3.56 \(\int \frac{\tan ^6(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=142 \[ \frac{3 i \tan ^4(c+d x)}{2 a^2 d (1+i \tan (c+d x))}-\frac{25 \tan ^3(c+d x)}{12 a^2 d}-\frac{3 i \tan ^2(c+d x)}{a^2 d}+\frac{25 \tan (c+d x)}{4 a^2 d}-\frac{6 i \log (\cos (c+d x))}{a^2 d}-\frac{25 x}{4 a^2}-\frac{\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

(-25*x)/(4*a^2) - ((6*I)*Log[Cos[c + d*x]])/(a^2*d) + (25*Tan[c + d*x])/(4*a^2*d) - ((3*I)*Tan[c + d*x]^2)/(a^
2*d) - (25*Tan[c + d*x]^3)/(12*a^2*d) + (((3*I)/2)*Tan[c + d*x]^4)/(a^2*d*(1 + I*Tan[c + d*x])) - Tan[c + d*x]
^5/(4*d*(a + I*a*Tan[c + d*x])^2)

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Rubi [A]  time = 0.226274, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3558, 3595, 3528, 3525, 3475} \[ \frac{3 i \tan ^4(c+d x)}{2 a^2 d (1+i \tan (c+d x))}-\frac{25 \tan ^3(c+d x)}{12 a^2 d}-\frac{3 i \tan ^2(c+d x)}{a^2 d}+\frac{25 \tan (c+d x)}{4 a^2 d}-\frac{6 i \log (\cos (c+d x))}{a^2 d}-\frac{25 x}{4 a^2}-\frac{\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(-25*x)/(4*a^2) - ((6*I)*Log[Cos[c + d*x]])/(a^2*d) + (25*Tan[c + d*x])/(4*a^2*d) - ((3*I)*Tan[c + d*x]^2)/(a^
2*d) - (25*Tan[c + d*x]^3)/(12*a^2*d) + (((3*I)/2)*Tan[c + d*x]^4)/(a^2*d*(1 + I*Tan[c + d*x])) - Tan[c + d*x]
^5/(4*d*(a + I*a*Tan[c + d*x])^2)

Rule 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*m), x] + Dist[1/(2*a^2*m), Int[(a
+ b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*n])

Rule 3595

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

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))^2} \, dx &=-\frac{\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\tan ^4(c+d x) (-5 a+7 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{3 i \tan ^4(c+d x)}{2 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan ^3(c+d x) \left (-48 i a^2-50 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{25 \tan ^3(c+d x)}{12 a^2 d}+\frac{3 i \tan ^4(c+d x)}{2 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan ^2(c+d x) \left (50 a^2-48 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{3 i \tan ^2(c+d x)}{a^2 d}-\frac{25 \tan ^3(c+d x)}{12 a^2 d}+\frac{3 i \tan ^4(c+d x)}{2 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan (c+d x) \left (48 i a^2+50 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{25 x}{4 a^2}+\frac{25 \tan (c+d x)}{4 a^2 d}-\frac{3 i \tan ^2(c+d x)}{a^2 d}-\frac{25 \tan ^3(c+d x)}{12 a^2 d}+\frac{3 i \tan ^4(c+d x)}{2 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{(6 i) \int \tan (c+d x) \, dx}{a^2}\\ &=-\frac{25 x}{4 a^2}-\frac{6 i \log (\cos (c+d x))}{a^2 d}+\frac{25 \tan (c+d x)}{4 a^2 d}-\frac{3 i \tan ^2(c+d x)}{a^2 d}-\frac{25 \tan ^3(c+d x)}{12 a^2 d}+\frac{3 i \tan ^4(c+d x)}{2 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^5(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 6.4384, size = 882, normalized size = 6.21 \[ \frac{i \sec (c) (\cos (d x)+i \sin (d x))^2 (-\cos (2 c-d x)+\cos (2 c+d x)-i \sin (2 c-d x)+i \sin (2 c+d x)) \sec ^5(c+d x)}{6 d (i \tan (c+d x) a+a)^2}+\frac{\sec (c) (3 \cos (c)-i \sin (c)) \left (\frac{1}{3} \sin (2 c)-\frac{1}{3} i \cos (2 c)\right ) (\cos (d x)+i \sin (d x))^2 \sec ^4(c+d x)}{d (i \tan (c+d x) a+a)^2}-\frac{13 i \sec (c) (\cos (d x)+i \sin (d x))^2 (-\cos (2 c-d x)+\cos (2 c+d x)-i \sin (2 c-d x)+i \sin (2 c+d x)) \sec ^3(c+d x)}{6 d (i \tan (c+d x) a+a)^2}-\frac{25 x \cos (2 c) (\cos (d x)+i \sin (d x))^2 \sec ^2(c+d x)}{4 (i \tan (c+d x) a+a)^2}-\frac{6 \tan ^{-1}(\tan (d x)) \cos (2 c) (\cos (d x)+i \sin (d x))^2 \sec ^2(c+d x)}{d (i \tan (c+d x) a+a)^2}+\frac{5 i \cos (2 d x) (\cos (d x)+i \sin (d x))^2 \sec ^2(c+d x)}{4 d (i \tan (c+d x) a+a)^2}-\frac{3 i \cos (2 c) \log \left (\cos ^2(c+d x)\right ) (\cos (d x)+i \sin (d x))^2 \sec ^2(c+d x)}{d (i \tan (c+d x) a+a)^2}+\frac{\cos (4 d x) \left (-\frac{1}{16} i \cos (2 c)-\frac{1}{16} \sin (2 c)\right ) (\cos (d x)+i \sin (d x))^2 \sec ^2(c+d x)}{d (i \tan (c+d x) a+a)^2}-\frac{25 i x \sin (2 c) (\cos (d x)+i \sin (d x))^2 \sec ^2(c+d x)}{4 (i \tan (c+d x) a+a)^2}-\frac{6 i \tan ^{-1}(\tan (d x)) \sin (2 c) (\cos (d x)+i \sin (d x))^2 \sec ^2(c+d x)}{d (i \tan (c+d x) a+a)^2}+\frac{3 \log \left (\cos ^2(c+d x)\right ) \sin (2 c) (\cos (d x)+i \sin (d x))^2 \sec ^2(c+d x)}{d (i \tan (c+d x) a+a)^2}+\frac{5 (\cos (d x)+i \sin (d x))^2 \sin (2 d x) \sec ^2(c+d x)}{4 d (i \tan (c+d x) a+a)^2}+\frac{\left (\frac{1}{16} i \sin (2 c)-\frac{1}{16} \cos (2 c)\right ) (\cos (d x)+i \sin (d x))^2 \sin (4 d x) \sec ^2(c+d x)}{d (i \tan (c+d x) a+a)^2}+\frac{x (\cos (d x)+i \sin (d x))^2 (i (6 \cos (2 c)+6 i \sin (2 c)) \tan (c)+6 i \tan (c)+6) \sec ^2(c+d x)}{(i \tan (c+d x) a+a)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(-25*x*Cos[2*c]*Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^2)/(4*(a + I*a*Tan[c + d*x])^2) - (6*ArcTan[Tan[d*x]]*C
os[2*c]*Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^2)/(d*(a + I*a*Tan[c + d*x])^2) + (((5*I)/4)*Cos[2*d*x]*Sec[c +
d*x]^2*(Cos[d*x] + I*Sin[d*x])^2)/(d*(a + I*a*Tan[c + d*x])^2) - ((3*I)*Cos[2*c]*Log[Cos[c + d*x]^2]*Sec[c +
d*x]^2*(Cos[d*x] + I*Sin[d*x])^2)/(d*(a + I*a*Tan[c + d*x])^2) + (Cos[4*d*x]*Sec[c + d*x]^2*((-I/16)*Cos[2*c]
- Sin[2*c]/16)*(Cos[d*x] + I*Sin[d*x])^2)/(d*(a + I*a*Tan[c + d*x])^2) + (Sec[c]*Sec[c + d*x]^4*(3*Cos[c] - I*
Sin[c])*((-I/3)*Cos[2*c] + Sin[2*c]/3)*(Cos[d*x] + I*Sin[d*x])^2)/(d*(a + I*a*Tan[c + d*x])^2) - (((25*I)/4)*x
*Sec[c + d*x]^2*Sin[2*c]*(Cos[d*x] + I*Sin[d*x])^2)/(a + I*a*Tan[c + d*x])^2 - ((6*I)*ArcTan[Tan[d*x]]*Sec[c +
d*x]^2*Sin[2*c]*(Cos[d*x] + I*Sin[d*x])^2)/(d*(a + I*a*Tan[c + d*x])^2) + (3*Log[Cos[c + d*x]^2]*Sec[c + d*x]
^2*Sin[2*c]*(Cos[d*x] + I*Sin[d*x])^2)/(d*(a + I*a*Tan[c + d*x])^2) + (5*Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x]
)^2*Sin[2*d*x])/(4*d*(a + I*a*Tan[c + d*x])^2) + (Sec[c + d*x]^2*(-Cos[2*c]/16 + (I/16)*Sin[2*c])*(Cos[d*x] +
I*Sin[d*x])^2*Sin[4*d*x])/(d*(a + I*a*Tan[c + d*x])^2) - (((13*I)/6)*Sec[c]*Sec[c + d*x]^3*(Cos[d*x] + I*Sin[d
*x])^2*(-Cos[2*c - d*x] + Cos[2*c + d*x] - I*Sin[2*c - d*x] + I*Sin[2*c + d*x]))/(d*(a + I*a*Tan[c + d*x])^2)
+ ((I/6)*Sec[c]*Sec[c + d*x]^5*(Cos[d*x] + I*Sin[d*x])^2*(-Cos[2*c - d*x] + Cos[2*c + d*x] - I*Sin[2*c - d*x]
+ I*Sin[2*c + d*x]))/(d*(a + I*a*Tan[c + d*x])^2) + (x*Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^2*(6 + (6*I)*Tan
[c] + I*(6*Cos[2*c] + (6*I)*Sin[2*c])*Tan[c]))/(a + I*a*Tan[c + d*x])^2

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Maple [A]  time = 0.027, size = 126, normalized size = 0.9 \begin{align*} 4\,{\frac{\tan \left ( dx+c \right ) }{{a}^{2}d}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,{a}^{2}d}}-{\frac{i \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{{a}^{2}d}}+{\frac{{\frac{i}{4}}}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{{\frac{49\,i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{{a}^{2}d}}+{\frac{11}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{2}d}} \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))^2,x)

[Out]

4*tan(d*x+c)/a^2/d-1/3*tan(d*x+c)^3/a^2/d-I/d/a^2*tan(d*x+c)^2+1/4*I/d/a^2/(tan(d*x+c)-I)^2+49/8*I/d/a^2*ln(ta
n(d*x+c)-I)+11/4/a^2/d/(tan(d*x+c)-I)-1/8*I/d/a^2*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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.23459, size = 618, normalized size = 4.35 \begin{align*} -\frac{588 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} +{\left (1764 \, d x - 348 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (1764 \, d x - 753 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (588 \, d x - 587 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (-288 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 864 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 864 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 288 i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 51 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i}{48 \,{\left (a^{2} d e^{\left (10 i \, d x + 10 i \, c\right )} + 3 \, a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 3 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 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))^2,x, algorithm="fricas")

[Out]

-1/48*(588*d*x*e^(10*I*d*x + 10*I*c) + (1764*d*x - 348*I)*e^(8*I*d*x + 8*I*c) + (1764*d*x - 753*I)*e^(6*I*d*x
+ 6*I*c) + (588*d*x - 587*I)*e^(4*I*d*x + 4*I*c) - (-288*I*e^(10*I*d*x + 10*I*c) - 864*I*e^(8*I*d*x + 8*I*c) -
864*I*e^(6*I*d*x + 6*I*c) - 288*I*e^(4*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 51*I*e^(2*I*d*x + 2*I*c
) + 3*I)/(a^2*d*e^(10*I*d*x + 10*I*c) + 3*a^2*d*e^(8*I*d*x + 8*I*c) + 3*a^2*d*e^(6*I*d*x + 6*I*c) + a^2*d*e^(4
*I*d*x + 4*I*c))

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Sympy [A]  time = 3.79549, size = 218, normalized size = 1.54 \begin{align*} \frac{\frac{6 i e^{- 2 i c} e^{4 i d x}}{a^{2} d} + \frac{12 i e^{- 4 i c} e^{2 i d x}}{a^{2} d} + \frac{26 i e^{- 6 i c}}{3 a^{2} d}}{e^{6 i d x} + 3 e^{- 2 i c} e^{4 i d x} + 3 e^{- 4 i c} e^{2 i d x} + e^{- 6 i c}} - \frac{\left (\begin{cases} 49 x e^{4 i c} - \frac{5 i e^{2 i c} e^{- 2 i d x}}{d} + \frac{i e^{- 4 i d x}}{4 d} & \text{for}\: d \neq 0 \\x \left (49 e^{4 i c} - 10 e^{2 i c} + 1\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i c}}{4 a^{2}} - \frac{6 i \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} 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))**2,x)

[Out]

(6*I*exp(-2*I*c)*exp(4*I*d*x)/(a**2*d) + 12*I*exp(-4*I*c)*exp(2*I*d*x)/(a**2*d) + 26*I*exp(-6*I*c)/(3*a**2*d))
/(exp(6*I*d*x) + 3*exp(-2*I*c)*exp(4*I*d*x) + 3*exp(-4*I*c)*exp(2*I*d*x) + exp(-6*I*c)) - Piecewise((49*x*exp(
4*I*c) - 5*I*exp(2*I*c)*exp(-2*I*d*x)/d + I*exp(-4*I*d*x)/(4*d), Ne(d, 0)), (x*(49*exp(4*I*c) - 10*exp(2*I*c)
+ 1), True))*exp(-4*I*c)/(4*a**2) - 6*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d)

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Giac [A]  time = 4.97304, size = 150, normalized size = 1.06 \begin{align*} -\frac{\frac{6 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} - \frac{294 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac{3 \,{\left (147 i \, \tan \left (d x + c\right )^{2} + 250 \, \tan \left (d x + c\right ) - 107 i\right )}}{a^{2}{\left (\tan \left (d x + c\right ) - i\right )}^{2}} + \frac{16 \,{\left (a^{4} \tan \left (d x + c\right )^{3} + 3 i \, a^{4} \tan \left (d x + c\right )^{2} - 12 \, a^{4} \tan \left (d x + c\right )\right )}}{a^{6}}}{48 \, 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))^2,x, algorithm="giac")

[Out]

-1/48*(6*I*log(tan(d*x + c) + I)/a^2 - 294*I*log(tan(d*x + c) - I)/a^2 + 3*(147*I*tan(d*x + c)^2 + 250*tan(d*x
+ c) - 107*I)/(a^2*(tan(d*x + c) - I)^2) + 16*(a^4*tan(d*x + c)^3 + 3*I*a^4*tan(d*x + c)^2 - 12*a^4*tan(d*x +
c))/a^6)/d