Integrand size = 24, antiderivative size = 109 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {8 i (a+i a \tan (c+d x))^7}{7 a^4 d}+\frac {3 i (a+i a \tan (c+d x))^8}{2 a^5 d}-\frac {2 i (a+i a \tan (c+d x))^9}{3 a^6 d}+\frac {i (a+i a \tan (c+d x))^{10}}{10 a^7 d} \] Output:
-8/7*I*(a+I*a*tan(d*x+c))^7/a^4/d+3/2*I*(a+I*a*tan(d*x+c))^8/a^5/d-2/3*I*( a+I*a*tan(d*x+c))^9/a^6/d+1/10*I*(a+I*a*tan(d*x+c))^10/a^7/d
Time = 0.61 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.72 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {a^3 \sec ^9(c+d x) (\cos (7 (c+d x))+i \sin (7 (c+d x))) (-66 i+242 i \cos (2 (c+d x))+119 \sec (c+d x) \sin (3 (c+d x))+35 \tan (c+d x))}{840 d} \] Input:
Integrate[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]
Output:
-1/840*(a^3*Sec[c + d*x]^9*(Cos[7*(c + d*x)] + I*Sin[7*(c + d*x)])*(-66*I + (242*I)*Cos[2*(c + d*x)] + 119*Sec[c + d*x]*Sin[3*(c + d*x)] + 35*Tan[c + d*x]))/d
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86, 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.
| \(\displaystyle \int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^8 (a+i a \tan (c+d x))^3dx\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^3 (i \tan (c+d x) a+a)^6d(i a \tan (c+d x))}{a^7 d}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle -\frac {i \int \left (-(i \tan (c+d x) a+a)^9+6 a (i \tan (c+d x) a+a)^8-12 a^2 (i \tan (c+d x) a+a)^7+8 a^3 (i \tan (c+d x) a+a)^6\right )d(i a \tan (c+d x))}{a^7 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {i \left (\frac {8}{7} a^3 (a+i a \tan (c+d x))^7-\frac {3}{2} a^2 (a+i a \tan (c+d x))^8-\frac {1}{10} (a+i a \tan (c+d x))^{10}+\frac {2}{3} a (a+i a \tan (c+d x))^9\right )}{a^7 d}\) |
Input:
Int[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]
Output:
((-I)*((8*a^3*(a + I*a*Tan[c + d*x])^7)/7 - (3*a^2*(a + I*a*Tan[c + d*x])^ 8)/2 + (2*a*(a + I*a*Tan[c + d*x])^9)/3 - (a + I*a*Tan[c + d*x])^10/10))/( a^7*d)
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 = 169.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83
| method | result | size |
| risch | \(\frac {128 i a^{3} \left (210 \,{\mathrm e}^{12 i \left (d x +c \right )}+252 \,{\mathrm e}^{10 i \left (d x +c \right )}+210 \,{\mathrm e}^{8 i \left (d x +c \right )}+120 \,{\mathrm e}^{6 i \left (d x +c \right )}+45 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}\) | \(91\) |
| derivativedivides | \(\frac {-i a^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \sin \left (d x +c \right )^{4}}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{40 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \sin \left (d x +c \right )^{3}}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \sin \left (d x +c \right )^{3}}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {3 i a^{3}}{8 \cos \left (d x +c \right )^{8}}-a^{3} \left (-\frac {16}{35}-\frac {\sec \left (d x +c \right )^{6}}{7}-\frac {6 \sec \left (d x +c \right )^{4}}{35}-\frac {8 \sec \left (d x +c \right )^{2}}{35}\right ) \tan \left (d x +c \right )}{d}\) | \(220\) |
| default | \(\frac {-i a^{3} \left (\frac {\sin \left (d x +c \right )^{4}}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \sin \left (d x +c \right )^{4}}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin \left (d x +c \right )^{4}}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin \left (d x +c \right )^{4}}{40 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin \left (d x +c \right )^{3}}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \sin \left (d x +c \right )^{3}}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \sin \left (d x +c \right )^{3}}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {3 i a^{3}}{8 \cos \left (d x +c \right )^{8}}-a^{3} \left (-\frac {16}{35}-\frac {\sec \left (d x +c \right )^{6}}{7}-\frac {6 \sec \left (d x +c \right )^{4}}{35}-\frac {8 \sec \left (d x +c \right )^{2}}{35}\right ) \tan \left (d x +c \right )}{d}\) | \(220\) |
Input:
int(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
128/105*I*a^3*(210*exp(12*I*(d*x+c))+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/(e xp(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. 215 vs. \(2 (85) = 170\).
Time = 0.09 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.97 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {128 \, {\left (-210 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} - 252 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 210 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 120 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 45 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 10 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )}}{105 \, {\left (d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
-128/105*(-210*I*a^3*e^(12*I*d*x + 12*I*c) - 252*I*a^3*e^(10*I*d*x + 10*I* c) - 210*I*a^3*e^(8*I*d*x + 8*I*c) - 120*I*a^3*e^(6*I*d*x + 6*I*c) - 45*I* a^3*e^(4*I*d*x + 4*I*c) - 10*I*a^3*e^(2*I*d*x + 2*I*c) - I*a^3)/(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) + 120*d*e^(6*I*d*x + 6*I*c) + 4 5*d*e^(4*I*d*x + 4*I*c) + 10*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=- i a^{3} \left (\int i \sec ^{8}{\left (c + d x \right )}\, dx + \int \left (- 3 \tan {\left (c + d x \right )} \sec ^{8}{\left (c + d x \right )}\right )\, dx + \int \tan ^{3}{\left (c + d x \right )} \sec ^{8}{\left (c + d x \right )}\, dx + \int \left (- 3 i \tan ^{2}{\left (c + d x \right )} \sec ^{8}{\left (c + d x \right )}\right )\, dx\right ) \] Input:
integrate(sec(d*x+c)**8*(a+I*a*tan(d*x+c))**3,x)
Output:
-I*a**3*(Integral(I*sec(c + d*x)**8, x) + Integral(-3*tan(c + d*x)*sec(c + d*x)**8, x) + Integral(tan(c + d*x)**3*sec(c + d*x)**8, x) + Integral(-3* I*tan(c + d*x)**2*sec(c + d*x)**8, x))
Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.99 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {21 i \, a^{3} \tan \left (d x + c\right )^{10} + 70 \, a^{3} \tan \left (d x + c\right )^{9} + 240 \, a^{3} \tan \left (d x + c\right )^{7} - 210 i \, a^{3} \tan \left (d x + c\right )^{6} + 252 \, a^{3} \tan \left (d x + c\right )^{5} - 420 i \, a^{3} \tan \left (d x + c\right )^{4} - 315 i \, a^{3} \tan \left (d x + c\right )^{2} - 210 \, a^{3} \tan \left (d x + c\right )}{210 \, d} \] Input:
integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/210*(21*I*a^3*tan(d*x + c)^10 + 70*a^3*tan(d*x + c)^9 + 240*a^3*tan(d*x + c)^7 - 210*I*a^3*tan(d*x + c)^6 + 252*a^3*tan(d*x + c)^5 - 420*I*a^3*ta n(d*x + c)^4 - 315*I*a^3*tan(d*x + c)^2 - 210*a^3*tan(d*x + c))/d
Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.99 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {21 i \, a^{3} \tan \left (d x + c\right )^{10} + 70 \, a^{3} \tan \left (d x + c\right )^{9} + 240 \, a^{3} \tan \left (d x + c\right )^{7} - 210 i \, a^{3} \tan \left (d x + c\right )^{6} + 252 \, a^{3} \tan \left (d x + c\right )^{5} - 420 i \, a^{3} \tan \left (d x + c\right )^{4} - 315 i \, a^{3} \tan \left (d x + c\right )^{2} - 210 \, a^{3} \tan \left (d x + c\right )}{210 \, d} \] Input:
integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
-1/210*(21*I*a^3*tan(d*x + c)^10 + 70*a^3*tan(d*x + c)^9 + 240*a^3*tan(d*x + c)^7 - 210*I*a^3*tan(d*x + c)^6 + 252*a^3*tan(d*x + c)^5 - 420*I*a^3*ta n(d*x + c)^4 - 315*I*a^3*tan(d*x + c)^2 - 210*a^3*tan(d*x + c))/d
Time = 0.38 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.39 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=-\frac {a^3\,\sin \left (c+d\,x\right )\,\left (-210\,{\cos \left (c+d\,x\right )}^9-{\cos \left (c+d\,x\right )}^8\,\sin \left (c+d\,x\right )\,315{}\mathrm {i}-{\cos \left (c+d\,x\right )}^6\,{\sin \left (c+d\,x\right )}^3\,420{}\mathrm {i}+252\,{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^4-{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^5\,210{}\mathrm {i}+240\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^6+70\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^8+{\sin \left (c+d\,x\right )}^9\,21{}\mathrm {i}\right )}{210\,d\,{\cos \left (c+d\,x\right )}^{10}} \] Input:
int((a + a*tan(c + d*x)*1i)^3/cos(c + d*x)^8,x)
Output:
-(a^3*sin(c + d*x)*(70*cos(c + d*x)*sin(c + d*x)^8 - cos(c + d*x)^8*sin(c + d*x)*315i - 210*cos(c + d*x)^9 + sin(c + d*x)^9*21i + 240*cos(c + d*x)^3 *sin(c + d*x)^6 - cos(c + d*x)^4*sin(c + d*x)^5*210i + 252*cos(c + d*x)^5* sin(c + d*x)^4 - cos(c + d*x)^6*sin(c + d*x)^3*420i))/(210*d*cos(c + d*x)^ 10)
Time = 0.17 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.76 \[ \int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {\sin \left (d x +c \right ) a^{3} \left (-128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8}+576 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}-1008 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+840 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-210 \cos \left (d x +c \right )-84 \sin \left (d x +c \right )^{9} i +420 \sin \left (d x +c \right )^{7} i -840 \sin \left (d x +c \right )^{5} i +840 \sin \left (d x +c \right )^{3} i -315 \sin \left (d x +c \right ) i \right )}{210 d \left (\sin \left (d x +c \right )^{10}-5 \sin \left (d x +c \right )^{8}+10 \sin \left (d x +c \right )^{6}-10 \sin \left (d x +c \right )^{4}+5 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x)
Output:
(sin(c + d*x)*a**3*( - 128*cos(c + d*x)*sin(c + d*x)**8 + 576*cos(c + d*x) *sin(c + d*x)**6 - 1008*cos(c + d*x)*sin(c + d*x)**4 + 840*cos(c + d*x)*si n(c + d*x)**2 - 210*cos(c + d*x) - 84*sin(c + d*x)**9*i + 420*sin(c + d*x) **7*i - 840*sin(c + d*x)**5*i + 840*sin(c + d*x)**3*i - 315*sin(c + d*x)*i ))/(210*d*(sin(c + d*x)**10 - 5*sin(c + d*x)**8 + 10*sin(c + d*x)**6 - 10* sin(c + d*x)**4 + 5*sin(c + d*x)**2 - 1))