Integrand size = 24, antiderivative size = 147 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {15 i x}{16 a^4}-\frac {\log (\cos (c+d x))}{a^4 d}+\frac {15}{16 a^4 d (1+i \tan (c+d x))}+\frac {7 \tan ^2(c+d x)}{16 a^4 d (1+i \tan (c+d x))^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac {i \tan ^3(c+d x)}{4 a d (a+i a \tan (c+d x))^3} \]
15/16*I*x/a^4-ln(cos(d*x+c))/a^4/d+15/16/a^4/d/(1+I*tan(d*x+c))+7/16*tan(d *x+c)^2/a^4/d/(1+I*tan(d*x+c))^2-1/8*tan(d*x+c)^4/d/(a+I*a*tan(d*x+c))^4+1 /4*I*tan(d*x+c)^3/a/d/(a+I*a*tan(d*x+c))^3
Time = 0.59 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) (-16+56 \cos (2 (c+d x))+2 \cos (4 (c+d x)) (-20+31 \log (i-\tan (c+d x))+\log (i+\tan (c+d x)))+48 i \sin (2 (c+d x))+i (-39+62 \log (i-\tan (c+d x))+2 \log (i+\tan (c+d x))) \sin (4 (c+d x)))}{64 a^4 d (-i+\tan (c+d x))^4} \]
(Sec[c + d*x]^4*(-16 + 56*Cos[2*(c + d*x)] + 2*Cos[4*(c + d*x)]*(-20 + 31* Log[I - Tan[c + d*x]] + Log[I + Tan[c + d*x]]) + (48*I)*Sin[2*(c + d*x)] + I*(-39 + 62*Log[I - Tan[c + d*x]] + 2*Log[I + Tan[c + d*x]])*Sin[4*(c + d *x)]))/(64*a^4*d*(-I + Tan[c + d*x])^4)
Time = 0.99 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.16, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4041, 27, 3042, 4078, 27, 3042, 4078, 27, 3042, 4072, 27, 3042, 3956, 4009, 24}
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.
| \(\displaystyle \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^5}{(a+i a \tan (c+d x))^4}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle -\frac {\int -\frac {4 \tan ^3(c+d x) (a-2 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^3}dx}{8 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\tan ^3(c+d x) (a-2 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^3}dx}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\tan (c+d x)^3 (a-2 i a \tan (c+d x))}{(i \tan (c+d x) a+a)^3}dx}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {3 \tan ^2(c+d x) \left (4 \tan (c+d x) a^2+3 i a^2\right )}{(i \tan (c+d x) a+a)^2}dx}{6 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan ^2(c+d x) \left (4 \tan (c+d x) a^2+3 i a^2\right )}{(i \tan (c+d x) a+a)^2}dx}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\int \frac {\tan (c+d x)^2 \left (4 \tan (c+d x) a^2+3 i a^2\right )}{(i \tan (c+d x) a+a)^2}dx}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 4078 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {-\frac {\int -\frac {2 \tan (c+d x) \left (7 a^3-8 i a^3 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{4 a^2}-\frac {7 \tan ^2(c+d x)}{4 d (1+i \tan (c+d x))^2}}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\frac {\int \frac {\tan (c+d x) \left (7 a^3-8 i a^3 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {7 \tan ^2(c+d x)}{4 d (1+i \tan (c+d x))^2}}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\frac {\int \frac {\tan (c+d x) \left (7 a^3-8 i a^3 \tan (c+d x)\right )}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {7 \tan ^2(c+d x)}{4 d (1+i \tan (c+d x))^2}}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 4072 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\frac {-8 a^2 \int \tan (c+d x)dx-\frac {i \int \frac {15 i a^4 \tan (c+d x)}{i \tan (c+d x) a+a}dx}{a}}{2 a^2}-\frac {7 \tan ^2(c+d x)}{4 d (1+i \tan (c+d x))^2}}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\frac {-8 a^2 \int \tan (c+d x)dx+15 a^3 \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {7 \tan ^2(c+d x)}{4 d (1+i \tan (c+d x))^2}}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\frac {-8 a^2 \int \tan (c+d x)dx+15 a^3 \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {7 \tan ^2(c+d x)}{4 d (1+i \tan (c+d x))^2}}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {8 a^2 \log (\cos (c+d x))}{d}+15 a^3 \int \frac {\tan (c+d x)}{i \tan (c+d x) a+a}dx}{2 a^2}-\frac {7 \tan ^2(c+d x)}{4 d (1+i \tan (c+d x))^2}}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 4009 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {8 a^2 \log (\cos (c+d x))}{d}+15 a^3 \left (-\frac {i \int 1dx}{2 a}-\frac {1}{2 d (a+i a \tan (c+d x))}\right )}{2 a^2}-\frac {7 \tan ^2(c+d x)}{4 d (1+i \tan (c+d x))^2}}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {i a \tan ^3(c+d x)}{2 d (a+i a \tan (c+d x))^3}-\frac {\frac {\frac {8 a^2 \log (\cos (c+d x))}{d}+15 a^3 \left (-\frac {1}{2 d (a+i a \tan (c+d x))}-\frac {i x}{2 a}\right )}{2 a^2}-\frac {7 \tan ^2(c+d x)}{4 d (1+i \tan (c+d x))^2}}{2 a^2}}{2 a^2}-\frac {\tan ^4(c+d x)}{8 d (a+i a \tan (c+d x))^4}\) |
-1/8*Tan[c + d*x]^4/(d*(a + I*a*Tan[c + d*x])^4) + (((I/2)*a*Tan[c + d*x]^ 3)/(d*(a + I*a*Tan[c + d*x])^3) - ((-7*Tan[c + d*x]^2)/(4*d*(1 + I*Tan[c + d*x])^2) + ((8*a^2*Log[Cos[c + d*x]])/d + 15*a^3*(((-1/2*I)*x)/a - 1/(2*d *(a + I*a*Tan[c + d*x]))))/(2*a^2))/(2*a^2))/(2*a^2)
3.1.76.3.1 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/(2*a *f*m)), x] + Simp[(b*c + a*d)/(2*a*b) Int[(a + b*Tan[e + f*x])^(m + 1), x ], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 , 0] && LtQ[m, 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_)])*((c_.) + (d_.)*tan[(e_.) + (f_ .)*(x_)]))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*(d/ b) Int[Tan[e + f*x], x], x] + Simp[1/b Int[Simp[A*b*c + (A*b*d + B*(b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d , e, f, A, B}, x] && NeQ[b*c - a*d, 0]
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 = 0.43 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.73
| method | result | size |
| risch | \(\frac {31 i x}{16 a^{4}}+\frac {13 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{4} d}-\frac {{\mathrm e}^{-4 i \left (d x +c \right )}}{4 a^{4} d}+\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{16 a^{4} d}-\frac {{\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}+\frac {2 i c}{a^{4} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{4} d}\) | \(107\) |
| derivativedivides | \(\frac {3 i}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {31}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}+\frac {15 i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}\) | \(115\) |
| default | \(\frac {3 i}{4 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {49 i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}+\frac {31}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}+\frac {15 i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}\) | \(115\) |
| norman | \(\frac {\frac {5 \left (\tan ^{6}\left (d x +c \right )\right )}{a d}+\frac {15 i x}{16 a}+\frac {7}{4 a d}+\frac {35 \left (\tan ^{4}\left (d x +c \right )\right )}{4 a d}-\frac {15 i \tan \left (d x +c \right )}{16 d a}-\frac {55 i \left (\tan ^{3}\left (d x +c \right )\right )}{16 d a}-\frac {73 i \left (\tan ^{5}\left (d x +c \right )\right )}{16 a d}-\frac {49 i \left (\tan ^{7}\left (d x +c \right )\right )}{16 a d}+\frac {15 i x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}+\frac {45 i x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}+\frac {15 i x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}+\frac {15 i x \left (\tan ^{8}\left (d x +c \right )\right )}{16 a}+\frac {13 \left (\tan ^{2}\left (d x +c \right )\right )}{2 a d}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{4} a^{3}}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{4}}\) | \(227\) |
31/16*I*x/a^4+13/16/a^4/d*exp(-2*I*(d*x+c))-1/4/a^4/d*exp(-4*I*(d*x+c))+1/ 16/a^4/d*exp(-6*I*(d*x+c))-1/128/a^4/d*exp(-8*I*(d*x+c))+2*I/a^4/d*c-1/a^4 /d*ln(exp(2*I*(d*x+c))+1)
Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.60 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (248 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 128 \, e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 104 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 32 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{128 \, a^{4} d} \]
1/128*(248*I*d*x*e^(8*I*d*x + 8*I*c) - 128*e^(8*I*d*x + 8*I*c)*log(e^(2*I* d*x + 2*I*c) + 1) + 104*e^(6*I*d*x + 6*I*c) - 32*e^(4*I*d*x + 4*I*c) + 8*e ^(2*I*d*x + 2*I*c) - 1)*e^(-8*I*d*x - 8*I*c)/(a^4*d)
Time = 1.35 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.49 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (106496 a^{12} d^{3} e^{18 i c} e^{- 2 i d x} - 32768 a^{12} d^{3} e^{16 i c} e^{- 4 i d x} + 8192 a^{12} d^{3} e^{14 i c} e^{- 6 i d x} - 1024 a^{12} d^{3} e^{12 i c} e^{- 8 i d x}\right ) e^{- 20 i c}}{131072 a^{16} d^{4}} & \text {for}\: a^{16} d^{4} e^{20 i c} \neq 0 \\x \left (\frac {\left (31 i e^{8 i c} - 26 i e^{6 i c} + 16 i e^{4 i c} - 6 i e^{2 i c} + i\right ) e^{- 8 i c}}{16 a^{4}} - \frac {31 i}{16 a^{4}}\right ) & \text {otherwise} \end {cases} + \frac {31 i x}{16 a^{4}} - \frac {\log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{4} d} \]
Piecewise(((106496*a**12*d**3*exp(18*I*c)*exp(-2*I*d*x) - 32768*a**12*d**3 *exp(16*I*c)*exp(-4*I*d*x) + 8192*a**12*d**3*exp(14*I*c)*exp(-6*I*d*x) - 1 024*a**12*d**3*exp(12*I*c)*exp(-8*I*d*x))*exp(-20*I*c)/(131072*a**16*d**4) , Ne(a**16*d**4*exp(20*I*c), 0)), (x*((31*I*exp(8*I*c) - 26*I*exp(6*I*c) + 16*I*exp(4*I*c) - 6*I*exp(2*I*c) + I)*exp(-8*I*c)/(16*a**4) - 31*I/(16*a* *4)), True)) + 31*I*x/(16*a**4) - log(exp(2*I*d*x) + exp(-2*I*c))/(a**4*d)
Exception generated. \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
Time = 2.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.61 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {12 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} + \frac {372 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac {775 \, \tan \left (d x + c\right )^{4} - 1924 i \, \tan \left (d x + c\right )^{3} - 1866 \, \tan \left (d x + c\right )^{2} + 772 i \, \tan \left (d x + c\right ) + 103}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]
1/384*(12*log(tan(d*x + c) + I)/a^4 + 372*log(tan(d*x + c) - I)/a^4 - (775 *tan(d*x + c)^4 - 1924*I*tan(d*x + c)^3 - 1866*tan(d*x + c)^2 + 772*I*tan( d*x + c) + 103)/(a^4*(tan(d*x + c) - I)^4))/d
Time = 4.58 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.87 \[ \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {31\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{32\,a^4\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{32\,a^4\,d}+\frac {\frac {\mathrm {tan}\left (c+d\,x\right )\,97{}\mathrm {i}}{16\,a^4}+\frac {7}{4\,a^4}-\frac {29\,{\mathrm {tan}\left (c+d\,x\right )}^2}{4\,a^4}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,49{}\mathrm {i}}{16\,a^4}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,4{}\mathrm {i}-6\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,4{}\mathrm {i}+1\right )} \]