\(\int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\) [149]

Optimal result

Integrand size = 24, antiderivative size = 55 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i (a-i a \tan (c+d x))^6}{3 a^{10} d}-\frac {i (a-i a \tan (c+d x))^7}{7 a^{11} d} \]

[Out]

1/3*I*(a-I*a*tan(d*x+c))^6/a^10/d-1/7*I*(a-I*a*tan(d*x+c))^7/a^11/d

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 55, 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} \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {i (a-i a \tan (c+d x))^6}{3 a^{10} d}-\frac {i (a-i a \tan (c+d x))^7}{7 a^{11} d} \]

[In]

Int[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I/3)*(a - I*a*Tan[c + d*x])^6)/(a^10*d) - ((I/7)*(a - I*a*Tan[c + d*x])^7)/(a^11*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)^5 (a+x) \, dx,x,i a \tan (c+d x)\right )}{a^{11} d} \\ & = -\frac {i \text {Subst}\left (\int \left (2 a (a-x)^5-(a-x)^6\right ) \, dx,x,i a \tan (c+d x)\right )}{a^{11} d} \\ & = \frac {i (a-i a \tan (c+d x))^6}{3 a^{10} d}-\frac {i (a-i a \tan (c+d x))^7}{7 a^{11} d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {(i+\tan (c+d x))^6 (-4 i+3 \tan (c+d x))}{21 a^4 d} \]

[In]

Integrate[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^4,x]

[Out]

((I + Tan[c + d*x])^6*(-4*I + 3*Tan[c + d*x]))/(21*a^4*d)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65

method	result	size
risch	\(\frac {64 i \left (7 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{21 d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{7}}\)	\(36\)
derivativedivides	\(-\frac {-\tan \left (d x +c \right )-\frac {\left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {2 i \left (\tan ^{6}\left (d x +c \right )\right )}{3}+\tan ^{5}\left (d x +c \right )+\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+2 i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{4} d}\)	\(68\)
default	\(-\frac {-\tan \left (d x +c \right )-\frac {\left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {2 i \left (\tan ^{6}\left (d x +c \right )\right )}{3}+\tan ^{5}\left (d x +c \right )+\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{3}+2 i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{4} d}\)	\(68\)

[In]

int(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

64/21*I*(7*exp(2*I*(d*x+c))+1)/d/a^4/(exp(2*I*(d*x+c))+1)^7

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. 127 vs. \(2 (43) = 86\).

Time = 0.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.31 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {64 \, {\left (-7 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{21 \, {\left (a^{4} d e^{\left (14 i \, d x + 14 i \, c\right )} + 7 \, a^{4} d e^{\left (12 i \, d x + 12 i \, c\right )} + 21 \, a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + 35 \, a^{4} d e^{\left (8 i \, d x + 8 i \, c\right )} + 35 \, a^{4} d e^{\left (6 i \, d x + 6 i \, c\right )} + 21 \, a^{4} d e^{\left (4 i \, d x + 4 i \, c\right )} + 7 \, a^{4} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{4} d\right )}} \]

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

-64/21*(-7*I*e^(2*I*d*x + 2*I*c) - I)/(a^4*d*e^(14*I*d*x + 14*I*c) + 7*a^4*d*e^(12*I*d*x + 12*I*c) + 21*a^4*d*
e^(10*I*d*x + 10*I*c) + 35*a^4*d*e^(8*I*d*x + 8*I*c) + 35*a^4*d*e^(6*I*d*x + 6*I*c) + 21*a^4*d*e^(4*I*d*x + 4*
I*c) + 7*a^4*d*e^(2*I*d*x + 2*I*c) + a^4*d)

Sympy [F]

\[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\int \frac {\sec ^{12}{\left (c + d x \right )}}{\tan ^{4}{\left (c + d x \right )} - 4 i \tan ^{3}{\left (c + d x \right )} - 6 \tan ^{2}{\left (c + d x \right )} + 4 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

[In]

integrate(sec(d*x+c)**12/(a+I*a*tan(d*x+c))**4,x)

[Out]

Integral(sec(c + d*x)**12/(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 = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {3 \, \tan \left (d x + c\right )^{7} + 14 i \, \tan \left (d x + c\right )^{6} - 21 \, \tan \left (d x + c\right )^{5} - 35 \, \tan \left (d x + c\right )^{3} - 42 i \, \tan \left (d x + c\right )^{2} + 21 \, \tan \left (d x + c\right )}{21 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

1/21*(3*tan(d*x + c)^7 + 14*I*tan(d*x + c)^6 - 21*tan(d*x + c)^5 - 35*tan(d*x + c)^3 - 42*I*tan(d*x + c)^2 + 2
1*tan(d*x + c))/(a^4*d)

Giac [A] (verification not implemented)

none

Time = 0.80 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.22 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {3 \, \tan \left (d x + c\right )^{7} + 14 i \, \tan \left (d x + c\right )^{6} - 21 \, \tan \left (d x + c\right )^{5} - 35 \, \tan \left (d x + c\right )^{3} - 42 i \, \tan \left (d x + c\right )^{2} + 21 \, \tan \left (d x + c\right )}{21 \, a^{4} d} \]

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/21*(3*tan(d*x + c)^7 + 14*I*tan(d*x + c)^6 - 21*tan(d*x + c)^5 - 35*tan(d*x + c)^3 - 42*I*tan(d*x + c)^2 + 2
1*tan(d*x + c))/(a^4*d)

Mupad [B] (verification not implemented)

Time = 4.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 2.05 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (21\,{\cos \left (c+d\,x\right )}^6-{\cos \left (c+d\,x\right )}^5\,\sin \left (c+d\,x\right )\,42{}\mathrm {i}-35\,{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^2-21\,{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^4+\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^5\,14{}\mathrm {i}+3\,{\sin \left (c+d\,x\right )}^6\right )}{21\,a^4\,d\,{\cos \left (c+d\,x\right )}^7} \]

[In]

int(1/(cos(c + d*x)^12*(a + a*tan(c + d*x)*1i)^4),x)

[Out]

(sin(c + d*x)*(cos(c + d*x)*sin(c + d*x)^5*14i - cos(c + d*x)^5*sin(c + d*x)*42i + 21*cos(c + d*x)^6 + 3*sin(c
+ d*x)^6 - 21*cos(c + d*x)^2*sin(c + d*x)^4 - 35*cos(c + d*x)^4*sin(c + d*x)^2))/(21*a^4*d*cos(c + d*x)^7)