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

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

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

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*sec(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*tan 
(d*x+c))/d
 

Mathematica [A] (verified)

Time = 2.01 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.69 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 (\cos (5 d x)+i \sin (5 d x)) \left (5040 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right )+i \sec ^5(c+d x) (1344+1920 \cos (2 (c+d x))+640 \cos (4 (c+d x))+450 i \sin (2 (c+d x))+325 i \sin (4 (c+d x)))\right )}{320 d (\cos (d x)+i \sin (d x))^5} \] Input:

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

Output:

(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*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)])))/(320*d*(Cos[d*x] 
 + I*Sin[d*x])^5)
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 3979, 3042, 3979, 3042, 3979, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (c+d x) (a+i a \tan (c+d x))^5dx\)

\(\Big \downarrow \) 3979

\(\displaystyle \frac {9}{5} a \int \sec (c+d x) (i \tan (c+d x) a+a)^4dx+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{5} a \int \sec (c+d x) (i \tan (c+d x) a+a)^4dx+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3979

\(\displaystyle \frac {9}{5} a \left (\frac {7}{4} a \int \sec (c+d x) (i \tan (c+d x) a+a)^3dx+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^3}{4 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{5} a \left (\frac {7}{4} a \int \sec (c+d x) (i \tan (c+d x) a+a)^3dx+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^3}{4 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3979

\(\displaystyle \frac {9}{5} a \left (\frac {7}{4} a \left (\frac {5}{3} a \int \sec (c+d x) (i \tan (c+d x) a+a)^2dx+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^3}{4 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{5} a \left (\frac {7}{4} a \left (\frac {5}{3} a \int \sec (c+d x) (i \tan (c+d x) a+a)^2dx+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^3}{4 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3979

\(\displaystyle \frac {9}{5} a \left (\frac {7}{4} a \left (\frac {5}{3} a \left (\frac {3}{2} a \int \sec (c+d x) (i \tan (c+d x) a+a)dx+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^3}{4 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{5} a \left (\frac {7}{4} a \left (\frac {5}{3} a \left (\frac {3}{2} a \int \sec (c+d x) (i \tan (c+d x) a+a)dx+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^3}{4 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3967

\(\displaystyle \frac {9}{5} a \left (\frac {7}{4} a \left (\frac {5}{3} a \left (\frac {3}{2} a \left (a \int \sec (c+d x)dx+\frac {i a \sec (c+d x)}{d}\right )+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^3}{4 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{5} a \left (\frac {7}{4} a \left (\frac {5}{3} a \left (\frac {3}{2} a \left (a \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {i a \sec (c+d x)}{d}\right )+\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^3}{4 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {9}{5} a \left (\frac {7}{4} a \left (\frac {5}{3} a \left (\frac {i \sec (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{2 d}+\frac {3}{2} a \left (\frac {a \text {arctanh}(\sin (c+d x))}{d}+\frac {i a \sec (c+d x)}{d}\right )\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^3}{4 d}\right )+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^4}{5 d}\)

Input:

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

Output:

((I/5)*a*Sec[c + d*x]*(a + I*a*Tan[c + d*x])^4)/d + (9*a*(((I/4)*a*Sec[c + 
 d*x]*(a + I*a*Tan[c + d*x])^3)/d + (7*a*(((I/3)*a*Sec[c + d*x]*(a + I*a*T 
an[c + d*x])^2)/d + (5*a*((3*a*((a*ArcTanh[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))/4)) 
/5
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3967
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])
 

rule 3979
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]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 
Maple [A] (verified)

Time = 7.19 (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 \(\frac {i a^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin \left (d x +c \right )^{6}}{15 \cos \left (d x +c \right )^{3}}+\frac {\sin \left (d x +c \right )^{6}}{5 \cos \left (d x +c \right )}+\frac {\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}\right )+5 a^{5} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-10 i a^{5} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )-10 a^{5} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {5 i a^{5}}{\cos \left (d x +c \right )}+a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) \(312\)
default \(\frac {i a^{5} \left (\frac {\sin \left (d x +c \right )^{6}}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin \left (d x +c \right )^{6}}{15 \cos \left (d x +c \right )^{3}}+\frac {\sin \left (d x +c \right )^{6}}{5 \cos \left (d x +c \right )}+\frac {\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}\right )+5 a^{5} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-10 i a^{5} \left (\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin \left (d x +c \right )^{4}}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}\right )-10 a^{5} \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {5 i a^{5}}{\cos \left (d x +c \right )}+a^{5} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) \(312\)

Input:

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

Output:

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 = 0.10 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.86 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {1930 i \, a^{5} e^{\left (9 i \, d x + 9 i \, c\right )} + 4740 i \, a^{5} e^{\left (7 i \, d x + 7 i \, c\right )} + 5376 i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )} + 2940 i \, a^{5} e^{\left (3 i \, d x + 3 i \, c\right )} + 630 i \, a^{5} e^{\left (i \, d x + i \, c\right )} + 315 \, {\left (a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 315 \, {\left (a^{5} e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{5}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{40 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

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

Output:

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) + 53 
76*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]

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

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

Output:

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)

Time = 0.04 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.29 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {75 \, a^{5} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 600 \, a^{5} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, a^{5} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac {1200 i \, a^{5}}{\cos \left (d x + c\right )} + \frac {800 i \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{5}}{\cos \left (d x + c\right )^{3}} + \frac {16 i \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 3\right )} a^{5}}{\cos \left (d x + c\right )^{5}}}{240 \, d} \] Input:

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

Output:

1/240*(75*a^5*(2*(5*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^4 - 2*s 
in(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) - log 
(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*(1 
5*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)

Time = 0.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.13 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {315 \, a^{5} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 315 \, a^{5} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {2 \, {\left (275 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 200 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 750 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1600 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 3280 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 750 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2240 i \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 275 \, a^{5} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 488 i \, a^{5}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{40 \, d} \] Input:

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

Output:

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 - 2 
240*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 = 4.23 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.37 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {63\,a^5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {55\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,10{}\mathrm {i}-\frac {75\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,80{}\mathrm {i}+a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,164{}\mathrm {i}+\frac {75\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,112{}\mathrm {i}-\frac {55\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {a^5\,122{}\mathrm {i}}{5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:

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

Output:

(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*ta 
n(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*t 
an(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)^1 
0 - 1))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.73 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {a^{5} \left (-315 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+630 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-315 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+315 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-630 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+315 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-488 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} i +325 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+976 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i -275 \cos \left (d x +c \right ) \sin \left (d x +c \right )-488 \cos \left (d x +c \right ) i +640 \sin \left (d x +c \right )^{4} i -1120 \sin \left (d x +c \right )^{2} i +488 i \right )}{40 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

(a**5*( - 315*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 630 
*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 315*cos(c + d*x) 
*log(tan((c + d*x)/2) - 1) + 315*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*si 
n(c + d*x)**4 - 630*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 
 + 315*cos(c + d*x)*log(tan((c + d*x)/2) + 1) - 488*cos(c + d*x)*sin(c + d 
*x)**4*i + 325*cos(c + d*x)*sin(c + d*x)**3 + 976*cos(c + d*x)*sin(c + d*x 
)**2*i - 275*cos(c + d*x)*sin(c + d*x) - 488*cos(c + d*x)*i + 640*sin(c + 
d*x)**4*i - 1120*sin(c + d*x)**2*i + 488*i))/(40*cos(c + d*x)*d*(sin(c + d 
*x)**4 - 2*sin(c + d*x)**2 + 1))