3.66 \(\int \frac{\tan ^6(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=161 \[ \frac{55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{7 i \tan ^2(c+d x)}{2 a^3 d}-\frac{55 \tan (c+d x)}{8 a^3 d}+\frac{7 i \log (\cos (c+d x))}{a^3 d}+\frac{55 x}{8 a^3}-\frac{\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2} \]

[Out]

(55*x)/(8*a^3) + ((7*I)*Log[Cos[c + d*x]])/(a^3*d) - (55*Tan[c + d*x])/(8*a^3*d) + (((7*I)/2)*Tan[c + d*x]^2)/
(a^3*d) - Tan[c + d*x]^5/(6*d*(a + I*a*Tan[c + d*x])^3) + (((13*I)/24)*Tan[c + d*x]^4)/(a*d*(a + I*a*Tan[c + d
*x])^2) + (55*Tan[c + d*x]^3)/(24*d*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.318401, antiderivative size = 161, 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} \[ \frac{55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{7 i \tan ^2(c+d x)}{2 a^3 d}-\frac{55 \tan (c+d x)}{8 a^3 d}+\frac{7 i \log (\cos (c+d x))}{a^3 d}+\frac{55 x}{8 a^3}-\frac{\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(55*x)/(8*a^3) + ((7*I)*Log[Cos[c + d*x]])/(a^3*d) - (55*Tan[c + d*x])/(8*a^3*d) + (((7*I)/2)*Tan[c + d*x]^2)/
(a^3*d) - Tan[c + d*x]^5/(6*d*(a + I*a*Tan[c + d*x])^3) + (((13*I)/24)*Tan[c + d*x]^4)/(a*d*(a + I*a*Tan[c + d
*x])^2) + (55*Tan[c + d*x]^3)/(24*d*(a^3 + I*a^3*Tan[c + d*x]))

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))^3} \, dx &=-\frac{\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac{\int \frac{\tan ^4(c+d x) (-5 a+8 i a \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=-\frac{\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\tan ^3(c+d x) \left (-52 i a^2-58 a^2 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=-\frac{\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\int \tan ^2(c+d x) \left (330 a^3-336 i a^3 \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=\frac{7 i \tan ^2(c+d x)}{2 a^3 d}-\frac{\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{\int \tan (c+d x) \left (336 i a^3+330 a^3 \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=\frac{55 x}{8 a^3}-\frac{55 \tan (c+d x)}{8 a^3 d}+\frac{7 i \tan ^2(c+d x)}{2 a^3 d}-\frac{\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{(7 i) \int \tan (c+d x) \, dx}{a^3}\\ &=\frac{55 x}{8 a^3}+\frac{7 i \log (\cos (c+d x))}{a^3 d}-\frac{55 \tan (c+d x)}{8 a^3 d}+\frac{7 i \tan ^2(c+d x)}{2 a^3 d}-\frac{\tan ^5(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{13 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{55 \tan ^3(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 3.94587, size = 264, normalized size = 1.64 \[ \frac{\sec ^3(c+d x) (\cos (d x)+i \sin (d x))^3 \left (660 i d x \sin (3 c)-234 i \sin (c) \sin (2 d x)-27 i \sin (c) \sin (4 d x)+2 i \sin (3 c) \sin (6 d x)+234 \sin (c) \cos (2 d x)+27 \sin (c) \cos (4 d x)-2 \sin (3 c) \cos (6 d x)+9 \cos (c) (29 \sin (d x)-23 i \cos (d x)) (\cos (3 d x)-i \sin (3 d x))-48 \sin (3 c) \sec ^2(c+d x)-288 i \sin (3 c) \sec (c) \sin (d x) \sec (c+d x)-672 \sin (3 c) \log (\cos (c+d x))+\cos (3 c) \left (48 i \sec ^2(c+d x)+672 i \log (\cos (c+d x))-288 \sec (c) \sin (d x) \sec (c+d x)+660 d x-2 \sin (6 d x)-2 i \cos (6 d x)\right )\right )}{96 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c + d*x]^3*(Cos[d*x] + I*Sin[d*x])^3*(234*Cos[2*d*x]*Sin[c] + 27*Cos[4*d*x]*Sin[c] + (660*I)*d*x*Sin[3*c]
- 2*Cos[6*d*x]*Sin[3*c] - 672*Log[Cos[c + d*x]]*Sin[3*c] - 48*Sec[c + d*x]^2*Sin[3*c] - (288*I)*Sec[c]*Sec[c
+ d*x]*Sin[3*c]*Sin[d*x] - (234*I)*Sin[c]*Sin[2*d*x] + 9*Cos[c]*((-23*I)*Cos[d*x] + 29*Sin[d*x])*(Cos[3*d*x] -
I*Sin[3*d*x]) - (27*I)*Sin[c]*Sin[4*d*x] + Cos[3*c]*(660*d*x - (2*I)*Cos[6*d*x] + (672*I)*Log[Cos[c + d*x]] +
(48*I)*Sec[c + d*x]^2 - 288*Sec[c]*Sec[c + d*x]*Sin[d*x] - 2*Sin[6*d*x]) + (2*I)*Sin[3*c]*Sin[6*d*x]))/(96*d*
(a + I*a*Tan[c + d*x])^3)

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Maple [A]  time = 0.028, size = 129, normalized size = 0.8 \begin{align*}{\frac{{\frac{i}{2}} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d{a}^{3}}}-3\,{\frac{\tan \left ( dx+c \right ) }{d{a}^{3}}}-{\frac{{\frac{111\,i}{16}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{d{a}^{3}}}-{\frac{{\frac{11\,i}{8}}}{d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{1}{6\,d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}-{\frac{49}{8\,d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{d{a}^{3}}} \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))^3,x)

[Out]

1/2*I/d/a^3*tan(d*x+c)^2-3*tan(d*x+c)/a^3/d-111/16*I/d/a^3*ln(tan(d*x+c)-I)-11/8*I/d/a^3/(tan(d*x+c)-I)^2+1/6/
d/a^3/(tan(d*x+c)-I)^3-49/8/a^3/d/(tan(d*x+c)-I)-1/16*I/d/a^3*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))^3,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.25928, size = 525, normalized size = 3.26 \begin{align*} \frac{1332 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} +{\left (2664 \, d x - 618 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} +{\left (1332 \, d x - 1017 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (672 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 1344 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 672 i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 182 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 23 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i}{96 \,{\left (a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 2 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + a^{3} d e^{\left (6 i \, d x + 6 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))^3,x, algorithm="fricas")

[Out]

1/96*(1332*d*x*e^(10*I*d*x + 10*I*c) + (2664*d*x - 618*I)*e^(8*I*d*x + 8*I*c) + (1332*d*x - 1017*I)*e^(6*I*d*x
+ 6*I*c) + (672*I*e^(10*I*d*x + 10*I*c) + 1344*I*e^(8*I*d*x + 8*I*c) + 672*I*e^(6*I*d*x + 6*I*c))*log(e^(2*I*
d*x + 2*I*c) + 1) - 182*I*e^(4*I*d*x + 4*I*c) + 23*I*e^(2*I*d*x + 2*I*c) - 2*I)/(a^3*d*e^(10*I*d*x + 10*I*c) +
2*a^3*d*e^(8*I*d*x + 8*I*c) + a^3*d*e^(6*I*d*x + 6*I*c))

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Sympy [A]  time = 4.71115, size = 209, normalized size = 1.3 \begin{align*} \frac{- \frac{4 i e^{- 2 i c} e^{2 i d x}}{a^{3} d} - \frac{6 i e^{- 4 i c}}{a^{3} d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \frac{\left (\begin{cases} 111 x e^{6 i c} - \frac{39 i e^{4 i c} e^{- 2 i d x}}{2 d} + \frac{9 i e^{2 i c} e^{- 4 i d x}}{4 d} - \frac{i e^{- 6 i d x}}{6 d} & \text{for}\: d \neq 0 \\x \left (111 e^{6 i c} - 39 e^{4 i c} + 9 e^{2 i c} - 1\right ) & \text{otherwise} \end{cases}\right ) e^{- 6 i c}}{8 a^{3}} + \frac{7 i \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} 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))**3,x)

[Out]

(-4*I*exp(-2*I*c)*exp(2*I*d*x)/(a**3*d) - 6*I*exp(-4*I*c)/(a**3*d))/(exp(4*I*d*x) + 2*exp(-2*I*c)*exp(2*I*d*x)
+ exp(-4*I*c)) + Piecewise((111*x*exp(6*I*c) - 39*I*exp(4*I*c)*exp(-2*I*d*x)/(2*d) + 9*I*exp(2*I*c)*exp(-4*I*
d*x)/(4*d) - I*exp(-6*I*d*x)/(6*d), Ne(d, 0)), (x*(111*exp(6*I*c) - 39*exp(4*I*c) + 9*exp(2*I*c) - 1), True))*
exp(-6*I*c)/(8*a**3) + 7*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**3*d)

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Giac [A]  time = 4.95758, size = 150, normalized size = 0.93 \begin{align*} -\frac{\frac{666 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac{6 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a^{3}} + \frac{48 \,{\left (-i \, a^{3} \tan \left (d x + c\right )^{2} + 6 \, a^{3} \tan \left (d x + c\right )\right )}}{a^{6}} - \frac{1221 i \, \tan \left (d x + c\right )^{3} + 3075 \, \tan \left (d x + c\right )^{2} - 2619 i \, \tan \left (d x + c\right ) - 749}{a^{3}{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, 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))^3,x, algorithm="giac")

[Out]

-1/96*(666*I*log(tan(d*x + c) - I)/a^3 + 6*I*log(I*tan(d*x + c) - 1)/a^3 + 48*(-I*a^3*tan(d*x + c)^2 + 6*a^3*t
an(d*x + c))/a^6 - (1221*I*tan(d*x + c)^3 + 3075*tan(d*x + c)^2 - 2619*I*tan(d*x + c) - 749)/(a^3*(tan(d*x + c
) - I)^3))/d