3.140 \(\int \frac {\sec ^9(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=119 \[ -\frac {7 i \sec ^5(c+d x)}{15 a^3 d}+\frac {7 \tanh ^{-1}(\sin (c+d x))}{8 a^3 d}+\frac {7 \tan (c+d x) \sec ^3(c+d x)}{12 a^3 d}+\frac {7 \tan (c+d x) \sec (c+d x)}{8 a^3 d}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2} \]

[Out]

7/8*arctanh(sin(d*x+c))/a^3/d-7/15*I*sec(d*x+c)^5/a^3/d+7/8*sec(d*x+c)*tan(d*x+c)/a^3/d+7/12*sec(d*x+c)^3*tan(
d*x+c)/a^3/d-2/3*I*sec(d*x+c)^7/a/d/(a+I*a*tan(d*x+c))^2

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Rubi [A]  time = 0.11, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3500, 3501, 3768, 3770} \[ -\frac {7 i \sec ^5(c+d x)}{15 a^3 d}+\frac {7 \tanh ^{-1}(\sin (c+d x))}{8 a^3 d}+\frac {7 \tan (c+d x) \sec ^3(c+d x)}{12 a^3 d}+\frac {7 \tan (c+d x) \sec (c+d x)}{8 a^3 d}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(7*ArcTanh[Sin[c + d*x]])/(8*a^3*d) - (((7*I)/15)*Sec[c + d*x]^5)/(a^3*d) + (7*Sec[c + d*x]*Tan[c + d*x])/(8*a
^3*d) + (7*Sec[c + d*x]^3*Tan[c + d*x])/(12*a^3*d) - (((2*I)/3)*Sec[c + d*x]^7)/(a*d*(a + I*a*Tan[c + d*x])^2)

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 3501

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

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))^3} \, dx &=-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}+\frac {7 \int \frac {\sec ^7(c+d x)}{a+i a \tan (c+d x)} \, dx}{3 a^2}\\ &=-\frac {7 i \sec ^5(c+d x)}{15 a^3 d}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}+\frac {7 \int \sec ^5(c+d x) \, dx}{3 a^3}\\ &=-\frac {7 i \sec ^5(c+d x)}{15 a^3 d}+\frac {7 \sec ^3(c+d x) \tan (c+d x)}{12 a^3 d}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}+\frac {7 \int \sec ^3(c+d x) \, dx}{4 a^3}\\ &=-\frac {7 i \sec ^5(c+d x)}{15 a^3 d}+\frac {7 \sec (c+d x) \tan (c+d x)}{8 a^3 d}+\frac {7 \sec ^3(c+d x) \tan (c+d x)}{12 a^3 d}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}+\frac {7 \int \sec (c+d x) \, dx}{8 a^3}\\ &=\frac {7 \tanh ^{-1}(\sin (c+d x))}{8 a^3 d}-\frac {7 i \sec ^5(c+d x)}{15 a^3 d}+\frac {7 \sec (c+d x) \tan (c+d x)}{8 a^3 d}+\frac {7 \sec ^3(c+d x) \tan (c+d x)}{12 a^3 d}-\frac {2 i \sec ^7(c+d x)}{3 a d (a+i a \tan (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 113, normalized size = 0.95 \[ \frac {\sec ^8(c+d x) (\cos (3 (c+d x))+i \sin (3 (c+d x))) \left (-150 i \sin (2 (c+d x))+105 i \sin (4 (c+d x))+640 \cos (2 (c+d x))+1680 i \cos ^5(c+d x) \tanh ^{-1}\left (\cos (c) \tan \left (\frac {d x}{2}\right )+\sin (c)\right )+448\right )}{960 a^3 d (\tan (c+d x)-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^8*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)])*(448 + (1680*I)*ArcTanh[Sin[c] + Cos[c]*Tan[(d*x)/2]]*
Cos[c + d*x]^5 + 640*Cos[2*(c + d*x)] - (150*I)*Sin[2*(c + d*x)] + (105*I)*Sin[4*(c + d*x)]))/(960*a^3*d*(-I +
Tan[c + d*x])^3)

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fricas [B]  time = 0.52, size = 278, normalized size = 2.34 \[ \frac {105 \, {\left (e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 105 \, {\left (e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right ) - 210 i \, e^{\left (9 i \, d x + 9 i \, c\right )} - 980 i \, e^{\left (7 i \, d x + 7 i \, c\right )} - 1792 i \, e^{\left (5 i \, d x + 5 i \, c\right )} - 1580 i \, e^{\left (3 i \, d x + 3 i \, c\right )} + 210 i \, e^{\left (i \, d x + i \, c\right )}}{120 \, {\left (a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 5 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/120*(105*(e^(10*I*d*x + 10*I*c) + 5*e^(8*I*d*x + 8*I*c) + 10*e^(6*I*d*x + 6*I*c) + 10*e^(4*I*d*x + 4*I*c) +
5*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) + I) - 105*(e^(10*I*d*x + 10*I*c) + 5*e^(8*I*d*x + 8*I*c) + 10*
e^(6*I*d*x + 6*I*c) + 10*e^(4*I*d*x + 4*I*c) + 5*e^(2*I*d*x + 2*I*c) + 1)*log(e^(I*d*x + I*c) - I) - 210*I*e^(
9*I*d*x + 9*I*c) - 980*I*e^(7*I*d*x + 7*I*c) - 1792*I*e^(5*I*d*x + 5*I*c) - 1580*I*e^(3*I*d*x + 3*I*c) + 210*I
*e^(I*d*x + I*c))/(a^3*d*e^(10*I*d*x + 10*I*c) + 5*a^3*d*e^(8*I*d*x + 8*I*c) + 10*a^3*d*e^(6*I*d*x + 6*I*c) +
10*a^3*d*e^(4*I*d*x + 4*I*c) + 5*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)

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giac [A]  time = 1.94, size = 164, normalized size = 1.38 \[ \frac {\frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3}} - \frac {105 \, \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}{a^{3}} + \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 390 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 960 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 400 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 390 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 320 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 136 i\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5} a^{3}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/120*(105*log(tan(1/2*d*x + 1/2*c) + 1)/a^3 - 105*log(tan(1/2*d*x + 1/2*c) - 1)/a^3 + 2*(15*tan(1/2*d*x + 1/2
*c)^9 + 360*I*tan(1/2*d*x + 1/2*c)^8 - 390*tan(1/2*d*x + 1/2*c)^7 - 960*I*tan(1/2*d*x + 1/2*c)^6 + 400*I*tan(1
/2*d*x + 1/2*c)^4 + 390*tan(1/2*d*x + 1/2*c)^3 - 320*I*tan(1/2*d*x + 1/2*c)^2 - 15*tan(1/2*d*x + 1/2*c) + 136*
I)/((tan(1/2*d*x + 1/2*c)^2 - 1)^5*a^3))/d

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maple [B]  time = 0.41, size = 430, normalized size = 3.61 \[ -\frac {i}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {7 i}{12 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {3}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {13 i}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {5}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {i}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {3}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {i}{5 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 a^{3} d}+\frac {11 i}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {i}{5 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}+\frac {3}{4 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {11 i}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {1}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {7 i}{12 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3}{2 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {13 i}{8 a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^3,x)

[Out]

-1/2*I/a^3/d/(tan(1/2*d*x+1/2*c)+1)^4+1/8/a^3/d/(tan(1/2*d*x+1/2*c)-1)+7/12*I/a^3/d/(tan(1/2*d*x+1/2*c)-1)^3-3
/4/a^3/d/(tan(1/2*d*x+1/2*c)-1)^4+13/8*I/a^3/d/(tan(1/2*d*x+1/2*c)-1)-5/8/a^3/d/(tan(1/2*d*x+1/2*c)-1)^2-1/2*I
/a^3/d/(tan(1/2*d*x+1/2*c)-1)^4-3/2/a^3/d/(tan(1/2*d*x+1/2*c)-1)^3+1/5*I/a^3/d/(tan(1/2*d*x+1/2*c)+1)^5-7/8/a^
3/d*ln(tan(1/2*d*x+1/2*c)-1)+11/8*I/a^3/d/(tan(1/2*d*x+1/2*c)+1)^2+5/8/a^3/d/(tan(1/2*d*x+1/2*c)+1)^2-1/5*I/a^
3/d/(tan(1/2*d*x+1/2*c)-1)^5+3/4/a^3/d/(tan(1/2*d*x+1/2*c)+1)^4+11/8*I/a^3/d/(tan(1/2*d*x+1/2*c)-1)^2+1/8/a^3/
d/(tan(1/2*d*x+1/2*c)+1)-7/12*I/a^3/d/(tan(1/2*d*x+1/2*c)+1)^3-3/2/a^3/d/(tan(1/2*d*x+1/2*c)+1)^3-13/8*I/a^3/d
/(tan(1/2*d*x+1/2*c)+1)+7/8/a^3/d*ln(tan(1/2*d*x+1/2*c)+1)

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maxima [B]  time = 0.45, size = 341, normalized size = 2.87 \[ \frac {\frac {16 \, {\left (-\frac {15 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {320 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {390 i \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {400 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {960 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {390 i \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {360 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {15 i \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - 136\right )}}{-120 i \, a^{3} + \frac {600 i \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1200 i \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1200 i \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {600 i \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {120 i \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {7 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} - \frac {7 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/8*(16*(-15*I*sin(d*x + c)/(cos(d*x + c) + 1) + 320*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 390*I*sin(d*x + c)^
3/(cos(d*x + c) + 1)^3 - 400*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 960*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 3
90*I*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 360*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 15*I*sin(d*x + c)^9/(cos(
d*x + c) + 1)^9 - 136)/(-120*I*a^3 + 600*I*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1200*I*a^3*sin(d*x + c)^4
/(cos(d*x + c) + 1)^4 + 1200*I*a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 600*I*a^3*sin(d*x + c)^8/(cos(d*x + c
) + 1)^8 + 120*I*a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) + 7*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 -
7*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d

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mupad [B]  time = 6.08, size = 150, normalized size = 1.26 \[ \frac {7\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,a^3\,d}+\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,6{}\mathrm {i}-\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,16{}\mathrm {i}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,20{}\mathrm {i}}{3}+\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,16{}\mathrm {i}}{3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}+\frac {34}{15}{}\mathrm {i}}{a^3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}^5} \]

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)^3),x)

[Out]

(7*atanh(tan(c/2 + (d*x)/2)))/(4*a^3*d) + ((13*tan(c/2 + (d*x)/2)^3)/2 - (tan(c/2 + (d*x)/2)^2*16i)/3 - tan(c/
2 + (d*x)/2)/4 + (tan(c/2 + (d*x)/2)^4*20i)/3 - tan(c/2 + (d*x)/2)^6*16i - (13*tan(c/2 + (d*x)/2)^7)/2 + tan(c
/2 + (d*x)/2)^8*6i + tan(c/2 + (d*x)/2)^9/4 + 34i/15)/(a^3*d*(tan(c/2 + (d*x)/2)^2 - 1)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \int \frac {\sec ^{9}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**9/(a+I*a*tan(d*x+c))**3,x)

[Out]

I*Integral(sec(c + d*x)**9/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan(c + d*x) + I), x)/a**3

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