3.1.45 \(\int \frac {\tan ^6(c+d x)}{a+i a \tan (c+d x)} \, dx\) [45]

Optimal. Leaf size=130 \[ \frac {5 x}{2 a}+\frac {3 i \log (\cos (c+d x))}{a d}-\frac {5 \tan (c+d x)}{2 a d}+\frac {3 i \tan ^2(c+d x)}{2 a d}+\frac {5 \tan ^3(c+d x)}{6 a d}-\frac {3 i \tan ^4(c+d x)}{4 a d}-\frac {\tan ^5(c+d x)}{2 d (a+i a \tan (c+d x))} \]

[Out]

5/2*x/a+3*I*ln(cos(d*x+c))/a/d-5/2*tan(d*x+c)/a/d+3/2*I*tan(d*x+c)^2/a/d+5/6*tan(d*x+c)^3/a/d-3/4*I*tan(d*x+c)
^4/a/d-1/2*tan(d*x+c)^5/d/(a+I*a*tan(d*x+c))

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Rubi [A]
time = 0.10, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3631, 3609, 3606, 3556} \begin {gather*} -\frac {\tan ^5(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {3 i \tan ^4(c+d x)}{4 a d}+\frac {5 \tan ^3(c+d x)}{6 a d}+\frac {3 i \tan ^2(c+d x)}{2 a d}-\frac {5 \tan (c+d x)}{2 a d}+\frac {3 i \log (\cos (c+d x))}{a d}+\frac {5 x}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x)/(2*a) + ((3*I)*Log[Cos[c + d*x]])/(a*d) - (5*Tan[c + d*x])/(2*a*d) + (((3*I)/2)*Tan[c + d*x]^2)/(a*d) +
(5*Tan[c + d*x]^3)/(6*a*d) - (((3*I)/4)*Tan[c + d*x]^4)/(a*d) - Tan[c + d*x]^5/(2*d*(a + I*a*Tan[c + d*x]))

Rule 3556

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

Rule 3606

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[b*d*(Tan[e + f*x]/f), x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*
((a + b*Tan[e + f*x])^m/(f*m)), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3631

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

Rubi steps

\begin {align*} \int \frac {\tan ^6(c+d x)}{a+i a \tan (c+d x)} \, dx &=-\frac {\tan ^5(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \tan ^4(c+d x) (5 a-6 i a \tan (c+d x)) \, dx}{2 a^2}\\ &=-\frac {3 i \tan ^4(c+d x)}{4 a d}-\frac {\tan ^5(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \tan ^3(c+d x) (6 i a+5 a \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac {5 \tan ^3(c+d x)}{6 a d}-\frac {3 i \tan ^4(c+d x)}{4 a d}-\frac {\tan ^5(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \tan ^2(c+d x) (-5 a+6 i a \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac {3 i \tan ^2(c+d x)}{2 a d}+\frac {5 \tan ^3(c+d x)}{6 a d}-\frac {3 i \tan ^4(c+d x)}{4 a d}-\frac {\tan ^5(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac {\int \tan (c+d x) (-6 i a-5 a \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac {5 x}{2 a}-\frac {5 \tan (c+d x)}{2 a d}+\frac {3 i \tan ^2(c+d x)}{2 a d}+\frac {5 \tan ^3(c+d x)}{6 a d}-\frac {3 i \tan ^4(c+d x)}{4 a d}-\frac {\tan ^5(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac {(3 i) \int \tan (c+d x) \, dx}{a}\\ &=\frac {5 x}{2 a}+\frac {3 i \log (\cos (c+d x))}{a d}-\frac {5 \tan (c+d x)}{2 a d}+\frac {3 i \tan ^2(c+d x)}{2 a d}+\frac {5 \tan ^3(c+d x)}{6 a d}-\frac {3 i \tan ^4(c+d x)}{4 a d}-\frac {\tan ^5(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(840\) vs. \(2(130)=260\).
time = 6.58, size = 840, normalized size = 6.46 \begin {gather*} \frac {5 x \cos (c) \sec (c+d x) (\cos (d x)+i \sin (d x))}{2 (a+i a \tan (c+d x))}+\frac {3 \text {ArcTan}(\tan (d x)) \cos (c) \sec (c+d x) (\cos (d x)+i \sin (d x))}{d (a+i a \tan (c+d x))}+\frac {3 i \cos (c) \log \left (\cos ^2(c+d x)\right ) \sec (c+d x) (\cos (d x)+i \sin (d x))}{2 d (a+i a \tan (c+d x))}+\frac {\cos (2 d x) \sec (c+d x) \left (-\frac {1}{4} i \cos (c)-\frac {\sin (c)}{4}\right ) (\cos (d x)+i \sin (d x))}{d (a+i a \tan (c+d x))}+\frac {\sec ^3(c+d x) \left (\frac {1}{6} i \cos (c)-\frac {\sin (c)}{6}\right ) (9 \cos (c)-2 i \sin (c)) (\cos (d x)+i \sin (d x))}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) (a+i a \tan (c+d x))}+\frac {\sec ^5(c+d x) \left (-\frac {1}{4} i \cos (c)+\frac {\sin (c)}{4}\right ) (\cos (d x)+i \sin (d x))}{d (a+i a \tan (c+d x))}+\frac {5 i x \sec (c+d x) \sin (c) (\cos (d x)+i \sin (d x))}{2 (a+i a \tan (c+d x))}+\frac {3 i \text {ArcTan}(\tan (d x)) \sec (c+d x) \sin (c) (\cos (d x)+i \sin (d x))}{d (a+i a \tan (c+d x))}-\frac {3 \log \left (\cos ^2(c+d x)\right ) \sec (c+d x) \sin (c) (\cos (d x)+i \sin (d x))}{2 d (a+i a \tan (c+d x))}+\frac {\sec (c+d x) \left (-\frac {\cos (c)}{4}+\frac {1}{4} i \sin (c)\right ) (\cos (d x)+i \sin (d x)) \sin (2 d x)}{d (a+i a \tan (c+d x))}+\frac {7 i \sec ^2(c+d x) (\cos (d x)+i \sin (d x)) (-\cos (c-d x)+\cos (c+d x)-i \sin (c-d x)+i \sin (c+d x))}{6 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) (a+i a \tan (c+d x))}-\frac {i \sec ^4(c+d x) (\cos (d x)+i \sin (d x)) (-\cos (c-d x)+\cos (c+d x)-i \sin (c-d x)+i \sin (c+d x))}{6 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) (a+i a \tan (c+d x))}+\frac {x \sec (c+d x) (\cos (d x)+i \sin (d x)) (-3 \sec (c)-i (3 \cos (c)+3 i \sin (c)) \tan (c))}{a+i a \tan (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x*Cos[c]*Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x]))/(2*(a + I*a*Tan[c + d*x])) + (3*ArcTan[Tan[d*x]]*Cos[c]*Sec[
c + d*x]*(Cos[d*x] + I*Sin[d*x]))/(d*(a + I*a*Tan[c + d*x])) + (((3*I)/2)*Cos[c]*Log[Cos[c + d*x]^2]*Sec[c + d
*x]*(Cos[d*x] + I*Sin[d*x]))/(d*(a + I*a*Tan[c + d*x])) + (Cos[2*d*x]*Sec[c + d*x]*((-1/4*I)*Cos[c] - Sin[c]/4
)*(Cos[d*x] + I*Sin[d*x]))/(d*(a + I*a*Tan[c + d*x])) + (Sec[c + d*x]^3*((I/6)*Cos[c] - Sin[c]/6)*(9*Cos[c] -
(2*I)*Sin[c])*(Cos[d*x] + I*Sin[d*x]))/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(a + I*a*Tan[c + d*x]))
+ (Sec[c + d*x]^5*((-1/4*I)*Cos[c] + Sin[c]/4)*(Cos[d*x] + I*Sin[d*x]))/(d*(a + I*a*Tan[c + d*x])) + (((5*I)/2
)*x*Sec[c + d*x]*Sin[c]*(Cos[d*x] + I*Sin[d*x]))/(a + I*a*Tan[c + d*x]) + ((3*I)*ArcTan[Tan[d*x]]*Sec[c + d*x]
*Sin[c]*(Cos[d*x] + I*Sin[d*x]))/(d*(a + I*a*Tan[c + d*x])) - (3*Log[Cos[c + d*x]^2]*Sec[c + d*x]*Sin[c]*(Cos[
d*x] + I*Sin[d*x]))/(2*d*(a + I*a*Tan[c + d*x])) + (Sec[c + d*x]*(-1/4*Cos[c] + (I/4)*Sin[c])*(Cos[d*x] + I*Si
n[d*x])*Sin[2*d*x])/(d*(a + I*a*Tan[c + d*x])) + (((7*I)/6)*Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])*(-Cos[c - d
*x] + Cos[c + d*x] - I*Sin[c - d*x] + I*Sin[c + d*x]))/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(a + I*a
*Tan[c + d*x])) - ((I/6)*Sec[c + d*x]^4*(Cos[d*x] + I*Sin[d*x])*(-Cos[c - d*x] + Cos[c + d*x] - I*Sin[c - d*x]
+ I*Sin[c + d*x]))/(d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(a + I*a*Tan[c + d*x])) + (x*Sec[c + d*x]*(
Cos[d*x] + I*Sin[d*x])*(-3*Sec[c] - I*(3*Cos[c] + (3*I)*Sin[c])*Tan[c]))/(a + I*a*Tan[c + d*x])

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Maple [A]
time = 0.13, size = 88, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {-2 \tan \left (d x +c \right )-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+i \left (\tan ^{2}\left (d x +c \right )\right )-\frac {11 i \ln \left (\tan \left (d x +c \right )-i\right )}{4}-\frac {1}{2 \left (\tan \left (d x +c \right )-i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{4}}{d a}\)	\(88\)
default	\(\frac {-2 \tan \left (d x +c \right )-\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+i \left (\tan ^{2}\left (d x +c \right )\right )-\frac {11 i \ln \left (\tan \left (d x +c \right )-i\right )}{4}-\frac {1}{2 \left (\tan \left (d x +c \right )-i\right )}-\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{4}}{d a}\)	\(88\)
risch	\(\frac {11 x}{2 a}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 d a}+\frac {6 c}{d a}-\frac {2 i \left (9 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+7\right )}{3 d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d a}\)	\(102\)
norman	\(\frac {\frac {5 x}{2 a}+\frac {\tan ^{5}\left (d x +c \right )}{3 d a}+\frac {5 x \left (\tan ^{2}\left (d x +c \right )\right )}{2 a}-\frac {3 i}{2 d a}-\frac {5 \tan \left (d x +c \right )}{2 d a}-\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{3 d a}+\frac {3 i \left (\tan ^{4}\left (d x +c \right )\right )}{4 d a}-\frac {i \left (\tan ^{6}\left (d x +c \right )\right )}{4 d a}}{1+\tan ^{2}\left (d x +c \right )}-\frac {3 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d a}\)	\(145\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-2*tan(d*x+c)-1/4*I*tan(d*x+c)^4+1/3*tan(d*x+c)^3+I*tan(d*x+c)^2-11/4*I*ln(tan(d*x+c)-I)-1/2/(tan(d*x+c
)-I)-1/4*I*ln(tan(d*x+c)+I))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (110) = 220\).
time = 0.38, size = 221, normalized size = 1.70 \begin {gather*} \frac {66 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} + 3 \, {\left (88 \, d x - i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 12 \, {\left (33 \, d x - 7 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (132 \, d x - 49 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (33 \, d x - 34 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 36 \, {\left (-i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 4 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 6 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 3 i}{12 \, {\left (a d e^{\left (10 i \, d x + 10 i \, c\right )} + 4 \, a d e^{\left (8 i \, d x + 8 i \, c\right )} + 6 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 4 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(66*d*x*e^(10*I*d*x + 10*I*c) + 3*(88*d*x - I)*e^(8*I*d*x + 8*I*c) + 12*(33*d*x - 7*I)*e^(6*I*d*x + 6*I*c
) + 2*(132*d*x - 49*I)*e^(4*I*d*x + 4*I*c) + 2*(33*d*x - 34*I)*e^(2*I*d*x + 2*I*c) - 36*(-I*e^(10*I*d*x + 10*I
*c) - 4*I*e^(8*I*d*x + 8*I*c) - 6*I*e^(6*I*d*x + 6*I*c) - 4*I*e^(4*I*d*x + 4*I*c) - I*e^(2*I*d*x + 2*I*c))*log
(e^(2*I*d*x + 2*I*c) + 1) - 3*I)/(a*d*e^(10*I*d*x + 10*I*c) + 4*a*d*e^(8*I*d*x + 8*I*c) + 6*a*d*e^(6*I*d*x + 6
*I*c) + 4*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

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Sympy [A]
time = 0.35, size = 219, normalized size = 1.68 \begin {gather*} \frac {- 18 i e^{4 i c} e^{4 i d x} - 20 i e^{2 i c} e^{2 i d x} - 14 i}{3 a d e^{8 i c} e^{8 i d x} + 12 a d e^{6 i c} e^{6 i d x} + 18 a d e^{4 i c} e^{4 i d x} + 12 a d e^{2 i c} e^{2 i d x} + 3 a d} + \begin {cases} - \frac {i e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text {for}\: a d e^{2 i c} \neq 0 \\x \left (\frac {\left (11 e^{2 i c} - 1\right ) e^{- 2 i c}}{2 a} - \frac {11}{2 a}\right ) & \text {otherwise} \end {cases} + \frac {11 x}{2 a} + \frac {3 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+I*a*tan(d*x+c)),x)

[Out]

(-18*I*exp(4*I*c)*exp(4*I*d*x) - 20*I*exp(2*I*c)*exp(2*I*d*x) - 14*I)/(3*a*d*exp(8*I*c)*exp(8*I*d*x) + 12*a*d*
exp(6*I*c)*exp(6*I*d*x) + 18*a*d*exp(4*I*c)*exp(4*I*d*x) + 12*a*d*exp(2*I*c)*exp(2*I*d*x) + 3*a*d) + Piecewise
((-I*exp(-2*I*c)*exp(-2*I*d*x)/(4*a*d), Ne(a*d*exp(2*I*c), 0)), (x*((11*exp(2*I*c) - 1)*exp(-2*I*c)/(2*a) - 11
/(2*a)), True)) + 11*x/(2*a) + 3*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a*d)

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Giac [A]
time = 2.43, size = 116, normalized size = 0.89 \begin {gather*} -\frac {\frac {33 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {3 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a} + \frac {3 \, {\left (-11 i \, \tan \left (d x + c\right ) - 9\right )}}{a {\left (\tan \left (d x + c\right ) - i\right )}} + \frac {3 i \, a^{3} \tan \left (d x + c\right )^{4} - 4 \, a^{3} \tan \left (d x + c\right )^{3} - 12 i \, a^{3} \tan \left (d x + c\right )^{2} + 24 \, a^{3} \tan \left (d x + c\right )}{a^{4}}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(33*I*log(tan(d*x + c) - I)/a + 3*I*log(I*tan(d*x + c) - 1)/a + 3*(-11*I*tan(d*x + c) - 9)/(a*(tan(d*x +
c) - I)) + (3*I*a^3*tan(d*x + c)^4 - 4*a^3*tan(d*x + c)^3 - 12*I*a^3*tan(d*x + c)^2 + 24*a^3*tan(d*x + c))/a^
4)/d

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Mupad [B]
time = 4.07, size = 125, normalized size = 0.96 \begin {gather*} \frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a\,d}-\frac {2\,\mathrm {tan}\left (c+d\,x\right )}{a\,d}-\frac {1{}\mathrm {i}}{2\,a\,d\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{a\,d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,11{}\mathrm {i}}{4\,a\,d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{4\,a\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6/(a + a*tan(c + d*x)*1i),x)

[Out]

(tan(c + d*x)^2*1i)/(a*d) - (log(tan(c + d*x) + 1i)*1i)/(4*a*d) - (2*tan(c + d*x))/(a*d) - 1i/(2*a*d*(tan(c +
d*x)*1i + 1)) - (log(tan(c + d*x) - 1i)*11i)/(4*a*d) + tan(c + d*x)^3/(3*a*d) - (tan(c + d*x)^4*1i)/(4*a*d)

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