\(\int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx\) [70]

Optimal result

Integrand size = 22, antiderivative size = 167 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {63 a^5 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {63 i a^5 \sec (c+d x)}{8 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {21 i a \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d} \]

[Out]

63/8*a^5*arctanh(sin(d*x+c))/d+63/8*I*a^5*sec(d*x+c)/d+9/20*I*a^2*sec(d*x+c)*(a+I*a*tan(d*x+c))^3/d+1/5*I*a*se
c(d*x+c)*(a+I*a*tan(d*x+c))^4/d+21/20*I*a*sec(d*x+c)*(a^2+I*a^2*tan(d*x+c))^2/d+21/8*I*sec(d*x+c)*(a^5+I*a^5*t
an(d*x+c))/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3579, 3567, 3855} \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {63 a^5 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {63 i a^5 \sec (c+d x)}{8 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {21 i a \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d} \]

[In]

Int[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(63*a^5*ArcTanh[Sin[c + d*x]])/(8*d) + (((63*I)/8)*a^5*Sec[c + d*x])/d + (((9*I)/20)*a^2*Sec[c + d*x]*(a + I*a
*Tan[c + d*x])^3)/d + ((I/5)*a*Sec[c + d*x]*(a + I*a*Tan[c + d*x])^4)/d + (((21*I)/20)*a*Sec[c + d*x]*(a^2 + I
*a^2*Tan[c + d*x])^2)/d + (((21*I)/8)*Sec[c + d*x]*(a^5 + I*a^5*Tan[c + d*x]))/d

Rule 3567

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 3579

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 3855

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

Rubi steps \begin{align*} \text {integral}& = \frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {1}{5} (9 a) \int \sec (c+d x) (a+i a \tan (c+d x))^4 \, dx \\ & = \frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {1}{20} \left (63 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx \\ & = \frac {21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {1}{4} \left (21 a^3\right ) \int \sec (c+d x) (a+i a \tan (c+d x))^2 \, dx \\ & = \frac {21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac {1}{8} \left (63 a^4\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx \\ & = \frac {63 i a^5 \sec (c+d x)}{8 d}+\frac {21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d}+\frac {1}{8} \left (63 a^5\right ) \int \sec (c+d x) \, dx \\ & = \frac {63 a^5 \text {arctanh}(\sin (c+d x))}{8 d}+\frac {63 i a^5 \sec (c+d x)}{8 d}+\frac {21 i a^3 \sec (c+d x) (a+i a \tan (c+d x))^2}{20 d}+\frac {9 i a^2 \sec (c+d x) (a+i a \tan (c+d x))^3}{20 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}+\frac {21 i \sec (c+d x) \left (a^5+i a^5 \tan (c+d x)\right )}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.11 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.69 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 (\cos (5 d x)+i \sin (5 d x)) \left (5040 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right )+i \sec ^5(c+d x) (1344+1920 \cos (2 (c+d x))+640 \cos (4 (c+d x))+450 i \sin (2 (c+d x))+325 i \sin (4 (c+d x)))\right )}{320 d (\cos (d x)+i \sin (d x))^5} \]

[In]

Integrate[Sec[c + d*x]*(a + I*a*Tan[c + d*x])^5,x]

[Out]

(a^5*(Cos[5*d*x] + I*Sin[5*d*x])*(5040*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]] + I*Sec[c + d*x]^5*(1344 + 1920*C
os[2*(c + d*x)] + 640*Cos[4*(c + d*x)] + (450*I)*Sin[2*(c + d*x)] + (325*I)*Sin[4*(c + d*x)])))/(320*d*(Cos[d*
x] + I*Sin[d*x])^5)

Maple [A] (verified)

Time = 8.78 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.73

method	result	size
risch	\(\frac {i a^{5} \left (965 \,{\mathrm e}^{9 i \left (d x +c \right )}+2370 \,{\mathrm e}^{7 i \left (d x +c \right )}+2688 \,{\mathrm e}^{5 i \left (d x +c \right )}+1470 \,{\mathrm e}^{3 i \left (d x +c \right )}+315 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{20 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {63 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {63 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\)	\(122\)
derivativedivides	\(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{6}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{6}\left (d x +c \right )}{5 \cos \left (d x +c \right )}+\frac {\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}\right )+5 a^{5} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-10 i a^{5} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )-10 a^{5} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {5 i a^{5}}{\cos \left (d x +c \right )}+a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(312\)
default	\(\frac {i a^{5} \left (\frac {\sin ^{6}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{6}\left (d x +c \right )}{15 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{6}\left (d x +c \right )}{5 \cos \left (d x +c \right )}+\frac {\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}\right )+5 a^{5} \left (\frac {\sin ^{5}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{5}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-10 i a^{5} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )-10 a^{5} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {5 i a^{5}}{\cos \left (d x +c \right )}+a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(312\)

[In]

int(sec(d*x+c)*(a+I*a*tan(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

1/20*I*a^5/d/(exp(2*I*(d*x+c))+1)^5*(965*exp(9*I*(d*x+c))+2370*exp(7*I*(d*x+c))+2688*exp(5*I*(d*x+c))+1470*exp
(3*I*(d*x+c))+315*exp(I*(d*x+c)))+63/8/d*a^5*ln(exp(I*(d*x+c))+I)-63/8/d*a^5*ln(exp(I*(d*x+c))-I)

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. 310 vs. \(2 (137) = 274\).

Time = 0.25 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.86 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {1930 i \, a^{5} e^{\left (9 i \, d x + 9 i \, c\right )} + 4740 i \, a^{5} e^{\left (7 i \, d x + 7 i \, c\right )} + 5376 i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )} + 2940 i \, a^{5} e^{\left (3 i \, d x + 3 i \, c\right )} + 630 i \, a^{5} e^{\left (i \, d x + i \, c\right )} + 315 \, {\left (a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 315 \, {\left (a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{40 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^5,x, algorithm="fricas")

[Out]

1/40*(1930*I*a^5*e^(9*I*d*x + 9*I*c) + 4740*I*a^5*e^(7*I*d*x + 7*I*c) + 5376*I*a^5*e^(5*I*d*x + 5*I*c) + 2940*
I*a^5*e^(3*I*d*x + 3*I*c) + 630*I*a^5*e^(I*d*x + I*c) + 315*(a^5*e^(10*I*d*x + 10*I*c) + 5*a^5*e^(8*I*d*x + 8*
I*c) + 10*a^5*e^(6*I*d*x + 6*I*c) + 10*a^5*e^(4*I*d*x + 4*I*c) + 5*a^5*e^(2*I*d*x + 2*I*c) + a^5)*log(e^(I*d*x
+ I*c) + I) - 315*(a^5*e^(10*I*d*x + 10*I*c) + 5*a^5*e^(8*I*d*x + 8*I*c) + 10*a^5*e^(6*I*d*x + 6*I*c) + 10*a^
5*e^(4*I*d*x + 4*I*c) + 5*a^5*e^(2*I*d*x + 2*I*c) + a^5)*log(e^(I*d*x + I*c) - I))/(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 [F]

\[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=i a^{5} \left (\int \left (- i \sec {\left (c + d x \right )}\right )\, dx + \int 5 \tan {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \left (- 10 \tan ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\right )\, dx + \int \tan ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 10 i \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \left (- 5 i \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\right )\, dx\right ) \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))**5,x)

[Out]

I*a**5*(Integral(-I*sec(c + d*x), x) + Integral(5*tan(c + d*x)*sec(c + d*x), x) + Integral(-10*tan(c + d*x)**3
*sec(c + d*x), x) + Integral(tan(c + d*x)**5*sec(c + d*x), x) + Integral(10*I*tan(c + d*x)**2*sec(c + d*x), x)
+ Integral(-5*I*tan(c + d*x)**4*sec(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.29 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {75 \, a^{5} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 600 \, a^{5} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, a^{5} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac {1200 i \, a^{5}}{\cos \left (d x + c\right )} + \frac {800 i \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{5}}{\cos \left (d x + c\right )^{3}} + \frac {16 i \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 3\right )} a^{5}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^5,x, algorithm="maxima")

[Out]

1/240*(75*a^5*(2*(5*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) + 3*log(sin(d*x +
c) + 1) - 3*log(sin(d*x + c) - 1)) + 600*a^5*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - l
og(sin(d*x + c) - 1)) + 240*a^5*log(sec(d*x + c) + tan(d*x + c)) + 1200*I*a^5/cos(d*x + c) + 800*I*(3*cos(d*x
+ c)^2 - 1)*a^5/cos(d*x + c)^3 + 16*I*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 + 3)*a^5/cos(d*x + c)^5)/d

Giac [A] (verification not implemented)

none

Time = 0.72 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.13 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {315 \, a^{5} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 315 \, a^{5} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {2 \, {\left (275 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 200 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 750 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1600 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3280 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 750 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2240 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 275 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 488 i \, a^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{40 \, d} \]

[In]

integrate(sec(d*x+c)*(a+I*a*tan(d*x+c))^5,x, algorithm="giac")

[Out]

1/40*(315*a^5*log(tan(1/2*d*x + 1/2*c) + 1) - 315*a^5*log(tan(1/2*d*x + 1/2*c) - 1) - 2*(275*a^5*tan(1/2*d*x +
1/2*c)^9 + 200*I*a^5*tan(1/2*d*x + 1/2*c)^8 - 750*a^5*tan(1/2*d*x + 1/2*c)^7 - 1600*I*a^5*tan(1/2*d*x + 1/2*c
)^6 + 3280*I*a^5*tan(1/2*d*x + 1/2*c)^4 + 750*a^5*tan(1/2*d*x + 1/2*c)^3 - 2240*I*a^5*tan(1/2*d*x + 1/2*c)^2 -
275*a^5*tan(1/2*d*x + 1/2*c) + 488*I*a^5)/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 7.80 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.37 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {63\,a^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {55\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,10{}\mathrm {i}-\frac {75\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,80{}\mathrm {i}+a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,164{}\mathrm {i}+\frac {75\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,112{}\mathrm {i}-\frac {55\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {a^5\,122{}\mathrm {i}}{5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

int((a + a*tan(c + d*x)*1i)^5/cos(c + d*x),x)

[Out]

(63*a^5*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((75*a^5*tan(c/2 + (d*x)/2)^3)/2 - a^5*tan(c/2 + (d*x)/2)^2*112i +
a^5*tan(c/2 + (d*x)/2)^4*164i - a^5*tan(c/2 + (d*x)/2)^6*80i - (75*a^5*tan(c/2 + (d*x)/2)^7)/2 + a^5*tan(c/2 +
(d*x)/2)^8*10i + (55*a^5*tan(c/2 + (d*x)/2)^9)/4 + (a^5*122i)/5 - (55*a^5*tan(c/2 + (d*x)/2))/4)/(d*(5*tan(c/
2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/
2)^10 - 1))