Integrand size = 24, antiderivative size = 159 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {65 x}{16 a^4}-\frac {65 \cot (c+d x)}{16 a^4 d}-\frac {4 i \log (\sin (c+d x))}{a^4 d}+\frac {31 \cot (c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}+\frac {2 \cot (c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 \cot (c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]
-65/16*x/a^4-65/16*cot(d*x+c)/a^4/d-4*I*ln(sin(d*x+c))/a^4/d+31/48*cot(d*x +c)/a^4/d/(1+I*tan(d*x+c))^2+2*cot(d*x+c)/a^4/d/(1+I*tan(d*x+c))+1/8*cot(d *x+c)/d/(a+I*a*tan(d*x+c))^4+7/24*cot(d*x+c)/a/d/(a+I*a*tan(d*x+c))^3
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.76 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {14 \cot ^4(c+d x)}{a^4 (i+\cot (c+d x))^3}+\frac {\frac {31 \cot ^3(c+d x)}{(i+\cot (c+d x))^2}+\frac {96 \cot ^2(c+d x)}{i+\cot (c+d x)}-195 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\tan ^2(c+d x)\right )-192 i (\log (\cos (c+d x))+\log (\tan (c+d x)))}{a^4}+\frac {6 \cot (c+d x)}{(a+i a \tan (c+d x))^4}}{48 d} \]
((14*Cot[c + d*x]^4)/(a^4*(I + Cot[c + d*x])^3) + ((31*Cot[c + d*x]^3)/(I + Cot[c + d*x])^2 + (96*Cot[c + d*x]^2)/(I + Cot[c + d*x]) - 195*Cot[c + d *x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] - (192*I)*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]]))/a^4 + (6*Cot[c + d*x])/(a + I*a*Tan[c + d*x] )^4)/(48*d)
Time = 1.29 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.14, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.792, Rules used = {3042, 4042, 3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4012, 25, 3042, 4014, 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.
\(\displaystyle \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^2 (a+i a \tan (c+d x))^4}dx\) |
\(\Big \downarrow \) 4042 |
\(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (9 a-5 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^3}dx}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {9 a-5 i a \tan (c+d x)}{\tan (c+d x)^2 (i \tan (c+d x) a+a)^3}dx}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {\frac {\int \frac {4 \cot ^2(c+d x) \left (17 a^2-14 i a^2 \tan (c+d x)\right )}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \int \frac {\cot ^2(c+d x) \left (17 a^2-14 i a^2 \tan (c+d x)\right )}{(i \tan (c+d x) a+a)^2}dx}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \int \frac {17 a^2-14 i a^2 \tan (c+d x)}{\tan (c+d x)^2 (i \tan (c+d x) a+a)^2}dx}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {\frac {2 \left (\frac {\int \frac {3 \cot ^2(c+d x) \left (33 a^3-31 i a^3 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \int \frac {\cot ^2(c+d x) \left (33 a^3-31 i a^3 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \int \frac {33 a^3-31 i a^3 \tan (c+d x)}{\tan (c+d x)^2 (i \tan (c+d x) a+a)}dx}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {\int 2 \cot ^2(c+d x) \left (65 a^4-64 i a^4 \tan (c+d x)\right )dx}{2 a^2}+\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {\int \cot ^2(c+d x) \left (65 a^4-64 i a^4 \tan (c+d x)\right )dx}{a^2}+\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {\int \frac {65 a^4-64 i a^4 \tan (c+d x)}{\tan (c+d x)^2}dx}{a^2}+\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {-\frac {65 a^4 \cot (c+d x)}{d}+\int -\cot (c+d x) \left (65 \tan (c+d x) a^4+64 i a^4\right )dx}{a^2}+\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {-\frac {65 a^4 \cot (c+d x)}{d}-\int \cot (c+d x) \left (65 \tan (c+d x) a^4+64 i a^4\right )dx}{a^2}+\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {-\frac {65 a^4 \cot (c+d x)}{d}-\int \frac {65 \tan (c+d x) a^4+64 i a^4}{\tan (c+d x)}dx}{a^2}+\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {-64 i a^4 \int \cot (c+d x)dx-\frac {65 a^4 \cot (c+d x)}{d}-65 a^4 x}{a^2}+\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {-64 i a^4 \int -\tan \left (c+d x+\frac {\pi }{2}\right )dx-\frac {65 a^4 \cot (c+d x)}{d}-65 a^4 x}{a^2}+\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {64 i a^4 \int \tan \left (\frac {1}{2} (2 c+\pi )+d x\right )dx-\frac {65 a^4 \cot (c+d x)}{d}-65 a^4 x}{a^2}+\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\frac {2 \left (\frac {3 \left (\frac {32 a^3 \cot (c+d x)}{d (a+i a \tan (c+d x))}+\frac {-\frac {65 a^4 \cot (c+d x)}{d}-\frac {64 i a^4 \log (-\sin (c+d x))}{d}-65 a^4 x}{a^2}\right )}{4 a^2}+\frac {31 \cot (c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}+\frac {7 a \cot (c+d x)}{3 d (a+i a \tan (c+d x))^3}}{8 a^2}+\frac {\cot (c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
Cot[c + d*x]/(8*d*(a + I*a*Tan[c + d*x])^4) + ((7*a*Cot[c + d*x])/(3*d*(a + I*a*Tan[c + d*x])^3) + (2*((31*Cot[c + d*x])/(4*d*(1 + I*Tan[c + d*x])^2 ) + (3*((-65*a^4*x - (65*a^4*Cot[c + d*x])/d - ((64*I)*a^4*Log[-Sin[c + d* x]])/d)/a^2 + (32*a^3*Cot[c + d*x])/(d*(a + I*a*Tan[c + d*x]))))/(4*a^2))) /(3*a^2))/(8*a^2)
3.1.83.3.1 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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N eQ[a*c + b*d, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) In t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 0]
Time = 0.56 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-\frac {129 x}{16 a^{4}}-\frac {9 i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{4} d}-\frac {15 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{4} d}-\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{12 a^{4} d}-\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}-\frac {8 c}{a^{4} d}-\frac {2 i}{d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}-\frac {4 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{4} d}\) | \(132\) |
derivativedivides | \(\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {17 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {5}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {4 i \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d}-\frac {1}{a^{4} d \tan \left (d x +c \right )}\) | \(147\) |
default | \(\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {17 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {5}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {4 i \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d}-\frac {1}{a^{4} d \tan \left (d x +c \right )}\) | \(147\) |
norman | \(\frac {-\frac {7 i \left (\tan ^{5}\left (d x +c \right )\right )}{a d}-\frac {1}{a d}-\frac {949 \left (\tan ^{4}\left (d x +c \right )\right )}{48 a d}-\frac {715 \left (\tan ^{6}\left (d x +c \right )\right )}{48 a d}-\frac {65 \left (\tan ^{8}\left (d x +c \right )\right )}{16 a d}-\frac {65 x \tan \left (d x +c \right )}{16 a}-\frac {65 x \left (\tan ^{3}\left (d x +c \right )\right )}{4 a}-\frac {195 x \left (\tan ^{5}\left (d x +c \right )\right )}{8 a}-\frac {65 x \left (\tan ^{7}\left (d x +c \right )\right )}{4 a}-\frac {65 x \left (\tan ^{9}\left (d x +c \right )\right )}{16 a}-\frac {175 \left (\tan ^{2}\left (d x +c \right )\right )}{16 a d}-\frac {2 i \left (\tan ^{7}\left (d x +c \right )\right )}{a d}-\frac {14 i \tan \left (d x +c \right )}{3 d a}-\frac {26 i \left (\tan ^{3}\left (d x +c \right )\right )}{3 d a}}{\tan \left (d x +c \right ) a^{3} \left (1+\tan ^{2}\left (d x +c \right )\right )^{4}}-\frac {4 i \ln \left (\tan \left (d x +c \right )\right )}{a^{4} d}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}\) | \(269\) |
-129/16*x/a^4-9/4*I/a^4/d*exp(-2*I*(d*x+c))-15/32*I/a^4/d*exp(-4*I*(d*x+c) )-1/12*I/a^4/d*exp(-6*I*(d*x+c))-1/128*I/a^4/d*exp(-8*I*(d*x+c))-8/a^4/d*c -2*I/d/a^4/(exp(2*I*(d*x+c))-1)-4*I/a^4/d*ln(exp(2*I*(d*x+c))-1)
Time = 0.24 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.86 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {3096 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} - 24 \, {\left (129 \, d x - 68 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 1536 \, {\left (i \, e^{\left (10 i \, d x + 10 i \, c\right )} - i \, e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 684 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 148 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 29 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i}{384 \, {\left (a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} - a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )}\right )}} \]
-1/384*(3096*d*x*e^(10*I*d*x + 10*I*c) - 24*(129*d*x - 68*I)*e^(8*I*d*x + 8*I*c) + 1536*(I*e^(10*I*d*x + 10*I*c) - I*e^(8*I*d*x + 8*I*c))*log(e^(2*I *d*x + 2*I*c) - 1) - 684*I*e^(6*I*d*x + 6*I*c) - 148*I*e^(4*I*d*x + 4*I*c) - 29*I*e^(2*I*d*x + 2*I*c) - 3*I)/(a^4*d*e^(10*I*d*x + 10*I*c) - a^4*d*e^ (8*I*d*x + 8*I*c))
Time = 0.36 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.58 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (- 442368 i a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 92160 i a^{12} d^{3} e^{16 i c} e^{- 4 i d x} - 16384 i a^{12} d^{3} e^{14 i c} e^{- 6 i d x} - 1536 i a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{196608 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (- 129 e^{8 i c} - 72 e^{6 i c} - 30 e^{4 i c} - 8 e^{2 i c} - 1\right ) e^{- 8 i c}}{16 a^{4}} + \frac {129}{16 a^{4}}\right ) & \text {otherwise} \end {cases} - \frac {2 i}{a^{4} d e^{2 i c} e^{2 i d x} - a^{4} d} - \frac {129 x}{16 a^{4}} - \frac {4 i \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a^{4} d} \]
Piecewise(((-442368*I*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) - 92160*I*a**12 *d**3*exp(16*I*c)*exp(-4*I*d*x) - 16384*I*a**12*d**3*exp(14*I*c)*exp(-6*I* d*x) - 1536*I*a**12*d**3*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(196608*a **16*d**4), Ne(a**16*d**4*exp(20*I*c), 0)), (x*((-129*exp(8*I*c) - 72*exp( 6*I*c) - 30*exp(4*I*c) - 8*exp(2*I*c) - 1)*exp(-8*I*c)/(16*a**4) + 129/(16 *a**4)), True)) - 2*I/(a**4*d*exp(2*I*c)*exp(2*I*d*x) - a**4*d) - 129*x/(1 6*a**4) - 4*I*log(exp(2*I*d*x) - exp(-2*I*c))/(a**4*d)
Exception generated. \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
Time = 1.15 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.77 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {12 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {1548 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} + \frac {1536 i \, \log \left (\tan \left (d x + c\right )\right )}{a^{4}} + \frac {384 \, {\left (-4 i \, \tan \left (d x + c\right ) + 1\right )}}{a^{4} \tan \left (d x + c\right )} + \frac {3225 i \, \tan \left (d x + c\right )^{4} + 14076 \, \tan \left (d x + c\right )^{3} - 23286 i \, \tan \left (d x + c\right )^{2} - 17404 \, \tan \left (d x + c\right ) + 5017 i}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]
-1/384*(12*I*log(tan(d*x + c) + I)/a^4 - 1548*I*log(tan(d*x + c) - I)/a^4 + 1536*I*log(tan(d*x + c))/a^4 + 384*(-4*I*tan(d*x + c) + 1)/(a^4*tan(d*x + c)) + (3225*I*tan(d*x + c)^4 + 14076*tan(d*x + c)^3 - 23286*I*tan(d*x + c)^2 - 17404*tan(d*x + c) + 5017*I)/(a^4*(tan(d*x + c) - I)^4))/d
Time = 4.62 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,129{}\mathrm {i}}{32\,a^4\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,a^4\,d}-\frac {\frac {1}{a^4}-\frac {851\,{\mathrm {tan}\left (c+d\,x\right )}^2}{48\,a^4}+\frac {65\,{\mathrm {tan}\left (c+d\,x\right )}^4}{16\,a^4}+\frac {\mathrm {tan}\left (c+d\,x\right )\,26{}\mathrm {i}}{3\,a^4}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,57{}\mathrm {i}}{4\,a^4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^5-{\mathrm {tan}\left (c+d\,x\right )}^4\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^3+{\mathrm {tan}\left (c+d\,x\right )}^2\,4{}\mathrm {i}+\mathrm {tan}\left (c+d\,x\right )\right )}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,4{}\mathrm {i}}{a^4\,d} \]
(log(tan(c + d*x) - 1i)*129i)/(32*a^4*d) - (log(tan(c + d*x) + 1i)*1i)/(32 *a^4*d) - ((tan(c + d*x)*26i)/(3*a^4) + 1/a^4 - (851*tan(c + d*x)^2)/(48*a ^4) - (tan(c + d*x)^3*57i)/(4*a^4) + (65*tan(c + d*x)^4)/(16*a^4))/(d*(tan (c + d*x) + tan(c + d*x)^2*4i - 6*tan(c + d*x)^3 - tan(c + d*x)^4*4i + tan (c + d*x)^5)) - (log(tan(c + d*x))*4i)/(a^4*d)