Integrand size = 24, antiderivative size = 171 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {65 x}{16 a^4}-\frac {4 i \log (\cos (c+d x))}{a^4 d}+\frac {65 \tan (c+d x)}{16 a^4 d}-\frac {2 i \tan ^2(c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac {31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]
-65/16*x/a^4-4*I*ln(cos(d*x+c))/a^4/d+65/16*tan(d*x+c)/a^4/d-2*I*tan(d*x+c )^2/a^4/d/(1+I*tan(d*x+c))+31/48*tan(d*x+c)^3/a^4/d/(1+I*tan(d*x+c))^2-1/8 *tan(d*x+c)^5/d/(a+I*a*tan(d*x+c))^4+7/24*I*tan(d*x+c)^4/a/d/(a+I*a*tan(d* x+c))^3
Time = 0.82 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.55 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i \sec ^5(c+d x) (268 \cos (c+d x)-550 \cos (5 (c+d x))+774 \cos (5 (c+d x)) \log (i-\tan (c+d x))+6 \cos (3 (c+d x)) (47+129 \log (i-\tan (c+d x))-\log (i+\tan (c+d x)))-6 \cos (5 (c+d x)) \log (i+\tan (c+d x))+416 i \sin (c+d x)+253 i \sin (3 (c+d x))+774 i \log (i-\tan (c+d x)) \sin (3 (c+d x))-6 i \log (i+\tan (c+d x)) \sin (3 (c+d x))-547 i \sin (5 (c+d x))+774 i \log (i-\tan (c+d x)) \sin (5 (c+d x))-6 i \log (i+\tan (c+d x)) \sin (5 (c+d x)))}{384 a^4 d (-i+\tan (c+d x))^4} \]
((I/384)*Sec[c + d*x]^5*(268*Cos[c + d*x] - 550*Cos[5*(c + d*x)] + 774*Cos [5*(c + d*x)]*Log[I - Tan[c + d*x]] + 6*Cos[3*(c + d*x)]*(47 + 129*Log[I - Tan[c + d*x]] - Log[I + Tan[c + d*x]]) - 6*Cos[5*(c + d*x)]*Log[I + Tan[c + d*x]] + (416*I)*Sin[c + d*x] + (253*I)*Sin[3*(c + d*x)] + (774*I)*Log[I - Tan[c + d*x]]*Sin[3*(c + d*x)] - (6*I)*Log[I + Tan[c + d*x]]*Sin[3*(c + d*x)] - (547*I)*Sin[5*(c + d*x)] + (774*I)*Log[I - Tan[c + d*x]]*Sin[5*(c + d*x)] - (6*I)*Log[I + Tan[c + d*x]]*Sin[5*(c + d*x)]))/(a^4*d*(-I + Tan [c + d*x])^4)
Time = 1.12 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.13, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4041, 25, 3042, 4078, 27, 3042, 4078, 27, 3042, 4078, 27, 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.
| \(\displaystyle \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^6}{(a+i a \tan (c+d x))^4}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {\tan ^4(c+d x) (5 a-9 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^3}dx}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\tan ^4(c+d x) (5 a-9 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^3}dx}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\tan (c+d x)^4 (5 a-9 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^3}dx}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {4 \tan ^3(c+d x) \left (17 \tan (c+d x) a^2+14 i a^2\right )}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \int \frac {\tan ^3(c+d x) \left (17 \tan (c+d x) a^2+14 i a^2\right )}{(i \tan (c+d x) a+a)^2}dx}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \int \frac {\tan (c+d x)^3 \left (17 \tan (c+d x) a^2+14 i a^2\right )}{(i \tan (c+d x) a+a)^2}dx}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \left (-\frac {\int -\frac {3 \tan ^2(c+d x) \left (31 a^3-33 i a^3 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {31 \tan ^3(c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \left (\frac {3 \int \frac {\tan ^2(c+d x) \left (31 a^3-33 i a^3 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {31 \tan ^3(c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \left (\frac {3 \int \frac {\tan (c+d x)^2 \left (31 a^3-33 i a^3 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {31 \tan ^3(c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \left (\frac {3 \left (\frac {32 i a^3 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int 2 \tan (c+d x) \left (65 \tan (c+d x) a^4+64 i a^4\right )dx}{2 a^2}\right )}{4 a^2}-\frac {31 \tan ^3(c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \left (\frac {3 \left (\frac {32 i a^3 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int \tan (c+d x) \left (65 \tan (c+d x) a^4+64 i a^4\right )dx}{a^2}\right )}{4 a^2}-\frac {31 \tan ^3(c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \left (\frac {3 \left (\frac {32 i a^3 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\int \tan (c+d x) \left (65 \tan (c+d x) a^4+64 i a^4\right )dx}{a^2}\right )}{4 a^2}-\frac {31 \tan ^3(c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 4008 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \left (\frac {3 \left (\frac {32 i a^3 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}-\frac {64 i a^4 \int \tan (c+d x)dx+\frac {65 a^4 \tan (c+d x)}{d}-65 a^4 x}{a^2}\right )}{4 a^2}-\frac {31 \tan ^3(c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \left (\frac {3 \left (\frac {32 i a^3 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}-\frac {64 i a^4 \int \tan (c+d x)dx+\frac {65 a^4 \tan (c+d x)}{d}-65 a^4 x}{a^2}\right )}{4 a^2}-\frac {31 \tan ^3(c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle \frac {\frac {7 i a \tan ^4(c+d x)}{3 d (a+i a \tan (c+d x))^3}-\frac {2 \left (\frac {3 \left (\frac {32 i a^3 \tan ^2(c+d x)}{d (a+i a \tan (c+d x))}-\frac {\frac {65 a^4 \tan (c+d x)}{d}-\frac {64 i a^4 \log (\cos (c+d x))}{d}-65 a^4 x}{a^2}\right )}{4 a^2}-\frac {31 \tan ^3(c+d x)}{4 d (1+i \tan (c+d x))^2}\right )}{3 a^2}}{8 a^2}-\frac {\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
-1/8*Tan[c + d*x]^5/(d*(a + I*a*Tan[c + d*x])^4) + ((((7*I)/3)*a*Tan[c + d *x]^4)/(d*(a + I*a*Tan[c + d*x])^3) - (2*((-31*Tan[c + d*x]^3)/(4*d*(1 + I *Tan[c + d*x])^2) + (3*(((32*I)*a^3*Tan[c + d*x]^2)/(d*(a + I*a*Tan[c + d* x])) - (-65*a^4*x - ((64*I)*a^4*Log[Cos[c + d*x]])/d + (65*a^4*Tan[c + d*x ])/d)/a^2))/(4*a^2)))/(3*a^2))/(8*a^2)
3.1.75.3.1 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_)])*((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_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && GtQ[n, 1] && (In tegerQ[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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f* x])^(n - 1)*Simp[A*(a*c*m + b*d*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a *A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 0]
Time = 0.39 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.75
| method | result | size |
| derivativedivides | \(\frac {\tan \left (d x +c \right )}{a^{4} d}+\frac {49 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {11}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {111}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}\) | \(128\) |
| default | \(\frac {\tan \left (d x +c \right )}{a^{4} d}+\frac {49 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}-\frac {65 \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}-\frac {11}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {111}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}\) | \(128\) |
| risch | \(-\frac {129 x}{16 a^{4}}+\frac {9 i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{4} d}-\frac {15 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{4} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{12 a^{4} d}-\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}-\frac {8 c}{a^{4} d}+\frac {2 i}{d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {4 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{4} d}\) | \(132\) |
| norman | \(\frac {\frac {\tan ^{9}\left (d x +c \right )}{a d}-\frac {65 x}{16 a}+\frac {949 \left (\tan ^{5}\left (d x +c \right )\right )}{48 a d}+\frac {175 \left (\tan ^{7}\left (d x +c \right )\right )}{16 a d}-\frac {65 x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}-\frac {195 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}-\frac {65 x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}-\frac {65 x \left (\tan ^{8}\left (d x +c \right )\right )}{16 a}+\frac {14 i}{3 a d}+\frac {65 \tan \left (d x +c \right )}{16 a d}+\frac {715 \left (\tan ^{3}\left (d x +c \right )\right )}{48 a d}+\frac {10 i \left (\tan ^{6}\left (d x +c \right )\right )}{a d}+\frac {21 i \left (\tan ^{4}\left (d x +c \right )\right )}{d a}+\frac {50 i \left (\tan ^{2}\left (d x +c \right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{4} a^{3}}+\frac {2 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d \,a^{4}}\) | \(238\) |
tan(d*x+c)/a^4/d+49/16*I/d/a^4/(tan(d*x+c)-I)^2-1/8*I/d/a^4/(tan(d*x+c)-I) ^4+2*I/d/a^4*ln(1+tan(d*x+c)^2)-65/16/d/a^4*arctan(tan(d*x+c))-11/12/d/a^4 /(tan(d*x+c)-I)^3+111/16/d/a^4/(tan(d*x+c)-I)
Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.79 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {3096 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} + 24 \, {\left (129 \, d x - 68 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 1536 \, {\left (i \, e^{\left (10 i \, d x + 10 i \, c\right )} + i \, e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 684 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 148 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 29 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i}{384 \, {\left (a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )}\right )}} \]
-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 = 0.40 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.45 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (442368 i a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 92160 i a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 16384 i a^{12} d^{3} e^{14 i c} e^{- 6 i d x} - 1536 i a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{196608 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (- 129 e^{8 i c} + 72 e^{6 i c} - 30 e^{4 i c} + 8 e^{2 i c} - 1\right ) e^{- 8 i c}}{16 a^{4}} + \frac {129}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {2 i}{a^{4} d e^{2 i c} e^{2 i d x} + a^{4} d} - \frac {129 x}{16 a^{4}} - \frac {4 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{4} d} \]
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/(16 *a**4) - 4*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**4*d)
Exception generated. \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
Time = 2.70 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.58 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {\frac {12 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {1548 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {384 \, \tan \left (d x + c\right )}{a^{4}} - \frac {-3225 i \, \tan \left (d x + c\right )^{4} - 10236 \, \tan \left (d x + c\right )^{3} + 12534 i \, \tan \left (d x + c\right )^{2} + 6908 \, \tan \left (d x + c\right ) - 1433 i}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]
-1/384*(12*I*log(tan(d*x + c) + I)/a^4 - 1548*I*log(tan(d*x + c) - I)/a^4 - 384*tan(d*x + c)/a^4 - (-3225*I*tan(d*x + c)^4 - 10236*tan(d*x + c)^3 + 12534*I*tan(d*x + c)^2 + 6908*tan(d*x + c) - 1433*I)/(a^4*(tan(d*x + c) - I)^4))/d
Time = 5.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.77 \[ \int \frac {\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )}{a^4\,d}-\frac {65\,x}{16\,a^4}+\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,2{}\mathrm {i}}{a^4\,d}-\frac {\frac {749\,\mathrm {tan}\left (c+d\,x\right )}{48\,a^4}-\frac {111\,{\mathrm {tan}\left (c+d\,x\right )}^3}{16\,a^4}-\frac {14{}\mathrm {i}}{3\,a^4}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,71{}\mathrm {i}}{4\,a^4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )} \]