3.45 \(\int \frac{\tan ^6(c+d x)}{a+i a \tan (c+d x)} \, dx\)

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

[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]))

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Rubi [A]  time = 0.151399, antiderivative size = 130, normalized size of antiderivative = 1., 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 = {3550, 3528, 3525, 3475} \[ -\frac{\tan ^5(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{3 i \tan ^4(c+d x)}{4 a d}+\frac{5 \tan ^3(c+d x)}{6 a d}+\frac{3 i \tan ^2(c+d x)}{2 a d}-\frac{5 \tan (c+d x)}{2 a d}+\frac{3 i \log (\cos (c+d x))}{a d}+\frac{5 x}{2 a} \]

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 3550

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]

Rule 3528

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 3525

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 3475

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

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*}

Mathematica [B]  time = 6.50613, size = 840, normalized size = 6.46 \[ \frac{\left (\frac{\sin (c)}{4}-\frac{1}{4} i \cos (c)\right ) (\cos (d x)+i \sin (d x)) \sec ^5(c+d x)}{d (i \tan (c+d x) a+a)}-\frac{i (\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)) \sec ^4(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 ) (i \tan (c+d x) a+a)}+\frac{\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)) \sec ^3(c+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 ) (i \tan (c+d x) a+a)}+\frac{7 i (\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)) \sec ^2(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 ) (i \tan (c+d x) a+a)}+\frac{5 x \cos (c) (\cos (d x)+i \sin (d x)) \sec (c+d x)}{2 (i \tan (c+d x) a+a)}+\frac{3 \tan ^{-1}(\tan (d x)) \cos (c) (\cos (d x)+i \sin (d x)) \sec (c+d x)}{d (i \tan (c+d x) a+a)}+\frac{3 i \cos (c) \log \left (\cos ^2(c+d x)\right ) (\cos (d x)+i \sin (d x)) \sec (c+d x)}{2 d (i \tan (c+d x) a+a)}+\frac{\cos (2 d x) \left (-\frac{1}{4} i \cos (c)-\frac{\sin (c)}{4}\right ) (\cos (d x)+i \sin (d x)) \sec (c+d x)}{d (i \tan (c+d x) a+a)}+\frac{5 i x \sin (c) (\cos (d x)+i \sin (d x)) \sec (c+d x)}{2 (i \tan (c+d x) a+a)}+\frac{3 i \tan ^{-1}(\tan (d x)) \sin (c) (\cos (d x)+i \sin (d x)) \sec (c+d x)}{d (i \tan (c+d x) a+a)}-\frac{3 \log \left (\cos ^2(c+d x)\right ) \sin (c) (\cos (d x)+i \sin (d x)) \sec (c+d x)}{2 d (i \tan (c+d x) a+a)}+\frac{\left (\frac{1}{4} i \sin (c)-\frac{\cos (c)}{4}\right ) (\cos (d x)+i \sin (d x)) \sin (2 d x) \sec (c+d x)}{d (i \tan (c+d x) a+a)}+\frac{x (\cos (d x)+i \sin (d x)) (-3 \sec (c)-i (3 \cos (c)+3 i \sin (c)) \tan (c)) \sec (c+d x)}{i \tan (c+d x) a+a} \]

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]*((-I/4)*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*((-I/4)*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]*(-Cos[c]/4 + (I/4)*Sin[c])*(Cos[d*x] + I*Sin[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*S
in[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 = 0.024, size = 123, normalized size = 1. \begin{align*} -2\,{\frac{\tan \left ( dx+c \right ) }{ad}}-{\frac{{\frac{i}{4}} \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{ad}}+{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,ad}}+{\frac{i \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{ad}}-{\frac{{\frac{11\,i}{4}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{ad}}-{\frac{1}{2\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{4}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{ad}} \end{align*}

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)

[Out]

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

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

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

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Fricas [A]  time = 2.29588, size = 683, normalized size = 5.25 \begin{align*} \frac{66 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} +{\left (264 \, d x - 3 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (396 \, d x - 84 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (264 \, d x - 98 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (66 \, d x - 68 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (36 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 144 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 216 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 144 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 36 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{align*}

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) + (264*d*x - 3*I)*e^(8*I*d*x + 8*I*c) + (396*d*x - 84*I)*e^(6*I*d*x + 6*I*c
) + (264*d*x - 98*I)*e^(4*I*d*x + 4*I*c) + (66*d*x - 68*I)*e^(2*I*d*x + 2*I*c) + (36*I*e^(10*I*d*x + 10*I*c) +
144*I*e^(8*I*d*x + 8*I*c) + 216*I*e^(6*I*d*x + 6*I*c) + 144*I*e^(4*I*d*x + 4*I*c) + 36*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 = 3.85989, size = 201, normalized size = 1.55 \begin{align*} \frac{- \frac{6 i e^{- 4 i c} e^{4 i d x}}{a d} - \frac{20 i e^{- 6 i c} e^{2 i d x}}{3 a d} - \frac{14 i e^{- 8 i c}}{3 a d}}{e^{8 i d x} + 4 e^{- 2 i c} e^{6 i d x} + 6 e^{- 4 i c} e^{4 i d x} + 4 e^{- 6 i c} e^{2 i d x} + e^{- 8 i c}} + \frac{\left (\begin{cases} 11 x e^{2 i c} - \frac{i e^{- 2 i d x}}{2 d} & \text{for}\: d \neq 0 \\x \left (11 e^{2 i c} - 1\right ) & \text{otherwise} \end{cases}\right ) e^{- 2 i c}}{2 a} + \frac{3 i \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} \end{align*}

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]

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

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Giac [A]  time = 5.36787, size = 157, normalized size = 1.21 \begin{align*} -\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{align*}

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