3.91 \(\int \cos (c+d x) (a+i a \tan (c+d x))^8 \, dx\)

Optimal. Leaf size=235 \[ -\frac{3003 i a^8 \sec (c+d x)}{16 d}-\frac{3003 a^8 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac{429 i a^2 \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^3}{40 d}-\frac{143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac{1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac{1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d} \]

[Out]

(-3003*a^8*ArcTanh[Sin[c + d*x]])/(16*d) - (((3003*I)/16)*a^8*Sec[c + d*x])/d - (((13*I)/6)*a^3*Sec[c + d*x]*(
a + I*a*Tan[c + d*x])^5)/d - ((2*I)*a*Cos[c + d*x]*(a + I*a*Tan[c + d*x])^7)/d - (((429*I)/40)*a^2*Sec[c + d*x
]*(a^2 + I*a^2*Tan[c + d*x])^3)/d - (((143*I)/30)*Sec[c + d*x]*(a^2 + I*a^2*Tan[c + d*x])^4)/d - (((1001*I)/40
)*Sec[c + d*x]*(a^4 + I*a^4*Tan[c + d*x])^2)/d - (((1001*I)/16)*Sec[c + d*x]*(a^8 + I*a^8*Tan[c + d*x]))/d

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Rubi [A]  time = 0.20371, antiderivative size = 235, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3496, 3498, 3486, 3770} \[ -\frac{3003 i a^8 \sec (c+d x)}{16 d}-\frac{3003 a^8 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac{429 i a^2 \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^3}{40 d}-\frac{143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac{1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac{1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3003*a^8*ArcTanh[Sin[c + d*x]])/(16*d) - (((3003*I)/16)*a^8*Sec[c + d*x])/d - (((13*I)/6)*a^3*Sec[c + d*x]*(
a + I*a*Tan[c + d*x])^5)/d - ((2*I)*a*Cos[c + d*x]*(a + I*a*Tan[c + d*x])^7)/d - (((429*I)/40)*a^2*Sec[c + d*x
]*(a^2 + I*a^2*Tan[c + d*x])^3)/d - (((143*I)/30)*Sec[c + d*x]*(a^2 + I*a^2*Tan[c + d*x])^4)/d - (((1001*I)/40
)*Sec[c + d*x]*(a^4 + I*a^4*Tan[c + d*x])^2)/d - (((1001*I)/16)*Sec[c + d*x]*(a^8 + I*a^8*Tan[c + d*x]))/d

Rule 3496

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] - Dist[(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]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rule 3498

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] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[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]

Rule 3486

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] + Dist[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])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+i a \tan (c+d x))^8 \, dx &=-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\left (13 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^6 \, dx\\ &=-\frac{13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac{1}{6} \left (143 a^3\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx\\ &=-\frac{13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac{143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac{1}{10} \left (429 a^4\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^4 \, dx\\ &=-\frac{429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac{13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac{143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac{1}{40} \left (3003 a^5\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx\\ &=-\frac{429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac{13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac{143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac{1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac{1}{8} \left (1001 a^6\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=-\frac{429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac{13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac{143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac{1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac{1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}-\frac{1}{16} \left (3003 a^7\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx\\ &=-\frac{3003 i a^8 \sec (c+d x)}{16 d}-\frac{429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac{13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac{143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac{1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac{1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}-\frac{1}{16} \left (3003 a^8\right ) \int \sec (c+d x) \, dx\\ &=-\frac{3003 a^8 \tanh ^{-1}(\sin (c+d x))}{16 d}-\frac{3003 i a^8 \sec (c+d x)}{16 d}-\frac{429 i a^5 \sec (c+d x) (a+i a \tan (c+d x))^3}{40 d}-\frac{13 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^5}{6 d}-\frac{2 i a \cos (c+d x) (a+i a \tan (c+d x))^7}{d}-\frac{143 i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^4}{30 d}-\frac{1001 i \sec (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )^2}{40 d}-\frac{1001 i \sec (c+d x) \left (a^8+i a^8 \tan (c+d x)\right )}{16 d}\\ \end{align*}

Mathematica [A]  time = 2.6415, size = 205, normalized size = 0.87 \[ \frac{a^8 (\cos (8 c)-i \sin (8 c)) \cos ^2(c+d x) (\tan (c+d x)-i)^8 \left (-658944 i \cos (c+d x)+5 (12870 \sin (c+d x)+22165 \sin (3 (c+d x))+10959 \sin (5 (c+d x))+1536 \sin (7 (c+d x))-73216 i \cos (3 (c+d x))-19968 i \cos (5 (c+d x))-1536 i \cos (7 (c+d x)))+720720 \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{3840 d (\cos (d x)+i \sin (d x))^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^8*Cos[c + d*x]^2*(Cos[8*c] - I*Sin[8*c])*((-658944*I)*Cos[c + d*x] + 720720*Cos[c + d*x]^6*(Log[Cos[(c + d*
x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 5*((-73216*I)*Cos[3*(c + d*x)] - (1996
8*I)*Cos[5*(c + d*x)] - (1536*I)*Cos[7*(c + d*x)] + 12870*Sin[c + d*x] + 22165*Sin[3*(c + d*x)] + 10959*Sin[5*
(c + d*x)] + 1536*Sin[7*(c + d*x)]))*(-I + Tan[c + d*x])^8)/(3840*d*(Cos[d*x] + I*Sin[d*x])^8)

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Maple [B]  time = 0.104, size = 464, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+I*a*tan(d*x+c))^8,x)

[Out]

5/16*a^8*sin(d*x+c)^7/d+175/16*a^8*sin(d*x+c)^5/d+2555/48*a^8*sin(d*x+c)^3/d-3003/16/d*a^8*ln(sec(d*x+c)+tan(d
*x+c))+3019/16*a^8*sin(d*x+c)/d+1/6/d*a^8*sin(d*x+c)^9/cos(d*x+c)^6-1/8/d*a^8*sin(d*x+c)^9/cos(d*x+c)^4+5/16/d
*a^8*sin(d*x+c)^9/cos(d*x+c)^2+56/3*I/d*a^8*sin(d*x+c)^6/cos(d*x+c)^3-8/5*I/d*a^8*sin(d*x+c)^8/cos(d*x+c)^5+8/
5*I/d*a^8*sin(d*x+c)^8/cos(d*x+c)^3-4424/15*I/d*a^8*cos(d*x+c)-328/5*I/d*a^8*cos(d*x+c)*sin(d*x+c)^4-8*I/d*a^8
*cos(d*x+c)*sin(d*x+c)^6-8*I/d*a^8*sin(d*x+c)^8/cos(d*x+c)-2152/15*I/d*a^8*cos(d*x+c)*sin(d*x+c)^2-56*I/d*a^8*
sin(d*x+c)^6/cos(d*x+c)-56*I/d*a^8*sin(d*x+c)^4/cos(d*x+c)-7/d*a^8*sin(d*x+c)^7/cos(d*x+c)^4+21/2/d*a^8*sin(d*
x+c)^7/cos(d*x+c)^2+35/d*a^8*sin(d*x+c)^5/cos(d*x+c)^2

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Maxima [B]  time = 1.18915, size = 535, normalized size = 2.28 \begin{align*} -\frac{5 \, a^{8}{\left (\frac{2 \,{\left (87 \, \sin \left (d x + c\right )^{5} - 136 \, \sin \left (d x + c\right )^{3} + 57 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + 105 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 96 \, \sin \left (d x + c\right )\right )} + 840 \, a^{8}{\left (\frac{2 \,{\left (9 \, \sin \left (d x + c\right )^{3} - 7 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, \sin \left (d x + c\right )\right )} + 8400 \, a^{8}{\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 )} + 26880 i \, a^{8}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} + 8960 i \, a^{8}{\left (\frac{6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} + 768 i \, a^{8}{\left (\frac{15 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 1}{\cos \left (d x + c\right )^{5}} + 5 \, \cos \left (d x + c\right )\right )} + 6720 \, a^{8}{\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 )} + 3840 i \, a^{8} \cos \left (d x + c\right ) - 480 \, a^{8} \sin \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(5*a^8*(2*(87*sin(d*x + c)^5 - 136*sin(d*x + c)^3 + 57*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4
+ 3*sin(d*x + c)^2 - 1) + 105*log(sin(d*x + c) + 1) - 105*log(sin(d*x + c) - 1) - 96*sin(d*x + c)) + 840*a^8*
(2*(9*sin(d*x + c)^3 - 7*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) + 15*log(sin(d*x + c) + 1) - 15
*log(sin(d*x + c) - 1) - 16*sin(d*x + c)) + 8400*a^8*(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)) + 26880*I*a^8*(1/cos(d*x + c) + cos(d*x + c)) + 8960*I*a^8*(
(6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c)) + 768*I*a^8*((15*cos(d*x + c)^4 - 5*cos(d*x + c)^2 + 1
)/cos(d*x + c)^5 + 5*cos(d*x + c)) + 6720*a^8*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c))
+ 3840*I*a^8*cos(d*x + c) - 480*a^8*sin(d*x + c))/d

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Fricas [A]  time = 1.79936, size = 1137, normalized size = 4.84 \begin{align*} \frac{-30720 i \, a^{8} e^{\left (13 i \, d x + 13 i \, c\right )} - 309270 i \, a^{8} e^{\left (11 i \, d x + 11 i \, c\right )} - 953810 i \, a^{8} e^{\left (9 i \, d x + 9 i \, c\right )} - 1446588 i \, a^{8} e^{\left (7 i \, d x + 7 i \, c\right )} - 1189188 i \, a^{8} e^{\left (5 i \, d x + 5 i \, c\right )} - 510510 i \, a^{8} e^{\left (3 i \, d x + 3 i \, c\right )} - 90090 i \, a^{8} e^{\left (i \, d x + i \, c\right )} - 45045 \,{\left (a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{8}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) + 45045 \,{\left (a^{8} e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{8} e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{8} e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{8} e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{8} e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{8} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{8}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{240 \,{\left (d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(-30720*I*a^8*e^(13*I*d*x + 13*I*c) - 309270*I*a^8*e^(11*I*d*x + 11*I*c) - 953810*I*a^8*e^(9*I*d*x + 9*I
*c) - 1446588*I*a^8*e^(7*I*d*x + 7*I*c) - 1189188*I*a^8*e^(5*I*d*x + 5*I*c) - 510510*I*a^8*e^(3*I*d*x + 3*I*c)
- 90090*I*a^8*e^(I*d*x + I*c) - 45045*(a^8*e^(12*I*d*x + 12*I*c) + 6*a^8*e^(10*I*d*x + 10*I*c) + 15*a^8*e^(8*
I*d*x + 8*I*c) + 20*a^8*e^(6*I*d*x + 6*I*c) + 15*a^8*e^(4*I*d*x + 4*I*c) + 6*a^8*e^(2*I*d*x + 2*I*c) + a^8)*lo
g(e^(I*d*x + I*c) + I) + 45045*(a^8*e^(12*I*d*x + 12*I*c) + 6*a^8*e^(10*I*d*x + 10*I*c) + 15*a^8*e^(8*I*d*x +
8*I*c) + 20*a^8*e^(6*I*d*x + 6*I*c) + 15*a^8*e^(4*I*d*x + 4*I*c) + 6*a^8*e^(2*I*d*x + 2*I*c) + a^8)*log(e^(I*d
*x + I*c) - I))/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*
d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [A]  time = 12.3875, size = 326, normalized size = 1.39 \begin{align*} \frac{3003 a^{8} \left (\frac{\log{\left (e^{i d x} - i e^{- i c} \right )}}{16} - \frac{\log{\left (e^{i d x} + i e^{- i c} \right )}}{16}\right )}{d} + \frac{- \frac{4165 i a^{8} e^{- i c} e^{11 i d x}}{8 d} - \frac{49301 i a^{8} e^{- 3 i c} e^{9 i d x}}{24 d} - \frac{69349 i a^{8} e^{- 5 i c} e^{7 i d x}}{20 d} - \frac{60699 i a^{8} e^{- 7 i c} e^{5 i d x}}{20 d} - \frac{10873 i a^{8} e^{- 9 i c} e^{3 i d x}}{8 d} - \frac{1979 i a^{8} e^{- 11 i c} e^{i d x}}{8 d}}{e^{12 i d x} + 6 e^{- 2 i c} e^{10 i d x} + 15 e^{- 4 i c} e^{8 i d x} + 20 e^{- 6 i c} e^{6 i d x} + 15 e^{- 8 i c} e^{4 i d x} + 6 e^{- 10 i c} e^{2 i d x} + e^{- 12 i c}} + \begin{cases} - \frac{128 i a^{8} e^{i c} e^{i d x}}{d} & \text{for}\: d \neq 0 \\128 a^{8} x e^{i c} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3003*a**8*(log(exp(I*d*x) - I*exp(-I*c))/16 - log(exp(I*d*x) + I*exp(-I*c))/16)/d + (-4165*I*a**8*exp(-I*c)*ex
p(11*I*d*x)/(8*d) - 49301*I*a**8*exp(-3*I*c)*exp(9*I*d*x)/(24*d) - 69349*I*a**8*exp(-5*I*c)*exp(7*I*d*x)/(20*d
) - 60699*I*a**8*exp(-7*I*c)*exp(5*I*d*x)/(20*d) - 10873*I*a**8*exp(-9*I*c)*exp(3*I*d*x)/(8*d) - 1979*I*a**8*e
xp(-11*I*c)*exp(I*d*x)/(8*d))/(exp(12*I*d*x) + 6*exp(-2*I*c)*exp(10*I*d*x) + 15*exp(-4*I*c)*exp(8*I*d*x) + 20*
exp(-6*I*c)*exp(6*I*d*x) + 15*exp(-8*I*c)*exp(4*I*d*x) + 6*exp(-10*I*c)*exp(2*I*d*x) + exp(-12*I*c)) + Piecewi
se((-128*I*a**8*exp(I*c)*exp(I*d*x)/d, Ne(d, 0)), (128*a**8*x*exp(I*c), True))

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Giac [B]  time = 2.30031, size = 1247, normalized size = 5.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/61440*(11512215*a^8*e^(12*I*d*x + 12*I*c)*log(I*e^(I*d*x + I*c) + 1) + 69073290*a^8*e^(10*I*d*x + 10*I*c)*lo
g(I*e^(I*d*x + I*c) + 1) + 172683225*a^8*e^(8*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 230244300*a^8*e^(6*I
*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 172683225*a^8*e^(4*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 6907
3290*a^8*e^(2*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) - 19305*a^8*e^(12*I*d*x + 12*I*c)*log(I*e^(I*d*x + I*c
) - 1) - 115830*a^8*e^(10*I*d*x + 10*I*c)*log(I*e^(I*d*x + I*c) - 1) - 289575*a^8*e^(8*I*d*x + 8*I*c)*log(I*e^
(I*d*x + I*c) - 1) - 386100*a^8*e^(6*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) - 1) - 289575*a^8*e^(4*I*d*x + 4*I*c
)*log(I*e^(I*d*x + I*c) - 1) - 115830*a^8*e^(2*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) - 1) - 11512215*a^8*e^(12*
I*d*x + 12*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 69073290*a^8*e^(10*I*d*x + 10*I*c)*log(-I*e^(I*d*x + I*c) + 1) -
172683225*a^8*e^(8*I*d*x + 8*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 230244300*a^8*e^(6*I*d*x + 6*I*c)*log(-I*e^(I
*d*x + I*c) + 1) - 172683225*a^8*e^(4*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 69073290*a^8*e^(2*I*d*x + 2
*I*c)*log(-I*e^(I*d*x + I*c) + 1) + 19305*a^8*e^(12*I*d*x + 12*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 115830*a^8*e
^(10*I*d*x + 10*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 289575*a^8*e^(8*I*d*x + 8*I*c)*log(-I*e^(I*d*x + I*c) - 1)
+ 386100*a^8*e^(6*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 289575*a^8*e^(4*I*d*x + 4*I*c)*log(-I*e^(I*d*x
+ I*c) - 1) + 115830*a^8*e^(2*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 7864320*I*a^8*e^(13*I*d*x + 13*I*c)
- 79173120*I*a^8*e^(11*I*d*x + 11*I*c) - 244175360*I*a^8*e^(9*I*d*x + 9*I*c) - 370326528*I*a^8*e^(7*I*d*x + 7
*I*c) - 304432128*I*a^8*e^(5*I*d*x + 5*I*c) - 130690560*I*a^8*e^(3*I*d*x + 3*I*c) - 23063040*I*a^8*e^(I*d*x +
I*c) + 11512215*a^8*log(I*e^(I*d*x + I*c) + 1) - 19305*a^8*log(I*e^(I*d*x + I*c) - 1) - 11512215*a^8*log(-I*e^
(I*d*x + I*c) + 1) + 19305*a^8*log(-I*e^(I*d*x + I*c) - 1))/(d*e^(12*I*d*x + 12*I*c) + 6*d*e^(10*I*d*x + 10*I*
c) + 15*d*e^(8*I*d*x + 8*I*c) + 20*d*e^(6*I*d*x + 6*I*c) + 15*d*e^(4*I*d*x + 4*I*c) + 6*d*e^(2*I*d*x + 2*I*c)
+ d)