Integrand size = 24, antiderivative size = 126 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=4 i a^3 x+\frac {4 a^3 \log (\cos (c+d x))}{d}-\frac {4 i a^3 \tan (c+d x)}{d}+\frac {2 a^3 \tan ^2(c+d x)}{d}+\frac {4 i a^3 \tan ^3(c+d x)}{3 d}-\frac {11 a^3 \tan ^4(c+d x)}{20 d}-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d} \]
4*I*a^3*x+4*a^3*ln(cos(d*x+c))/d-4*I*a^3*tan(d*x+c)/d+2*a^3*tan(d*x+c)^2/d +4/3*I*a^3*tan(d*x+c)^3/d-11/20*a^3*tan(d*x+c)^4/d-1/5*tan(d*x+c)^4*(a^3+I *a^3*tan(d*x+c))/d
Time = 0.55 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {4 a^3 \log (i+\tan (c+d x))}{d}-\frac {4 i a^3 \tan (c+d x)}{d}+\frac {2 a^3 \tan ^2(c+d x)}{d}+\frac {4 i a^3 \tan ^3(c+d x)}{3 d}-\frac {3 a^3 \tan ^4(c+d x)}{4 d}-\frac {i a^3 \tan ^5(c+d x)}{5 d} \]
(-4*a^3*Log[I + Tan[c + d*x]])/d - ((4*I)*a^3*Tan[c + d*x])/d + (2*a^3*Tan [c + d*x]^2)/d + (((4*I)/3)*a^3*Tan[c + d*x]^3)/d - (3*a^3*Tan[c + d*x]^4) /(4*d) - ((I/5)*a^3*Tan[c + d*x]^5)/d
Time = 0.79 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4039, 3042, 4075, 3042, 4011, 3042, 4011, 3042, 4008, 3042, 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.
\(\displaystyle \int \tan ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^3 (a+i a \tan (c+d x))^3dx\) |
\(\Big \downarrow \) 4039 |
\(\displaystyle \frac {1}{5} a \int \tan ^3(c+d x) (i \tan (c+d x) a+a) (11 i \tan (c+d x) a+9 a)dx-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} a \int \tan (c+d x)^3 (i \tan (c+d x) a+a) (11 i \tan (c+d x) a+9 a)dx-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 4075 |
\(\displaystyle \frac {1}{5} a \left (-\frac {11 a^2 \tan ^4(c+d x)}{4 d}+\int \tan ^3(c+d x) \left (20 i \tan (c+d x) a^2+20 a^2\right )dx\right )-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} a \left (-\frac {11 a^2 \tan ^4(c+d x)}{4 d}+\int \tan (c+d x)^3 \left (20 i \tan (c+d x) a^2+20 a^2\right )dx\right )-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {1}{5} a \left (\int \tan ^2(c+d x) \left (20 a^2 \tan (c+d x)-20 i a^2\right )dx-\frac {11 a^2 \tan ^4(c+d x)}{4 d}+\frac {20 i a^2 \tan ^3(c+d x)}{3 d}\right )-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} a \left (\int \tan (c+d x)^2 \left (20 a^2 \tan (c+d x)-20 i a^2\right )dx-\frac {11 a^2 \tan ^4(c+d x)}{4 d}+\frac {20 i a^2 \tan ^3(c+d x)}{3 d}\right )-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \frac {1}{5} a \left (\int \tan (c+d x) \left (-20 i \tan (c+d x) a^2-20 a^2\right )dx-\frac {11 a^2 \tan ^4(c+d x)}{4 d}+\frac {20 i a^2 \tan ^3(c+d x)}{3 d}+\frac {10 a^2 \tan ^2(c+d x)}{d}\right )-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} a \left (\int \tan (c+d x) \left (-20 i \tan (c+d x) a^2-20 a^2\right )dx-\frac {11 a^2 \tan ^4(c+d x)}{4 d}+\frac {20 i a^2 \tan ^3(c+d x)}{3 d}+\frac {10 a^2 \tan ^2(c+d x)}{d}\right )-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle \frac {1}{5} a \left (-20 a^2 \int \tan (c+d x)dx-\frac {11 a^2 \tan ^4(c+d x)}{4 d}+\frac {20 i a^2 \tan ^3(c+d x)}{3 d}+\frac {10 a^2 \tan ^2(c+d x)}{d}-\frac {20 i a^2 \tan (c+d x)}{d}+20 i a^2 x\right )-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} a \left (-20 a^2 \int \tan (c+d x)dx-\frac {11 a^2 \tan ^4(c+d x)}{4 d}+\frac {20 i a^2 \tan ^3(c+d x)}{3 d}+\frac {10 a^2 \tan ^2(c+d x)}{d}-\frac {20 i a^2 \tan (c+d x)}{d}+20 i a^2 x\right )-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {1}{5} a \left (-\frac {11 a^2 \tan ^4(c+d x)}{4 d}+\frac {20 i a^2 \tan ^3(c+d x)}{3 d}+\frac {10 a^2 \tan ^2(c+d x)}{d}-\frac {20 i a^2 \tan (c+d x)}{d}+\frac {20 a^2 \log (\cos (c+d x))}{d}+20 i a^2 x\right )-\frac {\tan ^4(c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{5 d}\) |
-1/5*(Tan[c + d*x]^4*(a^3 + I*a^3*Tan[c + d*x]))/d + (a*((20*I)*a^2*x + (2 0*a^2*Log[Cos[c + d*x]])/d - ((20*I)*a^2*Tan[c + d*x])/d + (10*a^2*Tan[c + d*x]^2)/d + (((20*I)/3)*a^2*Tan[c + d*x]^3)/d - (11*a^2*Tan[c + d*x]^4)/( 4*d)))/5
3.1.24.3.1 Defintions of rubi rules used
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_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(a*c - b*d)*x, x] + (Simp[b*d*(Tan[e + f*x]/f), x] + Simp[(b*c + a*d) Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*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] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 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_)]), x_Symbol] :> Simp[B *d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1]
Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66
method | result | size |
derivativedivides | \(\frac {a^{3} \left (-4 i \tan \left (d x +c \right )-\frac {i \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {3 \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {4 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+2 \left (\tan ^{2}\left (d x +c \right )\right )-2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+4 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(83\) |
default | \(\frac {a^{3} \left (-4 i \tan \left (d x +c \right )-\frac {i \left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {3 \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {4 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+2 \left (\tan ^{2}\left (d x +c \right )\right )-2 \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+4 i \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(83\) |
parallelrisch | \(\frac {-12 i a^{3} \left (\tan ^{5}\left (d x +c \right )\right )+80 i a^{3} \left (\tan ^{3}\left (d x +c \right )\right )-45 \left (\tan ^{4}\left (d x +c \right )\right ) a^{3}+240 i a^{3} x d -240 i a^{3} \tan \left (d x +c \right )+120 a^{3} \left (\tan ^{2}\left (d x +c \right )\right )-120 a^{3} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{60 d}\) | \(97\) |
risch | \(-\frac {8 i a^{3} c}{d}+\frac {2 a^{3} \left (240 \,{\mathrm e}^{8 i \left (d x +c \right )}+585 \,{\mathrm e}^{6 i \left (d x +c \right )}+695 \,{\mathrm e}^{4 i \left (d x +c \right )}+385 \,{\mathrm e}^{2 i \left (d x +c \right )}+83\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(99\) |
norman | \(4 i a^{3} x +\frac {2 a^{3} \left (\tan ^{2}\left (d x +c \right )\right )}{d}-\frac {3 a^{3} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {4 i a^{3} \tan \left (d x +c \right )}{d}+\frac {4 i a^{3} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {i a^{3} \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}-\frac {2 a^{3} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) | \(109\) |
parts | \(\frac {a^{3} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}-\frac {i a^{3} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 i a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}-\frac {3 a^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}\) | \(155\) |
1/d*a^3*(-4*I*tan(d*x+c)-1/5*I*tan(d*x+c)^5-3/4*tan(d*x+c)^4+4/3*I*tan(d*x +c)^3+2*tan(d*x+c)^2-2*ln(1+tan(d*x+c)^2)+4*I*arctan(tan(d*x+c)))
Time = 0.24 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.70 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {2 \, {\left (240 \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 585 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 695 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 385 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 83 \, a^{3} + 30 \, {\left (a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
2/15*(240*a^3*e^(8*I*d*x + 8*I*c) + 585*a^3*e^(6*I*d*x + 6*I*c) + 695*a^3* e^(4*I*d*x + 4*I*c) + 385*a^3*e^(2*I*d*x + 2*I*c) + 83*a^3 + 30*(a^3*e^(10 *I*d*x + 10*I*c) + 5*a^3*e^(8*I*d*x + 8*I*c) + 10*a^3*e^(6*I*d*x + 6*I*c) + 10*a^3*e^(4*I*d*x + 4*I*c) + 5*a^3*e^(2*I*d*x + 2*I*c) + a^3)*log(e^(2*I *d*x + 2*I*c) + 1))/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 1 0*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I* c) + d)
Time = 0.31 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.64 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {4 a^{3} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {480 a^{3} e^{8 i c} e^{8 i d x} + 1170 a^{3} e^{6 i c} e^{6 i d x} + 1390 a^{3} e^{4 i c} e^{4 i d x} + 770 a^{3} e^{2 i c} e^{2 i d x} + 166 a^{3}}{15 d e^{10 i c} e^{10 i d x} + 75 d e^{8 i c} e^{8 i d x} + 150 d e^{6 i c} e^{6 i d x} + 150 d e^{4 i c} e^{4 i d x} + 75 d e^{2 i c} e^{2 i d x} + 15 d} \]
4*a**3*log(exp(2*I*d*x) + exp(-2*I*c))/d + (480*a**3*exp(8*I*c)*exp(8*I*d* x) + 1170*a**3*exp(6*I*c)*exp(6*I*d*x) + 1390*a**3*exp(4*I*c)*exp(4*I*d*x) + 770*a**3*exp(2*I*c)*exp(2*I*d*x) + 166*a**3)/(15*d*exp(10*I*c)*exp(10*I *d*x) + 75*d*exp(8*I*c)*exp(8*I*d*x) + 150*d*exp(6*I*c)*exp(6*I*d*x) + 150 *d*exp(4*I*c)*exp(4*I*d*x) + 75*d*exp(2*I*c)*exp(2*I*d*x) + 15*d)
Time = 0.70 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.75 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {12 i \, a^{3} \tan \left (d x + c\right )^{5} + 45 \, a^{3} \tan \left (d x + c\right )^{4} - 80 i \, a^{3} \tan \left (d x + c\right )^{3} - 120 \, a^{3} \tan \left (d x + c\right )^{2} - 240 i \, {\left (d x + c\right )} a^{3} + 120 \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 240 i \, a^{3} \tan \left (d x + c\right )}{60 \, d} \]
-1/60*(12*I*a^3*tan(d*x + c)^5 + 45*a^3*tan(d*x + c)^4 - 80*I*a^3*tan(d*x + c)^3 - 120*a^3*tan(d*x + c)^2 - 240*I*(d*x + c)*a^3 + 120*a^3*log(tan(d* x + c)^2 + 1) + 240*I*a^3*tan(d*x + c))/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 274 vs. \(2 (112) = 224\).
Time = 0.77 (sec) , antiderivative size = 274, normalized size of antiderivative = 2.17 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {2 \, {\left (30 \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 150 \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 300 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 300 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 150 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 240 \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} + 585 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 695 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 385 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 30 \, a^{3} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 83 \, a^{3}\right )}}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
2/15*(30*a^3*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 150*a^3* e^(8*I*d*x + 8*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 300*a^3*e^(6*I*d*x + 6* I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 300*a^3*e^(4*I*d*x + 4*I*c)*log(e^(2*I *d*x + 2*I*c) + 1) + 150*a^3*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 240*a^3*e^(8*I*d*x + 8*I*c) + 585*a^3*e^(6*I*d*x + 6*I*c) + 695*a^3* e^(4*I*d*x + 4*I*c) + 385*a^3*e^(2*I*d*x + 2*I*c) + 30*a^3*log(e^(2*I*d*x + 2*I*c) + 1) + 83*a^3)/(d*e^(10*I*d*x + 10*I*c) + 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) + 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) + d)
Time = 4.50 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.69 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {4\,a^3\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )+a^3\,\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}-2\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^2-\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}}{3}+\frac {3\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}+\frac {a^3\,{\mathrm {tan}\left (c+d\,x\right )}^5\,1{}\mathrm {i}}{5}}{d} \]