\(\int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [148]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {4 i (a-i a \tan (c+d x))^7}{7 a^{11} d}-\frac {i (a-i a \tan (c+d x))^8}{2 a^{12} d}+\frac {i (a-i a \tan (c+d x))^9}{9 a^{13} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i (a-i a \tan (c+d x))^9}{9 a^{13} d}-\frac {i (a-i a \tan (c+d x))^8}{2 a^{12} d}+\frac {4 i (a-i a \tan (c+d x))^7}{7 a^{11} d} \]

[In]

Int[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(((4*I)/7)*(a - I*a*Tan[c + d*x])^7)/(a^11*d) - ((I/2)*(a - I*a*Tan[c + d*x])^8)/(a^12*d) + ((I/9)*(a - I*a*Ta
n[c + d*x])^9)/(a^13*d)

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int (a-x)^6 (a+x)^2 \, dx,x,i a \tan (c+d x)\right )}{a^{13} d} \\ & = -\frac {i \text {Subst}\left (\int \left (4 a^2 (a-x)^6-4 a (a-x)^7+(a-x)^8\right ) \, dx,x,i a \tan (c+d x)\right )}{a^{13} d} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.54 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {(i+\tan (c+d x))^7 \left (-23-35 i \tan (c+d x)+14 \tan ^2(c+d x)\right )}{126 a^4 d} \]

[In]

Integrate[Sec[c + d*x]^14/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I + Tan[c + d*x])^7*(-23 - (35*I)*Tan[c + d*x] + 14*Tan[c + d*x]^2))/(126*a^4*d)

Maple [A] (verified)

Time = 0.47 (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	\(-\frac {-\tan \left (d x +c \right )-\frac {\left (\tan ^{9}\left (d x +c \right )\right )}{9}-\frac {i \left (\tan ^{8}\left (d x +c \right )\right )}{2}+\frac {4 \left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {2 i \left (\tan ^{6}\left (d x +c \right )\right )}{3}+2 \left (\tan ^{5}\left (d x +c \right )\right )+i \left (\tan ^{4}\left (d x +c \right )\right )+\frac {4 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+2 i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{4} d}\)	\(102\)
default	\(-\frac {-\tan \left (d x +c \right )-\frac {\left (\tan ^{9}\left (d x +c \right )\right )}{9}-\frac {i \left (\tan ^{8}\left (d x +c \right )\right )}{2}+\frac {4 \left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {2 i \left (\tan ^{6}\left (d x +c \right )\right )}{3}+2 \left (\tan ^{5}\left (d x +c \right )\right )+i \left (\tan ^{4}\left (d x +c \right )\right )+\frac {4 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+2 i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{4} d}\)	\(102\)

[In]

int(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

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 = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.05 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {128 \, {\left (-36 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 9 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{63 \, {\left (a^{4} d e^{\left (18 i \, d x + 18 i \, c\right )} + 9 \, a^{4} d e^{\left (16 i \, d x + 16 i \, c\right )} + 36 \, a^{4} d e^{\left (14 i \, d x + 14 i \, c\right )} + 84 \, a^{4} d e^{\left (12 i \, d x + 12 i \, c\right )} + 126 \, a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + 126 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 84 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 36 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \]

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-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 + 1
0*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)

Sympy [F]

\[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{14}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate(sec(d*x+c)**14/(a+I*a*tan(d*x+c))**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {14 \, \tan \left (d x + c\right )^{9} + 63 i \, \tan \left (d x + c\right )^{8} - 72 \, \tan \left (d x + c\right )^{7} + 84 i \, \tan \left (d x + c\right )^{6} - 252 \, \tan \left (d x + c\right )^{5} - 126 i \, \tan \left (d x + c\right )^{4} - 168 \, \tan \left (d x + c\right )^{3} - 252 i \, \tan \left (d x + c\right )^{2} + 126 \, \tan \left (d x + c\right )}{126 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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)

Giac [A] (verification not implemented)

none

Time = 0.87 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {14 \, \tan \left (d x + c\right )^{9} + 63 i \, \tan \left (d x + c\right )^{8} - 72 \, \tan \left (d x + c\right )^{7} + 84 i \, \tan \left (d x + c\right )^{6} - 252 \, \tan \left (d x + c\right )^{5} - 126 i \, \tan \left (d x + c\right )^{4} - 168 \, \tan \left (d x + c\right )^{3} - 252 i \, \tan \left (d x + c\right )^{2} + 126 \, \tan \left (d x + c\right )}{126 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^14/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 4.06 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^{14}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\cos \left (c+d\,x\right )}^9\,105{}\mathrm {i}+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^8+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^6+48\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^4-{\cos \left (c+d\,x\right )}^3\,168{}\mathrm {i}-128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^2+\cos \left (c+d\,x\right )\,63{}\mathrm {i}+14\,\sin \left (c+d\,x\right )}{126\,a^4\,d\,{\cos \left (c+d\,x\right )}^9} \]

[In]

int(1/(cos(c + d*x)^14*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

(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*co
s(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)