Integrand size = 24, antiderivative size = 82 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {4 i (a+i a \tan (c+d x))^5}{5 a^3 d}+\frac {2 i (a+i a \tan (c+d x))^6}{3 a^4 d}-\frac {i (a+i a \tan (c+d x))^7}{7 a^5 d} \] Output:
-4/5*I*(a+I*a*tan(d*x+c))^5/a^3/d+2/3*I*(a+I*a*tan(d*x+c))^6/a^4/d-1/7*I*( a+I*a*tan(d*x+c))^7/a^5/d
Time = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.77 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 \sec ^7(c+d x) (7+22 \cos (2 (c+d x))-20 i \sin (2 (c+d x))) (-i \cos (5 (c+d x))+\sin (5 (c+d x)))}{105 d} \] Input:
Integrate[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]
Output:
(a^2*Sec[c + d*x]^7*(7 + 22*Cos[2*(c + d*x)] - (20*I)*Sin[2*(c + d*x)])*(( -I)*Cos[5*(c + d*x)] + Sin[5*(c + d*x)]))/(105*d)
Time = 0.27 (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.
\(\displaystyle \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (c+d x)^6 (a+i a \tan (c+d x))^2dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^2 (i \tan (c+d x) a+a)^4d(i a \tan (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left ((i \tan (c+d x) a+a)^6-4 a (i \tan (c+d x) a+a)^5+4 a^2 (i \tan (c+d x) a+a)^4\right )d(i a \tan (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (\frac {4}{5} a^2 (a+i a \tan (c+d x))^5+\frac {1}{7} (a+i a \tan (c+d x))^7-\frac {2}{3} a (a+i a \tan (c+d x))^6\right )}{a^5 d}\) |
Input:
Int[Sec[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]
Output:
((-I)*((4*a^2*(a + I*a*Tan[c + d*x])^5)/5 - (2*a*(a + I*a*Tan[c + d*x])^6) /3 + (a + I*a*Tan[c + d*x])^7/7))/(a^5*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 = 33.53 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {128 i a^{2} \left (35 \,{\mathrm e}^{8 i \left (d x +c \right )}+35 \,{\mathrm e}^{6 i \left (d x +c \right )}+21 \,{\mathrm e}^{4 i \left (d x +c \right )}+7 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\) | \(69\) |
derivativedivides | \(\frac {-a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \sin \left (d x +c \right )^{3}}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{3 \cos \left (d x +c \right )^{6}}-a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(113\) |
default | \(\frac {-a^{2} \left (\frac {\sin \left (d x +c \right )^{3}}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \sin \left (d x +c \right )^{3}}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \sin \left (d x +c \right )^{3}}{105 \cos \left (d x +c \right )^{3}}\right )+\frac {i a^{2}}{3 \cos \left (d x +c \right )^{6}}-a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(113\) |
Input:
int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
128/105*I*a^2*(35*exp(8*I*(d*x+c))+35*exp(6*I*(d*x+c))+21*exp(4*I*(d*x+c)) +7*exp(2*I*(d*x+c))+1)/d/(exp(2*I*(d*x+c))+1)^7
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (64) = 128\).
Time = 0.09 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.84 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {128 \, {\left (-35 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 35 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 21 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 7 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{2}\right )}}{105 \, {\left (d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
Output:
-128/105*(-35*I*a^2*e^(8*I*d*x + 8*I*c) - 35*I*a^2*e^(6*I*d*x + 6*I*c) - 2 1*I*a^2*e^(4*I*d*x + 4*I*c) - 7*I*a^2*e^(2*I*d*x + 2*I*c) - I*a^2)/(d*e^(1 4*I*d*x + 14*I*c) + 7*d*e^(12*I*d*x + 12*I*c) + 21*d*e^(10*I*d*x + 10*I*c) + 35*d*e^(8*I*d*x + 8*I*c) + 35*d*e^(6*I*d*x + 6*I*c) + 21*d*e^(4*I*d*x + 4*I*c) + 7*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=- a^{2} \left (\int \tan ^{2}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \left (- 2 i \tan {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\right )\, dx + \int \left (- \sec ^{6}{\left (c + d x \right )}\right )\, dx\right ) \] Input:
integrate(sec(d*x+c)**6*(a+I*a*tan(d*x+c))**2,x)
Output:
-a**2*(Integral(tan(c + d*x)**2*sec(c + d*x)**6, x) + Integral(-2*I*tan(c + d*x)*sec(c + d*x)**6, x) + Integral(-sec(c + d*x)**6, x))
Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {15 \, a^{2} \tan \left (d x + c\right )^{7} - 35 i \, a^{2} \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} - 105 i \, a^{2} \tan \left (d x + c\right )^{4} - 35 \, a^{2} \tan \left (d x + c\right )^{3} - 105 i \, a^{2} \tan \left (d x + c\right )^{2} - 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \] Input:
integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
Output:
-1/105*(15*a^2*tan(d*x + c)^7 - 35*I*a^2*tan(d*x + c)^6 + 21*a^2*tan(d*x + c)^5 - 105*I*a^2*tan(d*x + c)^4 - 35*a^2*tan(d*x + c)^3 - 105*I*a^2*tan(d *x + c)^2 - 105*a^2*tan(d*x + c))/d
Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {15 \, a^{2} \tan \left (d x + c\right )^{7} - 35 i \, a^{2} \tan \left (d x + c\right )^{6} + 21 \, a^{2} \tan \left (d x + c\right )^{5} - 105 i \, a^{2} \tan \left (d x + c\right )^{4} - 35 \, a^{2} \tan \left (d x + c\right )^{3} - 105 i \, a^{2} \tan \left (d x + c\right )^{2} - 105 \, a^{2} \tan \left (d x + c\right )}{105 \, d} \] Input:
integrate(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
Output:
-1/105*(15*a^2*tan(d*x + c)^7 - 35*I*a^2*tan(d*x + c)^6 + 21*a^2*tan(d*x + c)^5 - 105*I*a^2*tan(d*x + c)^4 - 35*a^2*tan(d*x + c)^3 - 105*I*a^2*tan(d *x + c)^2 - 105*a^2*tan(d*x + c))/d
Time = 0.34 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.61 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2\,\sin \left (c+d\,x\right )\,\left (105\,{\cos \left (c+d\,x\right )}^6+{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )\,105{}\mathrm {i}+35\,{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2+{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^3\,105{}\mathrm {i}-21\,{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^4+\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^5\,35{}\mathrm {i}-15\,{\sin \left (c+d\,x\right )}^6\right )}{105\,d\,{\cos \left (c+d\,x\right )}^7} \] Input:
int((a + a*tan(c + d*x)*1i)^2/cos(c + d*x)^6,x)
Output:
(a^2*sin(c + d*x)*(cos(c + d*x)*sin(c + d*x)^5*35i + cos(c + d*x)^5*sin(c + d*x)*105i + 105*cos(c + d*x)^6 - 15*sin(c + d*x)^6 - 21*cos(c + d*x)^2*s in(c + d*x)^4 + cos(c + d*x)^3*sin(c + d*x)^3*105i + 35*cos(c + d*x)^4*sin (c + d*x)^2))/(105*d*cos(c + d*x)^7)
Time = 0.16 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.65 \[ \int \sec ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {\sin \left (d x +c \right ) a^{2} \left (-35 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} i +105 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} i -105 \cos \left (d x +c \right ) \sin \left (d x +c \right ) i +64 \sin \left (d x +c \right )^{6}-224 \sin \left (d x +c \right )^{4}+280 \sin \left (d x +c \right )^{2}-105\right )}{105 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{6}-3 \sin \left (d x +c \right )^{4}+3 \sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(sec(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x)
Output:
(sin(c + d*x)*a**2*( - 35*cos(c + d*x)*sin(c + d*x)**5*i + 105*cos(c + d*x )*sin(c + d*x)**3*i - 105*cos(c + d*x)*sin(c + d*x)*i + 64*sin(c + d*x)**6 - 224*sin(c + d*x)**4 + 280*sin(c + d*x)**2 - 105))/(105*cos(c + d*x)*d*( sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3*sin(c + d*x)**2 - 1))