3.75 \(\int \frac{\tan ^6(c+d x)}{(a+i a \tan (c+d x))^4} \, dx\)

Optimal. Leaf size=171 \[ \frac{31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac{2 i \tan ^2(c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac{65 \tan (c+d x)}{16 a^4 d}-\frac{4 i \log (\cos (c+d x))}{a^4 d}-\frac{65 x}{16 a^4}-\frac{\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

[Out]

(-65*x)/(16*a^4) - ((4*I)*Log[Cos[c + d*x]])/(a^4*d) + (65*Tan[c + d*x])/(16*a^4*d) - ((2*I)*Tan[c + d*x]^2)/(
a^4*d*(1 + I*Tan[c + d*x])) + (31*Tan[c + d*x]^3)/(48*a^4*d*(1 + I*Tan[c + d*x])^2) - Tan[c + d*x]^5/(8*d*(a +
I*a*Tan[c + d*x])^4) + (((7*I)/24)*Tan[c + d*x]^4)/(a*d*(a + I*a*Tan[c + d*x])^3)

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Rubi [A]  time = 0.393183, antiderivative size = 171, 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 = {3558, 3595, 3525, 3475} \[ \frac{31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac{2 i \tan ^2(c+d x)}{a^4 d (1+i \tan (c+d x))}+\frac{65 \tan (c+d x)}{16 a^4 d}-\frac{4 i \log (\cos (c+d x))}{a^4 d}-\frac{65 x}{16 a^4}-\frac{\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^4,x]

[Out]

(-65*x)/(16*a^4) - ((4*I)*Log[Cos[c + d*x]])/(a^4*d) + (65*Tan[c + d*x])/(16*a^4*d) - ((2*I)*Tan[c + d*x]^2)/(
a^4*d*(1 + I*Tan[c + d*x])) + (31*Tan[c + d*x]^3)/(48*a^4*d*(1 + I*Tan[c + d*x])^2) - Tan[c + d*x]^5/(8*d*(a +
I*a*Tan[c + d*x])^4) + (((7*I)/24)*Tan[c + d*x]^4)/(a*d*(a + I*a*Tan[c + d*x])^3)

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 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))^4} \, dx &=-\frac{\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}-\frac{\int \frac{\tan ^4(c+d x) (-5 a+9 i a \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx}{8 a^2}\\ &=-\frac{\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}+\frac{\int \frac{\tan ^3(c+d x) \left (-56 i a^2-68 a^2 \tan (c+d x)\right )}{(a+i a \tan (c+d x))^2} \, dx}{48 a^4}\\ &=\frac{31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac{\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac{\int \frac{\tan ^2(c+d x) \left (372 a^3-396 i a^3 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{192 a^6}\\ &=\frac{31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac{\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac{2 i \tan ^2(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{\int \tan (c+d x) \left (1536 i a^4+1560 a^4 \tan (c+d x)\right ) \, dx}{384 a^8}\\ &=-\frac{65 x}{16 a^4}+\frac{65 \tan (c+d x)}{16 a^4 d}+\frac{31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac{\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac{2 i \tan ^2(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}+\frac{(4 i) \int \tan (c+d x) \, dx}{a^4}\\ &=-\frac{65 x}{16 a^4}-\frac{4 i \log (\cos (c+d x))}{a^4 d}+\frac{65 \tan (c+d x)}{16 a^4 d}+\frac{31 \tan ^3(c+d x)}{48 a^4 d (1+i \tan (c+d x))^2}-\frac{\tan ^5(c+d x)}{8 d (a+i a \tan (c+d x))^4}+\frac{7 i \tan ^4(c+d x)}{24 a d (a+i a \tan (c+d x))^3}-\frac{2 i \tan ^2(c+d x)}{d \left (a^4+i a^4 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.749966, size = 429, normalized size = 2.51 \[ -\frac{\sec (c) \sec ^5(c+d x) (832 \sin (2 c+d x)+1560 i d x \sin (2 c+3 d x)+835 \sin (2 c+3 d x)+1560 i d x \sin (4 c+3 d x)+1603 \sin (4 c+3 d x)+1560 i d x \sin (4 c+5 d x)-765 \sin (4 c+5 d x)+1560 i d x \sin (6 c+5 d x)+3 \sin (6 c+5 d x)-536 i \cos (2 c+d x)+1560 d x \cos (2 c+3 d x)-893 i \cos (2 c+3 d x)+1560 d x \cos (4 c+3 d x)-1661 i \cos (4 c+3 d x)+1560 d x \cos (4 c+5 d x)+771 i \cos (4 c+5 d x)+1560 d x \cos (6 c+5 d x)+3 i \cos (6 c+5 d x)+1536 i \cos (2 c+3 d x) \log (\cos (c+d x))+1536 i \cos (4 c+3 d x) \log (\cos (c+d x))+1536 i \cos (4 c+5 d x) \log (\cos (c+d x))+1536 i \cos (6 c+5 d x) \log (\cos (c+d x))-1536 \sin (2 c+3 d x) \log (\cos (c+d x))-1536 \sin (4 c+3 d x) \log (\cos (c+d x))-1536 \sin (4 c+5 d x) \log (\cos (c+d x))-1536 \sin (6 c+5 d x) \log (\cos (c+d x))+832 \sin (d x)-536 i \cos (d x))}{1536 a^4 d (\tan (c+d x)-i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + I*a*Tan[c + d*x])^4,x]

[Out]

-(Sec[c]*Sec[c + d*x]^5*((-536*I)*Cos[d*x] - (536*I)*Cos[2*c + d*x] - (893*I)*Cos[2*c + 3*d*x] + 1560*d*x*Cos[
2*c + 3*d*x] - (1661*I)*Cos[4*c + 3*d*x] + 1560*d*x*Cos[4*c + 3*d*x] + (771*I)*Cos[4*c + 5*d*x] + 1560*d*x*Cos
[4*c + 5*d*x] + (3*I)*Cos[6*c + 5*d*x] + 1560*d*x*Cos[6*c + 5*d*x] + (1536*I)*Cos[2*c + 3*d*x]*Log[Cos[c + d*x
]] + (1536*I)*Cos[4*c + 3*d*x]*Log[Cos[c + d*x]] + (1536*I)*Cos[4*c + 5*d*x]*Log[Cos[c + d*x]] + (1536*I)*Cos[
6*c + 5*d*x]*Log[Cos[c + d*x]] + 832*Sin[d*x] + 832*Sin[2*c + d*x] + 835*Sin[2*c + 3*d*x] + (1560*I)*d*x*Sin[2
*c + 3*d*x] - 1536*Log[Cos[c + d*x]]*Sin[2*c + 3*d*x] + 1603*Sin[4*c + 3*d*x] + (1560*I)*d*x*Sin[4*c + 3*d*x]
- 1536*Log[Cos[c + d*x]]*Sin[4*c + 3*d*x] - 765*Sin[4*c + 5*d*x] + (1560*I)*d*x*Sin[4*c + 5*d*x] - 1536*Log[Co
s[c + d*x]]*Sin[4*c + 5*d*x] + 3*Sin[6*c + 5*d*x] + (1560*I)*d*x*Sin[6*c + 5*d*x] - 1536*Log[Cos[c + d*x]]*Sin
[6*c + 5*d*x]))/(1536*a^4*d*(-I + Tan[c + d*x])^4)

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Maple [A]  time = 0.029, size = 131, normalized size = 0.8 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{{a}^{4}d}}+{\frac{{\frac{49\,i}{16}}}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{{\frac{i}{8}}}{{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{4}}}+{\frac{{\frac{129\,i}{32}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{{a}^{4}d}}-{\frac{11}{12\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}+{\frac{111}{16\,{a}^{4}d \left ( \tan \left ( dx+c \right ) -i \right ) }}-{\frac{{\frac{i}{32}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{{a}^{4}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))^4,x)

[Out]

tan(d*x+c)/a^4/d+49/16*I/d/a^4/(tan(d*x+c)-I)^2-1/8*I/d/a^4/(tan(d*x+c)-I)^4+129/32*I/d/a^4*ln(tan(d*x+c)-I)-1
1/12/d/a^4/(tan(d*x+c)-I)^3+111/16/d/a^4/(tan(d*x+c)-I)-1/32*I/d/a^4*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))^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.61925, size = 435, normalized size = 2.54 \begin{align*} -\frac{3096 \, d x e^{\left (10 i \, d x + 10 i \, c\right )} +{\left (3096 \, d x - 1632 i\right )} e^{\left (8 i \, d x + 8 i \, c\right )} -{\left (-1536 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 1536 i \, e^{\left (8 i \, d x + 8 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 684 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 148 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 29 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i}{384 \,{\left (a^{4} d e^{\left (10 i \, d x + 10 i \, c\right )} + a^{4} d e^{\left (8 i \, d x + 8 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))^4,x, algorithm="fricas")

[Out]

-1/384*(3096*d*x*e^(10*I*d*x + 10*I*c) + (3096*d*x - 1632*I)*e^(8*I*d*x + 8*I*c) - (-1536*I*e^(10*I*d*x + 10*I
*c) - 1536*I*e^(8*I*d*x + 8*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 684*I*e^(6*I*d*x + 6*I*c) + 148*I*e^(4*I*d*x
+ 4*I*c) - 29*I*e^(2*I*d*x + 2*I*c) + 3*I)/(a^4*d*e^(10*I*d*x + 10*I*c) + a^4*d*e^(8*I*d*x + 8*I*c))

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Sympy [A]  time = 10.7219, size = 196, normalized size = 1.15 \begin{align*} - \frac{\left (\begin{cases} 129 x e^{8 i c} - \frac{36 i e^{6 i c} e^{- 2 i d x}}{d} + \frac{15 i e^{4 i c} e^{- 4 i d x}}{2 d} - \frac{4 i e^{2 i c} e^{- 6 i d x}}{3 d} + \frac{i e^{- 8 i d x}}{8 d} & \text{for}\: d \neq 0 \\x \left (129 e^{8 i c} - 72 e^{6 i c} + 30 e^{4 i c} - 8 e^{2 i c} + 1\right ) & \text{otherwise} \end{cases}\right ) e^{- 8 i c}}{16 a^{4}} - \frac{4 i \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{4} d} + \frac{2 i e^{- 2 i c}}{a^{4} d \left (e^{2 i d x} + e^{- 2 i c}\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))**4,x)

[Out]

-Piecewise((129*x*exp(8*I*c) - 36*I*exp(6*I*c)*exp(-2*I*d*x)/d + 15*I*exp(4*I*c)*exp(-4*I*d*x)/(2*d) - 4*I*exp
(2*I*c)*exp(-6*I*d*x)/(3*d) + I*exp(-8*I*d*x)/(8*d), Ne(d, 0)), (x*(129*exp(8*I*c) - 72*exp(6*I*c) + 30*exp(4*
I*c) - 8*exp(2*I*c) + 1), True))*exp(-8*I*c)/(16*a**4) - 4*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**4*d) + 2*I*ex
p(-2*I*c)/(a**4*d*(exp(2*I*d*x) + exp(-2*I*c)))

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Giac [A]  time = 5.36575, size = 135, normalized size = 0.79 \begin{align*} -\frac{\frac{12 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac{1548 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} - \frac{384 \, \tan \left (d x + c\right )}{a^{4}} - \frac{-3225 i \, \tan \left (d x + c\right )^{4} - 10236 \, \tan \left (d x + c\right )^{3} + 12534 i \, \tan \left (d x + c\right )^{2} + 6908 \, \tan \left (d x + c\right ) - 1433 i}{a^{4}{\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, 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))^4,x, algorithm="giac")

[Out]

-1/384*(12*I*log(tan(d*x + c) + I)/a^4 - 1548*I*log(tan(d*x + c) - I)/a^4 - 384*tan(d*x + c)/a^4 - (-3225*I*ta
n(d*x + c)^4 - 10236*tan(d*x + c)^3 + 12534*I*tan(d*x + c)^2 + 6908*tan(d*x + c) - 1433*I)/(a^4*(tan(d*x + c)
- I)^4))/d