\(\int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\) [170]

Optimal result

Integrand size = 24, antiderivative size = 81 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {4 i}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac {i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {i}{d \left (a^2+i a^2 \tan (c+d x)\right )^4} \]

[Out]

4/5*I/a^3/d/(a+I*a*tan(d*x+c))^5+1/3*I/a^5/d/(a+I*a*tan(d*x+c))^3-I/d/(a^2+I*a^2*tan(d*x+c))^4

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 81, 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} \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {i}{3 a^5 d (a+i a \tan (c+d x))^3}+\frac {4 i}{5 a^3 d (a+i a \tan (c+d x))^5}-\frac {i}{d \left (a^2+i a^2 \tan (c+d x)\right )^4} \]

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^8,x]

[Out]

((4*I)/5)/(a^3*d*(a + I*a*Tan[c + d*x])^5) + (I/3)/(a^5*d*(a + I*a*Tan[c + d*x])^3) - I/(d*(a^2 + I*a^2*Tan[c
+ d*x])^4)

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 \frac {(a-x)^2}{(a+x)^6} \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = -\frac {i \text {Subst}\left (\int \left (\frac {4 a^2}{(a+x)^6}-\frac {4 a}{(a+x)^5}+\frac {1}{(a+x)^4}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d} \\ & = \frac {4 i}{5 a^3 d (a+i a \tan (c+d x))^5}+\frac {i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {i}{d \left (a^2+i a^2 \tan (c+d x)\right )^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.54 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {2-5 i \tan (c+d x)-5 \tan ^2(c+d x)}{15 a^8 d (-i+\tan (c+d x))^5} \]

[In]

Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(2 - (5*I)*Tan[c + d*x] - 5*Tan[c + d*x]^2)/(15*a^8*d*(-I + Tan[c + d*x])^5)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.62

method	result	size
derivativedivides	\(-\frac {\frac {i}{\left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {4}{5 \left (\tan \left (d x +c \right )-i\right )^{5}}+\frac {1}{3 \left (\tan \left (d x +c \right )-i\right )^{3}}}{a^{8} d}\)	\(50\)
default	\(-\frac {\frac {i}{\left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {4}{5 \left (\tan \left (d x +c \right )-i\right )^{5}}+\frac {1}{3 \left (\tan \left (d x +c \right )-i\right )^{3}}}{a^{8} d}\)	\(50\)
risch	\(\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{24 a^{8} d}+\frac {i {\mathrm e}^{-8 i \left (d x +c \right )}}{16 a^{8} d}+\frac {i {\mathrm e}^{-10 i \left (d x +c \right )}}{40 a^{8} d}\)	\(56\)

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^8,x,method=_RETURNVERBOSE)

[Out]

-1/a^8/d*(I/(tan(d*x+c)-I)^4-4/5/(tan(d*x+c)-I)^5+1/3/(tan(d*x+c)-I)^3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.51 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {{\left (10 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i\right )} e^{\left (-10 i \, d x - 10 i \, c\right )}}{240 \, a^{8} d} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/240*(10*I*e^(4*I*d*x + 4*I*c) + 15*I*e^(2*I*d*x + 2*I*c) + 6*I)*e^(-10*I*d*x - 10*I*c)/(a^8*d)

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 466 vs. \(2 (65) = 130\).

Time = 8.73 (sec) , antiderivative size = 466, normalized size of antiderivative = 5.75 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\begin {cases} - \frac {i \tan ^{2}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}}{240 a^{8} d \tan ^{8}{\left (c + d x \right )} - 1920 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{6}{\left (c + d x \right )} + 13440 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 16800 a^{8} d \tan ^{4}{\left (c + d x \right )} - 13440 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{2}{\left (c + d x \right )} + 1920 i a^{8} d \tan {\left (c + d x \right )} + 240 a^{8} d} - \frac {8 \tan {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}}{240 a^{8} d \tan ^{8}{\left (c + d x \right )} - 1920 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{6}{\left (c + d x \right )} + 13440 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 16800 a^{8} d \tan ^{4}{\left (c + d x \right )} - 13440 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{2}{\left (c + d x \right )} + 1920 i a^{8} d \tan {\left (c + d x \right )} + 240 a^{8} d} + \frac {31 i \sec ^{6}{\left (c + d x \right )}}{240 a^{8} d \tan ^{8}{\left (c + d x \right )} - 1920 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{6}{\left (c + d x \right )} + 13440 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 16800 a^{8} d \tan ^{4}{\left (c + d x \right )} - 13440 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{2}{\left (c + d x \right )} + 1920 i a^{8} d \tan {\left (c + d x \right )} + 240 a^{8} d} & \text {for}\: d \neq 0 \\\frac {x \sec ^{6}{\left (c \right )}}{\left (i a \tan {\left (c \right )} + a\right )^{8}} & \text {otherwise} \end {cases} \]

[In]

integrate(sec(d*x+c)**6/(a+I*a*tan(d*x+c))**8,x)

[Out]

Piecewise((-I*tan(c + d*x)**2*sec(c + d*x)**6/(240*a**8*d*tan(c + d*x)**8 - 1920*I*a**8*d*tan(c + d*x)**7 - 67
20*a**8*d*tan(c + d*x)**6 + 13440*I*a**8*d*tan(c + d*x)**5 + 16800*a**8*d*tan(c + d*x)**4 - 13440*I*a**8*d*tan
(c + d*x)**3 - 6720*a**8*d*tan(c + d*x)**2 + 1920*I*a**8*d*tan(c + d*x) + 240*a**8*d) - 8*tan(c + d*x)*sec(c +
d*x)**6/(240*a**8*d*tan(c + d*x)**8 - 1920*I*a**8*d*tan(c + d*x)**7 - 6720*a**8*d*tan(c + d*x)**6 + 13440*I*a
**8*d*tan(c + d*x)**5 + 16800*a**8*d*tan(c + d*x)**4 - 13440*I*a**8*d*tan(c + d*x)**3 - 6720*a**8*d*tan(c + d*
x)**2 + 1920*I*a**8*d*tan(c + d*x) + 240*a**8*d) + 31*I*sec(c + d*x)**6/(240*a**8*d*tan(c + d*x)**8 - 1920*I*a
**8*d*tan(c + d*x)**7 - 6720*a**8*d*tan(c + d*x)**6 + 13440*I*a**8*d*tan(c + d*x)**5 + 16800*a**8*d*tan(c + d*
x)**4 - 13440*I*a**8*d*tan(c + d*x)**3 - 6720*a**8*d*tan(c + d*x)**2 + 1920*I*a**8*d*tan(c + d*x) + 240*a**8*d
), Ne(d, 0)), (x*sec(c)**6/(I*a*tan(c) + a)**8, True))

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (65) = 130\).

Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {5 \, \tan \left (d x + c\right )^{4} - 5 i \, \tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )^{2} - i \, \tan \left (d x + c\right ) + 2}{15 \, {\left (a^{8} \tan \left (d x + c\right )^{7} - 7 i \, a^{8} \tan \left (d x + c\right )^{6} - 21 \, a^{8} \tan \left (d x + c\right )^{5} + 35 i \, a^{8} \tan \left (d x + c\right )^{4} + 35 \, a^{8} \tan \left (d x + c\right )^{3} - 21 i \, a^{8} \tan \left (d x + c\right )^{2} - 7 \, a^{8} \tan \left (d x + c\right ) + i \, a^{8}\right )} d} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

-1/15*(5*tan(d*x + c)^4 - 5*I*tan(d*x + c)^3 + 3*tan(d*x + c)^2 - I*tan(d*x + c) + 2)/((a^8*tan(d*x + c)^7 - 7
*I*a^8*tan(d*x + c)^6 - 21*a^8*tan(d*x + c)^5 + 35*I*a^8*tan(d*x + c)^4 + 35*a^8*tan(d*x + c)^3 - 21*I*a^8*tan
(d*x + c)^2 - 7*a^8*tan(d*x + c) + I*a^8)*d)

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (65) = 130\).

Time = 1.50 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.69 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=-\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 30 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 170 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 282 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 170 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{15 \, a^{8} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{10}} \]

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-2/15*(15*tan(1/2*d*x + 1/2*c)^9 - 30*I*tan(1/2*d*x + 1/2*c)^8 - 140*tan(1/2*d*x + 1/2*c)^7 + 170*I*tan(1/2*d*
x + 1/2*c)^6 + 282*tan(1/2*d*x + 1/2*c)^5 - 170*I*tan(1/2*d*x + 1/2*c)^4 - 140*tan(1/2*d*x + 1/2*c)^3 + 30*I*t
an(1/2*d*x + 1/2*c)^2 + 15*tan(1/2*d*x + 1/2*c))/(a^8*d*(tan(1/2*d*x + 1/2*c) - I)^10)

Mupad [B] (verification not implemented)

Time = 4.23 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.05 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx=\frac {-{\mathrm {tan}\left (c+d\,x\right )}^2\,5{}\mathrm {i}+5\,\mathrm {tan}\left (c+d\,x\right )+2{}\mathrm {i}}{15\,a^8\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}+1\right )} \]

[In]

int(1/(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^8),x)

[Out]

(5*tan(c + d*x) - tan(c + d*x)^2*5i + 2i)/(15*a^8*d*(tan(c + d*x)*5i - 10*tan(c + d*x)^2 - tan(c + d*x)^3*10i
+ 5*tan(c + d*x)^4 + tan(c + d*x)^5*1i + 1))