3.167 \(\int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=126 \[ \frac {\tan ^3(c+d x)}{3 a^8 d}+\frac {4 i \tan ^2(c+d x)}{a^8 d}-\frac {31 \tan (c+d x)}{a^8 d}-\frac {80 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac {80 i \log (\cos (c+d x))}{a^8 d}+\frac {80 x}{a^8}+\frac {16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2} \]

[Out]

80*x/a^8+80*I*ln(cos(d*x+c))/a^8/d-31*tan(d*x+c)/a^8/d+4*I*tan(d*x+c)^2/a^8/d+1/3*tan(d*x+c)^3/a^8/d+16*I/d/(a
^4+I*a^4*tan(d*x+c))^2-80*I/d/(a^8+I*a^8*tan(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3487, 43} \[ \frac {\tan ^3(c+d x)}{3 a^8 d}+\frac {4 i \tan ^2(c+d x)}{a^8 d}-\frac {31 \tan (c+d x)}{a^8 d}-\frac {80 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac {16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2}+\frac {80 i \log (\cos (c+d x))}{a^8 d}+\frac {80 x}{a^8} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(80*x)/a^8 + ((80*I)*Log[Cos[c + d*x]])/(a^8*d) - (31*Tan[c + d*x])/(a^8*d) + ((4*I)*Tan[c + d*x]^2)/(a^8*d) +
Tan[c + d*x]^3/(3*a^8*d) + (16*I)/(d*(a^4 + I*a^4*Tan[c + d*x])^2) - (80*I)/(d*(a^8 + I*a^8*Tan[c + d*x]))

Rule 43

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 3487

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*} \int \frac {\sec ^{12}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=-\frac {i \operatorname {Subst}\left (\int \frac {(a-x)^5}{(a+x)^3} \, dx,x,i a \tan (c+d x)\right )}{a^{11} d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (-31 a^2+8 a x-x^2+\frac {32 a^5}{(a+x)^3}-\frac {80 a^4}{(a+x)^2}+\frac {80 a^3}{a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^{11} d}\\ &=\frac {80 x}{a^8}+\frac {80 i \log (\cos (c+d x))}{a^8 d}-\frac {31 \tan (c+d x)}{a^8 d}+\frac {4 i \tan ^2(c+d x)}{a^8 d}+\frac {\tan ^3(c+d x)}{3 a^8 d}+\frac {16 i}{d \left (a^4+i a^4 \tan (c+d x)\right )^2}-\frac {80 i}{d \left (a^8+i a^8 \tan (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.55, size = 537, normalized size = 4.26 \[ \frac {\sec (c) \sec ^{11}(c+d x) (\cos (6 (c+d x))+i \sin (6 (c+d x))) (120 i d x \sin (2 c+d x)+87 \sin (2 c+d x)+180 i d x \sin (2 c+3 d x)-96 \sin (2 c+3 d x)+180 i d x \sin (4 c+3 d x)+45 \sin (4 c+3 d x)+60 i d x \sin (4 c+5 d x)-44 \sin (4 c+5 d x)+60 i d x \sin (6 c+5 d x)+3 \sin (6 c+5 d x)+180 d x \cos (2 c+3 d x)+66 i \cos (2 c+3 d x)+180 d x \cos (4 c+3 d x)-75 i \cos (4 c+3 d x)+60 d x \cos (4 c+5 d x)+50 i \cos (4 c+5 d x)+60 d x \cos (6 c+5 d x)+3 i \cos (6 c+5 d x)+180 i \cos (2 c+3 d x) \log (\cos (c+d x))+3 \cos (2 c+d x) (80 i \log (\cos (c+d x))+80 d x-71 i)+\cos (d x) (240 i \log (\cos (c+d x))+240 d x-119 i)+180 i \cos (4 c+3 d x) \log (\cos (c+d x))+60 i \cos (4 c+5 d x) \log (\cos (c+d x))+60 i \cos (6 c+5 d x) \log (\cos (c+d x))-120 \sin (d x) \log (\cos (c+d x))-120 \sin (2 c+d x) \log (\cos (c+d x))-180 \sin (2 c+3 d x) \log (\cos (c+d x))-180 \sin (4 c+3 d x) \log (\cos (c+d x))-60 \sin (4 c+5 d x) \log (\cos (c+d x))-60 \sin (6 c+5 d x) \log (\cos (c+d x))+120 i d x \sin (d x)-101 \sin (d x))}{12 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^12/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(Sec[c]*Sec[c + d*x]^11*(Cos[6*(c + d*x)] + I*Sin[6*(c + d*x)])*((66*I)*Cos[2*c + 3*d*x] + 180*d*x*Cos[2*c + 3
*d*x] - (75*I)*Cos[4*c + 3*d*x] + 180*d*x*Cos[4*c + 3*d*x] + (50*I)*Cos[4*c + 5*d*x] + 60*d*x*Cos[4*c + 5*d*x]
+ (3*I)*Cos[6*c + 5*d*x] + 60*d*x*Cos[6*c + 5*d*x] + 3*Cos[2*c + d*x]*(-71*I + 80*d*x + (80*I)*Log[Cos[c + d*
x]]) + Cos[d*x]*(-119*I + 240*d*x + (240*I)*Log[Cos[c + d*x]]) + (180*I)*Cos[2*c + 3*d*x]*Log[Cos[c + d*x]] +
(180*I)*Cos[4*c + 3*d*x]*Log[Cos[c + d*x]] + (60*I)*Cos[4*c + 5*d*x]*Log[Cos[c + d*x]] + (60*I)*Cos[6*c + 5*d*
x]*Log[Cos[c + d*x]] - 101*Sin[d*x] + (120*I)*d*x*Sin[d*x] - 120*Log[Cos[c + d*x]]*Sin[d*x] + 87*Sin[2*c + d*x
] + (120*I)*d*x*Sin[2*c + d*x] - 120*Log[Cos[c + d*x]]*Sin[2*c + d*x] - 96*Sin[2*c + 3*d*x] + (180*I)*d*x*Sin[
2*c + 3*d*x] - 180*Log[Cos[c + d*x]]*Sin[2*c + 3*d*x] + 45*Sin[4*c + 3*d*x] + (180*I)*d*x*Sin[4*c + 3*d*x] - 1
80*Log[Cos[c + d*x]]*Sin[4*c + 3*d*x] - 44*Sin[4*c + 5*d*x] + (60*I)*d*x*Sin[4*c + 5*d*x] - 60*Log[Cos[c + d*x
]]*Sin[4*c + 5*d*x] + 3*Sin[6*c + 5*d*x] + (60*I)*d*x*Sin[6*c + 5*d*x] - 60*Log[Cos[c + d*x]]*Sin[6*c + 5*d*x]
))/(12*a^8*d*(-I + Tan[c + d*x])^8)

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 195, normalized size = 1.55 \[ \frac {480 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} + {\left (1440 \, d x - 240 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + {\left (1440 \, d x - 600 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (480 \, d x - 440 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (240 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 720 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 720 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 240 i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 60 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 12 i}{3 \, {\left (a^{8} d e^{\left (10 i \, d x + 10 i \, c\right )} + 3 \, a^{8} d e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, a^{8} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{8} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/3*(480*d*x*e^(10*I*d*x + 10*I*c) + (1440*d*x - 240*I)*e^(8*I*d*x + 8*I*c) + (1440*d*x - 600*I)*e^(6*I*d*x +
6*I*c) + (480*d*x - 440*I)*e^(4*I*d*x + 4*I*c) + (240*I*e^(10*I*d*x + 10*I*c) + 720*I*e^(8*I*d*x + 8*I*c) + 72
0*I*e^(6*I*d*x + 6*I*c) + 240*I*e^(4*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 60*I*e^(2*I*d*x + 2*I*c) +
12*I)/(a^8*d*e^(10*I*d*x + 10*I*c) + 3*a^8*d*e^(8*I*d*x + 8*I*c) + 3*a^8*d*e^(6*I*d*x + 6*I*c) + a^8*d*e^(4*I
*d*x + 4*I*c))

________________________________________________________________________________________

giac [A]  time = 4.97, size = 223, normalized size = 1.77 \[ -\frac {2 \, {\left (-\frac {120 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{8}} + \frac {240 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}{a^{8}} - \frac {120 i \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{8}} + \frac {220 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 93 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 684 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 190 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 684 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 93 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 220 i}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{8}} + \frac {-500 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2144 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3384 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2144 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 500 i}{a^{8} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{4}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-2/3*(-120*I*log(tan(1/2*d*x + 1/2*c) + 1)/a^8 + 240*I*log(tan(1/2*d*x + 1/2*c) - I)/a^8 - 120*I*log(tan(1/2*d
*x + 1/2*c) - 1)/a^8 + (220*I*tan(1/2*d*x + 1/2*c)^6 - 93*tan(1/2*d*x + 1/2*c)^5 - 684*I*tan(1/2*d*x + 1/2*c)^
4 + 190*tan(1/2*d*x + 1/2*c)^3 + 684*I*tan(1/2*d*x + 1/2*c)^2 - 93*tan(1/2*d*x + 1/2*c) - 220*I)/((tan(1/2*d*x
+ 1/2*c)^2 - 1)^3*a^8) + (-500*I*tan(1/2*d*x + 1/2*c)^4 - 2144*tan(1/2*d*x + 1/2*c)^3 + 3384*I*tan(1/2*d*x +
1/2*c)^2 + 2144*tan(1/2*d*x + 1/2*c) - 500*I)/(a^8*(tan(1/2*d*x + 1/2*c) - I)^4))/d

________________________________________________________________________________________

maple [A]  time = 0.45, size = 107, normalized size = 0.85 \[ -\frac {31 \tan \left (d x +c \right )}{a^{8} d}+\frac {\tan ^{3}\left (d x +c \right )}{3 a^{8} d}+\frac {4 i \left (\tan ^{2}\left (d x +c \right )\right )}{a^{8} d}-\frac {16 i}{a^{8} d \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {80}{a^{8} d \left (\tan \left (d x +c \right )-i\right )}-\frac {80 i \ln \left (\tan \left (d x +c \right )-i\right )}{a^{8} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^8,x)

[Out]

-31*tan(d*x+c)/a^8/d+1/3*tan(d*x+c)^3/a^8/d+4*I*tan(d*x+c)^2/a^8/d-16*I/a^8/d/(tan(d*x+c)-I)^2-80/a^8/d/(tan(d
*x+c)-I)-80*I/a^8/d*ln(tan(d*x+c)-I)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 213, normalized size = 1.69 \[ -\frac {\frac {3 \, {\left (1680 \, \tan \left (d x + c\right )^{6} - 9744 i \, \tan \left (d x + c\right )^{5} - 23520 \, \tan \left (d x + c\right )^{4} + 30240 i \, \tan \left (d x + c\right )^{3} + 21840 \, \tan \left (d x + c\right )^{2} - 8400 i \, \tan \left (d x + c\right ) - 1344\right )}}{21 \, a^{8} \tan \left (d x + c\right )^{7} - 147 i \, a^{8} \tan \left (d x + c\right )^{6} - 441 \, a^{8} \tan \left (d x + c\right )^{5} + 735 i \, a^{8} \tan \left (d x + c\right )^{4} + 735 \, a^{8} \tan \left (d x + c\right )^{3} - 441 i \, a^{8} \tan \left (d x + c\right )^{2} - 147 \, a^{8} \tan \left (d x + c\right ) + 21 i \, a^{8}} - \frac {\tan \left (d x + c\right )^{3} + 12 i \, \tan \left (d x + c\right )^{2} - 93 \, \tan \left (d x + c\right )}{a^{8}} + \frac {240 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a^{8}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^12/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

-1/3*(3*(1680*tan(d*x + c)^6 - 9744*I*tan(d*x + c)^5 - 23520*tan(d*x + c)^4 + 30240*I*tan(d*x + c)^3 + 21840*t
an(d*x + c)^2 - 8400*I*tan(d*x + c) - 1344)/(21*a^8*tan(d*x + c)^7 - 147*I*a^8*tan(d*x + c)^6 - 441*a^8*tan(d*
x + c)^5 + 735*I*a^8*tan(d*x + c)^4 + 735*a^8*tan(d*x + c)^3 - 441*I*a^8*tan(d*x + c)^2 - 147*a^8*tan(d*x + c)
+ 21*I*a^8) - (tan(d*x + c)^3 + 12*I*tan(d*x + c)^2 - 93*tan(d*x + c))/a^8 + 240*I*log(I*tan(d*x + c) + 1)/a^
8)/d

________________________________________________________________________________________

mupad [B]  time = 3.44, size = 114, normalized size = 0.90 \[ \frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,a^8\,d}-\frac {31\,\mathrm {tan}\left (c+d\,x\right )}{a^8\,d}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,4{}\mathrm {i}}{a^8\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,80{}\mathrm {i}}{a^8\,d}-\frac {\frac {64}{a^8}+\frac {\mathrm {tan}\left (c+d\,x\right )\,80{}\mathrm {i}}{a^8}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^12*(a + a*tan(c + d*x)*1i)^8),x)

[Out]

(tan(c + d*x)^2*4i)/(a^8*d) - (31*tan(c + d*x))/(a^8*d) - (log(tan(c + d*x) - 1i)*80i)/(a^8*d) + tan(c + d*x)^
3/(3*a^8*d) - ((tan(c + d*x)*80i)/a^8 + 64/a^8)/(d*(2*tan(c + d*x) + tan(c + d*x)^2*1i - 1i))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{12}{\left (c + d x \right )}}{\tan ^{8}{\left (c + d x \right )} - 8 i \tan ^{7}{\left (c + d x \right )} - 28 \tan ^{6}{\left (c + d x \right )} + 56 i \tan ^{5}{\left (c + d x \right )} + 70 \tan ^{4}{\left (c + d x \right )} - 56 i \tan ^{3}{\left (c + d x \right )} - 28 \tan ^{2}{\left (c + d x \right )} + 8 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**12/(a+I*a*tan(d*x+c))**8,x)

[Out]

Integral(sec(c + d*x)**12/(tan(c + d*x)**8 - 8*I*tan(c + d*x)**7 - 28*tan(c + d*x)**6 + 56*I*tan(c + d*x)**5 +
70*tan(c + d*x)**4 - 56*I*tan(c + d*x)**3 - 28*tan(c + d*x)**2 + 8*I*tan(c + d*x) + 1), x)/a**8

________________________________________________________________________________________