Integrand size = 24, antiderivative size = 134 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {192 x}{a^8}-\frac {192 i \log (\cos (c+d x))}{a^8 d}+\frac {129 \tan (c+d x)}{a^8 d}-\frac {36 i \tan ^2(c+d x)}{a^8 d}-\frac {10 \tan ^3(c+d x)}{a^8 d}+\frac {2 i \tan ^4(c+d x)}{a^8 d}+\frac {\tan ^5(c+d x)}{5 a^8 d}+\frac {64 i}{d \left (a^8+i a^8 \tan (c+d x)\right )} \]
-192*x/a^8-192*I*ln(cos(d*x+c))/a^8/d+129*tan(d*x+c)/a^8/d-36*I*tan(d*x+c) ^2/a^8/d-10*tan(d*x+c)^3/a^8/d+2*I*tan(d*x+c)^4/a^8/d+1/5*tan(d*x+c)^5/a^8 /d+64*I/d/(a^8+I*a^8*tan(d*x+c))
Time = 0.84 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.73 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {i \left (-960 \log (i-\tan (c+d x))+645 i \tan (c+d x)+180 \tan ^2(c+d x)-50 i \tan ^3(c+d x)-10 \tan ^4(c+d x)+i \tan ^5(c+d x)+\frac {320 i}{-i+\tan (c+d x)}\right )}{5 a^8 d} \]
((-1/5*I)*(-960*Log[I - Tan[c + d*x]] + (645*I)*Tan[c + d*x] + 180*Tan[c + d*x]^2 - (50*I)*Tan[c + d*x]^3 - 10*Tan[c + d*x]^4 + I*Tan[c + d*x]^5 + ( 320*I)/(-I + Tan[c + d*x])))/(a^8*d)
Time = 0.29 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3968, 49, 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.
| \(\displaystyle \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{14}}{(a+i a \tan (c+d x))^8}dx\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle -\frac {i \int \frac {(a-i a \tan (c+d x))^6}{(i \tan (c+d x) a+a)^2}d(i a \tan (c+d x))}{a^{13} d}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle -\frac {i \int \left (\frac {64 a^6}{(i \tan (c+d x) a+a)^2}-\frac {192 a^5}{i \tan (c+d x) a+a}+\tan ^4(c+d x) a^4+8 i \tan ^3(c+d x) a^4-30 \tan ^2(c+d x) a^4-72 i \tan (c+d x) a^4+129 a^4\right )d(i a \tan (c+d x))}{a^{13} d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {i \left (-\frac {64 a^6}{a+i a \tan (c+d x)}+\frac {1}{5} i a^5 \tan ^5(c+d x)-2 a^5 \tan ^4(c+d x)-10 i a^5 \tan ^3(c+d x)+36 a^5 \tan ^2(c+d x)+129 i a^5 \tan (c+d x)-192 a^5 \log (a+i a \tan (c+d x))\right )}{a^{13} d}\) |
((-I)*(-192*a^5*Log[a + I*a*Tan[c + d*x]] + (129*I)*a^5*Tan[c + d*x] + 36* a^5*Tan[c + d*x]^2 - (10*I)*a^5*Tan[c + d*x]^3 - 2*a^5*Tan[c + d*x]^4 + (I /5)*a^5*Tan[c + d*x]^5 - (64*a^6)/(a + I*a*Tan[c + d*x])))/(a^13*d)
3.2.66.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 0.56 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93
| method | result | size |
| risch | \(\frac {32 i {\mathrm e}^{-2 i \left (d x +c \right )}}{a^{8} d}-\frac {384 x}{a^{8}}-\frac {384 c}{a^{8} d}+\frac {16 i \left (50 \,{\mathrm e}^{8 i \left (d x +c \right )}+220 \,{\mathrm e}^{6 i \left (d x +c \right )}+370 \,{\mathrm e}^{4 i \left (d x +c \right )}+285 \,{\mathrm e}^{2 i \left (d x +c \right )}+87\right )}{5 d \,a^{8} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {192 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{8} d}\) | \(124\) |
| derivativedivides | \(\frac {129 \tan \left (d x +c \right )}{a^{8} d}+\frac {\tan ^{5}\left (d x +c \right )}{5 a^{8} d}+\frac {2 i \left (\tan ^{4}\left (d x +c \right )\right )}{a^{8} d}-\frac {10 \left (\tan ^{3}\left (d x +c \right )\right )}{a^{8} d}-\frac {36 i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{8} d}+\frac {64}{a^{8} d \left (\tan \left (d x +c \right )-i\right )}-\frac {192 \arctan \left (\tan \left (d x +c \right )\right )}{a^{8} d}+\frac {96 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{8} d}\) | \(136\) |
| default | \(\frac {129 \tan \left (d x +c \right )}{a^{8} d}+\frac {\tan ^{5}\left (d x +c \right )}{5 a^{8} d}+\frac {2 i \left (\tan ^{4}\left (d x +c \right )\right )}{a^{8} d}-\frac {10 \left (\tan ^{3}\left (d x +c \right )\right )}{a^{8} d}-\frac {36 i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{8} d}+\frac {64}{a^{8} d \left (\tan \left (d x +c \right )-i\right )}-\frac {192 \arctan \left (\tan \left (d x +c \right )\right )}{a^{8} d}+\frac {96 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{a^{8} d}\) | \(136\) |
32*I/a^8/d*exp(-2*I*(d*x+c))-384*x/a^8-384/a^8/d*c+16/5*I*(50*exp(8*I*(d*x +c))+220*exp(6*I*(d*x+c))+370*exp(4*I*(d*x+c))+285*exp(2*I*(d*x+c))+87)/d/ a^8/(exp(2*I*(d*x+c))+1)^5-192*I/a^8/d*ln(exp(2*I*(d*x+c))+1)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (122) = 244\).
Time = 0.30 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.04 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {16 \, {\left (120 \, d x e^{\left (12 i \, d x + 12 i \, c\right )} + 60 \, {\left (10 \, d x - i\right )} e^{\left (10 i \, d x + 10 i \, c\right )} + 30 \, {\left (40 \, d x - 9 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, {\left (120 \, d x - 47 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, {\left (120 \, d x - 77 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (120 \, d x - 137 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 60 \, {\left (i \, e^{\left (12 i \, d x + 12 i \, c\right )} + 5 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 10 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 10 i\right )}}{5 \, {\left (a^{8} d e^{\left (12 i \, d x + 12 i \, c\right )} + 5 \, a^{8} d e^{\left (10 i \, d x + 10 i \, c\right )} + 10 \, a^{8} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{8} d e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, a^{8} d e^{\left (4 i \, d x + 4 i \, c\right )} + a^{8} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]
-16/5*(120*d*x*e^(12*I*d*x + 12*I*c) + 60*(10*d*x - I)*e^(10*I*d*x + 10*I* c) + 30*(40*d*x - 9*I)*e^(8*I*d*x + 8*I*c) + 10*(120*d*x - 47*I)*e^(6*I*d* x + 6*I*c) + 5*(120*d*x - 77*I)*e^(4*I*d*x + 4*I*c) + (120*d*x - 137*I)*e^ (2*I*d*x + 2*I*c) + 60*(I*e^(12*I*d*x + 12*I*c) + 5*I*e^(10*I*d*x + 10*I*c ) + 10*I*e^(8*I*d*x + 8*I*c) + 10*I*e^(6*I*d*x + 6*I*c) + 5*I*e^(4*I*d*x + 4*I*c) + I*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 10*I)/(a^8 *d*e^(12*I*d*x + 12*I*c) + 5*a^8*d*e^(10*I*d*x + 10*I*c) + 10*a^8*d*e^(8*I *d*x + 8*I*c) + 10*a^8*d*e^(6*I*d*x + 6*I*c) + 5*a^8*d*e^(4*I*d*x + 4*I*c) + a^8*d*e^(2*I*d*x + 2*I*c))
\[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\int \frac {\sec ^{14}{\left (c + d x \right )}}{\tan ^{8}{\left (c + d x \right )} - 8 i \tan ^{7}{\left (c + d x \right )} - 28 \tan ^{6}{\left (c + d x \right )} + 56 i \tan ^{5}{\left (c + d x \right )} + 70 \tan ^{4}{\left (c + d x \right )} - 56 i \tan ^{3}{\left (c + d x \right )} - 28 \tan ^{2}{\left (c + d x \right )} + 8 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{8}} \]
Integral(sec(c + d*x)**14/(tan(c + d*x)**8 - 8*I*tan(c + d*x)**7 - 28*tan( c + d*x)**6 + 56*I*tan(c + d*x)**5 + 70*tan(c + d*x)**4 - 56*I*tan(c + d*x )**3 - 28*tan(c + d*x)**2 + 8*I*tan(c + d*x) + 1), x)/a**8
Time = 0.21 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.71 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\frac {320 \, {\left (\tan \left (d x + c\right )^{6} - 6 i \, \tan \left (d x + c\right )^{5} - 15 \, \tan \left (d x + c\right )^{4} + 20 i \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )^{2} - 6 i \, \tan \left (d x + c\right ) - 1\right )}}{a^{8} \tan \left (d x + c\right )^{7} - 7 i \, a^{8} \tan \left (d x + c\right )^{6} - 21 \, a^{8} \tan \left (d x + c\right )^{5} + 35 i \, a^{8} \tan \left (d x + c\right )^{4} + 35 \, a^{8} \tan \left (d x + c\right )^{3} - 21 i \, a^{8} \tan \left (d x + c\right )^{2} - 7 \, a^{8} \tan \left (d x + c\right ) + i \, a^{8}} + \frac {\tan \left (d x + c\right )^{5} + 10 i \, \tan \left (d x + c\right )^{4} - 50 \, \tan \left (d x + c\right )^{3} - 180 i \, \tan \left (d x + c\right )^{2} + 645 \, \tan \left (d x + c\right )}{a^{8}} + \frac {960 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{8}}}{5 \, d} \]
1/5*(320*(tan(d*x + c)^6 - 6*I*tan(d*x + c)^5 - 15*tan(d*x + c)^4 + 20*I*t an(d*x + c)^3 + 15*tan(d*x + c)^2 - 6*I*tan(d*x + c) - 1)/(a^8*tan(d*x + c )^7 - 7*I*a^8*tan(d*x + c)^6 - 21*a^8*tan(d*x + c)^5 + 35*I*a^8*tan(d*x + c)^4 + 35*a^8*tan(d*x + c)^3 - 21*I*a^8*tan(d*x + c)^2 - 7*a^8*tan(d*x + c ) + I*a^8) + (tan(d*x + c)^5 + 10*I*tan(d*x + c)^4 - 50*tan(d*x + c)^3 - 1 80*I*tan(d*x + c)^2 + 645*tan(d*x + c))/a^8 + 960*I*log(I*tan(d*x + c) + 1 )/a^8)/d
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (122) = 244\).
Time = 1.93 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.87 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {2 \, {\left (\frac {480 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{8}} - \frac {960 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{8}} + \frac {480 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{8}} + \frac {160 \, {\left (9 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 i\right )}}{a^{8} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{2}} + \frac {-1096 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 645 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 5840 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 2780 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12120 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 4286 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12120 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2780 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5840 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 645 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1096 i}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5} a^{8}}\right )}}{5 \, d} \]
-2/5*(480*I*log(tan(1/2*d*x + 1/2*c) + 1)/a^8 - 960*I*log(tan(1/2*d*x + 1/ 2*c) - I)/a^8 + 480*I*log(tan(1/2*d*x + 1/2*c) - 1)/a^8 + 160*(9*I*tan(1/2 *d*x + 1/2*c)^2 + 20*tan(1/2*d*x + 1/2*c) - 9*I)/(a^8*(tan(1/2*d*x + 1/2*c ) - I)^2) + (-1096*I*tan(1/2*d*x + 1/2*c)^10 + 645*tan(1/2*d*x + 1/2*c)^9 + 5840*I*tan(1/2*d*x + 1/2*c)^8 - 2780*tan(1/2*d*x + 1/2*c)^7 - 12120*I*ta n(1/2*d*x + 1/2*c)^6 + 4286*tan(1/2*d*x + 1/2*c)^5 + 12120*I*tan(1/2*d*x + 1/2*c)^4 - 2780*tan(1/2*d*x + 1/2*c)^3 - 5840*I*tan(1/2*d*x + 1/2*c)^2 + 645*tan(1/2*d*x + 1/2*c) + 1096*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^5*a^8))/d
Time = 4.13 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {\frac {129\,\mathrm {tan}\left (c+d\,x\right )}{a^8}-\frac {10\,{\mathrm {tan}\left (c+d\,x\right )}^3}{a^8}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{5\,a^8}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,192{}\mathrm {i}}{a^8}+\frac {64{}\mathrm {i}}{a^8\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,36{}\mathrm {i}}{a^8}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,2{}\mathrm {i}}{a^8}}{d} \]