Integrand size = 22, antiderivative size = 76 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {i a \sec ^5(c+d x)}{5 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:
3/8*a*arctanh(sin(d*x+c))/d+1/5*I*a*sec(d*x+c)^5/d+3/8*a*sec(d*x+c)*tan(d* x+c)/d+1/4*a*sec(d*x+c)^3*tan(d*x+c)/d
Time = 0.02 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {3 a \text {arctanh}(\sin (c+d x))}{8 d}+\frac {i a \sec ^5(c+d x)}{5 d}+\frac {3 a \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Input:
Integrate[Sec[c + d*x]^5*(a + I*a*Tan[c + d*x]),x]
Output:
(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + ((I/5)*a*Sec[c + d*x]^5)/d + (3*a*Sec[ c + d*x]*Tan[c + d*x])/(8*d) + (a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)
Time = 0.44 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3967, 3042, 4255, 3042, 4255, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^5 (a+i a \tan (c+d x))dx\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle a \int \sec ^5(c+d x)dx+\frac {i a \sec ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx+\frac {i a \sec ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle a \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {i a \sec ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {i a \sec ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle a \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {i a \sec ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {i a \sec ^5(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle a \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {i a \sec ^5(c+d x)}{5 d}\) |
Input:
Int[Sec[c + d*x]^5*(a + I*a*Tan[c + d*x]),x]
Output:
((I/5)*a*Sec[c + d*x]^5)/d + a*((Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*( ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/4)
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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 6.93 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84
| method | result | size |
| derivativedivides | \(\frac {\frac {i a}{5 \cos \left (d x +c \right )^{5}}+a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(64\) |
| default | \(\frac {\frac {i a}{5 \cos \left (d x +c \right )^{5}}+a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(64\) |
| risch | \(-\frac {i a \left (15 \,{\mathrm e}^{9 i \left (d x +c \right )}+70 \,{\mathrm e}^{7 i \left (d x +c \right )}-128 \,{\mathrm e}^{5 i \left (d x +c \right )}-70 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{20 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}\) | \(116\) |
Input:
int(sec(d*x+c)^5*(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/5*I*a/cos(d*x+c)^5+a*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c )+3/8*ln(sec(d*x+c)+tan(d*x+c))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (66) = 132\).
Time = 0.09 (sec) , antiderivative size = 276, normalized size of antiderivative = 3.63 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {-30 i \, a e^{\left (9 i \, d x + 9 i \, c\right )} - 140 i \, a e^{\left (7 i \, d x + 7 i \, c\right )} + 256 i \, a e^{\left (5 i \, d x + 5 i \, c\right )} + 140 i \, a e^{\left (3 i \, d x + 3 i \, c\right )} + 30 i \, a e^{\left (i \, d x + i \, c\right )} + 15 \, {\left (a e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \, {\left (a e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{40 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
1/40*(-30*I*a*e^(9*I*d*x + 9*I*c) - 140*I*a*e^(7*I*d*x + 7*I*c) + 256*I*a* e^(5*I*d*x + 5*I*c) + 140*I*a*e^(3*I*d*x + 3*I*c) + 30*I*a*e^(I*d*x + I*c) + 15*(a*e^(10*I*d*x + 10*I*c) + 5*a*e^(8*I*d*x + 8*I*c) + 10*a*e^(6*I*d*x + 6*I*c) + 10*a*e^(4*I*d*x + 4*I*c) + 5*a*e^(2*I*d*x + 2*I*c) + a)*log(e^ (I*d*x + I*c) + I) - 15*(a*e^(10*I*d*x + 10*I*c) + 5*a*e^(8*I*d*x + 8*I*c) + 10*a*e^(6*I*d*x + 6*I*c) + 10*a*e^(4*I*d*x + 4*I*c) + 5*a*e^(2*I*d*x + 2*I*c) + a)*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)
\[ \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx=i a \left (\int \left (- i \sec ^{5}{\left (c + d x \right )}\right )\, dx + \int \tan {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(sec(d*x+c)**5*(a+I*a*tan(d*x+c)),x)
Output:
I*a*(Integral(-I*sec(c + d*x)**5, x) + Integral(tan(c + d*x)*sec(c + d*x)* *5, x))
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.13 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {5 \, a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - \frac {16 i \, a}{\cos \left (d x + c\right )^{5}}}{80 \, d} \] Input:
integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
-1/80*(5*a*(2*(3*sin(d*x + c)^3 - 5*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)) - 16* I*a/cos(d*x + c)^5)/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (66) = 132\).
Time = 0.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.83 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {15 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 15 \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) + \frac {2 \, {\left (25 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 40 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 80 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 10 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 25 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 i \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{40 \, d} \] Input:
integrate(sec(d*x+c)^5*(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
1/40*(15*a*log(tan(1/2*d*x + 1/2*c) + 1) - 15*a*log(tan(1/2*d*x + 1/2*c) - 1) + 2*(25*a*tan(1/2*d*x + 1/2*c)^9 - 40*I*a*tan(1/2*d*x + 1/2*c)^8 - 10* a*tan(1/2*d*x + 1/2*c)^7 - 80*I*a*tan(1/2*d*x + 1/2*c)^4 + 10*a*tan(1/2*d* x + 1/2*c)^3 - 25*a*tan(1/2*d*x + 1/2*c) - 8*I*a)/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 4.06 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.34 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {-\frac {5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+2{}\mathrm {i}\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+4{}\mathrm {i}\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {5\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {a\,2{}\mathrm {i}}{5}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int((a + a*tan(c + d*x)*1i)/cos(c + d*x)^5,x)
Output:
(3*a*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((a*2i)/5 + (5*a*tan(c/2 + (d*x)/2 ))/4 - (a*tan(c/2 + (d*x)/2)^3)/2 + a*tan(c/2 + (d*x)/2)^4*4i + (a*tan(c/2 + (d*x)/2)^7)/2 + a*tan(c/2 + (d*x)/2)^8*2i - (5*a*tan(c/2 + (d*x)/2)^9)/ 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))
Time = 0.17 (sec) , antiderivative size = 265, normalized size of antiderivative = 3.49 \[ \int \sec ^5(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \left (-15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+30 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-30 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} i -15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i +25 \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 \cos \left (d x +c \right ) i +8 i \right )}{40 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^5*(a+I*a*tan(d*x+c)),x)
Output:
(a*( - 15*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 30*cos( c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 15*cos(c + d*x)*log(t an((c + d*x)/2) - 1) + 15*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d *x)**4 - 30*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 15*co s(c + d*x)*log(tan((c + d*x)/2) + 1) - 8*cos(c + d*x)*sin(c + d*x)**4*i - 15*cos(c + d*x)*sin(c + d*x)**3 + 16*cos(c + d*x)*sin(c + d*x)**2*i + 25*c os(c + d*x)*sin(c + d*x) - 8*cos(c + d*x)*i + 8*i))/(40*cos(c + d*x)*d*(si n(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))