Integrand size = 22, antiderivative size = 94 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {i a \sec ^{10}(c+d x)}{10 d}+\frac {a \tan (c+d x)}{d}+\frac {4 a \tan ^3(c+d x)}{3 d}+\frac {6 a \tan ^5(c+d x)}{5 d}+\frac {4 a \tan ^7(c+d x)}{7 d}+\frac {a \tan ^9(c+d x)}{9 d} \] Output:
1/10*I*a*sec(d*x+c)^10/d+a*tan(d*x+c)/d+4/3*a*tan(d*x+c)^3/d+6/5*a*tan(d*x +c)^5/d+4/7*a*tan(d*x+c)^7/d+1/9*a*tan(d*x+c)^9/d
Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.84 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {i a \sec ^{10}(c+d x)}{10 d}+\frac {a \left (\tan (c+d x)+\frac {4}{3} \tan ^3(c+d x)+\frac {6}{5} \tan ^5(c+d x)+\frac {4}{7} \tan ^7(c+d x)+\frac {1}{9} \tan ^9(c+d x)\right )}{d} \] Input:
Integrate[Sec[c + d*x]^10*(a + I*a*Tan[c + d*x]),x]
Output:
((I/10)*a*Sec[c + d*x]^10)/d + (a*(Tan[c + d*x] + (4*Tan[c + d*x]^3)/3 + ( 6*Tan[c + d*x]^5)/5 + (4*Tan[c + d*x]^7)/7 + Tan[c + d*x]^9/9))/d
Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {3042, 3967, 3042, 4254, 2009}
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.
\(\displaystyle \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^{10} (a+i a \tan (c+d x))dx\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle a \int \sec ^{10}(c+d x)dx+\frac {i a \sec ^{10}(c+d x)}{10 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^{10}dx+\frac {i a \sec ^{10}(c+d x)}{10 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle -\frac {a \int \left (\tan ^8(c+d x)+4 \tan ^6(c+d x)+6 \tan ^4(c+d x)+4 \tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}+\frac {i a \sec ^{10}(c+d x)}{10 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \left (-\frac {1}{9} \tan ^9(c+d x)-\frac {4}{7} \tan ^7(c+d x)-\frac {6}{5} \tan ^5(c+d x)-\frac {4}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}+\frac {i a \sec ^{10}(c+d x)}{10 d}\) |
Input:
Int[Sec[c + d*x]^10*(a + I*a*Tan[c + d*x]),x]
Output:
((I/10)*a*Sec[c + d*x]^10)/d - (a*(-Tan[c + d*x] - (4*Tan[c + d*x]^3)/3 - (6*Tan[c + d*x]^5)/5 - (4*Tan[c + d*x]^7)/7 - Tan[c + d*x]^9/9))/d
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_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Time = 162.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.83
method | result | size |
risch | \(\frac {256 i a \left (252 \,{\mathrm e}^{10 i \left (d x +c \right )}+210 \,{\mathrm e}^{8 i \left (d x +c \right )}+120 \,{\mathrm e}^{6 i \left (d x +c \right )}+45 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{315 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}\) | \(78\) |
derivativedivides | \(\frac {a \left (\tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{10}}{10}+\frac {\tan \left (d x +c \right )^{9}}{9}+\frac {i \tan \left (d x +c \right )^{8}}{2}+\frac {4 \tan \left (d x +c \right )^{7}}{7}+i \tan \left (d x +c \right )^{6}+\frac {6 \tan \left (d x +c \right )^{5}}{5}+i \tan \left (d x +c \right )^{4}+\frac {4 \tan \left (d x +c \right )^{3}}{3}+\frac {i \tan \left (d x +c \right )^{2}}{2}\right )}{d}\) | \(108\) |
default | \(\frac {a \left (\tan \left (d x +c \right )+\frac {i \tan \left (d x +c \right )^{10}}{10}+\frac {\tan \left (d x +c \right )^{9}}{9}+\frac {i \tan \left (d x +c \right )^{8}}{2}+\frac {4 \tan \left (d x +c \right )^{7}}{7}+i \tan \left (d x +c \right )^{6}+\frac {6 \tan \left (d x +c \right )^{5}}{5}+i \tan \left (d x +c \right )^{4}+\frac {4 \tan \left (d x +c \right )^{3}}{3}+\frac {i \tan \left (d x +c \right )^{2}}{2}\right )}{d}\) | \(108\) |
Input:
int(sec(d*x+c)^10*(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
256/315*I*a*(252*exp(10*I*(d*x+c))+210*exp(8*I*(d*x+c))+120*exp(6*I*(d*x+c ))+45*exp(4*I*(d*x+c))+10*exp(2*I*(d*x+c))+1)/d/(exp(2*I*(d*x+c))+1)^10
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (82) = 164\).
Time = 0.08 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.01 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {256 \, {\left (-252 i \, a e^{\left (10 i \, d x + 10 i \, c\right )} - 210 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 120 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 45 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 10 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a\right )}}{315 \, {\left (d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(sec(d*x+c)^10*(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
-256/315*(-252*I*a*e^(10*I*d*x + 10*I*c) - 210*I*a*e^(8*I*d*x + 8*I*c) - 1 20*I*a*e^(6*I*d*x + 6*I*c) - 45*I*a*e^(4*I*d*x + 4*I*c) - 10*I*a*e^(2*I*d* x + 2*I*c) - I*a)/(d*e^(20*I*d*x + 20*I*c) + 10*d*e^(18*I*d*x + 18*I*c) + 45*d*e^(16*I*d*x + 16*I*c) + 120*d*e^(14*I*d*x + 14*I*c) + 210*d*e^(12*I*d *x + 12*I*c) + 252*d*e^(10*I*d*x + 10*I*c) + 210*d*e^(8*I*d*x + 8*I*c) + 1 20*d*e^(6*I*d*x + 6*I*c) + 45*d*e^(4*I*d*x + 4*I*c) + 10*d*e^(2*I*d*x + 2* I*c) + d)
Time = 4.34 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.88 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\begin {cases} \frac {a \left (\frac {\tan ^{9}{\left (c + d x \right )}}{9} + \frac {4 \tan ^{7}{\left (c + d x \right )}}{7} + \frac {6 \tan ^{5}{\left (c + d x \right )}}{5} + \frac {4 \tan ^{3}{\left (c + d x \right )}}{3} + \tan {\left (c + d x \right )}\right ) + \frac {i a \sec ^{10}{\left (c + d x \right )}}{10}}{d} & \text {for}\: d \neq 0 \\x \left (i a \tan {\left (c \right )} + a\right ) \sec ^{10}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(sec(d*x+c)**10*(a+I*a*tan(d*x+c)),x)
Output:
Piecewise(((a*(tan(c + d*x)**9/9 + 4*tan(c + d*x)**7/7 + 6*tan(c + d*x)**5 /5 + 4*tan(c + d*x)**3/3 + tan(c + d*x)) + I*a*sec(c + d*x)**10/10)/d, Ne( d, 0)), (x*(I*a*tan(c) + a)*sec(c)**10, True))
Time = 0.04 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {63 i \, a \tan \left (d x + c\right )^{10} + 70 \, a \tan \left (d x + c\right )^{9} + 315 i \, a \tan \left (d x + c\right )^{8} + 360 \, a \tan \left (d x + c\right )^{7} + 630 i \, a \tan \left (d x + c\right )^{6} + 756 \, a \tan \left (d x + c\right )^{5} + 630 i \, a \tan \left (d x + c\right )^{4} + 840 \, a \tan \left (d x + c\right )^{3} + 315 i \, a \tan \left (d x + c\right )^{2} + 630 \, a \tan \left (d x + c\right )}{630 \, d} \] Input:
integrate(sec(d*x+c)^10*(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
1/630*(63*I*a*tan(d*x + c)^10 + 70*a*tan(d*x + c)^9 + 315*I*a*tan(d*x + c) ^8 + 360*a*tan(d*x + c)^7 + 630*I*a*tan(d*x + c)^6 + 756*a*tan(d*x + c)^5 + 630*I*a*tan(d*x + c)^4 + 840*a*tan(d*x + c)^3 + 315*I*a*tan(d*x + c)^2 + 630*a*tan(d*x + c))/d
Time = 0.17 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.21 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {-63 i \, a \tan \left (d x + c\right )^{10} - 70 \, a \tan \left (d x + c\right )^{9} - 315 i \, a \tan \left (d x + c\right )^{8} - 360 \, a \tan \left (d x + c\right )^{7} - 630 i \, a \tan \left (d x + c\right )^{6} - 756 \, a \tan \left (d x + c\right )^{5} - 630 i \, a \tan \left (d x + c\right )^{4} - 840 \, a \tan \left (d x + c\right )^{3} - 315 i \, a \tan \left (d x + c\right )^{2} - 630 \, a \tan \left (d x + c\right )}{630 \, d} \] Input:
integrate(sec(d*x+c)^10*(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
-1/630*(-63*I*a*tan(d*x + c)^10 - 70*a*tan(d*x + c)^9 - 315*I*a*tan(d*x + c)^8 - 360*a*tan(d*x + c)^7 - 630*I*a*tan(d*x + c)^6 - 756*a*tan(d*x + c)^ 5 - 630*I*a*tan(d*x + c)^4 - 840*a*tan(d*x + c)^3 - 315*I*a*tan(d*x + c)^2 - 630*a*tan(d*x + c))/d
Time = 0.67 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.13 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a\,\left (-{\cos \left (c+d\,x\right )}^{10}\,63{}\mathrm {i}+256\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^9+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^7+96\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^5+80\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3+70\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )+63{}\mathrm {i}\right )}{630\,d\,{\cos \left (c+d\,x\right )}^{10}} \] Input:
int((a + a*tan(c + d*x)*1i)/cos(c + d*x)^10,x)
Output:
(a*(70*cos(c + d*x)*sin(c + d*x) + 80*cos(c + d*x)^3*sin(c + d*x) + 96*cos (c + d*x)^5*sin(c + d*x) + 128*cos(c + d*x)^7*sin(c + d*x) + 256*cos(c + d *x)^9*sin(c + d*x) - cos(c + d*x)^10*63i + 63i))/(630*d*cos(c + d*x)^10)
Time = 0.16 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.02 \[ \int \sec ^{10}(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) a \left (-256 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8}+1152 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}-2016 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+1680 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-630 \cos \left (d x +c \right )-63 \sin \left (d x +c \right )^{9} i +315 \sin \left (d x +c \right )^{7} i -630 \sin \left (d x +c \right )^{5} i +630 \sin \left (d x +c \right )^{3} i -315 \sin \left (d x +c \right ) i \right )}{630 d \left (\sin \left (d x +c \right )^{10}-5 \sin \left (d x +c \right )^{8}+10 \sin \left (d x +c \right )^{6}-10 \sin \left (d x +c \right )^{4}+5 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^10*(a+I*a*tan(d*x+c)),x)
Output:
(sin(c + d*x)*a*( - 256*cos(c + d*x)*sin(c + d*x)**8 + 1152*cos(c + d*x)*s in(c + d*x)**6 - 2016*cos(c + d*x)*sin(c + d*x)**4 + 1680*cos(c + d*x)*sin (c + d*x)**2 - 630*cos(c + d*x) - 63*sin(c + d*x)**9*i + 315*sin(c + d*x)* *7*i - 630*sin(c + d*x)**5*i + 630*sin(c + d*x)**3*i - 315*sin(c + d*x)*i) )/(630*d*(sin(c + d*x)**10 - 5*sin(c + d*x)**8 + 10*sin(c + d*x)**6 - 10*s in(c + d*x)**4 + 5*sin(c + d*x)**2 - 1))