a) - sin(b*x + a))*sin(b*x + a) + 2*(b*d^2*x + b*c*d)*log(cos(b*x + a) 
 + I*sin(b*x + a) + 1)*sin(b*x + a) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log( 
cos(b*x + a) + I*sin(b*x + a) + I)*sin(b*x + a) + 2*(b*d^2*x + b*c*d)*log( 
cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a) + (b^2*c^2 - 2*a*b*c*d + a 
^2*d^2)*log(cos(b*x + a) - I*sin(b*x + a) + I)*sin(b*x + a) - (b^2*d^2*x^2 
 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(I*cos(b*x + a) + sin(b*x + a) + 
1)*sin(b*x + a) + (b^2*d^2*x^2 + 2*b^2*c*d*x + 2*a*b*c*d - a^2*d^2)*log(I* 
cos(b*x + a) - sin(b*x + a) + 1)*sin(b*x + a) - (b^2*d^2*x^2 + 2*b^2*c*d*x 
 + 2*a*b*c*d - a^2*d^2)*log(-I*cos(b*x + a) + sin(b*x + a) + 1)*sin(b*x...
 

Sympy [F]

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

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

Output:

Integral((c + d*x)**2*csc(a + b*x)**2*sec(a + b*x), 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. 1638 vs. \(2 (188) = 376\).

Time = 0.46 (sec) , antiderivative size = 1638, normalized size of antiderivative = 7.25 \[ \int (c+d x)^2 \csc ^2(a+b x) \sec (a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(c^2*(2/sin(b*x + a) - log(sin(b*x + a) + 1) + log(sin(b*x + a) - 1)) 
 - 2*a*c*d*(2/sin(b*x + a) - log(sin(b*x + a) + 1) + log(sin(b*x + a) - 1) 
)/b + a^2*d^2*(2/sin(b*x + a) - log(sin(b*x + a) + 1) + log(sin(b*x + a) - 
 1))/b^2 - 2*(2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) - ((b*x + a 
)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*cos(2*b*x + 2*a) - (I*(b*x + a)^2*d 
^2 + 2*(I*b*c*d - I*a*d^2)*(b*x + a))*sin(2*b*x + 2*a))*arctan2(cos(b*x + 
a), sin(b*x + a) + 1) + 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a) - 
 ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)*(b*x + a))*cos(2*b*x + 2*a) - (I*(b* 
x + a)^2*d^2 + 2*(I*b*c*d - I*a*d^2)*(b*x + a))*sin(2*b*x + 2*a))*arctan2( 
cos(b*x + a), -sin(b*x + a) + 1) + 4*(b*c*d + (b*x + a)*d^2 - a*d^2 - (b*c 
*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) - (I*b*c*d + I*(b*x + a)*d^2 
- I*a*d^2)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - 4*( 
b*c*d - a*d^2 - (b*c*d - a*d^2)*cos(2*b*x + 2*a) + (-I*b*c*d + I*a*d^2)*si 
n(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) - 1) - 4*((b*x + a)*d^2 
*cos(2*b*x + 2*a) + I*(b*x + a)*d^2*sin(2*b*x + 2*a) - (b*x + a)*d^2)*arct 
an2(sin(b*x + a), -cos(b*x + a) + 1) - 4*((b*x + a)^2*d^2 + 2*(b*c*d - a*d 
^2)*(b*x + a))*cos(b*x + a) + 4*(b*c*d + (b*x + a)*d^2 - a*d^2 - (b*c*d + 
(b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) - (I*b*c*d + I*(b*x + a)*d^2 - I*a 
*d^2)*sin(2*b*x + 2*a))*dilog(I*e^(I*b*x + I*a)) - 4*(b*c*d + (b*x + a)*d^ 
2 - a*d^2 - (b*c*d + (b*x + a)*d^2 - a*d^2)*cos(2*b*x + 2*a) + (-I*b*c*...
 

Giac [F(-2)]

Exception generated. \[ \int (c+d x)^2 \csc ^2(a+b x) \sec (a+b x) \, dx=\text {Exception raised: AttributeError} \] Input:

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

Output:

Exception raised: AttributeError >> type
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^2 \csc ^2(a+b x) \sec (a+b x) \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x)^2 \csc ^2(a+b x) \sec (a+b x) \, dx=\frac {8 \cos \left (b x +a \right ) b c d x -4 \left (\int \frac {x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}d x \right ) \sin \left (b x +a \right ) b^{2} c d +\left (\int \csc \left (b x +a \right )^{2} \sec \left (b x +a \right ) x^{2}d x \right ) \sin \left (b x +a \right ) b^{2} d^{2}+8 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right ) \sin \left (b x +a \right ) c d -\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right ) b \,c^{2}+\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right ) b \,c^{2}-14 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right ) c d +\sin \left (b x +a \right ) b^{2} c d \,x^{2}-b \,c^{2}+6 b c d x}{\sin \left (b x +a \right ) b^{2}} \] Input:

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

Output:

(8*cos(a + b*x)*b*c*d*x - 4*int(x/(tan((a + b*x)/2)**4 - tan((a + b*x)/2)* 
*2),x)*sin(a + b*x)*b**2*c*d + int(csc(a + b*x)**2*sec(a + b*x)*x**2,x)*si 
n(a + b*x)*b**2*d**2 + 8*log(tan((a + b*x)/2)**2 + 1)*sin(a + b*x)*c*d - l 
og(tan((a + b*x)/2) - 1)*sin(a + b*x)*b*c**2 + log(tan((a + b*x)/2) + 1)*s 
in(a + b*x)*b*c**2 - 14*log(tan((a + b*x)/2))*sin(a + b*x)*c*d + sin(a + b 
*x)*b**2*c*d*x**2 - b*c**2 + 6*b*c*d*x)/(sin(a + b*x)*b**2)