3.122 \(\int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=124 \[ \frac {7 \tanh ^{-1}(\sin (c+d x))}{16 a^2 d}-\frac {2 i \sec ^7(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {7 \tan (c+d x) \sec ^5(c+d x)}{30 a^2 d}+\frac {7 \tan (c+d x) \sec ^3(c+d x)}{24 a^2 d}+\frac {7 \tan (c+d x) \sec (c+d x)}{16 a^2 d} \]

[Out]

7/16*arctanh(sin(d*x+c))/a^2/d+7/16*sec(d*x+c)*tan(d*x+c)/a^2/d+7/24*sec(d*x+c)^3*tan(d*x+c)/a^2/d+7/30*sec(d*
x+c)^5*tan(d*x+c)/a^2/d-2/5*I*sec(d*x+c)^7/d/(a^2+I*a^2*tan(d*x+c))

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Rubi [A]  time = 0.08, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3500, 3768, 3770} \[ \frac {7 \tanh ^{-1}(\sin (c+d x))}{16 a^2 d}-\frac {2 i \sec ^7(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {7 \tan (c+d x) \sec ^5(c+d x)}{30 a^2 d}+\frac {7 \tan (c+d x) \sec ^3(c+d x)}{24 a^2 d}+\frac {7 \tan (c+d x) \sec (c+d x)}{16 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(7*ArcTanh[Sin[c + d*x]])/(16*a^2*d) + (7*Sec[c + d*x]*Tan[c + d*x])/(16*a^2*d) + (7*Sec[c + d*x]^3*Tan[c + d*
x])/(24*a^2*d) + (7*Sec[c + d*x]^5*Tan[c + d*x])/(30*a^2*d) - (((2*I)/5)*Sec[c + d*x]^7)/(d*(a^2 + I*a^2*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 ^9(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac {2 i \sec ^7(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {7 \int \sec ^7(c+d x) \, dx}{5 a^2}\\ &=\frac {7 \sec ^5(c+d x) \tan (c+d x)}{30 a^2 d}-\frac {2 i \sec ^7(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {7 \int \sec ^5(c+d x) \, dx}{6 a^2}\\ &=\frac {7 \sec ^3(c+d x) \tan (c+d x)}{24 a^2 d}+\frac {7 \sec ^5(c+d x) \tan (c+d x)}{30 a^2 d}-\frac {2 i \sec ^7(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {7 \int \sec ^3(c+d x) \, dx}{8 a^2}\\ &=\frac {7 \sec (c+d x) \tan (c+d x)}{16 a^2 d}+\frac {7 \sec ^3(c+d x) \tan (c+d x)}{24 a^2 d}+\frac {7 \sec ^5(c+d x) \tan (c+d x)}{30 a^2 d}-\frac {2 i \sec ^7(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {7 \int \sec (c+d x) \, dx}{16 a^2}\\ &=\frac {7 \tanh ^{-1}(\sin (c+d x))}{16 a^2 d}+\frac {7 \sec (c+d x) \tan (c+d x)}{16 a^2 d}+\frac {7 \sec ^3(c+d x) \tan (c+d x)}{24 a^2 d}+\frac {7 \sec ^5(c+d x) \tan (c+d x)}{30 a^2 d}-\frac {2 i \sec ^7(c+d x)}{5 d \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 2.15, size = 294, normalized size = 2.37 \[ -\frac {\sec ^6(c+d x) \left (5 \left (60 \sin (c+d x)-238 \sin (3 (c+d x))-42 \sin (5 (c+d x))+21 \cos (6 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+210 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+315 \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+126 \cos (4 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-21 \cos (6 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-210 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+3072 i \cos (c+d x)\right )}{7680 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^2,x]

[Out]

-1/7680*(Sec[c + d*x]^6*((3072*I)*Cos[c + d*x] + 5*(210*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 21*Cos[6*(c
+ d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 315*Cos[2*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)
/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 126*Cos[4*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/
2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 210*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 21*Cos[6*(c +
d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 60*Sin[c + d*x] - 238*Sin[3*(c + d*x)] - 42*Sin[5*(c + d*x)]
)))/(a^2*d)

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fricas [B]  time = 0.67, size = 326, normalized size = 2.63 \[ \frac {105 \, {\left (e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \, {\left (e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 210 i \, e^{\left (11 i \, d x + 11 i \, c\right )} - 1190 i \, e^{\left (9 i \, d x + 9 i \, c\right )} - 2772 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 3372 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 1190 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 210 i \, e^{\left (i \, d x + i \, c\right )}}{240 \, {\left (a^{2} d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/240*(105*(e^(12*I*d*x + 12*I*c) + 6*e^(10*I*d*x + 10*I*c) + 15*e^(8*I*d*x + 8*I*c) + 20*e^(6*I*d*x + 6*I*c)
+ 15*e^(4*I*d*x + 4*I*c) + 6*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) + I) - 105*(e^(12*I*d*x + 12*I*c) +
6*e^(10*I*d*x + 10*I*c) + 15*e^(8*I*d*x + 8*I*c) + 20*e^(6*I*d*x + 6*I*c) + 15*e^(4*I*d*x + 4*I*c) + 6*e^(2*I*
d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) - I) - 210*I*e^(11*I*d*x + 11*I*c) - 1190*I*e^(9*I*d*x + 9*I*c) - 2772*I
*e^(7*I*d*x + 7*I*c) - 3372*I*e^(5*I*d*x + 5*I*c) + 1190*I*e^(3*I*d*x + 3*I*c) + 210*I*e^(I*d*x + I*c))/(a^2*d
*e^(12*I*d*x + 12*I*c) + 6*a^2*d*e^(10*I*d*x + 10*I*c) + 15*a^2*d*e^(8*I*d*x + 8*I*c) + 20*a^2*d*e^(6*I*d*x +
6*I*c) + 15*a^2*d*e^(4*I*d*x + 4*I*c) + 6*a^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)

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giac [A]  time = 1.12, size = 203, normalized size = 1.64 \[ \frac {\frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{2}} - \frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{2}} + \frac {2 \, {\left (135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 480 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} - 445 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 480 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 330 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 330 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 960 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 445 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 96 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 135 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 96 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6} a^{2}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/240*(105*log(tan(1/2*d*x + 1/2*c) + 1)/a^2 - 105*log(tan(1/2*d*x + 1/2*c) - 1)/a^2 + 2*(135*tan(1/2*d*x + 1/
2*c)^11 + 480*I*tan(1/2*d*x + 1/2*c)^10 - 445*tan(1/2*d*x + 1/2*c)^9 - 480*I*tan(1/2*d*x + 1/2*c)^8 - 330*tan(
1/2*d*x + 1/2*c)^7 + 960*I*tan(1/2*d*x + 1/2*c)^6 - 330*tan(1/2*d*x + 1/2*c)^5 - 960*I*tan(1/2*d*x + 1/2*c)^4
- 445*tan(1/2*d*x + 1/2*c)^3 + 96*I*tan(1/2*d*x + 1/2*c)^2 + 135*tan(1/2*d*x + 1/2*c) - 96*I)/((tan(1/2*d*x +
1/2*c)^2 - 1)^6*a^2))/d

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maple [B]  time = 0.38, size = 514, normalized size = 4.15 \[ -\frac {1}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {5 i}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {9}{16 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3 i}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {9}{16 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 i}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{6 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3 i}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}+\frac {i}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {1}{6 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{6}}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 a^{2} d}+\frac {1}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {2 i}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {9}{16 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {i}{a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{6 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3 i}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{2 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {5 i}{4 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {9}{16 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {2 i}{5 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}+\frac {1}{6 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^2,x)

[Out]

-1/2/a^2/d/(tan(1/2*d*x+1/2*c)-1)^4+5/4*I/a^2/d/(tan(1/2*d*x+1/2*c)+1)^2+9/16/d/a^2/(tan(1/2*d*x+1/2*c)-1)^2+3
/4*I/a^2/d/(tan(1/2*d*x+1/2*c)-1)+9/16/a^2/d/(tan(1/2*d*x+1/2*c)-1)-3/4*I/a^2/d/(tan(1/2*d*x+1/2*c)+1)-1/6/d/a
^2/(tan(1/2*d*x+1/2*c)-1)^3-3/2*I/a^2/d/(tan(1/2*d*x+1/2*c)+1)^3-1/2/a^2/d/(tan(1/2*d*x+1/2*c)-1)^5+I/a^2/d/(t
an(1/2*d*x+1/2*c)-1)^4-1/6/a^2/d/(tan(1/2*d*x+1/2*c)-1)^6-7/16/a^2/d*ln(tan(1/2*d*x+1/2*c)-1)+1/2/a^2/d/(tan(1
/2*d*x+1/2*c)+1)^4-2/5*I/a^2/d/(tan(1/2*d*x+1/2*c)+1)^5+9/16/a^2/d/(tan(1/2*d*x+1/2*c)+1)+I/a^2/d/(tan(1/2*d*x
+1/2*c)+1)^4-1/6/d/a^2/(tan(1/2*d*x+1/2*c)+1)^3+3/2*I/a^2/d/(tan(1/2*d*x+1/2*c)-1)^3-1/2/a^2/d/(tan(1/2*d*x+1/
2*c)+1)^5+5/4*I/a^2/d/(tan(1/2*d*x+1/2*c)-1)^2-9/16/d/a^2/(tan(1/2*d*x+1/2*c)+1)^2+2/5*I/a^2/d/(tan(1/2*d*x+1/
2*c)-1)^5+1/6/a^2/d/(tan(1/2*d*x+1/2*c)+1)^6+7/16/a^2/d*ln(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.36, size = 421, normalized size = 3.40 \[ \frac {\frac {2 \, {\left (\frac {135 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {96 i \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {445 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {960 i \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {330 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {960 i \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {330 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {480 i \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {445 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {480 i \, \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {135 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - 96 i\right )}}{a^{2} - \frac {6 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {20 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {6 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} + \frac {105 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} - \frac {105 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/240*(2*(135*sin(d*x + c)/(cos(d*x + c) + 1) + 96*I*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 445*sin(d*x + c)^3/
(cos(d*x + c) + 1)^3 - 960*I*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 330*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 9
60*I*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 330*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 480*I*sin(d*x + c)^8/(cos
(d*x + c) + 1)^8 - 445*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 480*I*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 135
*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - 96*I)/(a^2 - 6*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^2*sin(d
*x + c)^4/(cos(d*x + c) + 1)^4 - 20*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^2*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8 - 6*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) + 105*lo
g(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 - 105*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2)/d

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mupad [B]  time = 6.31, size = 191, normalized size = 1.54 \[ \frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{8\,a^2\,d}-\frac {-\frac {9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,4{}\mathrm {i}+\frac {89\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,4{}\mathrm {i}+\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,8{}\mathrm {i}+\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,8{}\mathrm {i}+\frac {89\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,4{}\mathrm {i}}{5}-\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {4}{5}{}\mathrm {i}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^9*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

(7*atanh(tan(c/2 + (d*x)/2)))/(8*a^2*d) - ((89*tan(c/2 + (d*x)/2)^3)/24 - (tan(c/2 + (d*x)/2)^2*4i)/5 - (9*tan
(c/2 + (d*x)/2))/8 + tan(c/2 + (d*x)/2)^4*8i + (11*tan(c/2 + (d*x)/2)^5)/4 - tan(c/2 + (d*x)/2)^6*8i + (11*tan
(c/2 + (d*x)/2)^7)/4 + tan(c/2 + (d*x)/2)^8*4i + (89*tan(c/2 + (d*x)/2)^9)/24 - tan(c/2 + (d*x)/2)^10*4i - (9*
tan(c/2 + (d*x)/2)^11)/8 + 4i/5)/(a^2*d*(tan(c/2 + (d*x)/2)^2 - 1)^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\sec ^{9}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**9/(a+I*a*tan(d*x+c))**2,x)

[Out]

-Integral(sec(c + d*x)**9/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x)/a**2

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