\(\int (c+d x)^2 \cot ^2(a+b x) \, dx\) [107]

Optimal result

Integrand size = 16, antiderivative size = 97 \[ \int (c+d x)^2 \cot ^2(a+b x) \, dx=-\frac {i (c+d x)^2}{b}-\frac {(c+d x)^3}{3 d}-\frac {(c+d x)^2 \cot (a+b x)}{b}+\frac {2 d (c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b^2}-\frac {i d^2 \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^3} \] Output:

-I*(d*x+c)^2/b-1/3*(d*x+c)^3/d-(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. \(198\) vs. \(2(97)=194\).

Time = 4.12 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.04 \[ \int (c+d x)^2 \cot ^2(a+b x) \, dx=-\frac {1}{3} x \left (3 c^2+3 c d x+d^2 x^2\right )-\frac {2 c d (b x \cot (a)-\log (\sin (a+b x)))}{b^2}+\frac {d^2 \left (i b x (\pi -2 \arctan (\tan (a)))+\pi \log \left (1+e^{-2 i b x}\right )+2 (b x+\arctan (\tan (a))) \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )-\pi \log (\cos (b x))-2 \arctan (\tan (a)) \log (\sin (b x+\arctan (\tan (a))))-i \operatorname {PolyLog}\left (2,e^{2 i (b x+\arctan (\tan (a)))}\right )-b^2 e^{i \arctan (\tan (a))} x^2 \cot (a) \sqrt {\sec ^2(a)}\right )}{b^3}+\frac {(c+d x)^2 \csc (a) \csc (a+b x) \sin (b x)}{b} \] Input:

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

Output:

-1/3*(x*(3*c^2 + 3*c*d*x + d^2*x^2)) - (2*c*d*(b*x*Cot[a] - Log[Sin[a + b* 
x]]))/b^2 + (d^2*(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*Pol 
yLog[2, E^((2*I)*(b*x + ArcTan[Tan[a]]))] - b^2*E^(I*ArcTan[Tan[a]])*x^2*C 
ot[a]*Sqrt[Sec[a]^2]))/b^3 + ((c + d*x)^2*Csc[a]*Csc[a + b*x]*Sin[b*x])/b
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.24, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3042, 4203, 17, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4203

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

\(\Big \downarrow \) 17

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

-1/3*(c + d*x)^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
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

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 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 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. 304 vs. \(2 (89 ) = 178\).

Time = 0.44 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.14

method	result	size
risch	\(-\frac {d^{2} x^{3}}{3}-c d \,x^{2}-c^{2} x -\frac {c^{3}}{3 d}-\frac {2 i \left (x^{2} d^{2}+2 c d x +c^{2}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}-\frac {4 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {4 i d^{2} a x}{b^{2}}-\frac {2 i d^{2} a^{2}}{b^{3}}+\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {2 i d^{2} x^{2}}{b}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 i d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {4 d^{2} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 d^{2} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}\)	\(305\)

Input:

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

Output:

-1/3*d^2*x^3-c*d*x^2-c^2*x-1/3/d*c^3-2*I*(d^2*x^2+2*c*d*x+c^2)/b/(exp(2*I* 
(b*x+a))-1)-4*d/b^2*c*ln(exp(I*(b*x+a)))+2*d/b^2*c*ln(exp(I*(b*x+a))-1)+2* 
d/b^2*c*ln(exp(I*(b*x+a))+1)-2*I*d^2*polylog(2,exp(I*(b*x+a)))/b^3-4*I*d^2 
/b^2*a*x-2*I*d^2/b^3*a^2+2*d^2/b^2*ln(exp(I*(b*x+a))+1)*x-2*I*d^2/b*x^2+2* 
d^2/b^2*ln(1-exp(I*(b*x+a)))*x+2*d^2/b^3*ln(1-exp(I*(b*x+a)))*a-2*I*d^2/b^ 
3*polylog(2,-exp(I*(b*x+a)))+4*d^2/b^3*a*ln(exp(I*(b*x+a)))-2*d^2/b^3*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. 384 vs. \(2 (86) = 172\).

Time = 0.08 (sec) , antiderivative size = 384, normalized size of antiderivative = 3.96 \[ \int (c+d x)^2 \cot ^2(a+b x) \, dx=-\frac {6 \, b^{2} d^{2} x^{2} + 12 \, b^{2} c d x + 6 \, b^{2} c^{2} + 3 i \, d^{2} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 3 i \, d^{2} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b c d - a d^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (b d^{2} x + a d^{2}\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) \sin \left (2 \, b x + 2 \, a\right ) + 6 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}\right )} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (b^{3} d^{2} x^{3} + 3 \, b^{3} c d x^{2} + 3 \, b^{3} c^{2} x\right )} \sin \left (2 \, b x + 2 \, a\right )}{6 \, b^{3} \sin \left (2 \, b x + 2 \, a\right )} \] Input:

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

Output:

-1/6*(6*b^2*d^2*x^2 + 12*b^2*c*d*x + 6*b^2*c^2 + 3*I*d^2*dilog(cos(2*b*x + 
 2*a) + I*sin(2*b*x + 2*a))*sin(2*b*x + 2*a) - 3*I*d^2*dilog(cos(2*b*x + 2 
*a) - I*sin(2*b*x + 2*a))*sin(2*b*x + 2*a) - 6*(b*c*d - a*d^2)*log(-1/2*co 
s(2*b*x + 2*a) + 1/2*I*sin(2*b*x + 2*a) + 1/2)*sin(2*b*x + 2*a) - 6*(b*c*d 
 - a*d^2)*log(-1/2*cos(2*b*x + 2*a) - 1/2*I*sin(2*b*x + 2*a) + 1/2)*sin(2* 
b*x + 2*a) - 6*(b*d^2*x + a*d^2)*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a 
) + 1)*sin(2*b*x + 2*a) - 6*(b*d^2*x + a*d^2)*log(-cos(2*b*x + 2*a) - I*si 
n(2*b*x + 2*a) + 1)*sin(2*b*x + 2*a) + 6*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2* 
c^2)*cos(2*b*x + 2*a) + 2*(b^3*d^2*x^3 + 3*b^3*c*d*x^2 + 3*b^3*c^2*x)*sin( 
2*b*x + 2*a))/(b^3*sin(2*b*x + 2*a))
 

Sympy [F]

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

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

Output:

Integral((c + d*x)**2*cot(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. 645 vs. \(2 (86) = 172\).

Time = 0.22 (sec) , antiderivative size = 645, normalized size of antiderivative = 6.65 \[ \int (c+d x)^2 \cot ^2(a+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

(-I*b^3*d^2*x^3 - 3*I*b^3*c*d*x^2 - 3*I*b^3*c^2*x - 6*b^2*c^2 - 6*(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) + 6*(b*c*d*cos(2*b* 
x + 2*a) + I*b*c*d*sin(2*b*x + 2*a) - b*c*d)*arctan2(sin(b*x + a), cos(b*x 
 + a) - 1) - 6*(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) + (I*b^3*d^2*x^3 - 3*(-I*b 
^3*c*d + 2*b^2*d^2)*x^2 - 3*(-I*b^3*c^2 + 4*b^2*c*d)*x)*cos(2*b*x + 2*a) - 
 6*(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)) - 6*(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)) - 3*(-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) - 3*(-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) - (b^3*d^2*x^3 + 3*(b^3*c*d + 
2*I*b^2*d^2)*x^2 + 3*(b^3*c^2 + 4*I*b^2*c*d)*x)*sin(2*b*x + 2*a))/(-3*I*b^ 
3*cos(2*b*x + 2*a) + 3*b^3*sin(2*b*x + 2*a) + 3*I*b^3)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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