\(\int (c+d x)^2 \csc ^2(a+b x) \, dx\) [29]

Optimal result

Integrand size = 16, antiderivative size = 83 \[ \int (c+d x)^2 \csc ^2(a+b x) \, dx=-\frac {i (c+d x)^2}{b}-\frac {(c+d x)^2 \cot (a+b x)}{b}+\frac {2 d (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3} \] Output:

-I*(d*x+c)^2/b-(d*x+c)^2*cot(b*x+a)/b+2*d*(d*x+c)*ln(1-exp(2*I*(b*x+a)))/b 
^2-I*d^2*polylog(2,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. \(181\) vs. \(2(83)=166\).

Time = 6.12 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.18 \[ \int (c+d x)^2 \csc ^2(a+b x) \, dx=\frac {\csc (a) \left (-2 b c d (b x \cos (a)-\log (\sin (a+b x)) \sin (a))+d^2 \left (-b^2 e^{i \arctan (\tan (a))} x^2 \cos (a) \sqrt {\sec ^2(a)}-\left (-i b x (\pi -2 \arctan (\tan (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x+\arctan (\tan (a))) \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )+\pi \log (\cos (b x))+2 \arctan (\tan (a)) \log (\sin (b x+\arctan (\tan (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x+\arctan (\tan (a)))}\right )\right ) \sin (a)\right )+b^2 (c+d x)^2 \csc (a+b x) \sin (b x)\right )}{b^3} \] Input:

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

Output:

(Csc[a]*(-2*b*c*d*(b*x*Cos[a] - Log[Sin[a + b*x]]*Sin[a]) + d^2*(-(b^2*E^( 
I*ArcTan[Tan[a]])*x^2*Cos[a]*Sqrt[Sec[a]^2]) - ((-I)*b*x*(Pi - 2*ArcTan[Ta 
n[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[S 
in[b*x + ArcTan[Tan[a]]]] + I*PolyLog[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))] 
)*Sin[a]) + b^2*(c + d*x)^2*Csc[a + b*x]*Sin[b*x]))/b^3
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.28, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4672, 3042, 25, 4202, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4672

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

-(((c + d*x)^2*Cot[a + b*x])/b) - (2*d*(((I/2)*(c + d*x)^2)/d - (2*I)*(((- 
1/2*I)*(c + d*x)*Log[1 + E^(I*(2*a + Pi + 2*b*x))])/b - (d*PolyLog[2, -E^( 
I*(2*a + Pi + 2*b*x))])/(4*b^2))))/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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (77 ) = 154\).

Time = 1.09 (sec) , antiderivative size = 276, normalized size of antiderivative = 3.33

method	result	size
risch	\(-\frac {2 i \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {4 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}-\frac {2 i d^{2} x^{2}}{b}-\frac {4 i d^{2} a x}{b^{2}}-\frac {2 i d^{2} a^{2}}{b^{3}}+\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {4 d^{2} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 d^{2} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}\)	\(276\)

Input:

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

Output:

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

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. 379 vs. \(2 (74) = 148\).

Time = 0.10 (sec) , antiderivative size = 379, normalized size of antiderivative = 4.57 \[ \int (c+d x)^2 \csc ^2(a+b x) \, dx=\frac {-i \, d^{2} {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + i \, d^{2} {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + i \, d^{2} {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) - i \, d^{2} {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) \sin \left (b x + a\right ) + {\left (b d^{2} x + b c d\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + {\left (b d^{2} x + b c d\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) + {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) \sin \left (b x + a\right ) + {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (b x + a\right )}{b^{3} \sin \left (b x + a\right )} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((c + d*x)**2*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. 552 vs. \(2 (74) = 148\).

Time = 0.13 (sec) , antiderivative size = 552, normalized size of antiderivative = 6.65 \[ \int (c+d x)^2 \csc ^2(a+b x) \, dx=-\frac {2 \, b^{2} c^{2} + 2 \, {\left (b d^{2} x + b c d - {\left (b d^{2} x + b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (-i \, b d^{2} x - i \, b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - 2 \, {\left (b c d \cos \left (2 \, b x + 2 \, a\right ) + i \, b c d \sin \left (2 \, b x + 2 \, a\right ) - b c d\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) + 2 \, {\left (b d^{2} x \cos \left (2 \, b x + 2 \, a\right ) + i \, b d^{2} x \sin \left (2 \, b x + 2 \, a\right ) - b d^{2} x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + 2 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x\right )} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + 2 \, {\left (d^{2} \cos \left (2 \, b x + 2 \, a\right ) + i \, d^{2} \sin \left (2 \, b x + 2 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) - {\left (i \, b d^{2} x + i \, b c d + {\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (i \, b d^{2} x + i \, b c d + {\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (2 \, b x + 2 \, a\right ) + {\left (b d^{2} x + b c d\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) + 2 \, {\left (i \, b^{2} d^{2} x^{2} + 2 i \, b^{2} c d x\right )} \sin \left (2 \, b x + 2 \, a\right )}{-i \, b^{3} \cos \left (2 \, b x + 2 \, a\right ) + b^{3} \sin \left (2 \, b x + 2 \, a\right ) + i \, b^{3}} \] Input:

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

Output:

-(2*b^2*c^2 + 2*(b*d^2*x + b*c*d - (b*d^2*x + b*c*d)*cos(2*b*x + 2*a) + (- 
I*b*d^2*x - I*b*c*d)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), cos(b*x + a) 
+ 1) - 2*(b*c*d*cos(2*b*x + 2*a) + I*b*c*d*sin(2*b*x + 2*a) - b*c*d)*arcta 
n2(sin(b*x + a), cos(b*x + a) - 1) + 2*(b*d^2*x*cos(2*b*x + 2*a) + I*b*d^2 
*x*sin(2*b*x + 2*a) - b*d^2*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 
2*(b^2*d^2*x^2 + 2*b^2*c*d*x)*cos(2*b*x + 2*a) + 2*(d^2*cos(2*b*x + 2*a) + 
 I*d^2*sin(2*b*x + 2*a) - d^2)*dilog(-e^(I*b*x + I*a)) + 2*(d^2*cos(2*b*x 
+ 2*a) + I*d^2*sin(2*b*x + 2*a) - d^2)*dilog(e^(I*b*x + I*a)) - (I*b*d^2*x 
 + I*b*c*d + (-I*b*d^2*x - I*b*c*d)*cos(2*b*x + 2*a) + (b*d^2*x + b*c*d)*s 
in(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) 
 - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*d)*cos(2*b*x + 2*a) + (b*d^2 
*x + b*c*d)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos( 
b*x + a) + 1) + 2*(I*b^2*d^2*x^2 + 2*I*b^2*c*d*x)*sin(2*b*x + 2*a))/(-I*b^ 
3*cos(2*b*x + 2*a) + b^3*sin(2*b*x + 2*a) + I*b^3)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 2*cos(a + b*x)*tan((a + b*x)/2)*b*c**2 - 4*cos(a + b*x)*tan((a + b*x)/ 
2)*b*c*d*x - cos(a + b*x)*tan((a + b*x)/2)*b*d**2*x**2 + 2*int(x/tan((a + 
b*x)/2),x)*sin(a + b*x)*tan((a + b*x)/2)*b*d**2 - 2*int(tan((a + b*x)/2)*x 
,x)*sin(a + b*x)*tan((a + b*x)/2)*b*d**2 - 4*log(tan((a + b*x)/2)**2 + 1)* 
sin(a + b*x)*tan((a + b*x)/2)*c*d + 4*log(tan((a + b*x)/2))*sin(a + b*x)*t 
an((a + b*x)/2)*c*d - sin(a + b*x)*b*d**2*x**2 + tan((a + b*x)/2)*b*d**2*x 
**2)/(2*sin(a + b*x)*tan((a + b*x)/2)*b**2)