\(\int (c+d x)^2 \csc ^3(a+b x) \sec (a+b x) \, dx\) [242]

Optimal result

Integrand size = 22, antiderivative size = 194 \[ \int (c+d x)^2 \csc ^3(a+b x) \sec (a+b x) \, dx=-\frac {(c+d x)^2}{2 b}-\frac {2 (c+d x)^2 \text {arctanh}\left (e^{2 i (a+b x)}\right )}{b}-\frac {d (c+d x) \cot (a+b x)}{b^2}-\frac {(c+d x)^2 \cot ^2(a+b x)}{2 b}+\frac {d^2 \log (\sin (a+b x))}{b^3}+\frac {i d (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d (c+d x) \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac {d^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {d^2 \operatorname {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3} \] Output:

-1/2*(d*x+c)^2/b-2*(d*x+c)^2*arctanh(exp(2*I*(b*x+a)))/b-d*(d*x+c)*cot(b*x 
+a)/b^2-1/2*(d*x+c)^2*cot(b*x+a)^2/b+d^2*ln(sin(b*x+a))/b^3+I*d*(d*x+c)*po 
lylog(2,-exp(2*I*(b*x+a)))/b^2-I*d*(d*x+c)*polylog(2,exp(2*I*(b*x+a)))/b^2 
-1/2*d^2*polylog(3,-exp(2*I*(b*x+a)))/b^3+1/2*d^2*polylog(3,exp(2*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. \(880\) vs. \(2(194)=388\).

Time = 6.52 (sec) , antiderivative size = 880, normalized size of antiderivative = 4.54 \[ \int (c+d x)^2 \csc ^3(a+b x) \sec (a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*((c + d*x)^2*Csc[a + b*x]^2)/b - (d^2*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*P 
olyLog[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))])) 
/(6*b^3) + (x*(3*c^2 + 3*c*d*x + d^2*x^2)*Csc[a]*Sec[a])/3 - ((I/12)*d^2*( 
2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Log[1 + E^((-2*I)*(a + b*x))]) 
+ 6*b*(1 + E^((2*I)*a))*x*PolyLog[2, -E^((-2*I)*(a + b*x))] - (3*I)*(1 + E 
^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/(b^3*E^(I*a)) - (c^ 
2*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a]*Sin[b*x]] + b*x*Sin[a]))/(b* 
(Cos[a]^2 + Sin[a]^2)) + (c^2*Csc[a]*(-(b*x*Cos[a]) + Log[Cos[b*x]*Sin[a] 
+ Cos[a]*Sin[b*x]]*Sin[a]))/(b*(Cos[a]^2 + Sin[a]^2)) + (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)) - (c*d*Csc[a]*((b^2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b* 
x*(-Pi - 2*ArcTan[Cot[a]]) - Pi*Log[1 + E^((-2*I)*b*x)] - 2*(b*x - ArcTan[ 
Cot[a]])*Log[1 - E^((2*I)*(b*x - ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2* 
ArcTan[Cot[a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x 
 - ArcTan[Cot[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(b^2*Sqrt[Csc[a]^2*(Cos 
[a]^2 + Sin[a]^2)]) + (Csc[a]*Csc[a + b*x]*(c*d*Sin[b*x] + d^2*x*Sin[b*...
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4920, 27, 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.

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

\(\Big \downarrow \) 4920

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-1/2*((c + d*x)^2*Cot[a + b*x]^2)/b + ((c + d*x)^2*Log[Tan[a + b*x]])/b + 
(d*(-1/2*(c + d*x)^2/d - (2*(c + d*x)^2*ArcTanh[E^((2*I)*(a + b*x))])/d - 
((c + d*x)*Cot[a + b*x])/b + (d*Log[Sin[a + b*x]])/b^2 - ((c + d*x)^2*Log[ 
Tan[a + b*x]])/d + (I*(c + d*x)*PolyLog[2, -E^((2*I)*(a + b*x))])/b - (I*( 
c + d*x)*PolyLog[2, E^((2*I)*(a + b*x))])/b - (d*PolyLog[3, -E^((2*I)*(a + 
 b*x))])/(2*b^2) + (d*PolyLog[3, E^((2*I)*(a + b*x))])/(2*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. 631 vs. \(2 (174 ) = 348\).

Time = 0.53 (sec) , antiderivative size = 632, normalized size of antiderivative = 3.26

method	result	size
risch	\(-\frac {2 i c d \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 c d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {d^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}-\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}+\frac {d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {2 i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 i c d \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 c d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}+\frac {2 c d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {2 d^{2} \operatorname {polylog}\left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 b \,d^{2} x^{2} {\mathrm e}^{2 i \left (b x +a \right )}+4 b c d x \,{\mathrm e}^{2 i \left (b x +a \right )}+2 b \,c^{2} {\mathrm e}^{2 i \left (b x +a \right )}-2 i d^{2} x \,{\mathrm e}^{2 i \left (b x +a \right )}-2 i c d \,{\mathrm e}^{2 i \left (b x +a \right )}+2 i d^{2} x +2 i d c}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}-\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}-\frac {d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}+\frac {2 d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}-\frac {c^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {i d c \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}-\frac {2 d c \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b}\)	\(632\)

Input:

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

Output:

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

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. 1995 vs. \(2 (170) = 340\).

Time = 0.18 (sec) , antiderivative size = 1995, normalized size of antiderivative = 10.28 \[ \int (c+d x)^2 \csc ^3(a+b x) \sec (a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + 2*(b*d^2*x + b*c*d)*cos(b*x + a 
)*sin(b*x + a) - 2*(-I*b*d^2*x - I*b*c*d + (I*b*d^2*x + I*b*c*d)*cos(b*x + 
 a)^2)*dilog(cos(b*x + a) + I*sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d + (-I 
*b*d^2*x - I*b*c*d)*cos(b*x + a)^2)*dilog(cos(b*x + a) - I*sin(b*x + a)) - 
 2*(-I*b*d^2*x - I*b*c*d + (I*b*d^2*x + I*b*c*d)*cos(b*x + a)^2)*dilog(I*c 
os(b*x + a) + sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c 
*d)*cos(b*x + a)^2)*dilog(I*cos(b*x + a) - sin(b*x + a)) - 2*(I*b*d^2*x + 
I*b*c*d + (-I*b*d^2*x - I*b*c*d)*cos(b*x + a)^2)*dilog(-I*cos(b*x + a) + s 
in(b*x + a)) - 2*(-I*b*d^2*x - I*b*c*d + (I*b*d^2*x + I*b*c*d)*cos(b*x + a 
)^2)*dilog(-I*cos(b*x + a) - sin(b*x + a)) - 2*(I*b*d^2*x + I*b*c*d + (-I* 
b*d^2*x - I*b*c*d)*cos(b*x + a)^2)*dilog(-cos(b*x + a) + I*sin(b*x + a)) - 
 2*(-I*b*d^2*x - I*b*c*d + (I*b*d^2*x + I*b*c*d)*cos(b*x + a)^2)*dilog(-co 
s(b*x + a) - I*sin(b*x + a)) - (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - (b^2 
*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + d^2)*cos(b*x + a)^2 + d^2)*log(cos(b*x 
+ a) + I*sin(b*x + a) + 1) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2 - (b^2*c^2 - 2 
*a*b*c*d + a^2*d^2)*cos(b*x + a)^2)*log(cos(b*x + a) + I*sin(b*x + a) + I) 
 - (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2 
*c^2 + d^2)*cos(b*x + a)^2 + d^2)*log(cos(b*x + a) - I*sin(b*x + a) + 1) + 
 (b^2*c^2 - 2*a*b*c*d + a^2*d^2 - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cos(b*x 
+ a)^2)*log(cos(b*x + a) - I*sin(b*x + a) + I) + (b^2*d^2*x^2 + 2*b^2*c...
 

Sympy [F]

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

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

Output:

Integral((c + d*x)**2*csc(a + b*x)**3*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. 2528 vs. \(2 (170) = 340\).

Time = 0.62 (sec) , antiderivative size = 2528, normalized size of antiderivative = 13.03 \[ \int (c+d x)^2 \csc ^3(a+b x) \sec (a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(c^2*(1/sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1) - log(sin(b*x + a)^2 
)) - 2*a*c*d*(1/sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1) - log(sin(b*x + a 
)^2))/b + a^2*d^2*(1/sin(b*x + a)^2 + log(sin(b*x + a)^2 - 1) - log(sin(b* 
x + a)^2))/b^2 + 2*(4*(b*x + a)*d^2*cos(4*b*x + 4*a) + 4*I*(b*x + a)*d^2*s 
in(4*b*x + 4*a) - 4*b*c*d + 4*a*d^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(4*b*x + 
 4*a) - 2*((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(4*b*x + 4*a 
) - 2*(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(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 2*((b*x + a)^2*d^2 + 
 2*(b*c*d - a*d^2)*(b*x + a) + d^2 + ((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2)* 
(b*x + a) + d^2)*cos(4*b*x + 4*a) - 2*((b*x + a)^2*d^2 + 2*(b*c*d - a*d^2) 
*(b*x + a) + d^2)*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) + I*d^2)*sin(4*b*x + 4*a) + 2*(-I*(b*x + a)^2*d^2 + 2*(-I* 
b*c*d + I*a*d^2)*(b*x + a) - I*d^2)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a) 
, cos(b*x + a) + 1) - 2*(d^2*cos(4*b*x + 4*a) - 2*d^2*cos(2*b*x + 2*a) + I 
*d^2*sin(4*b*x + 4*a) - 2*I*d^2*sin(2*b*x + 2*a) + d^2)*arctan2(sin(b*x + 
a), cos(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(4*b*x + 4*a) - 2*((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...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^2 \csc ^3(a+b x) \sec (a+b x) \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x)^2 \csc ^3(a+b x) \sec (a+b x) \, dx=\frac {4 \left (\int \csc \left (b x +a \right )^{3} \sec \left (b x +a \right ) x^{2}d x \right ) \sin \left (b x +a \right )^{2} b \,d^{2}+8 \left (\int \csc \left (b x +a \right )^{3} \sec \left (b x +a \right ) x d x \right ) \sin \left (b x +a \right )^{2} b c d -4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{2} c^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{2} c^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2} c^{2}+\sin \left (b x +a \right )^{2} c^{2}-2 c^{2}}{4 \sin \left (b x +a \right )^{2} b} \] Input:

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

Output:

(4*int(csc(a + b*x)**3*sec(a + b*x)*x**2,x)*sin(a + b*x)**2*b*d**2 + 8*int 
(csc(a + b*x)**3*sec(a + b*x)*x,x)*sin(a + b*x)**2*b*c*d - 4*log(tan((a + 
b*x)/2) - 1)*sin(a + b*x)**2*c**2 - 4*log(tan((a + b*x)/2) + 1)*sin(a + b* 
x)**2*c**2 + 4*log(tan((a + b*x)/2))*sin(a + b*x)**2*c**2 + sin(a + b*x)** 
2*c**2 - 2*c**2)/(4*sin(a + b*x)**2*b)