\(\int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx\) [47]

Optimal result

Integrand size = 22, antiderivative size = 115 \[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=-\frac {3 i d (c+d x)^2}{2 b^2}-\frac {3 d (c+d x)^2 \cot (a+b x)}{2 b^2}-\frac {(c+d x)^3 \csc ^2(a+b x)}{2 b}+\frac {3 d^2 (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^3}-\frac {3 i d^3 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^4} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 4.97 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.76 \[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=\frac {-b^3 (c+d x)^3 \csc ^2(a+b x)-6 b c d^2 (b x \cot (a)-\log (\sin (a+b x)))+3 d^3 \left (i b x (\pi -2 \arctan (\tan (a)))+\pi \log \left (1+e^{-2 i b x}\right )+2 (b x+\arctan (\tan (a))) \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )-\pi \log (\cos (b x))-2 \arctan (\tan (a)) \log (\sin (b x+\arctan (\tan (a))))-i \operatorname {PolyLog}\left (2,e^{2 i (b x+\arctan (\tan (a)))}\right )-b^2 e^{i \arctan (\tan (a))} x^2 \cot (a) \sqrt {\sec ^2(a)}\right )+3 b^2 d (c+d x)^2 \csc (a) \csc (a+b x) \sin (b x)}{2 b^4} \] Input:

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

Output:

(-(b^3*(c + d*x)^3*Csc[a + b*x]^2) - 6*b*c*d^2*(b*x*Cot[a] - Log[Sin[a + b 
*x]]) + 3*d^3*(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*PolyLo 
g[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))] - b^2*E^(I*ArcTan[Tan[a]])*x^2*Cot[ 
a]*Sqrt[Sec[a]^2]) + 3*b^2*d*(c + d*x)^2*Csc[a]*Csc[a + b*x]*Sin[b*x])/(2* 
b^4)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {4910, 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.

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

\(\Big \downarrow \) 4910

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4672

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

-1/2*((c + d*x)^3*Csc[a + b*x]^2)/b + (3*d*(-(((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))/(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]
 

rule 4910

Int[Cot[(a_.) + (b_.)*(x_)]^(p_.)*Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d 
_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Csc[a + b*x]^n/(b*n)), x 
] + Simp[d*(m/(b*n))   Int[(c + d*x)^(m - 1)*Csc[a + b*x]^n, x], x] /; Free 
Q[{a, b, c, d, n}, x] && EqQ[p, 1] && 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. 408 vs. \(2 (101 ) = 202\).

Time = 0.51 (sec) , antiderivative size = 409, normalized size of antiderivative = 3.56

method	result	size
risch	\(\frac {2 b \,d^{3} x^{3} {\mathrm e}^{2 i \left (b x +a \right )}-3 i d^{3} x^{2} {\mathrm e}^{2 i \left (b x +a \right )}+6 b c \,d^{2} x^{2} {\mathrm e}^{2 i \left (b x +a \right )}-6 i c \,d^{2} x \,{\mathrm e}^{2 i \left (b x +a \right )}+6 b \,c^{2} d x \,{\mathrm e}^{2 i \left (b x +a \right )}-3 i c^{2} d \,{\mathrm e}^{2 i \left (b x +a \right )}+3 i d^{3} x^{2}+2 b \,c^{3} {\mathrm e}^{2 i \left (b x +a \right )}+6 i c \,d^{2} x +3 i c^{2} d}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {6 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {3 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {3 d^{2} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}-\frac {3 i d^{3} x^{2}}{b^{2}}-\frac {6 i d^{3} a x}{b^{3}}-\frac {3 i d^{3} a^{2}}{b^{4}}+\frac {3 d^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{3}}-\frac {3 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {3 d^{3} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{4}}-\frac {3 i d^{3} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 d^{3} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 d^{3} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{4}}\)	\(409\)

Input:

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

Output:

(2*b*d^3*x^3*exp(2*I*(b*x+a))-3*I*d^3*x^2*exp(2*I*(b*x+a))+6*b*c*d^2*x^2*e 
xp(2*I*(b*x+a))-6*I*c*d^2*x*exp(2*I*(b*x+a))+6*b*c^2*d*x*exp(2*I*(b*x+a))- 
3*I*c^2*d*exp(2*I*(b*x+a))+3*I*d^3*x^2+2*b*c^3*exp(2*I*(b*x+a))+6*I*c*d^2* 
x+3*I*c^2*d)/b^2/(exp(2*I*(b*x+a))-1)^2-6*d^2/b^3*c*ln(exp(I*(b*x+a)))+3*d 
^2/b^3*c*ln(exp(I*(b*x+a))-1)+3*d^2/b^3*c*ln(exp(I*(b*x+a))+1)-3*I*d^3/b^2 
*x^2-6*I*d^3/b^3*a*x-3*I*d^3/b^4*a^2+3*d^3/b^3*ln(exp(I*(b*x+a))+1)*x-3*I* 
d^3/b^4*polylog(2,-exp(I*(b*x+a)))+3*d^3/b^3*ln(1-exp(I*(b*x+a)))*x+3*d^3/ 
b^4*ln(1-exp(I*(b*x+a)))*a-3*I*d^3*polylog(2,exp(I*(b*x+a)))/b^4+6*d^3/b^4 
*a*ln(exp(I*(b*x+a)))-3*d^3/b^4*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. 591 vs. \(2 (98) = 196\).

Time = 0.10 (sec) , antiderivative size = 591, normalized size of antiderivative = 5.14 \[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((c + d*x)**3*cot(a + b*x)*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. 1044 vs. \(2 (98) = 196\).

Time = 0.33 (sec) , antiderivative size = 1044, normalized size of antiderivative = 9.08 \[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

(6*b^2*c^2*d + 6*(b*d^3*x + b*c*d^2 + (b*d^3*x + b*c*d^2)*cos(4*b*x + 4*a) 
 - 2*(b*d^3*x + b*c*d^2)*cos(2*b*x + 2*a) - (-I*b*d^3*x - I*b*c*d^2)*sin(4 
*b*x + 4*a) - 2*(I*b*d^3*x + I*b*c*d^2)*sin(2*b*x + 2*a))*arctan2(sin(b*x 
+ a), cos(b*x + a) + 1) + 6*(b*c*d^2*cos(4*b*x + 4*a) - 2*b*c*d^2*cos(2*b* 
x + 2*a) + I*b*c*d^2*sin(4*b*x + 4*a) - 2*I*b*c*d^2*sin(2*b*x + 2*a) + b*c 
*d^2)*arctan2(sin(b*x + a), cos(b*x + a) - 1) - 6*(b*d^3*x*cos(4*b*x + 4*a 
) - 2*b*d^3*x*cos(2*b*x + 2*a) + I*b*d^3*x*sin(4*b*x + 4*a) - 2*I*b*d^3*x* 
sin(2*b*x + 2*a) + b*d^3*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - 6*( 
b^2*d^3*x^2 + 2*b^2*c*d^2*x)*cos(4*b*x + 4*a) - 2*(2*I*b^3*d^3*x^3 + 2*I*b 
^3*c^3 + 3*b^2*c^2*d + 3*(2*I*b^3*c*d^2 - b^2*d^3)*x^2 + 6*(I*b^3*c^2*d - 
b^2*c*d^2)*x)*cos(2*b*x + 2*a) - 6*(d^3*cos(4*b*x + 4*a) - 2*d^3*cos(2*b*x 
 + 2*a) + I*d^3*sin(4*b*x + 4*a) - 2*I*d^3*sin(2*b*x + 2*a) + d^3)*dilog(- 
e^(I*b*x + I*a)) - 6*(d^3*cos(4*b*x + 4*a) - 2*d^3*cos(2*b*x + 2*a) + I*d^ 
3*sin(4*b*x + 4*a) - 2*I*d^3*sin(2*b*x + 2*a) + d^3)*dilog(e^(I*b*x + I*a) 
) - 3*(I*b*d^3*x + I*b*c*d^2 + (I*b*d^3*x + I*b*c*d^2)*cos(4*b*x + 4*a) + 
2*(-I*b*d^3*x - I*b*c*d^2)*cos(2*b*x + 2*a) - (b*d^3*x + b*c*d^2)*sin(4*b* 
x + 4*a) + 2*(b*d^3*x + b*c*d^2)*sin(2*b*x + 2*a))*log(cos(b*x + a)^2 + si 
n(b*x + a)^2 + 2*cos(b*x + a) + 1) - 3*(I*b*d^3*x + I*b*c*d^2 + (I*b*d^3*x 
 + I*b*c*d^2)*cos(4*b*x + 4*a) + 2*(-I*b*d^3*x - I*b*c*d^2)*cos(2*b*x + 2* 
a) - (b*d^3*x + b*c*d^2)*sin(4*b*x + 4*a) + 2*(b*d^3*x + b*c*d^2)*sin(2...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^3 \cot (a+b x) \csc ^2(a+b x) \, dx=\frac {-6 \cos \left (b x +a \right ) \sin \left (b x +a \right ) b^{2} c^{2} d -12 \cos \left (b x +a \right ) \sin \left (b x +a \right ) b^{2} c \,d^{2} x -6 \cos \left (b x +a \right ) \sin \left (b x +a \right ) b^{2} d^{3} x^{2}-12 \cos \left (b x +a \right ) \sin \left (b x +a \right ) d^{3}-12 \cos \left (b x +a \right ) b \,d^{3} x -6 \left (\int \frac {x}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}d x \right ) \sin \left (b x +a \right )^{2} b^{2} d^{3}-6 \left (\int \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) x d x \right ) \sin \left (b x +a \right )^{2} b^{2} d^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1\right ) \sin \left (b x +a \right )^{2} b c \,d^{2}+12 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \sin \left (b x +a \right )^{2} b c \,d^{2}+\sin \left (b x +a \right )^{2} b^{3} c^{3}-12 \sin \left (b x +a \right ) d^{3}-2 b^{3} c^{3}-6 b^{3} c^{2} d x -6 b^{3} c \,d^{2} x^{2}-2 b^{3} d^{3} x^{3}-12 b \,d^{3} x}{4 \sin \left (b x +a \right )^{2} b^{4}} \] Input:

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

Output:

( - 6*cos(a + b*x)*sin(a + b*x)*b**2*c**2*d - 12*cos(a + b*x)*sin(a + b*x) 
*b**2*c*d**2*x - 6*cos(a + b*x)*sin(a + b*x)*b**2*d**3*x**2 - 12*cos(a + b 
*x)*sin(a + b*x)*d**3 - 12*cos(a + b*x)*b*d**3*x - 6*int(x/tan((a + b*x)/2 
)**3,x)*sin(a + b*x)**2*b**2*d**3 - 6*int(tan((a + b*x)/2)*x,x)*sin(a + b* 
x)**2*b**2*d**3 - 12*log(tan((a + b*x)/2)**2 + 1)*sin(a + b*x)**2*b*c*d**2 
 + 12*log(tan((a + b*x)/2))*sin(a + b*x)**2*b*c*d**2 + sin(a + b*x)**2*b** 
3*c**3 - 12*sin(a + b*x)*d**3 - 2*b**3*c**3 - 6*b**3*c**2*d*x - 6*b**3*c*d 
**2*x**2 - 2*b**3*d**3*x**3 - 12*b*d**3*x)/(4*sin(a + b*x)**2*b**4)