Integrand size = 24, antiderivative size = 82 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {2 i (a+i a \tan (c+d x))^6}{3 a^3 d}+\frac {4 i (a+i a \tan (c+d x))^7}{7 a^4 d}-\frac {i (a+i a \tan (c+d x))^8}{8 a^5 d} \]
-2/3*I*(a+I*a*tan(d*x+c))^6/a^3/d+4/7*I*(a+I*a*tan(d*x+c))^7/a^4/d-1/8*I*( a+I*a*tan(d*x+c))^8/a^5/d
Time = 0.37 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 \sec ^8(c+d x) (8+29 \cos (2 (c+d x))-27 i \sin (2 (c+d x))) (-i \cos (6 (c+d x))+\sin (6 (c+d x)))}{168 d} \]
(a^3*Sec[c + d*x]^8*(8 + 29*Cos[2*(c + d*x)] - (27*I)*Sin[2*(c + d*x)])*(( -I)*Cos[6*(c + d*x)] + Sin[6*(c + d*x)]))/(168*d)
Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.88, 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.
\(\displaystyle \int \sec ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^6 (a+i a \tan (c+d x))^3dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^5d(i a \tan (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left ((i \tan (c+d x) a+a)^7-4 a (i \tan (c+d x) a+a)^6+4 a^2 (i \tan (c+d x) a+a)^5\right )d(i a \tan (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (\frac {2}{3} a^2 (a+i a \tan (c+d x))^6+\frac {1}{8} (a+i a \tan (c+d x))^8-\frac {4}{7} a (a+i a \tan (c+d x))^7\right )}{a^5 d}\) |
((-I)*((2*a^2*(a + I*a*Tan[c + d*x])^6)/3 - (4*a*(a + I*a*Tan[c + d*x])^7) /7 + (a + I*a*Tan[c + d*x])^8/8))/(a^5*d)
3.1.37.3.1 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 = 64.76 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.98
method | result | size |
risch | \(\frac {32 i a^{3} \left (56 \,{\mathrm e}^{10 i \left (d x +c \right )}+70 \,{\mathrm e}^{8 i \left (d x +c \right )}+56 \,{\mathrm e}^{6 i \left (d x +c \right )}+28 \,{\mathrm e}^{4 i \left (d x +c \right )}+8 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{21 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{8}}\) | \(80\) |
derivativedivides | \(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{3}}{2 \cos \left (d x +c \right )^{6}}-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(174\) |
default | \(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{3}}{2 \cos \left (d x +c \right )^{6}}-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(174\) |
32/21*I*a^3*(56*exp(10*I*(d*x+c))+70*exp(8*I*(d*x+c))+56*exp(6*I*(d*x+c))+ 28*exp(4*I*(d*x+c))+8*exp(2*I*(d*x+c))+1)/d/(exp(2*I*(d*x+c))+1)^8
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (64) = 128\).
Time = 0.24 (sec) , antiderivative size = 177, normalized size of antiderivative = 2.16 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {32 \, {\left (-56 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 70 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 56 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 28 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )}}{21 \, {\left (d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
-32/21*(-56*I*a^3*e^(10*I*d*x + 10*I*c) - 70*I*a^3*e^(8*I*d*x + 8*I*c) - 5 6*I*a^3*e^(6*I*d*x + 6*I*c) - 28*I*a^3*e^(4*I*d*x + 4*I*c) - 8*I*a^3*e^(2* I*d*x + 2*I*c) - I*a^3)/(d*e^(16*I*d*x + 16*I*c) + 8*d*e^(14*I*d*x + 14*I* c) + 28*d*e^(12*I*d*x + 12*I*c) + 56*d*e^(10*I*d*x + 10*I*c) + 70*d*e^(8*I *d*x + 8*I*c) + 56*d*e^(6*I*d*x + 6*I*c) + 28*d*e^(4*I*d*x + 4*I*c) + 8*d* e^(2*I*d*x + 2*I*c) + d)
\[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=- i a^{3} \left (\int i \sec ^{6}{\left (c + d x \right )}\, dx + \int \left (- 3 \tan {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx + \int \tan ^{3}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \left (- 3 i \tan ^{2}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx\right ) \]
-I*a**3*(Integral(I*sec(c + d*x)**6, x) + Integral(-3*tan(c + d*x)*sec(c + d*x)**6, x) + Integral(tan(c + d*x)**3*sec(c + d*x)**6, x) + Integral(-3* I*tan(c + d*x)**2*sec(c + d*x)**6, x))
Time = 0.39 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.32 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {21 i \, a^{3} \tan \left (d x + c\right )^{8} + 72 \, a^{3} \tan \left (d x + c\right )^{7} - 28 i \, a^{3} \tan \left (d x + c\right )^{6} + 168 \, a^{3} \tan \left (d x + c\right )^{5} - 210 i \, a^{3} \tan \left (d x + c\right )^{4} + 56 \, a^{3} \tan \left (d x + c\right )^{3} - 252 i \, a^{3} \tan \left (d x + c\right )^{2} - 168 \, a^{3} \tan \left (d x + c\right )}{168 \, d} \]
-1/168*(21*I*a^3*tan(d*x + c)^8 + 72*a^3*tan(d*x + c)^7 - 28*I*a^3*tan(d*x + c)^6 + 168*a^3*tan(d*x + c)^5 - 210*I*a^3*tan(d*x + c)^4 + 56*a^3*tan(d *x + c)^3 - 252*I*a^3*tan(d*x + c)^2 - 168*a^3*tan(d*x + c))/d
Time = 0.54 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.32 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {21 i \, a^{3} \tan \left (d x + c\right )^{8} + 72 \, a^{3} \tan \left (d x + c\right )^{7} - 28 i \, a^{3} \tan \left (d x + c\right )^{6} + 168 \, a^{3} \tan \left (d x + c\right )^{5} - 210 i \, a^{3} \tan \left (d x + c\right )^{4} + 56 \, a^{3} \tan \left (d x + c\right )^{3} - 252 i \, a^{3} \tan \left (d x + c\right )^{2} - 168 \, a^{3} \tan \left (d x + c\right )}{168 \, d} \]
-1/168*(21*I*a^3*tan(d*x + c)^8 + 72*a^3*tan(d*x + c)^7 - 28*I*a^3*tan(d*x + c)^6 + 168*a^3*tan(d*x + c)^5 - 210*I*a^3*tan(d*x + c)^4 + 56*a^3*tan(d *x + c)^3 - 252*I*a^3*tan(d*x + c)^2 - 168*a^3*tan(d*x + c))/d
Time = 4.17 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {a^3\,\sin \left (c+d\,x\right )\,\left (-168\,{\cos \left (c+d\,x\right )}^7-{\cos \left (c+d\,x\right )}^6\,\sin \left (c+d\,x\right )\,252{}\mathrm {i}+56\,{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^2-{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^3\,210{}\mathrm {i}+168\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^4-{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^5\,28{}\mathrm {i}+72\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^6+{\sin \left (c+d\,x\right )}^7\,21{}\mathrm {i}\right )}{168\,d\,{\cos \left (c+d\,x\right )}^8} \]
-(a^3*sin(c + d*x)*(72*cos(c + d*x)*sin(c + d*x)^6 - cos(c + d*x)^6*sin(c + d*x)*252i - 168*cos(c + d*x)^7 + sin(c + d*x)^7*21i - cos(c + d*x)^2*sin (c + d*x)^5*28i + 168*cos(c + d*x)^3*sin(c + d*x)^4 - cos(c + d*x)^4*sin(c + d*x)^3*210i + 56*cos(c + d*x)^5*sin(c + d*x)^2))/(168*d*cos(c + d*x)^8)