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

3.3.36.1 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
 

3.3.36.2 Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(593\) vs. \(2(226)=452\).

Time = 6.28 (sec) , antiderivative size = 593, normalized size of antiderivative = 2.62 \[ \int (c+d x)^2 \csc ^2(a+b x) \sec (a+b x) \, dx=-\frac {(c+d x)^2 \csc (a)}{b}+\frac {-2 i b^2 c^2 \arctan \left (e^{i (a+b x)}\right )+2 b^2 c d x \log \left (1-i e^{i (a+b x)}\right )+b^2 d^2 x^2 \log \left (1-i e^{i (a+b x)}\right )-2 b^2 c d x \log \left (1+i e^{i (a+b x)}\right )-b^2 d^2 x^2 \log \left (1+i e^{i (a+b x)}\right )+2 i b d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )-2 i b d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )-2 d^2 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )+2 d^2 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^3}+\frac {4 i c d \arctan \left (\frac {i \cos (a)-i \sin (a) \tan \left (\frac {b x}{2}\right )}{\sqrt {\cos ^2(a)+\sin ^2(a)}}\right )}{b^2 \sqrt {\cos ^2(a)+\sin ^2(a)}}+\frac {\sec \left (\frac {a}{2}\right ) \sec \left (\frac {a}{2}+\frac {b x}{2}\right ) \left (-c^2 \sin \left (\frac {b x}{2}\right )-2 c d x \sin \left (\frac {b x}{2}\right )-d^2 x^2 \sin \left (\frac {b x}{2}\right )\right )}{2 b}+\frac {\csc \left (\frac {a}{2}\right ) \csc \left (\frac {a}{2}+\frac {b x}{2}\right ) \left (c^2 \sin \left (\frac {b x}{2}\right )+2 c d x \sin \left (\frac {b x}{2}\right )+d^2 x^2 \sin \left (\frac {b x}{2}\right )\right )}{2 b}+\frac {2 d^2 \left (-\frac {2 \arctan (\tan (a)) \text {arctanh}\left (\frac {-\cos (a)+\sin (a) \tan \left (\frac {b x}{2}\right )}{\sqrt {\cos ^2(a)+\sin ^2(a)}}\right )}{\sqrt {\cos ^2(a)+\sin ^2(a)}}+\frac {\left ((b x+\arctan (\tan (a))) \left (\log \left (1-e^{i (b x+\arctan (\tan (a)))}\right )-\log \left (1+e^{i (b x+\arctan (\tan (a)))}\right )\right )+i \left (\operatorname {PolyLog}\left (2,-e^{i (b x+\arctan (\tan (a)))}\right )-\operatorname {PolyLog}\left (2,e^{i (b x+\arctan (\tan (a)))}\right )\right )\right ) \sec (a)}{\sqrt {1+\tan ^2(a)}}\right )}{b^3} \]

input

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

output

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

3.3.36.3 Rubi [A] (verified)

Time = 0.70 (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
 

3.3.36.3.1 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] 
]
 

3.3.36.4 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 = 1.61 (sec) , antiderivative size = 556, normalized size of antiderivative = 2.46

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

input

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

output

2*I/b^2*d^2*polylog(2,-I*exp(I*(b*x+a)))*x-2/b^2*c*d*ln(1+I*exp(I*(b*x+a)) 
)*a+2/b*c*d*ln(1-I*exp(I*(b*x+a)))*x-2*d^2/b^3*a*ln(exp(I*(b*x+a))-1)-2*d^ 
2/b^2*ln(exp(I*(b*x+a))+1)*x+2/b^2*c*d*ln(1-I*exp(I*(b*x+a)))*a+2*I/b^2*c* 
d*polylog(2,-I*exp(I*(b*x+a)))+2*d/b^2*c*ln(exp(I*(b*x+a))-1)-2*I/b*c^2*ar 
ctan(exp(I*(b*x+a)))+1/b*d^2*ln(1-I*exp(I*(b*x+a)))*x^2-2*d^2*polylog(3,-I 
*exp(I*(b*x+a)))/b^3-1/b*d^2*ln(1+I*exp(I*(b*x+a)))*x^2-2*I/b^2*c*d*polylo 
g(2,I*exp(I*(b*x+a)))+1/b^3*a^2*d^2*ln(1+I*exp(I*(b*x+a)))-2/b*c*d*ln(1+I* 
exp(I*(b*x+a)))*x-1/b^3*a^2*d^2*ln(1-I*exp(I*(b*x+a)))-2*d/b^2*c*ln(exp(I* 
(b*x+a))+1)+2*d^2*polylog(3,I*exp(I*(b*x+a)))/b^3+2*I/b^3*d^2*dilog(exp(I* 
(b*x+a))+1)-2*I/b^3*d^2*a^2*arctan(exp(I*(b*x+a)))+4*I/b^2*c*d*a*arctan(ex 
p(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)+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
 

3.3.36.5 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 = 0.32 (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...
 

3.3.36.6 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)
 

3.3.36.7 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 = 0.50 (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*...
 

3.3.36.8 Giac [F]

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

input

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

output

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

3.3.36.9 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}