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

Optimal. Leaf size=81 \[ \frac {i}{3 a^5 d (a+i a \tan (c+d x))^3}+\frac {4 i}{5 a^3 d (a+i a \tan (c+d x))^5}-\frac {i}{d \left (a^2+i a^2 \tan (c+d x)\right )^4} \]

[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

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Rubi [A]  time = 0.06, antiderivative size = 81, 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} \[ \frac {i}{3 a^5 d (a+i a \tan (c+d x))^3}-\frac {i}{d \left (a^2+i a^2 \tan (c+d x)\right )^4}+\frac {4 i}{5 a^3 d (a+i a \tan (c+d x))^5} \]

Antiderivative was successfully verified.

[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 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 ^6(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=-\frac {i \operatorname {Subst}\left (\int \frac {(a-x)^2}{(a+x)^6} \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac {i \operatorname {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*}

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Mathematica [A]  time = 0.28, size = 56, normalized size = 0.69 \[ \frac {i \sec ^8(c+d x) (4 i \sin (2 (c+d x))+16 \cos (2 (c+d x))+15)}{240 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/240)*Sec[c + d*x]^8*(15 + 16*Cos[2*(c + d*x)] + (4*I)*Sin[2*(c + d*x)]))/(a^8*d*(-I + Tan[c + d*x])^8)

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fricas [A]  time = 0.66, size = 41, normalized size = 0.51 \[ \frac {{\left (10 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i\right )} e^{\left (-10 i \, d x - 10 i \, c\right )}}{240 \, a^{8} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 5.21, size = 137, normalized size = 1.69 \[ -\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 30 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 170 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 282 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 170 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{15 \, a^{8} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.52, size = 49, normalized size = 0.60 \[ \frac {-\frac {i}{\left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {1}{3 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {4}{5 \left (\tan \left (d x +c \right )-i\right )^{5}}}{d \,a^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 142, normalized size = 1.75 \[ -\frac {35 \, \tan \left (d x + c\right )^{4} - 35 i \, \tan \left (d x + c\right )^{3} + 21 \, \tan \left (d x + c\right )^{2} - 7 i \, \tan \left (d x + c\right ) + 14}{{\left (105 \, a^{8} \tan \left (d x + c\right )^{7} - 735 i \, a^{8} \tan \left (d x + c\right )^{6} - 2205 \, a^{8} \tan \left (d x + c\right )^{5} + 3675 i \, a^{8} \tan \left (d x + c\right )^{4} + 3675 \, a^{8} \tan \left (d x + c\right )^{3} - 2205 i \, a^{8} \tan \left (d x + c\right )^{2} - 735 \, a^{8} \tan \left (d x + c\right ) + 105 i \, a^{8}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(35*tan(d*x + c)^4 - 35*I*tan(d*x + c)^3 + 21*tan(d*x + c)^2 - 7*I*tan(d*x + c) + 14)/((105*a^8*tan(d*x + c)^
7 - 735*I*a^8*tan(d*x + c)^6 - 2205*a^8*tan(d*x + c)^5 + 3675*I*a^8*tan(d*x + c)^4 + 3675*a^8*tan(d*x + c)^3 -
2205*I*a^8*tan(d*x + c)^2 - 735*a^8*tan(d*x + c) + 105*I*a^8)*d)

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mupad [B]  time = 3.50, size = 85, normalized size = 1.05 \[ \frac {-{\mathrm {tan}\left (c+d\,x\right )}^2\,5{}\mathrm {i}+5\,\mathrm {tan}\left (c+d\,x\right )+2{}\mathrm {i}}{15\,a^8\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (c+d\,x\right )}^4-{\mathrm {tan}\left (c+d\,x\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,5{}\mathrm {i}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 36.10, size = 466, normalized size = 5.75 \[ \begin {cases} - \frac {i \tan ^{2}{\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}}{240 a^{8} d \tan ^{8}{\left (c + d x \right )} - 1920 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{6}{\left (c + d x \right )} + 13440 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 16800 a^{8} d \tan ^{4}{\left (c + d x \right )} - 13440 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{2}{\left (c + d x \right )} + 1920 i a^{8} d \tan {\left (c + d x \right )} + 240 a^{8} d} - \frac {8 \tan {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}}{240 a^{8} d \tan ^{8}{\left (c + d x \right )} - 1920 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{6}{\left (c + d x \right )} + 13440 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 16800 a^{8} d \tan ^{4}{\left (c + d x \right )} - 13440 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{2}{\left (c + d x \right )} + 1920 i a^{8} d \tan {\left (c + d x \right )} + 240 a^{8} d} + \frac {31 i \sec ^{6}{\left (c + d x \right )}}{240 a^{8} d \tan ^{8}{\left (c + d x \right )} - 1920 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{6}{\left (c + d x \right )} + 13440 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 16800 a^{8} d \tan ^{4}{\left (c + d x \right )} - 13440 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 6720 a^{8} d \tan ^{2}{\left (c + d x \right )} + 1920 i a^{8} d \tan {\left (c + d x \right )} + 240 a^{8} d} & \text {for}\: d \neq 0 \\\frac {x \sec ^{6}{\relax (c )}}{\left (i a \tan {\relax (c )} + a\right )^{8}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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