\(\int (c+d x) \cot ^3(a+b x) \, dx\) [181]

Optimal result

Integrand size = 14, antiderivative size = 109 \[ \int (c+d x) \cot ^3(a+b x) \, dx=-\frac {d x}{2 b}+\frac {i (c+d x)^2}{2 d}-\frac {d \cot (a+b x)}{2 b^2}-\frac {(c+d x) \cot ^2(a+b x)}{2 b}-\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}+\frac {i d \operatorname {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(234\) vs. \(2(109)=218\).

Time = 5.37 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.15 \[ \int (c+d x) \cot ^3(a+b x) \, dx=-\frac {1}{2} d x^2 \cot (a)-\frac {c \csc ^2(a+b x)}{2 b}-\frac {d x \csc ^2(a+b x)}{2 b}-\frac {c \log (\sin (a+b x))}{b}+\frac {d \csc (a) \csc (a+b x) \sin (b x)}{2 b^2}+\frac {d \csc (a) \sec (a) \left (b^2 e^{i \arctan (\tan (a))} x^2+\frac {\left (i b x (-\pi +2 \arctan (\tan (a)))-\pi \log \left (1+e^{-2 i b x}\right )-2 (b x+\arctan (\tan (a))) \log \left (1-e^{2 i (b x+\arctan (\tan (a)))}\right )+\pi \log (\cos (b x))+2 \arctan (\tan (a)) \log (\sin (b x+\arctan (\tan (a))))+i \operatorname {PolyLog}\left (2,e^{2 i (b x+\arctan (\tan (a)))}\right )\right ) \tan (a)}{\sqrt {1+\tan ^2(a)}}\right )}{2 b^2 \sqrt {\sec ^2(a) \left (\cos ^2(a)+\sin ^2(a)\right )}} \] Input:

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

Output:

-1/2*(d*x^2*Cot[a]) - (c*Csc[a + b*x]^2)/(2*b) - (d*x*Csc[a + b*x]^2)/(2*b 
) - (c*Log[Sin[a + b*x]])/b + (d*Csc[a]*Csc[a + b*x]*Sin[b*x])/(2*b^2) + ( 
d*Csc[a]*Sec[a]*(b^2*E^(I*ArcTan[Tan[a]])*x^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]]))] 
)*Tan[a])/Sqrt[1 + Tan[a]^2]))/(2*b^2*Sqrt[Sec[a]^2*(Cos[a]^2 + Sin[a]^2)] 
)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3042, 25, 4203, 25, 3042, 25, 3954, 24, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4203

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

((I/2)*(c + d*x)^2)/d - ((c + d*x)*Cot[a + b*x]^2)/(2*b) + (d*(-x - Cot[a 
+ b*x]/b))/(2*b) - (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))
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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 3954

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d 
*x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2   Int[(b*Tan[c + d*x])^(n - 2), x] 
, x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
 

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. 280 vs. \(2 (93 ) = 186\).

Time = 0.26 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.58

method	result	size
risch	\(\frac {i d \,x^{2}}{2}-i x c +\frac {2 b d x \,{\mathrm e}^{2 i \left (b x +a \right )}-i d \,{\mathrm e}^{2 i \left (b x +a \right )}+2 b c \,{\mathrm e}^{2 i \left (b x +a \right )}+i d}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}+\frac {2 c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}-\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}+\frac {i d \operatorname {polylog}\left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {i d \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {i d \,a^{2}}{b^{2}}-\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}+\frac {2 i d a x}{b}-\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {2 d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}\)	\(281\)

Input:

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

Output:

1/2*I*d*x^2-I*x*c+(2*b*d*x*exp(2*I*(b*x+a))-I*d*exp(2*I*(b*x+a))+2*b*c*exp 
(2*I*(b*x+a))+I*d)/b^2/(exp(2*I*(b*x+a))-1)^2+2/b*c*ln(exp(I*(b*x+a)))-1/b 
*c*ln(exp(I*(b*x+a))-1)-1/b*c*ln(exp(I*(b*x+a))+1)+I/b^2*d*polylog(2,exp(I 
*(b*x+a)))+I*d*polylog(2,-exp(I*(b*x+a)))/b^2+I/b^2*d*a^2-1/b*d*ln(exp(I*( 
b*x+a))+1)*x+2*I/b*d*a*x-1/b*d*ln(1-exp(I*(b*x+a)))*x-1/b^2*d*ln(1-exp(I*( 
b*x+a)))*a-2/b^2*d*a*ln(exp(I*(b*x+a)))+1/b^2*d*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. 339 vs. \(2 (90) = 180\).

Time = 0.08 (sec) , antiderivative size = 339, normalized size of antiderivative = 3.11 \[ \int (c+d x) \cot ^3(a+b x) \, dx=\frac {4 \, b d x + 4 \, b c + {\left (i \, d \cos \left (2 \, b x + 2 \, a\right ) - i \, d\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + {\left (-i \, d \cos \left (2 \, b x + 2 \, a\right ) + i \, d\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 2 \, {\left (b c - a d - {\left (b c - a d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b c - a d - {\left (b c - a d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b d x + a d - {\left (b d x + a d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, {\left (b d x + a d - {\left (b d x + a d\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, d \sin \left (2 \, b x + 2 \, a\right )}{4 \, {\left (b^{2} \cos \left (2 \, b x + 2 \, a\right ) - b^{2}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((c + d*x)*cot(a + b*x)**3, 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. 830 vs. \(2 (90) = 180\).

Time = 0.19 (sec) , antiderivative size = 830, normalized size of antiderivative = 7.61 \[ \int (c+d x) \cot ^3(a+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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