\(\int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx\) [82]

Optimal result

Integrand size = 24, antiderivative size = 133 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx=-192 a^8 x+\frac {192 i a^8 \log (\cos (c+d x))}{d}+\frac {129 a^8 \tan (c+d x)}{d}+\frac {36 i a^8 \tan ^2(c+d x)}{d}-\frac {10 a^8 \tan ^3(c+d x)}{d}-\frac {2 i a^8 \tan ^4(c+d x)}{d}+\frac {a^8 \tan ^5(c+d x)}{5 d}-\frac {64 i a^9}{d (a-i a \tan (c+d x))} \]

[Out]

-192*a^8*x+192*I*a^8*ln(cos(d*x+c))/d+129*a^8*tan(d*x+c)/d+36*I*a^8*tan(d*x+c)^2/d-10*a^8*tan(d*x+c)^3/d-2*I*a
^8*tan(d*x+c)^4/d+1/5*a^8*tan(d*x+c)^5/d-64*I*a^9/d/(a-I*a*tan(d*x+c))

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {64 i a^9}{d (a-i a \tan (c+d x))}+\frac {a^8 \tan ^5(c+d x)}{5 d}-\frac {2 i a^8 \tan ^4(c+d x)}{d}-\frac {10 a^8 \tan ^3(c+d x)}{d}+\frac {36 i a^8 \tan ^2(c+d x)}{d}+\frac {129 a^8 \tan (c+d x)}{d}+\frac {192 i a^8 \log (\cos (c+d x))}{d}-192 a^8 x \]

[In]

Int[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^8,x]

[Out]

-192*a^8*x + ((192*I)*a^8*Log[Cos[c + d*x]])/d + (129*a^8*Tan[c + d*x])/d + ((36*I)*a^8*Tan[c + d*x]^2)/d - (1
0*a^8*Tan[c + d*x]^3)/d - ((2*I)*a^8*Tan[c + d*x]^4)/d + (a^8*Tan[c + d*x]^5)/(5*d) - ((64*I)*a^9)/(d*(a - I*a
*Tan[c + d*x]))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (i a^3\right ) \text {Subst}\left (\int \frac {(a+x)^6}{(a-x)^2} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (i a^3\right ) \text {Subst}\left (\int \left (129 a^4+\frac {64 a^6}{(a-x)^2}-\frac {192 a^5}{a-x}+72 a^3 x+30 a^2 x^2+8 a x^3+x^4\right ) \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -192 a^8 x+\frac {192 i a^8 \log (\cos (c+d x))}{d}+\frac {129 a^8 \tan (c+d x)}{d}+\frac {36 i a^8 \tan ^2(c+d x)}{d}-\frac {10 a^8 \tan ^3(c+d x)}{d}-\frac {2 i a^8 \tan ^4(c+d x)}{d}+\frac {a^8 \tan ^5(c+d x)}{5 d}-\frac {64 i a^9}{d (a-i a \tan (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.72 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {i a^8 \left (960 \log (i+\tan (c+d x))+645 i \tan (c+d x)-180 \tan ^2(c+d x)-50 i \tan ^3(c+d x)+10 \tan ^4(c+d x)+i \tan ^5(c+d x)+\frac {320 i}{i+\tan (c+d x)}\right )}{5 d} \]

[In]

Integrate[Cos[c + d*x]^2*(a + I*a*Tan[c + d*x])^8,x]

[Out]

((-1/5*I)*a^8*(960*Log[I + Tan[c + d*x]] + (645*I)*Tan[c + d*x] - 180*Tan[c + d*x]^2 - (50*I)*Tan[c + d*x]^3 +
10*Tan[c + d*x]^4 + I*Tan[c + d*x]^5 + (320*I)/(I + Tan[c + d*x])))/d

Maple [A] (verified)

Time = 84.15 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.89

method	result	size
risch	\(-\frac {32 i a^{8} {\mathrm e}^{2 i \left (d x +c \right )}}{d}+\frac {384 a^{8} c}{d}+\frac {16 i a^{8} \left (150 \,{\mathrm e}^{8 i \left (d x +c \right )}+500 \,{\mathrm e}^{6 i \left (d x +c \right )}+650 \,{\mathrm e}^{4 i \left (d x +c \right )}+385 \,{\mathrm e}^{2 i \left (d x +c \right )}+87\right )}{5 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {192 i a^{8} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(118\)
derivativedivides	\(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {4 \left (\sin ^{9}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}+\frac {8 \left (\sin ^{9}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )}+\frac {8 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )}{5}-\frac {7 d x}{2}-\frac {7 c}{2}\right )-56 i a^{8} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )-4 i a^{8} \left (\cos ^{2}\left (d x +c \right )\right )+70 a^{8} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )-8 i a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )-28 a^{8} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+56 i a^{8} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+a^{8} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(482\)
default	\(\frac {a^{8} \left (\frac {\sin ^{9}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {4 \left (\sin ^{9}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}+\frac {8 \left (\sin ^{9}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )}+\frac {8 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )}{5}-\frac {7 d x}{2}-\frac {7 c}{2}\right )-56 i a^{8} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-28 a^{8} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )-4 i a^{8} \left (\cos ^{2}\left (d x +c \right )\right )+70 a^{8} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )-8 i a^{8} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )-28 a^{8} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+56 i a^{8} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+a^{8} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(482\)

[In]

int(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)

[Out]

-32*I/d*a^8*exp(2*I*(d*x+c))+384/d*a^8*c+16/5*I*a^8*(150*exp(8*I*(d*x+c))+500*exp(6*I*(d*x+c))+650*exp(4*I*(d*
x+c))+385*exp(2*I*(d*x+c))+87)/d/(exp(2*I*(d*x+c))+1)^5+192*I/d*a^8*ln(exp(2*I*(d*x+c))+1)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (121) = 242\).

Time = 0.26 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.84 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {16 \, {\left (10 i \, a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} + 50 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} - 50 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} - 400 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} - 600 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} - 375 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} - 87 i \, a^{8} + 60 \, {\left (-i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} - 5 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} - 10 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} - 10 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} - 5 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{8}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{5 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

-16/5*(10*I*a^8*e^(12*I*d*x + 12*I*c) + 50*I*a^8*e^(10*I*d*x + 10*I*c) - 50*I*a^8*e^(8*I*d*x + 8*I*c) - 400*I*
a^8*e^(6*I*d*x + 6*I*c) - 600*I*a^8*e^(4*I*d*x + 4*I*c) - 375*I*a^8*e^(2*I*d*x + 2*I*c) - 87*I*a^8 + 60*(-I*a^
8*e^(10*I*d*x + 10*I*c) - 5*I*a^8*e^(8*I*d*x + 8*I*c) - 10*I*a^8*e^(6*I*d*x + 6*I*c) - 10*I*a^8*e^(4*I*d*x + 4
*I*c) - 5*I*a^8*e^(2*I*d*x + 2*I*c) - I*a^8)*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) + 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)

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.93 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {192 i a^{8} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {2400 i a^{8} e^{8 i c} e^{8 i d x} + 8000 i a^{8} e^{6 i c} e^{6 i d x} + 10400 i a^{8} e^{4 i c} e^{4 i d x} + 6160 i a^{8} e^{2 i c} e^{2 i d x} + 1392 i a^{8}}{5 d e^{10 i c} e^{10 i d x} + 25 d e^{8 i c} e^{8 i d x} + 50 d e^{6 i c} e^{6 i d x} + 50 d e^{4 i c} e^{4 i d x} + 25 d e^{2 i c} e^{2 i d x} + 5 d} + \begin {cases} - \frac {32 i a^{8} e^{2 i c} e^{2 i d x}}{d} & \text {for}\: d \neq 0 \\64 a^{8} x e^{2 i c} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**2*(a+I*a*tan(d*x+c))**8,x)

[Out]

192*I*a**8*log(exp(2*I*d*x) + exp(-2*I*c))/d + (2400*I*a**8*exp(8*I*c)*exp(8*I*d*x) + 8000*I*a**8*exp(6*I*c)*e
xp(6*I*d*x) + 10400*I*a**8*exp(4*I*c)*exp(4*I*d*x) + 6160*I*a**8*exp(2*I*c)*exp(2*I*d*x) + 1392*I*a**8)/(5*d*e
xp(10*I*c)*exp(10*I*d*x) + 25*d*exp(8*I*c)*exp(8*I*d*x) + 50*d*exp(6*I*c)*exp(6*I*d*x) + 50*d*exp(4*I*c)*exp(4
*I*d*x) + 25*d*exp(2*I*c)*exp(2*I*d*x) + 5*d) + Piecewise((-32*I*a**8*exp(2*I*c)*exp(2*I*d*x)/d, Ne(d, 0)), (6
4*a**8*x*exp(2*I*c), True))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {a^{8} \tan \left (d x + c\right )^{5} - 10 i \, a^{8} \tan \left (d x + c\right )^{4} - 50 \, a^{8} \tan \left (d x + c\right )^{3} + 180 i \, a^{8} \tan \left (d x + c\right )^{2} - 960 \, {\left (d x + c\right )} a^{8} - 480 i \, a^{8} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 645 \, a^{8} \tan \left (d x + c\right ) + \frac {320 \, {\left (a^{8} \tan \left (d x + c\right ) - i \, a^{8}\right )}}{\tan \left (d x + c\right )^{2} + 1}}{5 \, d} \]

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

1/5*(a^8*tan(d*x + c)^5 - 10*I*a^8*tan(d*x + c)^4 - 50*a^8*tan(d*x + c)^3 + 180*I*a^8*tan(d*x + c)^2 - 960*(d*
x + c)*a^8 - 480*I*a^8*log(tan(d*x + c)^2 + 1) + 645*a^8*tan(d*x + c) + 320*(a^8*tan(d*x + c) - I*a^8)/(tan(d*
x + c)^2 + 1))/d

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 302 vs. \(2 (121) = 242\).

Time = 0.91 (sec) , antiderivative size = 302, normalized size of antiderivative = 2.27 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx=-\frac {16 \, {\left (-60 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 300 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 600 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 600 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 300 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 10 i \, a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} + 50 i \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} - 50 i \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} - 400 i \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} - 600 i \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} - 375 i \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} - 60 i \, a^{8} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 87 i \, a^{8}\right )}}{5 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(cos(d*x+c)^2*(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-16/5*(-60*I*a^8*e^(10*I*d*x + 10*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 300*I*a^8*e^(8*I*d*x + 8*I*c)*log(e^(2*I
*d*x + 2*I*c) + 1) - 600*I*a^8*e^(6*I*d*x + 6*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) - 600*I*a^8*e^(4*I*d*x + 4*I*c
)*log(e^(2*I*d*x + 2*I*c) + 1) - 300*I*a^8*e^(2*I*d*x + 2*I*c)*log(e^(2*I*d*x + 2*I*c) + 1) + 10*I*a^8*e^(12*I
*d*x + 12*I*c) + 50*I*a^8*e^(10*I*d*x + 10*I*c) - 50*I*a^8*e^(8*I*d*x + 8*I*c) - 400*I*a^8*e^(6*I*d*x + 6*I*c)
- 600*I*a^8*e^(4*I*d*x + 4*I*c) - 375*I*a^8*e^(2*I*d*x + 2*I*c) - 60*I*a^8*log(e^(2*I*d*x + 2*I*c) + 1) - 87*
I*a^8)/(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)

Mupad [B] (verification not implemented)

Time = 3.80 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.77 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^8 \, dx=\frac {\frac {64\,a^8}{\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}}+129\,a^8\,\mathrm {tan}\left (c+d\,x\right )-10\,a^8\,{\mathrm {tan}\left (c+d\,x\right )}^3+\frac {a^8\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}-a^8\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,192{}\mathrm {i}+a^8\,{\mathrm {tan}\left (c+d\,x\right )}^2\,36{}\mathrm {i}-a^8\,{\mathrm {tan}\left (c+d\,x\right )}^4\,2{}\mathrm {i}}{d} \]

[In]

int(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^8,x)

[Out]

((64*a^8)/(tan(c + d*x) + 1i) - a^8*log(tan(c + d*x) + 1i)*192i + 129*a^8*tan(c + d*x) + a^8*tan(c + d*x)^2*36
i - 10*a^8*tan(c + d*x)^3 - a^8*tan(c + d*x)^4*2i + (a^8*tan(c + d*x)^5)/5)/d