Integrand size = 31, antiderivative size = 127 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\left (\left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x\right )-\frac {a^2 (2 A b+a B) \cot (c+d x)}{d}-\frac {b^3 B \log (\cos (c+d x))}{d}-\frac {a \left (a^2 A-3 A b^2-3 a b B\right ) \log (\sin (c+d x))}{d}-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d} \]
-(3*A*a^2*b-A*b^3+B*a^3-3*B*a*b^2)*x-a^2*(2*A*b+B*a)*cot(d*x+c)/d-b^3*B*ln (cos(d*x+c))/d-a*(A*a^2-3*A*b^2-3*B*a*b)*ln(sin(d*x+c))/d-1/2*a*A*cot(d*x+ c)^2*(a+b*tan(d*x+c))^2/d
Result contains complex when optimal does not.
Time = 0.45 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.99 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {-2 a^2 (3 A b+a B) \cot (c+d x)-a^3 A \cot ^2(c+d x)+(a+i b)^3 (A+i B) \log (i-\tan (c+d x))-2 a \left (a^2 A-3 A b^2-3 a b B\right ) \log (\tan (c+d x))+(a-i b)^3 (A-i B) \log (i+\tan (c+d x))}{2 d} \]
(-2*a^2*(3*A*b + a*B)*Cot[c + d*x] - a^3*A*Cot[c + d*x]^2 + (a + I*b)^3*(A + I*B)*Log[I - Tan[c + d*x]] - 2*a*(a^2*A - 3*A*b^2 - 3*a*b*B)*Log[Tan[c + d*x]] + (a - I*b)^3*(A - I*B)*Log[I + Tan[c + d*x]])/(2*d)
Time = 0.77 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4088, 27, 3042, 4118, 25, 3042, 4107, 3042, 25, 3956}
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.
| \(\displaystyle \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan (c+d x)^3}dx\) |
| \(\Big \downarrow \) 4088 |
| \(\displaystyle \frac {1}{2} \int 2 \cot ^2(c+d x) (a+b \tan (c+d x)) \left (b^2 B \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)\right )dx-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \cot ^2(c+d x) (a+b \tan (c+d x)) \left (b^2 B \tan ^2(c+d x)-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)\right )dx-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+b \tan (c+d x)) \left (b^2 B \tan (c+d x)^2-\left (A a^2-2 b B a-A b^2\right ) \tan (c+d x)+a (2 A b+a B)\right )}{\tan (c+d x)^2}dx-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 4118 |
| \(\displaystyle \int -\cot (c+d x) \left (-B \tan ^2(c+d x) b^3+a \left (A a^2-3 b B a-3 A b^2\right )+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)\right )dx-\frac {a^2 (a B+2 A b) \cot (c+d x)}{d}-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \cot (c+d x) \left (-B \tan ^2(c+d x) b^3+a \left (A a^2-3 b B a-3 A b^2\right )+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)\right )dx-\frac {a^2 (a B+2 A b) \cot (c+d x)}{d}-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\int \frac {-B \tan (c+d x)^2 b^3+a \left (A a^2-3 b B a-3 A b^2\right )+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\tan (c+d x)}dx-\frac {a^2 (a B+2 A b) \cot (c+d x)}{d}-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 4107 |
| \(\displaystyle -a \left (a^2 A-3 a b B-3 A b^2\right ) \int \cot (c+d x)dx+b^3 B \int \tan (c+d x)dx-\frac {a^2 (a B+2 A b) \cot (c+d x)}{d}-x \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -a \left (a^2 A-3 a b B-3 A b^2\right ) \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx+b^3 B \int \tan (c+d x)dx-\frac {a^2 (a B+2 A b) \cot (c+d x)}{d}-x \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle a \left (a^2 A-3 a b B-3 A b^2\right ) \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx+b^3 B \int \tan (c+d x)dx-\frac {a^2 (a B+2 A b) \cot (c+d x)}{d}-x \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle -\frac {a \left (a^2 A-3 a b B-3 A b^2\right ) \log (-\sin (c+d x))}{d}-\frac {a^2 (a B+2 A b) \cot (c+d x)}{d}-x \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right )-\frac {a A \cot ^2(c+d x) (a+b \tan (c+d x))^2}{2 d}-\frac {b^3 B \log (\cos (c+d x))}{d}\) |
-((3*a^2*A*b - A*b^3 + a^3*B - 3*a*b^2*B)*x) - (a^2*(2*A*b + a*B)*Cot[c + d*x])/d - (b^3*B*Log[Cos[c + d*x]])/d - (a*(a^2*A - 3*A*b^2 - 3*a*b*B)*Log [-Sin[c + d*x]])/d - (a*A*Cot[c + d*x]^2*(a + b*Tan[c + d*x])^2)/(2*d)
3.3.53.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x ])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(d*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* (b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] & & LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2 )/tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[B*x, x] + (Simp[A Int[1/Tan[ e + f*x], x], x] + Simp[C Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A, B, C}, x] && NeQ[A, C]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_. )*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(c^2*C - B*c*d + A*d^2)*((c + d*Tan[e + f*x])^(n + 1)/(d^2*f*(n + 1)*(c^2 + d^2))), x] + Simp[1/(d*(c^2 + d^2)) Int[(c + d*Tan[e + f*x])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b* (c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d + a*C*d) *Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n , -1]
Time = 0.21 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.07
| method | result | size |
| parallelrisch | \(\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (\sec ^{2}\left (d x +c \right )\right )+\left (-2 A \,a^{3}+6 A a \,b^{2}+6 B \,a^{2} b \right ) \ln \left (\tan \left (d x +c \right )\right )-A \left (\cot ^{2}\left (d x +c \right )\right ) a^{3}+\left (-6 A \,a^{2} b -2 B \,a^{3}\right ) \cot \left (d x +c \right )-6 \left (A \,a^{2} b -\frac {1}{3} A \,b^{3}+\frac {1}{3} B \,a^{3}-B a \,b^{2}\right ) d x}{2 d}\) | \(136\) |
| derivativedivides | \(\frac {\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{3}}{2 \tan \left (d x +c \right )^{2}}-\frac {a^{2} \left (3 A b +B a \right )}{\tan \left (d x +c \right )}-a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\) | \(140\) |
| default | \(\frac {\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}+\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) \arctan \left (\tan \left (d x +c \right )\right )-\frac {A \,a^{3}}{2 \tan \left (d x +c \right )^{2}}-\frac {a^{2} \left (3 A b +B a \right )}{\tan \left (d x +c \right )}-a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\) | \(140\) |
| norman | \(\frac {\left (-3 A \,a^{2} b +A \,b^{3}-B \,a^{3}+3 B a \,b^{2}\right ) x \left (\tan ^{2}\left (d x +c \right )\right )-\frac {A \,a^{3}}{2 d}-\frac {a^{2} \left (3 A b +B a \right ) \tan \left (d x +c \right )}{d}}{\tan \left (d x +c \right )^{2}}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \left (A \,a^{2}-3 A \,b^{2}-3 B a b \right ) \ln \left (\tan \left (d x +c \right )\right )}{d}\) | \(150\) |
| risch | \(-\frac {6 i B \,a^{2} b c}{d}-3 i B \,a^{2} b x +\frac {2 i B \,b^{3} c}{d}-\frac {6 i A a \,b^{2} c}{d}-3 A \,a^{2} b x +A \,b^{3} x -B \,a^{3} x +3 B a \,b^{2} x -\frac {2 i a^{2} \left (i A a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 A b \,{\mathrm e}^{2 i \left (d x +c \right )}+B a \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A b -B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}+i A \,a^{3} x -3 i A a \,b^{2} x +i B \,b^{3} x +\frac {2 i a^{3} A c}{d}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) A \,b^{2}}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) B b}{d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B \,b^{3}}{d}\) | \(267\) |
1/2*((A*a^3-3*A*a*b^2-3*B*a^2*b+B*b^3)*ln(sec(d*x+c)^2)+(-2*A*a^3+6*A*a*b^ 2+6*B*a^2*b)*ln(tan(d*x+c))-A*cot(d*x+c)^2*a^3+(-6*A*a^2*b-2*B*a^3)*cot(d* x+c)-6*(A*a^2*b-1/3*A*b^3+1/3*B*a^3-B*a*b^2)*d*x)/d
Time = 0.28 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.28 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {B b^{3} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + A a^{3} + {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )^{2} + {\left (A a^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} d x\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )^{2}} \]
-1/2*(B*b^3*log(1/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 + A*a^3 + (A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 + (A*a^3 + 2*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*d*x)*tan(d*x + c )^2 + 2*(B*a^3 + 3*A*a^2*b)*tan(d*x + c))/(d*tan(d*x + c)^2)
Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (121) = 242\).
Time = 1.08 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.06 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\begin {cases} \tilde {\infty } A a^{3} x & \text {for}\: c = 0 \wedge d = 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right )^{3} \cot ^{3}{\left (c \right )} & \text {for}\: d = 0 \\\tilde {\infty } A a^{3} x & \text {for}\: c = - d x \\\frac {A a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - \frac {A a^{3} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} - \frac {A a^{3}}{2 d \tan ^{2}{\left (c + d x \right )}} - 3 A a^{2} b x - \frac {3 A a^{2} b}{d \tan {\left (c + d x \right )}} - \frac {3 A a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 A a b^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + A b^{3} x - B a^{3} x - \frac {B a^{3}}{d \tan {\left (c + d x \right )}} - \frac {3 B a^{2} b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 B a^{2} b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 3 B a b^{2} x + \frac {B b^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {otherwise} \end {cases} \]
Piecewise((zoo*A*a**3*x, Eq(c, 0) & Eq(d, 0)), (x*(A + B*tan(c))*(a + b*ta n(c))**3*cot(c)**3, Eq(d, 0)), (zoo*A*a**3*x, Eq(c, -d*x)), (A*a**3*log(ta n(c + d*x)**2 + 1)/(2*d) - A*a**3*log(tan(c + d*x))/d - A*a**3/(2*d*tan(c + d*x)**2) - 3*A*a**2*b*x - 3*A*a**2*b/(d*tan(c + d*x)) - 3*A*a*b**2*log(t an(c + d*x)**2 + 1)/(2*d) + 3*A*a*b**2*log(tan(c + d*x))/d + A*b**3*x - B* a**3*x - B*a**3/(d*tan(c + d*x)) - 3*B*a**2*b*log(tan(c + d*x)**2 + 1)/(2* d) + 3*B*a**2*b*log(tan(c + d*x))/d + 3*B*a*b**2*x + B*b**3*log(tan(c + d* x)**2 + 1)/(2*d), True))
Time = 0.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.12 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} {\left (d x + c\right )} - {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {A a^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]
-1/2*(2*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*(d*x + c) - (A*a^3 - 3*B*a ^2*b - 3*A*a*b^2 + B*b^3)*log(tan(d*x + c)^2 + 1) + 2*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*log(tan(d*x + c)) + (A*a^3 + 2*(B*a^3 + 3*A*a^2*b)*tan(d*x + c ))/tan(d*x + c)^2)/d
Time = 1.40 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.52 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=-\frac {2 \, {\left (B a^{3} + 3 \, A a^{2} b - 3 \, B a b^{2} - A b^{3}\right )} {\left (d x + c\right )} - {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2} + B b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (A a^{3} - 3 \, B a^{2} b - 3 \, A a b^{2}\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) - \frac {3 \, A a^{3} \tan \left (d x + c\right )^{2} - 9 \, B a^{2} b \tan \left (d x + c\right )^{2} - 9 \, A a b^{2} \tan \left (d x + c\right )^{2} - 2 \, B a^{3} \tan \left (d x + c\right ) - 6 \, A a^{2} b \tan \left (d x + c\right ) - A a^{3}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]
-1/2*(2*(B*a^3 + 3*A*a^2*b - 3*B*a*b^2 - A*b^3)*(d*x + c) - (A*a^3 - 3*B*a ^2*b - 3*A*a*b^2 + B*b^3)*log(tan(d*x + c)^2 + 1) + 2*(A*a^3 - 3*B*a^2*b - 3*A*a*b^2)*log(abs(tan(d*x + c))) - (3*A*a^3*tan(d*x + c)^2 - 9*B*a^2*b*t an(d*x + c)^2 - 9*A*a*b^2*tan(d*x + c)^2 - 2*B*a^3*tan(d*x + c) - 6*A*a^2* b*tan(d*x + c) - A*a^3)/tan(d*x + c)^2)/d
Time = 8.15 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.06 \[ \int \cot ^3(c+d x) (a+b \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (-A\,a^3+3\,B\,a^2\,b+3\,A\,a\,b^2\right )}{d}-\frac {{\mathrm {cot}\left (c+d\,x\right )}^2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (B\,a^3+3\,A\,b\,a^2\right )+\frac {A\,a^3}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (A+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^3\,1{}\mathrm {i}}{2\,d} \]