\(\int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [131]

Optimal result

Integrand size = 24, antiderivative size = 82 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {2 i (a-i a \tan (c+d x))^6}{3 a^9 d}-\frac {4 i (a-i a \tan (c+d x))^7}{7 a^{10} d}+\frac {i (a-i a \tan (c+d x))^8}{8 a^{11} d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (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} \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {i (a-i a \tan (c+d x))^8}{8 a^{11} d}-\frac {4 i (a-i a \tan (c+d x))^7}{7 a^{10} d}+\frac {2 i (a-i a \tan (c+d x))^6}{3 a^9 d} \]

[In]

Int[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((2*I)/3)*(a - I*a*Tan[c + d*x])^6)/(a^9*d) - (((4*I)/7)*(a - I*a*Tan[c + d*x])^7)/(a^10*d) + ((I/8)*(a - I*a
*Tan[c + d*x])^8)/(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)^2 \, dx,x,i a \tan (c+d x)\right )}{a^{11} d} \\ & = -\frac {i \text {Subst}\left (\int \left (4 a^2 (a-x)^5-4 a (a-x)^6+(a-x)^7\right ) \, dx,x,i a \tan (c+d x)\right )}{a^{11} d} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.56 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {(i+\tan (c+d x))^6 \left (-37 i+54 \tan (c+d x)+21 i \tan ^2(c+d x)\right )}{168 a^3 d} \]

[In]

Integrate[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^3,x]

[Out]

((I + Tan[c + d*x])^6*(-37*I + 54*Tan[c + d*x] + (21*I)*Tan[c + d*x]^2))/(168*a^3*d)

Maple [A] (verified)

Time = 0.46 (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	\(\frac {i \left (\frac {\left (\tan ^{8}\left (d x +c \right )\right )}{8}-\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}+\frac {3 i \left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {5 \left (\tan ^{4}\left (d x +c \right )\right )}{4}+i \left (\tan ^{5}\left (d x +c \right )\right )-\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {i \left (\tan ^{3}\left (d x +c \right )\right )}{3}-i \tan \left (d x +c \right )\right )}{a^{3} d}\)	\(93\)
default	\(\frac {i \left (\frac {\left (\tan ^{8}\left (d x +c \right )\right )}{8}-\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}+\frac {3 i \left (\tan ^{7}\left (d x +c \right )\right )}{7}-\frac {5 \left (\tan ^{4}\left (d x +c \right )\right )}{4}+i \left (\tan ^{5}\left (d x +c \right )\right )-\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {i \left (\tan ^{3}\left (d x +c \right )\right )}{3}-i \tan \left (d x +c \right )\right )}{a^{3} d}\)	\(93\)

[In]

int(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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. 153 vs. \(2 (64) = 128\).

Time = 0.24 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.87 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {32 \, {\left (-28 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 8 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{21 \, {\left (a^{3} d e^{\left (16 i \, d x + 16 i \, c\right )} + 8 \, a^{3} d e^{\left (14 i \, d x + 14 i \, c\right )} + 28 \, a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 56 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 70 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 56 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 28 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-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)

Sympy [F]

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

[In]

integrate(sec(d*x+c)**12/(a+I*a*tan(d*x+c))**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-21 i \, \tan \left (d x + c\right )^{8} + 72 \, \tan \left (d x + c\right )^{7} + 28 i \, \tan \left (d x + c\right )^{6} + 168 \, \tan \left (d x + c\right )^{5} + 210 i \, \tan \left (d x + c\right )^{4} + 56 \, \tan \left (d x + c\right )^{3} + 252 i \, \tan \left (d x + c\right )^{2} - 168 \, \tan \left (d x + c\right )}{168 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/168*(-21*I*tan(d*x + c)^8 + 72*tan(d*x + c)^7 + 28*I*tan(d*x + c)^6 + 168*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)

Giac [A] (verification not implemented)

none

Time = 0.77 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-21 i \, \tan \left (d x + c\right )^{8} + 72 \, \tan \left (d x + c\right )^{7} + 28 i \, \tan \left (d x + c\right )^{6} + 168 \, \tan \left (d x + c\right )^{5} + 210 i \, \tan \left (d x + c\right )^{4} + 56 \, \tan \left (d x + c\right )^{3} + 252 i \, \tan \left (d x + c\right )^{2} - 168 \, \tan \left (d x + c\right )}{168 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/168*(-21*I*tan(d*x + c)^8 + 72*tan(d*x + c)^7 + 28*I*tan(d*x + c)^6 + 168*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)

Mupad [B] (verification not implemented)

Time = 4.26 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\cos \left (c+d\,x\right )}^8\,91{}\mathrm {i}+128\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^7+64\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^5+48\,\sin \left (c+d\,x\right )\,{\cos \left (c+d\,x\right )}^3-{\cos \left (c+d\,x\right )}^2\,112{}\mathrm {i}-72\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )+21{}\mathrm {i}}{168\,a^3\,d\,{\cos \left (c+d\,x\right )}^8} \]

[In]

int(1/(cos(c + d*x)^12*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

(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*112i + cos(c + d*x)^8*91i + 21i)/(168*a^3*d*cos(c + d*x)^8)