3.176 \(\int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx\)

Optimal. Leaf size=205 \[ \frac {1155 \tanh ^{-1}(\sin (c+d x))}{8 a^8 d}-\frac {154 i \sec ^5(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac {385 \tan (c+d x) \sec ^3(c+d x)}{4 a^8 d}+\frac {1155 \tan (c+d x) \sec (c+d x)}{8 a^8 d}-\frac {22 i \sec ^9(c+d x)}{3 a^3 d (a+i a \tan (c+d x))^5}-\frac {66 i \sec ^7(c+d x)}{a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}+\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7} \]

[Out]

1155/8*arctanh(sin(d*x+c))/a^8/d+1155/8*sec(d*x+c)*tan(d*x+c)/a^8/d+385/4*sec(d*x+c)^3*tan(d*x+c)/a^8/d+2/3*I*
sec(d*x+c)^11/a/d/(a+I*a*tan(d*x+c))^7-22/3*I*sec(d*x+c)^9/a^3/d/(a+I*a*tan(d*x+c))^5-66*I*sec(d*x+c)^7/a^2/d/
(a^2+I*a^2*tan(d*x+c))^3-154*I*sec(d*x+c)^5/d/(a^8+I*a^8*tan(d*x+c))

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Rubi [A]  time = 0.22, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3500, 3768, 3770} \[ \frac {1155 \tanh ^{-1}(\sin (c+d x))}{8 a^8 d}-\frac {22 i \sec ^9(c+d x)}{3 a^3 d (a+i a \tan (c+d x))^5}-\frac {66 i \sec ^7(c+d x)}{a^2 d \left (a^2+i a^2 \tan (c+d x)\right )^3}-\frac {154 i \sec ^5(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac {385 \tan (c+d x) \sec ^3(c+d x)}{4 a^8 d}+\frac {1155 \tan (c+d x) \sec (c+d x)}{8 a^8 d}+\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(1155*ArcTanh[Sin[c + d*x]])/(8*a^8*d) + (1155*Sec[c + d*x]*Tan[c + d*x])/(8*a^8*d) + (385*Sec[c + d*x]^3*Tan[
c + d*x])/(4*a^8*d) + (((2*I)/3)*Sec[c + d*x]^11)/(a*d*(a + I*a*Tan[c + d*x])^7) - (((22*I)/3)*Sec[c + d*x]^9)
/(a^3*d*(a + I*a*Tan[c + d*x])^5) - ((66*I)*Sec[c + d*x]^7)/(a^2*d*(a^2 + I*a^2*Tan[c + d*x])^3) - ((154*I)*Se
c[c + d*x]^5)/(d*(a^8 + I*a^8*Tan[c + d*x]))

Rule 3500

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*d^2
*(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + 2*n)), x] - Dist[(d^2*(m - 2))/(b^2*(m + 2*n
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\sec ^{13}(c+d x)}{(a+i a \tan (c+d x))^8} \, dx &=\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {11 \int \frac {\sec ^{11}(c+d x)}{(a+i a \tan (c+d x))^6} \, dx}{3 a^2}\\ &=\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {22 i \sec ^9(c+d x)}{3 a^3 d (a+i a \tan (c+d x))^5}+\frac {33 \int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^4} \, dx}{a^4}\\ &=\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {22 i \sec ^9(c+d x)}{3 a^3 d (a+i a \tan (c+d x))^5}-\frac {66 i \sec ^7(c+d x)}{a^5 d (a+i a \tan (c+d x))^3}+\frac {231 \int \frac {\sec ^7(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{a^6}\\ &=\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {22 i \sec ^9(c+d x)}{3 a^3 d (a+i a \tan (c+d x))^5}-\frac {66 i \sec ^7(c+d x)}{a^5 d (a+i a \tan (c+d x))^3}-\frac {154 i \sec ^5(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac {385 \int \sec ^5(c+d x) \, dx}{a^8}\\ &=\frac {385 \sec ^3(c+d x) \tan (c+d x)}{4 a^8 d}+\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {22 i \sec ^9(c+d x)}{3 a^3 d (a+i a \tan (c+d x))^5}-\frac {66 i \sec ^7(c+d x)}{a^5 d (a+i a \tan (c+d x))^3}-\frac {154 i \sec ^5(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac {1155 \int \sec ^3(c+d x) \, dx}{4 a^8}\\ &=\frac {1155 \sec (c+d x) \tan (c+d x)}{8 a^8 d}+\frac {385 \sec ^3(c+d x) \tan (c+d x)}{4 a^8 d}+\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {22 i \sec ^9(c+d x)}{3 a^3 d (a+i a \tan (c+d x))^5}-\frac {66 i \sec ^7(c+d x)}{a^5 d (a+i a \tan (c+d x))^3}-\frac {154 i \sec ^5(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}+\frac {1155 \int \sec (c+d x) \, dx}{8 a^8}\\ &=\frac {1155 \tanh ^{-1}(\sin (c+d x))}{8 a^8 d}+\frac {1155 \sec (c+d x) \tan (c+d x)}{8 a^8 d}+\frac {385 \sec ^3(c+d x) \tan (c+d x)}{4 a^8 d}+\frac {2 i \sec ^{11}(c+d x)}{3 a d (a+i a \tan (c+d x))^7}-\frac {22 i \sec ^9(c+d x)}{3 a^3 d (a+i a \tan (c+d x))^5}-\frac {66 i \sec ^7(c+d x)}{a^5 d (a+i a \tan (c+d x))^3}-\frac {154 i \sec ^5(c+d x)}{d \left (a^8+i a^8 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 6.40, size = 1704, normalized size = 8.31 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^13/(a + I*a*Tan[c + d*x])^8,x]

[Out]

(-1155*Cos[8*c]*Log[Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*(Cos[d*x] + I*Sin[d*x])^8)/(8*d*(a
+ I*a*Tan[c + d*x])^8) + (1155*Cos[8*c]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*(Cos[d*x]
+ I*Sin[d*x])^8)/(8*d*(a + I*a*Tan[c + d*x])^8) + (Cos[3*d*x]*Sec[c + d*x]^8*(((32*I)/3)*Cos[5*c] - (32*Sin[5
*c])/3)*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (Cos[d*x]*Sec[c + d*x]^8*((-160*I)*Cos[7*c]
+ 160*Sin[7*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) - (((1155*I)/8)*Log[Cos[c/2 + (d*x)/2]
- Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*Sin[8*c]*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (((11
55*I)/8)*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c + d*x]^8*Sin[8*c]*(Cos[d*x] + I*Sin[d*x])^8)/(d*(a
+ I*a*Tan[c + d*x])^8) + (Sec[c]*Sec[c + d*x]^8*(((-236*I)/3)*Cos[8*c] + (236*Sin[8*c])/3)*(Cos[d*x] + I*Sin[
d*x])^8)/(d*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*(-160*Cos[7*c] - (160*I)*Sin[7*c])*(Cos[d*x] + I*Sin[d
*x])^8*Sin[d*x])/(d*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*((32*Cos[5*c])/3 + ((32*I)/3)*Sin[5*c])*(Cos[d
*x] + I*Sin[d*x])^8*Sin[3*d*x])/(d*(a + I*a*Tan[c + d*x])^8) + (Sec[c + d*x]^8*(Cos[8*c]/16 + (I/16)*Sin[8*c])
*(Cos[d*x] + I*Sin[d*x])^8)/(d*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^4*(a + I*a*Tan[c + d*x])^8) - ((1/96
+ I/96)*Sec[c + d*x]^8*((-407*I)*Cos[(15*c)/2] + 343*Cos[(17*c)/2] + 407*Sin[(15*c)/2] + (343*I)*Sin[(17*c)/2]
)*(Cos[d*x] + I*Sin[d*x])^8)/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2*(a + I*a*Tan
[c + d*x])^8) + (Sec[c + d*x]^8*(-1/16*Cos[8*c] - (I/16)*Sin[8*c])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(Cos[c/2 + (d
*x)/2] + Sin[c/2 + (d*x)/2])^4*(a + I*a*Tan[c + d*x])^8) + ((1/96 + I/96)*Sec[c + d*x]^8*(407*Cos[(15*c)/2] -
(343*I)*Cos[(17*c)/2] + (407*I)*Sin[(15*c)/2] + 343*Sin[(17*c)/2])*(Cos[d*x] + I*Sin[d*x])^8)/(d*(Cos[c/2] + S
in[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2*(a + I*a*Tan[c + d*x])^8) + (236*Sec[c + d*x]^8*(Cos[d*x]
+ I*Sin[d*x])^8*(Cos[8*c - (d*x)/2]/2 - Cos[8*c + (d*x)/2]/2 + (I/2)*Sin[8*c - (d*x)/2] - (I/2)*Sin[8*c + (d*
x)/2]))/(3*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])*(a + I*a*Tan[c + d*x])^8) + (4*Se
c[c + d*x]^8*(Cos[d*x] + I*Sin[d*x])^8*(Cos[8*c - (d*x)/2]/2 - Cos[8*c + (d*x)/2]/2 + (I/2)*Sin[8*c - (d*x)/2]
- (I/2)*Sin[8*c + (d*x)/2]))/(3*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^3*(a + I*a*
Tan[c + d*x])^8) + (4*Sec[c + d*x]^8*(Cos[d*x] + I*Sin[d*x])^8*(-1/2*Cos[8*c - (d*x)/2] + Cos[8*c + (d*x)/2]/2
- (I/2)*Sin[8*c - (d*x)/2] + (I/2)*Sin[8*c + (d*x)/2]))/(3*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[
c/2 + (d*x)/2])^3*(a + I*a*Tan[c + d*x])^8) + (236*Sec[c + d*x]^8*(Cos[d*x] + I*Sin[d*x])^8*(-1/2*Cos[8*c - (d
*x)/2] + Cos[8*c + (d*x)/2]/2 - (I/2)*Sin[8*c - (d*x)/2] + (I/2)*Sin[8*c + (d*x)/2]))/(3*d*(Cos[c/2] + Sin[c/2
])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])*(a + I*a*Tan[c + d*x])^8)

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fricas [A]  time = 0.82, size = 267, normalized size = 1.30 \[ \frac {3465 \, {\left (e^{\left (11 i \, d x + 11 i \, c\right )} + 4 \, e^{\left (9 i \, d x + 9 i \, c\right )} + 6 \, e^{\left (7 i \, d x + 7 i \, c\right )} + 4 \, e^{\left (5 i \, d x + 5 i \, c\right )} + e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3465 \, {\left (e^{\left (11 i \, d x + 11 i \, c\right )} + 4 \, e^{\left (9 i \, d x + 9 i \, c\right )} + 6 \, e^{\left (7 i \, d x + 7 i \, c\right )} + 4 \, e^{\left (5 i \, d x + 5 i \, c\right )} + e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 6930 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 25410 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 33726 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 18414 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 2816 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 256 i}{24 \, {\left (a^{8} d e^{\left (11 i \, d x + 11 i \, c\right )} + 4 \, a^{8} d e^{\left (9 i \, d x + 9 i \, c\right )} + 6 \, a^{8} d e^{\left (7 i \, d x + 7 i \, c\right )} + 4 \, a^{8} d e^{\left (5 i \, d x + 5 i \, c\right )} + a^{8} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^8,x, algorithm="fricas")

[Out]

1/24*(3465*(e^(11*I*d*x + 11*I*c) + 4*e^(9*I*d*x + 9*I*c) + 6*e^(7*I*d*x + 7*I*c) + 4*e^(5*I*d*x + 5*I*c) + e^
(3*I*d*x + 3*I*c))*log(e^(I*d*x + I*c) + I) - 3465*(e^(11*I*d*x + 11*I*c) + 4*e^(9*I*d*x + 9*I*c) + 6*e^(7*I*d
*x + 7*I*c) + 4*e^(5*I*d*x + 5*I*c) + e^(3*I*d*x + 3*I*c))*log(e^(I*d*x + I*c) - I) - 6930*I*e^(10*I*d*x + 10*
I*c) - 25410*I*e^(8*I*d*x + 8*I*c) - 33726*I*e^(6*I*d*x + 6*I*c) - 18414*I*e^(4*I*d*x + 4*I*c) - 2816*I*e^(2*I
*d*x + 2*I*c) + 256*I)/(a^8*d*e^(11*I*d*x + 11*I*c) + 4*a^8*d*e^(9*I*d*x + 9*I*c) + 6*a^8*d*e^(7*I*d*x + 7*I*c
) + 4*a^8*d*e^(5*I*d*x + 5*I*c) + a^8*d*e^(3*I*d*x + 3*I*c))

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giac [A]  time = 6.77, size = 195, normalized size = 0.95 \[ \frac {\frac {3465 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{8}} - \frac {3465 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{8}} - \frac {1024 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7\right )}}{a^{8} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{3}} - \frac {2 \, {\left (369 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1728 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 393 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5568 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 393 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5696 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 369 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1856 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4} a^{8}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^8,x, algorithm="giac")

[Out]

1/24*(3465*log(tan(1/2*d*x + 1/2*c) + 1)/a^8 - 3465*log(tan(1/2*d*x + 1/2*c) - 1)/a^8 - 1024*(6*tan(1/2*d*x +
1/2*c)^2 - 15*I*tan(1/2*d*x + 1/2*c) - 7)/(a^8*(tan(1/2*d*x + 1/2*c) - I)^3) - 2*(369*tan(1/2*d*x + 1/2*c)^7 -
1728*I*tan(1/2*d*x + 1/2*c)^6 - 393*tan(1/2*d*x + 1/2*c)^5 + 5568*I*tan(1/2*d*x + 1/2*c)^4 - 393*tan(1/2*d*x
+ 1/2*c)^3 - 5696*I*tan(1/2*d*x + 1/2*c)^2 + 369*tan(1/2*d*x + 1/2*c) + 1856*I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^
4*a^8))/d

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maple [B]  time = 0.47, size = 409, normalized size = 2.00 \[ \frac {1}{2 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {4 i}{a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {121}{8 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {76 i}{a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {123}{8 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {4 i}{a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{4 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {1155 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 a^{8} d}+\frac {121}{8 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {76 i}{a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {1}{2 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {8 i}{3 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {123}{8 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {8 i}{3 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{4 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1155 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 a^{8} d}+\frac {128 i}{a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {256}{3 a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}-\frac {256}{a^{8} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^8,x)

[Out]

1/2/a^8/d/(tan(1/2*d*x+1/2*c)-1)^3-4*I/a^8/d/(tan(1/2*d*x+1/2*c)+1)^2-121/8/a^8/d/(tan(1/2*d*x+1/2*c)-1)^2-76*
I/a^8/d/(tan(1/2*d*x+1/2*c)+1)-123/8/a^8/d/(tan(1/2*d*x+1/2*c)-1)-4*I/a^8/d/(tan(1/2*d*x+1/2*c)-1)^2+1/4/a^8/d
/(tan(1/2*d*x+1/2*c)-1)^4-1155/8/a^8/d*ln(tan(1/2*d*x+1/2*c)-1)+121/8/a^8/d/(tan(1/2*d*x+1/2*c)+1)^2+76*I/a^8/
d/(tan(1/2*d*x+1/2*c)-1)+1/2/a^8/d/(tan(1/2*d*x+1/2*c)+1)^3-8/3*I/a^8/d/(tan(1/2*d*x+1/2*c)-1)^3-123/8/a^8/d/(
tan(1/2*d*x+1/2*c)+1)+8/3*I/a^8/d/(tan(1/2*d*x+1/2*c)+1)^3-1/4/a^8/d/(tan(1/2*d*x+1/2*c)+1)^4+1155/8/a^8/d*ln(
tan(1/2*d*x+1/2*c)+1)+128*I/a^8/d/(tan(1/2*d*x+1/2*c)-I)^2-256/3/a^8/d/(tan(1/2*d*x+1/2*c)-I)^3-256/a^8/d/(tan
(1/2*d*x+1/2*c)-I)

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maxima [B]  time = 1.24, size = 796, normalized size = 3.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^13/(a+I*a*tan(d*x+c))^8,x, algorithm="maxima")

[Out]

-((6930*cos(11*d*x + 11*c) + 27720*cos(9*d*x + 9*c) + 41580*cos(7*d*x + 7*c) + 27720*cos(5*d*x + 5*c) + 6930*c
os(3*d*x + 3*c) + 6930*I*sin(11*d*x + 11*c) + 27720*I*sin(9*d*x + 9*c) + 41580*I*sin(7*d*x + 7*c) + 27720*I*si
n(5*d*x + 5*c) + 6930*I*sin(3*d*x + 3*c))*arctan2(cos(d*x + c), sin(d*x + c) + 1) + (6930*cos(11*d*x + 11*c) +
27720*cos(9*d*x + 9*c) + 41580*cos(7*d*x + 7*c) + 27720*cos(5*d*x + 5*c) + 6930*cos(3*d*x + 3*c) + 6930*I*sin
(11*d*x + 11*c) + 27720*I*sin(9*d*x + 9*c) + 41580*I*sin(7*d*x + 7*c) + 27720*I*sin(5*d*x + 5*c) + 6930*I*sin(
3*d*x + 3*c))*arctan2(cos(d*x + c), -sin(d*x + c) + 1) - (-3465*I*cos(11*d*x + 11*c) - 13860*I*cos(9*d*x + 9*c
) - 20790*I*cos(7*d*x + 7*c) - 13860*I*cos(5*d*x + 5*c) - 3465*I*cos(3*d*x + 3*c) + 3465*sin(11*d*x + 11*c) +
13860*sin(9*d*x + 9*c) + 20790*sin(7*d*x + 7*c) + 13860*sin(5*d*x + 5*c) + 3465*sin(3*d*x + 3*c))*log(cos(d*x
+ c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - (3465*I*cos(11*d*x + 11*c) + 13860*I*cos(9*d*x + 9*c) + 20790*
I*cos(7*d*x + 7*c) + 13860*I*cos(5*d*x + 5*c) + 3465*I*cos(3*d*x + 3*c) - 3465*sin(11*d*x + 11*c) - 13860*sin(
9*d*x + 9*c) - 20790*sin(7*d*x + 7*c) - 13860*sin(5*d*x + 5*c) - 3465*sin(3*d*x + 3*c))*log(cos(d*x + c)^2 + s
in(d*x + c)^2 - 2*sin(d*x + c) + 1) + 13860*cos(10*d*x + 10*c) + 50820*cos(8*d*x + 8*c) + 67452*cos(6*d*x + 6*
c) + 36828*cos(4*d*x + 4*c) + 5632*cos(2*d*x + 2*c) + 13860*I*sin(10*d*x + 10*c) + 50820*I*sin(8*d*x + 8*c) +
67452*I*sin(6*d*x + 6*c) + 36828*I*sin(4*d*x + 4*c) + 5632*I*sin(2*d*x + 2*c) - 512)/((-48*I*a^8*cos(11*d*x +
11*c) - 192*I*a^8*cos(9*d*x + 9*c) - 288*I*a^8*cos(7*d*x + 7*c) - 192*I*a^8*cos(5*d*x + 5*c) - 48*I*a^8*cos(3*
d*x + 3*c) + 48*a^8*sin(11*d*x + 11*c) + 192*a^8*sin(9*d*x + 9*c) + 288*a^8*sin(7*d*x + 7*c) + 192*a^8*sin(5*d
*x + 5*c) + 48*a^8*sin(3*d*x + 3*c))*d)

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mupad [B]  time = 7.57, size = 344, normalized size = 1.68 \[ \frac {\frac {33847\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{6\,a^8}-\frac {12041\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3\,a^8}-\frac {3585\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{a^8}+\frac {3505\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4\,a^8}+\frac {4293\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^8}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,27565{}\mathrm {i}}{12\,a^8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,4575{}\mathrm {i}}{a^8}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,25993{}\mathrm {i}}{6\,a^8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,5639{}\mathrm {i}}{3\,a^8}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,1147{}\mathrm {i}}{4\,a^8}-\frac {1360{}\mathrm {i}}{3\,a^8}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,7{}\mathrm {i}+13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,18{}\mathrm {i}-22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,22{}\mathrm {i}+18\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,13{}\mathrm {i}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,3{}\mathrm {i}+1\right )}+\frac {1155\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^13*(a + a*tan(c + d*x)*1i)^8),x)

[Out]

((tan(c/2 + (d*x)/2)^2*27565i)/(12*a^8) - (12041*tan(c/2 + (d*x)/2)^3)/(3*a^8) - (tan(c/2 + (d*x)/2)^4*4575i)/
a^8 + (33847*tan(c/2 + (d*x)/2)^5)/(6*a^8) + (tan(c/2 + (d*x)/2)^6*25993i)/(6*a^8) - (3585*tan(c/2 + (d*x)/2)^
7)/a^8 - (tan(c/2 + (d*x)/2)^8*5639i)/(3*a^8) + (3505*tan(c/2 + (d*x)/2)^9)/(4*a^8) + (tan(c/2 + (d*x)/2)^10*1
147i)/(4*a^8) - 1360i/(3*a^8) + (4293*tan(c/2 + (d*x)/2))/(4*a^8))/(d*(tan(c/2 + (d*x)/2)*3i - 7*tan(c/2 + (d*
x)/2)^2 - tan(c/2 + (d*x)/2)^3*13i + 18*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^5*22i - 22*tan(c/2 + (d*x)/2
)^6 - tan(c/2 + (d*x)/2)^7*18i + 13*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^9*7i - 3*tan(c/2 + (d*x)/2)^10 -
tan(c/2 + (d*x)/2)^11*1i + 1)) + (1155*atanh(tan(c/2 + (d*x)/2)))/(4*a^8*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sec ^{13}{\left (c + d x \right )}}{\tan ^{8}{\left (c + d x \right )} - 8 i \tan ^{7}{\left (c + d x \right )} - 28 \tan ^{6}{\left (c + d x \right )} + 56 i \tan ^{5}{\left (c + d x \right )} + 70 \tan ^{4}{\left (c + d x \right )} - 56 i \tan ^{3}{\left (c + d x \right )} - 28 \tan ^{2}{\left (c + d x \right )} + 8 i \tan {\left (c + d x \right )} + 1}\, dx}{a^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**13/(a+I*a*tan(d*x+c))**8,x)

[Out]

Integral(sec(c + d*x)**13/(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

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