Integrand size = 24, antiderivative size = 82 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 i (a-i a \tan (c+d x))^6}{3 a^9 d}-\frac {4 i (a-i a \tan (c+d x))^7}{7 a^{10} d}+\frac {i (a-i a \tan (c+d x))^8}{8 a^{11} d} \] Output:
2/3*I*(a-I*a*tan(d*x+c))^6/a^9/d-4/7*I*(a-I*a*tan(d*x+c))^7/a^10/d+1/8*I*( a-I*a*tan(d*x+c))^8/a^11/d
Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.56 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {(i+\tan (c+d x))^6 \left (-37 i+54 \tan (c+d x)+21 i \tan ^2(c+d x)\right )}{168 a^3 d} \] Input:
Integrate[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^3,x]
Output:
((I + Tan[c + d*x])^6*(-37*I + 54*Tan[c + d*x] + (21*I)*Tan[c + d*x]^2))/( 168*a^3*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.
\(\displaystyle \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^{12}}{(a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^5 (i \tan (c+d x) a+a)^2d(i a \tan (c+d x))}{a^{11} d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left ((a-i a \tan (c+d x))^7-4 a (a-i a \tan (c+d x))^6+4 a^2 (a-i a \tan (c+d x))^5\right )d(i a \tan (c+d x))}{a^{11} d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-\frac {2}{3} a^2 (a-i a \tan (c+d x))^6-\frac {1}{8} (a-i a \tan (c+d x))^8+\frac {4}{7} a (a-i a \tan (c+d x))^7\right )}{a^{11} d}\) |
Input:
Int[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^3,x]
Output:
((-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^11*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 = 0.51 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.57
method | result | size |
risch | \(\frac {32 i \left (28 \,{\mathrm e}^{4 i \left (d x +c \right )}+8 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{21 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{8}}\) | \(47\) |
derivativedivides | \(\frac {i \left (\frac {\tan \left (d x +c \right )^{8}}{8}-\frac {\tan \left (d x +c \right )^{6}}{6}+\frac {3 i \tan \left (d x +c \right )^{7}}{7}-\frac {5 \tan \left (d x +c \right )^{4}}{4}+i \tan \left (d x +c \right )^{5}-\frac {3 \tan \left (d x +c \right )^{2}}{2}+\frac {i \tan \left (d x +c \right )^{3}}{3}-i \tan \left (d x +c \right )\right )}{a^{3} d}\) | \(93\) |
default | \(\frac {i \left (\frac {\tan \left (d x +c \right )^{8}}{8}-\frac {\tan \left (d x +c \right )^{6}}{6}+\frac {3 i \tan \left (d x +c \right )^{7}}{7}-\frac {5 \tan \left (d x +c \right )^{4}}{4}+i \tan \left (d x +c \right )^{5}-\frac {3 \tan \left (d x +c \right )^{2}}{2}+\frac {i \tan \left (d x +c \right )^{3}}{3}-i \tan \left (d x +c \right )\right )}{a^{3} d}\) | \(93\) |
Input:
int(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
32/21*I*(28*exp(4*I*(d*x+c))+8*exp(2*I*(d*x+c))+1)/d/a^3/(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. 153 vs. \(2 (64) = 128\).
Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.87 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {32 \, {\left (-28 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{21 \, {\left (a^{3} d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \] Input:
integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
-32/21*(-28*I*e^(4*I*d*x + 4*I*c) - 8*I*e^(2*I*d*x + 2*I*c) - I)/(a^3*d*e^ (16*I*d*x + 16*I*c) + 8*a^3*d*e^(14*I*d*x + 14*I*c) + 28*a^3*d*e^(12*I*d*x + 12*I*c) + 56*a^3*d*e^(10*I*d*x + 10*I*c) + 70*a^3*d*e^(8*I*d*x + 8*I*c) + 56*a^3*d*e^(6*I*d*x + 6*I*c) + 28*a^3*d*e^(4*I*d*x + 4*I*c) + 8*a^3*d*e ^(2*I*d*x + 2*I*c) + a^3*d)
\[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\sec ^{12}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)**12/(a+I*a*tan(d*x+c))**3,x)
Output:
I*Integral(sec(c + d*x)**12/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan (c + d*x) + I), x)/a**3
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-21 i \, \tan \left (d x + c\right )^{8} + 72 \, \tan \left (d x + c\right )^{7} + 28 i \, \tan \left (d x + c\right )^{6} + 168 \, \tan \left (d x + c\right )^{5} + 210 i \, \tan \left (d x + c\right )^{4} + 56 \, \tan \left (d x + c\right )^{3} + 252 i \, \tan \left (d x + c\right )^{2} - 168 \, \tan \left (d x + c\right )}{168 \, a^{3} d} \] Input:
integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
-1/168*(-21*I*tan(d*x + c)^8 + 72*tan(d*x + c)^7 + 28*I*tan(d*x + c)^6 + 1 68*tan(d*x + c)^5 + 210*I*tan(d*x + c)^4 + 56*tan(d*x + c)^3 + 252*I*tan(d *x + c)^2 - 168*tan(d*x + c))/(a^3*d)
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-21 i \, \tan \left (d x + c\right )^{8} + 72 \, \tan \left (d x + c\right )^{7} + 28 i \, \tan \left (d x + c\right )^{6} + 168 \, \tan \left (d x + c\right )^{5} + 210 i \, \tan \left (d x + c\right )^{4} + 56 \, \tan \left (d x + c\right )^{3} + 252 i \, \tan \left (d x + c\right )^{2} - 168 \, \tan \left (d x + c\right )}{168 \, a^{3} d} \] Input:
integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
-1/168*(-21*I*tan(d*x + c)^8 + 72*tan(d*x + c)^7 + 28*I*tan(d*x + c)^6 + 1 68*tan(d*x + c)^5 + 210*I*tan(d*x + c)^4 + 56*tan(d*x + c)^3 + 252*I*tan(d *x + c)^2 - 168*tan(d*x + c))/(a^3*d)
Time = 0.57 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\cos \left (c+d\,x\right )}^8\,91{}\mathrm {i}+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^7+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^5+48\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3-{\cos \left (c+d\,x\right )}^2\,112{}\mathrm {i}-72\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )+21{}\mathrm {i}}{168\,a^3\,d\,{\cos \left (c+d\,x\right )}^8} \] Input:
int(1/(cos(c + d*x)^12*(a + a*tan(c + d*x)*1i)^3),x)
Output:
(48*cos(c + d*x)^3*sin(c + d*x) - 72*cos(c + d*x)*sin(c + d*x) + 64*cos(c + d*x)^5*sin(c + d*x) + 128*cos(c + d*x)^7*sin(c + d*x) - cos(c + d*x)^2*1 12i + cos(c + d*x)^8*91i + 21i)/(168*a^3*d*cos(c + d*x)^8)
Time = 0.22 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.95 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {-128 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{7}+448 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-560 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+168 \cos \left (d x +c \right ) \sin \left (d x +c \right )-35 \sin \left (d x +c \right )^{8} i +140 \sin \left (d x +c \right )^{6} i -210 \sin \left (d x +c \right )^{4} i +252 \sin \left (d x +c \right )^{2} i -126 i}{168 a^{3} d \left (\sin \left (d x +c \right )^{8}-4 \sin \left (d x +c \right )^{6}+6 \sin \left (d x +c \right )^{4}-4 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^3,x)
Output:
( - 128*cos(c + d*x)*sin(c + d*x)**7 + 448*cos(c + d*x)*sin(c + d*x)**5 - 560*cos(c + d*x)*sin(c + d*x)**3 + 168*cos(c + d*x)*sin(c + d*x) - 35*sin( c + d*x)**8*i + 140*sin(c + d*x)**6*i - 210*sin(c + d*x)**4*i + 252*sin(c + d*x)**2*i - 126*i)/(168*a**3*d*(sin(c + d*x)**8 - 4*sin(c + d*x)**6 + 6* sin(c + d*x)**4 - 4*sin(c + d*x)**2 + 1))