Integrand size = 24, antiderivative size = 82 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {4 i (a-i a \tan (c+d x))^7}{7 a^{11} d}-\frac {i (a-i a \tan (c+d x))^8}{2 a^{12} d}+\frac {i (a-i a \tan (c+d x))^9}{9 a^{13} d} \] Output:
4/7*I*(a-I*a*tan(d*x+c))^7/a^11/d-1/2*I*(a-I*a*tan(d*x+c))^8/a^12/d+1/9*I* (a-I*a*tan(d*x+c))^9/a^13/d
Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.54 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {(i+\tan (c+d x))^7 \left (-23-35 i \tan (c+d x)+14 \tan ^2(c+d x)\right )}{126 a^4 d} \] Input:
Integrate[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^4,x]
Output:
((I + Tan[c + d*x])^7*(-23 - (35*I)*Tan[c + d*x] + 14*Tan[c + d*x]^2))/(12 6*a^4*d)
Time = 0.28 (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.
\(\displaystyle \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^{14}}{(a+i a \tan (c+d x))^4}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int (a-i a \tan (c+d x))^6 (i \tan (c+d x) a+a)^2d(i a \tan (c+d x))}{a^{13} d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {i \int \left ((a-i a \tan (c+d x))^8-4 a (a-i a \tan (c+d x))^7+4 a^2 (a-i a \tan (c+d x))^6\right )d(i a \tan (c+d x))}{a^{13} d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-\frac {4}{7} a^2 (a-i a \tan (c+d x))^7-\frac {1}{9} (a-i a \tan (c+d x))^9+\frac {1}{2} a (a-i a \tan (c+d x))^8\right )}{a^{13} d}\) |
Input:
Int[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^4,x]
Output:
((-I)*((-4*a^2*(a - I*a*Tan[c + d*x])^7)/7 + (a*(a - I*a*Tan[c + d*x])^8)/ 2 - (a - I*a*Tan[c + d*x])^9/9))/(a^13*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 = 0.65 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.57
method | result | size |
risch | \(\frac {128 i \left (36 \,{\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{63 d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{9}}\) | \(47\) |
derivativedivides | \(\frac {\tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )^{9}}{9}+\frac {i \tan \left (d x +c \right )^{8}}{2}-\frac {4 \tan \left (d x +c \right )^{7}}{7}+\frac {2 i \tan \left (d x +c \right )^{6}}{3}-2 \tan \left (d x +c \right )^{5}-i \tan \left (d x +c \right )^{4}-\frac {4 \tan \left (d x +c \right )^{3}}{3}-2 i \tan \left (d x +c \right )^{2}}{a^{4} d}\) | \(99\) |
default | \(\frac {\tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )^{9}}{9}+\frac {i \tan \left (d x +c \right )^{8}}{2}-\frac {4 \tan \left (d x +c \right )^{7}}{7}+\frac {2 i \tan \left (d x +c \right )^{6}}{3}-2 \tan \left (d x +c \right )^{5}-i \tan \left (d x +c \right )^{4}-\frac {4 \tan \left (d x +c \right )^{3}}{3}-2 i \tan \left (d x +c \right )^{2}}{a^{4} d}\) | \(99\) |
Input:
int(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
128/63*I*(36*exp(4*I*(d*x+c))+9*exp(2*I*(d*x+c))+1)/d/a^4/(exp(2*I*(d*x+c) )+1)^9
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (64) = 128\).
Time = 0.09 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.05 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {128 \, {\left (-36 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{63 \, {\left (a^{4} d e^{\left (18 i \, d x + 18 i \, c\right )} + 9 \, a^{4} d e^{\left (16 i \, d x + 16 i \, c\right )} + 36 \, a^{4} d e^{\left (14 i \, d x + 14 i \, c\right )} + 84 \, a^{4} d e^{\left (12 i \, d x + 12 i \, c\right )} + 126 \, a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + 126 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 84 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \] Input:
integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
Output:
-128/63*(-36*I*e^(4*I*d*x + 4*I*c) - 9*I*e^(2*I*d*x + 2*I*c) - I)/(a^4*d*e ^(18*I*d*x + 18*I*c) + 9*a^4*d*e^(16*I*d*x + 16*I*c) + 36*a^4*d*e^(14*I*d* x + 14*I*c) + 84*a^4*d*e^(12*I*d*x + 12*I*c) + 126*a^4*d*e^(10*I*d*x + 10* I*c) + 126*a^4*d*e^(8*I*d*x + 8*I*c) + 84*a^4*d*e^(6*I*d*x + 6*I*c) + 36*a ^4*d*e^(4*I*d*x + 4*I*c) + 9*a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)
\[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{14}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(sec(d*x+c)**14/(a+I*a*tan(d*x+c))**4,x)
Output:
Integral(sec(c + d*x)**14/(tan(c + d*x)**4 - 4*I*tan(c + d*x)**3 - 6*tan(c + d*x)**2 + 4*I*tan(c + d*x) + 1), x)/a**4
Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {14 \, \tan \left (d x + c\right )^{9} + 63 i \, \tan \left (d x + c\right )^{8} - 72 \, \tan \left (d x + c\right )^{7} + 84 i \, \tan \left (d x + c\right )^{6} - 252 \, \tan \left (d x + c\right )^{5} - 126 i \, \tan \left (d x + c\right )^{4} - 168 \, \tan \left (d x + c\right )^{3} - 252 i \, \tan \left (d x + c\right )^{2} + 126 \, \tan \left (d x + c\right )}{126 \, a^{4} d} \] Input:
integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
Output:
1/126*(14*tan(d*x + c)^9 + 63*I*tan(d*x + c)^8 - 72*tan(d*x + c)^7 + 84*I* tan(d*x + c)^6 - 252*tan(d*x + c)^5 - 126*I*tan(d*x + c)^4 - 168*tan(d*x + c)^3 - 252*I*tan(d*x + c)^2 + 126*tan(d*x + c))/(a^4*d)
Time = 0.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {14 \, \tan \left (d x + c\right )^{9} + 63 i \, \tan \left (d x + c\right )^{8} - 72 \, \tan \left (d x + c\right )^{7} + 84 i \, \tan \left (d x + c\right )^{6} - 252 \, \tan \left (d x + c\right )^{5} - 126 i \, \tan \left (d x + c\right )^{4} - 168 \, \tan \left (d x + c\right )^{3} - 252 i \, \tan \left (d x + c\right )^{2} + 126 \, \tan \left (d x + c\right )}{126 \, a^{4} d} \] Input:
integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
Output:
1/126*(14*tan(d*x + c)^9 + 63*I*tan(d*x + c)^8 - 72*tan(d*x + c)^7 + 84*I* tan(d*x + c)^6 - 252*tan(d*x + c)^5 - 126*I*tan(d*x + c)^4 - 168*tan(d*x + c)^3 - 252*I*tan(d*x + c)^2 + 126*tan(d*x + c))/(a^4*d)
Time = 0.66 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\cos \left (c+d\,x\right )}^9\,105{}\mathrm {i}+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^8+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^6+48\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^4-{\cos \left (c+d\,x\right )}^3\,168{}\mathrm {i}-128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^2+\cos \left (c+d\,x\right )\,63{}\mathrm {i}+14\,\sin \left (c+d\,x\right )}{126\,a^4\,d\,{\cos \left (c+d\,x\right )}^9} \] Input:
int(1/(cos(c + d*x)^14*(a + a*tan(c + d*x)*1i)^4),x)
Output:
(cos(c + d*x)*63i + 14*sin(c + d*x) - 128*cos(c + d*x)^2*sin(c + d*x) + 48 *cos(c + d*x)^4*sin(c + d*x) + 64*cos(c + d*x)^6*sin(c + d*x) + 128*cos(c + d*x)^8*sin(c + d*x) - cos(c + d*x)^3*168i + cos(c + d*x)^9*105i)/(126*a^ 4*d*cos(c + d*x)^9)
Time = 0.20 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.24 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {-7 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{8} i +28 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} i -42 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} i +196 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} i -112 \cos \left (d x +c \right ) i +128 \sin \left (d x +c \right )^{9}-576 \sin \left (d x +c \right )^{7}+1008 \sin \left (d x +c \right )^{5}-672 \sin \left (d x +c \right )^{3}+126 \sin \left (d x +c \right )}{126 \cos \left (d x +c \right ) a^{4} d \left (\sin \left (d x +c \right )^{8}-4 \sin \left (d x +c \right )^{6}+6 \sin \left (d x +c \right )^{4}-4 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^4,x)
Output:
( - 7*cos(c + d*x)*sin(c + d*x)**8*i + 28*cos(c + d*x)*sin(c + d*x)**6*i - 42*cos(c + d*x)*sin(c + d*x)**4*i + 196*cos(c + d*x)*sin(c + d*x)**2*i - 112*cos(c + d*x)*i + 128*sin(c + d*x)**9 - 576*sin(c + d*x)**7 + 1008*sin( c + d*x)**5 - 672*sin(c + d*x)**3 + 126*sin(c + d*x))/(126*cos(c + d*x)*a* *4*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))