Integrand size = 22, antiderivative size = 97 \[ \int \cos (c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {15 a^4 \text {arctanh}(\sin (c+d x))}{2 d}-\frac {15 i a^4 \sec (c+d x)}{2 d}-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}-\frac {5 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{2 d} \]
-15/2*a^4*arctanh(sin(d*x+c))/d-15/2*I*a^4*sec(d*x+c)/d-2*I*a*cos(d*x+c)*( a+I*a*tan(d*x+c))^3/d-5/2*I*sec(d*x+c)*(a^4+I*a^4*tan(d*x+c))/d
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(906\) vs. \(2(97)=194\).
Time = 6.77 (sec) , antiderivative size = 906, normalized size of antiderivative = 9.34 \[ \int \cos (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {15 \cos (4 c) \cos ^4(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) (a+i a \tan (c+d x))^4}{2 d (\cos (d x)+i \sin (d x))^4}-\frac {15 \cos (4 c) \cos ^4(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) (a+i a \tan (c+d x))^4}{2 d (\cos (d x)+i \sin (d x))^4}+\frac {\cos (d x) \cos ^4(c+d x) (-8 i \cos (3 c)-8 \sin (3 c)) (a+i a \tan (c+d x))^4}{d (\cos (d x)+i \sin (d x))^4}+\frac {\cos ^4(c+d x) \sec (c) (-4 i \cos (4 c)-4 \sin (4 c)) (a+i a \tan (c+d x))^4}{d (\cos (d x)+i \sin (d x))^4}-\frac {15 i \cos ^4(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sin (4 c) (a+i a \tan (c+d x))^4}{2 d (\cos (d x)+i \sin (d x))^4}+\frac {15 i \cos ^4(c+d x) \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right ) \sin (4 c) (a+i a \tan (c+d x))^4}{2 d (\cos (d x)+i \sin (d x))^4}+\frac {\cos ^4(c+d x) (8 \cos (3 c)-8 i \sin (3 c)) \sin (d x) (a+i a \tan (c+d x))^4}{d (\cos (d x)+i \sin (d x))^4}+\frac {\cos ^4(c+d x) \left (\frac {1}{4} \cos (4 c)-\frac {1}{4} i \sin (4 c)\right ) (a+i a \tan (c+d x))^4}{d (\cos (d x)+i \sin (d x))^4 \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}-\frac {i \cos ^4(c+d x) (4 \cos (4 c)-4 i \sin (4 c)) \sin \left (\frac {d x}{2}\right ) (a+i a \tan (c+d x))^4}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) (\cos (d x)+i \sin (d x))^4 \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\cos ^4(c+d x) \left (-\frac {1}{4} \cos (4 c)+\frac {1}{4} i \sin (4 c)\right ) (a+i a \tan (c+d x))^4}{d (\cos (d x)+i \sin (d x))^4 \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {i \cos ^4(c+d x) (4 \cos (4 c)-4 i \sin (4 c)) \sin \left (\frac {d x}{2}\right ) (a+i a \tan (c+d x))^4}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) (\cos (d x)+i \sin (d x))^4 \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )} \]
(15*Cos[4*c]*Cos[c + d*x]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*( a + I*a*Tan[c + d*x])^4)/(2*d*(Cos[d*x] + I*Sin[d*x])^4) - (15*Cos[4*c]*Co s[c + d*x]^4*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(a + I*a*Tan[c + d*x])^4)/(2*d*(Cos[d*x] + I*Sin[d*x])^4) + (Cos[d*x]*Cos[c + d*x]^4*((-8* I)*Cos[3*c] - 8*Sin[3*c])*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d *x])^4) + (Cos[c + d*x]^4*Sec[c]*((-4*I)*Cos[4*c] - 4*Sin[4*c])*(a + I*a*T an[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4) - (((15*I)/2)*Cos[c + d*x]^4 *Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sin[4*c]*(a + I*a*Tan[c + d* x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4) + (((15*I)/2)*Cos[c + d*x]^4*Log[Cos[ c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sin[4*c]*(a + I*a*Tan[c + d*x])^4)/(d *(Cos[d*x] + I*Sin[d*x])^4) + (Cos[c + d*x]^4*(8*Cos[3*c] - (8*I)*Sin[3*c] )*Sin[d*x]*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4) + (Cos[ c + d*x]^4*(Cos[4*c]/4 - (I/4)*Sin[4*c])*(a + I*a*Tan[c + d*x])^4)/(d*(Cos [d*x] + I*Sin[d*x])^4*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) - (I*Co s[c + d*x]^4*(4*Cos[4*c] - (4*I)*Sin[4*c])*Sin[(d*x)/2]*(a + I*a*Tan[c + d *x])^4)/(d*(Cos[c/2] - Sin[c/2])*(Cos[d*x] + I*Sin[d*x])^4*(Cos[c/2 + (d*x )/2] - Sin[c/2 + (d*x)/2])) + (Cos[c + d*x]^4*(-1/4*Cos[4*c] + (I/4)*Sin[4 *c])*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[d*x] + I*Sin[d*x])^4*(Cos[c/2 + (d* x)/2] + Sin[c/2 + (d*x)/2])^2) + (I*Cos[c + d*x]^4*(4*Cos[4*c] - (4*I)*Sin [4*c])*Sin[(d*x)/2]*(a + I*a*Tan[c + d*x])^4)/(d*(Cos[c/2] + Sin[c/2])*...
Time = 0.49 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3977, 3042, 3979, 3042, 3967, 3042, 4257}
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.
| \(\displaystyle \int \cos (c+d x) (a+i a \tan (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (c+d x))^4}{\sec (c+d x)}dx\) |
| \(\Big \downarrow \) 3977 |
| \(\displaystyle -5 a^2 \int \sec (c+d x) (i \tan (c+d x) a+a)^2dx-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -5 a^2 \int \sec (c+d x) (i \tan (c+d x) a+a)^2dx-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle -5 a^2 \left (\frac {3}{2} a \int \sec (c+d x) (i \tan (c+d x) a+a)dx+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -5 a^2 \left (\frac {3}{2} a \int \sec (c+d x) (i \tan (c+d x) a+a)dx+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle -5 a^2 \left (\frac {3}{2} a \left (a \int \sec (c+d x)dx+\frac {i a \sec (c+d x)}{d}\right )+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -5 a^2 \left (\frac {3}{2} a \left (a \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {i a \sec (c+d x)}{d}\right )+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -5 a^2 \left (\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}+\frac {3}{2} a \left (\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d}\right )\right )-\frac {2 i a \cos (c+d x) (a+i a \tan (c+d x))^3}{d}\) |
((-2*I)*a*Cos[c + d*x]*(a + I*a*Tan[c + d*x])^3)/d - 5*a^2*((3*a*((a*ArcTa nh[Sin[c + d*x]])/d + (I*a*Sec[c + d*x])/d))/2 + ((I/2)*Sec[c + d*x]*(a^2 + I*a^2*Tan[c + d*x]))/d)
3.1.54.3.1 Defintions of rubi rules used
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m)) Int[(d*Sec[e + f*x]) ^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & & LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Simp[a*((m + 2*n - 2)/(m + n - 1)) Int[(d*Se c[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ [2*m, 2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 3.19 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.10
| method | result | size |
| risch | \(-\frac {8 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i a^{4} \left (9 \,{\mathrm e}^{3 i \left (d x +c \right )}+7 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {15 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {15 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(107\) |
| derivativedivides | \(\frac {a^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-4 i a^{4} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )-6 a^{4} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-4 i a^{4} \cos \left (d x +c \right )+a^{4} \sin \left (d x +c \right )}{d}\) | \(154\) |
| default | \(\frac {a^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-4 i a^{4} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )-6 a^{4} \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )-4 i a^{4} \cos \left (d x +c \right )+a^{4} \sin \left (d x +c \right )}{d}\) | \(154\) |
-8*I*a^4/d*exp(I*(d*x+c))-I*a^4/d/(exp(2*I*(d*x+c))+1)^2*(9*exp(3*I*(d*x+c ))+7*exp(I*(d*x+c)))-15/2*a^4/d*ln(exp(I*(d*x+c))+I)+15/2*a^4/d*ln(exp(I*( d*x+c))-I)
Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.67 \[ \int \cos (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {-16 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 50 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 30 i \, a^{4} e^{\left (i \, d x + i \, c\right )} - 15 \, {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) + 15 \, {\left (a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{2 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
1/2*(-16*I*a^4*e^(5*I*d*x + 5*I*c) - 50*I*a^4*e^(3*I*d*x + 3*I*c) - 30*I*a ^4*e^(I*d*x + I*c) - 15*(a^4*e^(4*I*d*x + 4*I*c) + 2*a^4*e^(2*I*d*x + 2*I* c) + a^4)*log(e^(I*d*x + I*c) + I) + 15*(a^4*e^(4*I*d*x + 4*I*c) + 2*a^4*e ^(2*I*d*x + 2*I*c) + a^4)*log(e^(I*d*x + I*c) - I))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
Time = 0.24 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.58 \[ \int \cos (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {15 a^{4} \left (\frac {\log {\left (e^{i d x} - i e^{- i c} \right )}}{2} - \frac {\log {\left (e^{i d x} + i e^{- i c} \right )}}{2}\right )}{d} + \frac {- 9 i a^{4} e^{3 i c} e^{3 i d x} - 7 i a^{4} e^{i c} e^{i d x}}{d e^{4 i c} e^{4 i d x} + 2 d e^{2 i c} e^{2 i d x} + d} + \begin {cases} - \frac {8 i a^{4} e^{i c} e^{i d x}}{d} & \text {for}\: d \neq 0 \\8 a^{4} x e^{i c} & \text {otherwise} \end {cases} \]
15*a**4*(log(exp(I*d*x) - I*exp(-I*c))/2 - log(exp(I*d*x) + I*exp(-I*c))/2 )/d + (-9*I*a**4*exp(3*I*c)*exp(3*I*d*x) - 7*I*a**4*exp(I*c)*exp(I*d*x))/( d*exp(4*I*c)*exp(4*I*d*x) + 2*d*exp(2*I*c)*exp(2*I*d*x) + d) + Piecewise(( -8*I*a**4*exp(I*c)*exp(I*d*x)/d, Ne(d, 0)), (8*a**4*x*exp(I*c), True))
Time = 0.27 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.41 \[ \int \cos (c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, \sin \left (d x + c\right )\right )} + 16 i \, a^{4} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 12 \, a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )} + 16 i \, a^{4} \cos \left (d x + c\right ) - 4 \, a^{4} \sin \left (d x + c\right )}{4 \, d} \]
-1/4*(a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + 3*log(sin(d*x + c) + 1) - 3*log(sin(d*x + c) - 1) - 4*sin(d*x + c)) + 16*I*a^4*(1/cos(d*x + c) + co s(d*x + c)) + 12*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*si n(d*x + c)) + 16*I*a^4*cos(d*x + c) - 4*a^4*sin(d*x + c))/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (81) = 162\).
Time = 0.67 (sec) , antiderivative size = 372, normalized size of antiderivative = 3.84 \[ \int \cos (c+d x) (a+i a \tan (c+d x))^4 \, dx=\frac {235 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 470 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 5 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 10 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 235 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 470 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 5 \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 10 \, a^{4} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 256 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 800 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 480 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + 235 \, a^{4} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 5 \, a^{4} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 235 \, a^{4} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 5 \, a^{4} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right )}{32 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
1/32*(235*a^4*e^(4*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 470*a^4*e^( 2*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) - 5*a^4*e^(4*I*d*x + 4*I*c)*lo g(I*e^(I*d*x + I*c) - 1) - 10*a^4*e^(2*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c ) - 1) - 235*a^4*e^(4*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 470*a^4 *e^(2*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) + 5*a^4*e^(4*I*d*x + 4*I* c)*log(-I*e^(I*d*x + I*c) - 1) + 10*a^4*e^(2*I*d*x + 2*I*c)*log(-I*e^(I*d* x + I*c) - 1) - 256*I*a^4*e^(5*I*d*x + 5*I*c) - 800*I*a^4*e^(3*I*d*x + 3*I *c) - 480*I*a^4*e^(I*d*x + I*c) + 235*a^4*log(I*e^(I*d*x + I*c) + 1) - 5*a ^4*log(I*e^(I*d*x + I*c) - 1) - 235*a^4*log(-I*e^(I*d*x + I*c) + 1) + 5*a^ 4*log(-I*e^(I*d*x + I*c) - 1))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2 *I*c) + d)
Time = 6.34 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.64 \[ \int \cos (c+d x) (a+i a \tan (c+d x))^4 \, dx=-\frac {15\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {17\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,9{}\mathrm {i}-39\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,7{}\mathrm {i}+24\,a^4}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,1{}\mathrm {i}-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,2{}\mathrm {i}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]
(a^4*tan(c/2 + (d*x)/2)^3*9i - 39*a^4*tan(c/2 + (d*x)/2)^2 + 17*a^4*tan(c/ 2 + (d*x)/2)^4 + 24*a^4 - a^4*tan(c/2 + (d*x)/2)*7i)/(d*(tan(c/2 + (d*x)/2 ) - tan(c/2 + (d*x)/2)^2*2i - 2*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^ 4*1i + tan(c/2 + (d*x)/2)^5 + 1i)) - (15*a^4*atanh(tan(c/2 + (d*x)/2)))/d