3.171 \(\int \frac {\sec ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=55 \[ \frac {i}{3 a^2 d (a+i a \tan (c+d x))^6}-\frac {i}{5 a^3 d (a+i a \tan (c+d x))^5} \]

[Out]

1/3*I/a^2/d/(a+I*a*tan(d*x+c))^6-1/5*I/a^3/d/(a+I*a*tan(d*x+c))^5

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 55, 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} \[ \frac {i}{3 a^2 d (a+i a \tan (c+d x))^6}-\frac {i}{5 a^3 d (a+i a \tan (c+d x))^5} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(I/3)/(a^2*d*(a + I*a*Tan[c + d*x])^6) - (I/5)/(a^3*d*(a + I*a*Tan[c + d*x])^5)

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 ^4(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=-\frac {i \operatorname {Subst}\left (\int \frac {a-x}{(a+x)^7} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac {i \operatorname {Subst}\left (\int \left (\frac {2 a}{(a+x)^7}-\frac {1}{(a+x)^6}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=\frac {i}{3 a^2 d (a+i a \tan (c+d x))^6}-\frac {i}{5 a^3 d (a+i a \tan (c+d x))^5}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.27, size = 78, normalized size = 1.42 \[ \frac {i \sec ^8(c+d x) (16 i \sin (2 (c+d x))+10 i \sin (4 (c+d x))+64 \cos (2 (c+d x))+20 \cos (4 (c+d x))+45)}{960 a^8 d (\tan (c+d x)-i)^8} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + I*a*Tan[c + d*x])^8,x]

[Out]

((I/960)*Sec[c + d*x]^8*(45 + 64*Cos[2*(c + d*x)] + 20*Cos[4*(c + d*x)] + (16*I)*Sin[2*(c + d*x)] + (10*I)*Sin
[4*(c + d*x)]))/(a^8*d*(-I + Tan[c + d*x])^8)

________________________________________________________________________________________

fricas [A]  time = 0.70, size = 63, normalized size = 1.15 \[ \frac {{\left (15 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 40 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 45 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 24 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-12 i \, d x - 12 i \, c\right )}}{960 \, a^{8} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/960*(15*I*e^(8*I*d*x + 8*I*c) + 40*I*e^(6*I*d*x + 6*I*c) + 45*I*e^(4*I*d*x + 4*I*c) + 24*I*e^(2*I*d*x + 2*I*
c) + 5*I)*e^(-12*I*d*x - 12*I*c)/(a^8*d)

________________________________________________________________________________________

giac [B]  time = 3.39, size = 163, normalized size = 2.96 \[ -\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 60 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 235 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 480 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 822 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 904 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 822 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 480 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 235 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{15 \, a^{8} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

-2/15*(15*tan(1/2*d*x + 1/2*c)^11 - 60*I*tan(1/2*d*x + 1/2*c)^10 - 235*tan(1/2*d*x + 1/2*c)^9 + 480*I*tan(1/2*
d*x + 1/2*c)^8 + 822*tan(1/2*d*x + 1/2*c)^7 - 904*I*tan(1/2*d*x + 1/2*c)^6 - 822*tan(1/2*d*x + 1/2*c)^5 + 480*
I*tan(1/2*d*x + 1/2*c)^4 + 235*tan(1/2*d*x + 1/2*c)^3 - 60*I*tan(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)^12)

________________________________________________________________________________________

maple [A]  time = 0.49, size = 36, normalized size = 0.65 \[ \frac {-\frac {i}{3 \left (\tan \left (d x +c \right )-i\right )^{6}}-\frac {1}{5 \left (\tan \left (d x +c \right )-i\right )^{5}}}{d \,a^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x)

[Out]

1/d/a^8*(-1/3*I/(tan(d*x+c)-I)^6-1/5/(tan(d*x+c)-I)^5)

________________________________________________________________________________________

maxima [B]  time = 0.48, size = 122, normalized size = 2.22 \[ -\frac {7 \, {\left (3 \, \tan \left (d x + c\right )^{2} - i \, \tan \left (d x + c\right ) + 2\right )}}{{\left (105 \, a^{8} \tan \left (d x + c\right )^{7} - 735 i \, a^{8} \tan \left (d x + c\right )^{6} - 2205 \, a^{8} \tan \left (d x + c\right )^{5} + 3675 i \, a^{8} \tan \left (d x + c\right )^{4} + 3675 \, a^{8} \tan \left (d x + c\right )^{3} - 2205 i \, a^{8} \tan \left (d x + c\right )^{2} - 735 \, a^{8} \tan \left (d x + c\right ) + 105 i \, a^{8}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

-7*(3*tan(d*x + c)^2 - I*tan(d*x + c) + 2)/((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)

________________________________________________________________________________________

mupad [B]  time = 3.50, size = 85, normalized size = 1.55 \[ -\frac {-2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}}{15\,a^8\,d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^6\,1{}\mathrm {i}+6\,{\mathrm {tan}\left (c+d\,x\right )}^5-{\mathrm {tan}\left (c+d\,x\right )}^4\,15{}\mathrm {i}-20\,{\mathrm {tan}\left (c+d\,x\right )}^3+{\mathrm {tan}\left (c+d\,x\right )}^2\,15{}\mathrm {i}+6\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^8),x)

[Out]

-(tan(c + d*x)*3i - 2)/(15*a^8*d*(6*tan(c + d*x) + tan(c + d*x)^2*15i - 20*tan(c + d*x)^3 - tan(c + d*x)^4*15i
+ 6*tan(c + d*x)^5 + tan(c + d*x)^6*1i - 1i))

________________________________________________________________________________________

sympy [A]  time = 36.70, size = 774, normalized size = 14.07 \[ \begin {cases} \frac {i \tan ^{4}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{960 a^{8} d \tan ^{8}{\left (c + d x \right )} - 7680 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{6}{\left (c + d x \right )} + 53760 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 67200 a^{8} d \tan ^{4}{\left (c + d x \right )} - 53760 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{2}{\left (c + d x \right )} + 7680 i a^{8} d \tan {\left (c + d x \right )} + 960 a^{8} d} + \frac {8 \tan ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{960 a^{8} d \tan ^{8}{\left (c + d x \right )} - 7680 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{6}{\left (c + d x \right )} + 53760 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 67200 a^{8} d \tan ^{4}{\left (c + d x \right )} - 53760 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{2}{\left (c + d x \right )} + 7680 i a^{8} d \tan {\left (c + d x \right )} + 960 a^{8} d} - \frac {30 i \tan ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{960 a^{8} d \tan ^{8}{\left (c + d x \right )} - 7680 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{6}{\left (c + d x \right )} + 53760 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 67200 a^{8} d \tan ^{4}{\left (c + d x \right )} - 53760 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{2}{\left (c + d x \right )} + 7680 i a^{8} d \tan {\left (c + d x \right )} + 960 a^{8} d} - \frac {72 \tan {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}}{960 a^{8} d \tan ^{8}{\left (c + d x \right )} - 7680 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{6}{\left (c + d x \right )} + 53760 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 67200 a^{8} d \tan ^{4}{\left (c + d x \right )} - 53760 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{2}{\left (c + d x \right )} + 7680 i a^{8} d \tan {\left (c + d x \right )} + 960 a^{8} d} + \frac {129 i \sec ^{4}{\left (c + d x \right )}}{960 a^{8} d \tan ^{8}{\left (c + d x \right )} - 7680 i a^{8} d \tan ^{7}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{6}{\left (c + d x \right )} + 53760 i a^{8} d \tan ^{5}{\left (c + d x \right )} + 67200 a^{8} d \tan ^{4}{\left (c + d x \right )} - 53760 i a^{8} d \tan ^{3}{\left (c + d x \right )} - 26880 a^{8} d \tan ^{2}{\left (c + d x \right )} + 7680 i a^{8} d \tan {\left (c + d x \right )} + 960 a^{8} d} & \text {for}\: d \neq 0 \\\frac {x \sec ^{4}{\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)**4/(a+I*a*tan(d*x+c))**8,x)

[Out]

Piecewise((I*tan(c + d*x)**4*sec(c + d*x)**4/(960*a**8*d*tan(c + d*x)**8 - 7680*I*a**8*d*tan(c + d*x)**7 - 268
80*a**8*d*tan(c + d*x)**6 + 53760*I*a**8*d*tan(c + d*x)**5 + 67200*a**8*d*tan(c + d*x)**4 - 53760*I*a**8*d*tan
(c + d*x)**3 - 26880*a**8*d*tan(c + d*x)**2 + 7680*I*a**8*d*tan(c + d*x) + 960*a**8*d) + 8*tan(c + d*x)**3*sec
(c + d*x)**4/(960*a**8*d*tan(c + d*x)**8 - 7680*I*a**8*d*tan(c + d*x)**7 - 26880*a**8*d*tan(c + d*x)**6 + 5376
0*I*a**8*d*tan(c + d*x)**5 + 67200*a**8*d*tan(c + d*x)**4 - 53760*I*a**8*d*tan(c + d*x)**3 - 26880*a**8*d*tan(
c + d*x)**2 + 7680*I*a**8*d*tan(c + d*x) + 960*a**8*d) - 30*I*tan(c + d*x)**2*sec(c + d*x)**4/(960*a**8*d*tan(
c + d*x)**8 - 7680*I*a**8*d*tan(c + d*x)**7 - 26880*a**8*d*tan(c + d*x)**6 + 53760*I*a**8*d*tan(c + d*x)**5 +
67200*a**8*d*tan(c + d*x)**4 - 53760*I*a**8*d*tan(c + d*x)**3 - 26880*a**8*d*tan(c + d*x)**2 + 7680*I*a**8*d*t
an(c + d*x) + 960*a**8*d) - 72*tan(c + d*x)*sec(c + d*x)**4/(960*a**8*d*tan(c + d*x)**8 - 7680*I*a**8*d*tan(c
+ d*x)**7 - 26880*a**8*d*tan(c + d*x)**6 + 53760*I*a**8*d*tan(c + d*x)**5 + 67200*a**8*d*tan(c + d*x)**4 - 537
60*I*a**8*d*tan(c + d*x)**3 - 26880*a**8*d*tan(c + d*x)**2 + 7680*I*a**8*d*tan(c + d*x) + 960*a**8*d) + 129*I*
sec(c + d*x)**4/(960*a**8*d*tan(c + d*x)**8 - 7680*I*a**8*d*tan(c + d*x)**7 - 26880*a**8*d*tan(c + d*x)**6 + 5
3760*I*a**8*d*tan(c + d*x)**5 + 67200*a**8*d*tan(c + d*x)**4 - 53760*I*a**8*d*tan(c + d*x)**3 - 26880*a**8*d*t
an(c + d*x)**2 + 7680*I*a**8*d*tan(c + d*x) + 960*a**8*d), Ne(d, 0)), (x*sec(c)**4/(I*a*tan(c) + a)**8, True))

________________________________________________________________________________________