3.57 \(\int \frac{\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=124 \[ \frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{2 \tan ^2(c+d x)}{a^2 d}-\frac{15 i \tan (c+d x)}{4 a^2 d}-\frac{4 \log (\cos (c+d x))}{a^2 d}+\frac{15 i x}{4 a^2}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

(((15*I)/4)*x)/a^2 - (4*Log[Cos[c + d*x]])/(a^2*d) - (((15*I)/4)*Tan[c + d*x])/(a^2*d) - (2*Tan[c + d*x]^2)/(a
^2*d) + (((5*I)/4)*Tan[c + d*x]^3)/(a^2*d*(1 + I*Tan[c + d*x])) - Tan[c + d*x]^4/(4*d*(a + I*a*Tan[c + d*x])^2
)

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Rubi [A]  time = 0.194879, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ \frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{2 \tan ^2(c+d x)}{a^2 d}-\frac{15 i \tan (c+d x)}{4 a^2 d}-\frac{4 \log (\cos (c+d x))}{a^2 d}+\frac{15 i x}{4 a^2}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(((15*I)/4)*x)/a^2 - (4*Log[Cos[c + d*x]])/(a^2*d) - (((15*I)/4)*Tan[c + d*x])/(a^2*d) - (2*Tan[c + d*x]^2)/(a
^2*d) + (((5*I)/4)*Tan[c + d*x]^3)/(a^2*d*(1 + I*Tan[c + d*x])) - Tan[c + d*x]^4/(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 ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\tan ^3(c+d x) (-4 a+6 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan ^2(c+d x) \left (-30 i a^2-32 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac{2 \tan ^2(c+d x)}{a^2 d}+\frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan (c+d x) \left (32 a^2-30 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac{15 i x}{4 a^2}-\frac{15 i \tan (c+d x)}{4 a^2 d}-\frac{2 \tan ^2(c+d x)}{a^2 d}+\frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{4 \int \tan (c+d x) \, dx}{a^2}\\ &=\frac{15 i x}{4 a^2}-\frac{4 \log (\cos (c+d x))}{a^2 d}-\frac{15 i \tan (c+d x)}{4 a^2 d}-\frac{2 \tan ^2(c+d x)}{a^2 d}+\frac{5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac{\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 1.61437, size = 300, normalized size = 2.42 \[ \frac{\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-128 i d x \sin ^2(c)+60 d x \sin (2 c)-\sin (2 c) \sin (4 d x)-64 d x \tan (c)-i \sin (2 c) \cos (4 d x)-16 \sec (c) \cos (2 c-d x) \sec (c+d x)+16 \sec (c) \cos (2 c+d x) \sec (c+d x)+8 i \sin (2 c) \sec ^2(c+d x)-16 i \sec (c) \sin (2 c-d x) \sec (c+d x)+16 i \sec (c) \sin (2 c+d x) \sec (c+d x)+32 i \sin (2 c) \log \left (\cos ^2(c+d x)\right )+64 (\sin (2 c)-i \cos (2 c)) \tan ^{-1}(\tan (d x))+\cos (2 c) \left (-64 d x \tan (c)+8 \sec ^2(c+d x)+32 \log \left (\cos ^2(c+d x)\right )-60 i d x-i \sin (4 d x)+\cos (4 d x)\right )+64 i d x+16 i \sin (2 d x)-16 \cos (2 d x)\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^2*((64*I)*d*x - 16*Cos[2*d*x] - 16*Cos[2*c - d*x]*Sec[c]*Sec[c + d*x]
+ 16*Cos[2*c + d*x]*Sec[c]*Sec[c + d*x] - (128*I)*d*x*Sin[c]^2 + 60*d*x*Sin[2*c] - I*Cos[4*d*x]*Sin[2*c] + (32
*I)*Log[Cos[c + d*x]^2]*Sin[2*c] + (8*I)*Sec[c + d*x]^2*Sin[2*c] + 64*ArcTan[Tan[d*x]]*((-I)*Cos[2*c] + Sin[2*
c]) + (16*I)*Sin[2*d*x] - Sin[2*c]*Sin[4*d*x] - (16*I)*Sec[c]*Sec[c + d*x]*Sin[2*c - d*x] + (16*I)*Sec[c]*Sec[
c + d*x]*Sin[2*c + d*x] - 64*d*x*Tan[c] + Cos[2*c]*((-60*I)*d*x + Cos[4*d*x] + 32*Log[Cos[c + d*x]^2] + 8*Sec[
c + d*x]^2 - I*Sin[4*d*x] - 64*d*x*Tan[c])))/(16*a^2*d*(-I + Tan[c + d*x])^2)

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Maple [A]  time = 0.024, size = 108, normalized size = 0.9 \begin{align*}{\frac{-2\,i\tan \left ( dx+c \right ) }{{a}^{2}d}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,{a}^{2}d}}-{\frac{{\frac{9\,i}{4}}}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{1}{4\,{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{31\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{8\,{a}^{2}d}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{8\,{a}^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x)

[Out]

-2*I/d/a^2*tan(d*x+c)-1/2*tan(d*x+c)^2/a^2/d-9/4*I/d/a^2/(tan(d*x+c)-I)+1/4/d/a^2/(tan(d*x+c)-I)^2+31/8/d/a^2*
ln(tan(d*x+c)-I)+1/8/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)^5/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.24751, size = 451, normalized size = 3.64 \begin{align*} \frac{124 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 8 \,{\left (-31 i \, d x - 6\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (124 i \, d x + 95\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 64 \,{\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 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 ) + 14 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1}{16 \,{\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, 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)^5/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(124*I*d*x*e^(8*I*d*x + 8*I*c) - 8*(-31*I*d*x - 6)*e^(6*I*d*x + 6*I*c) + (124*I*d*x + 95)*e^(4*I*d*x + 4*
I*c) - 64*(e^(8*I*d*x + 8*I*c) + 2*e^(6*I*d*x + 6*I*c) + e^(4*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) + 1
4*e^(2*I*d*x + 2*I*c) - 1)/(a^2*d*e^(8*I*d*x + 8*I*c) + 2*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 = 1.98362, size = 219, normalized size = 1.77 \begin{align*} \frac{\frac{2 e^{- 2 i c} e^{2 i d x}}{a^{2} d} + \frac{4 e^{- 4 i c}}{a^{2} d}}{e^{4 i d x} + 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} + \begin{cases} \frac{\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} - a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{16 a^{4} d^{2}} & \text{for}\: 16 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac{\left (31 i e^{4 i c} - 8 i e^{2 i c} + i\right ) e^{- 4 i c}}{4 a^{2}} - \frac{31 i}{4 a^{2}}\right ) & \text{otherwise} \end{cases} + \frac{31 i x}{4 a^{2}} - \frac{4 \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)**5/(a+I*a*tan(d*x+c))**2,x)

[Out]

(2*exp(-2*I*c)*exp(2*I*d*x)/(a**2*d) + 4*exp(-4*I*c)/(a**2*d))/(exp(4*I*d*x) + 2*exp(-2*I*c)*exp(2*I*d*x) + ex
p(-4*I*c)) + Piecewise(((16*a**2*d*exp(4*I*c)*exp(-2*I*d*x) - a**2*d*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(16
*a**4*d**2), Ne(16*a**4*d**2*exp(6*I*c), 0)), (x*((31*I*exp(4*I*c) - 8*I*exp(2*I*c) + I)*exp(-4*I*c)/(4*a**2)
- 31*I/(4*a**2)), True)) + 31*I*x/(4*a**2) - 4*log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d)

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Giac [A]  time = 3.62507, size = 132, normalized size = 1.06 \begin{align*} \frac{\frac{2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{62 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac{8 \,{\left (a^{2} \tan \left (d x + c\right )^{2} + 4 i \, a^{2} \tan \left (d x + c\right )\right )}}{a^{4}} - \frac{93 \, \tan \left (d x + c\right )^{2} - 150 i \, \tan \left (d x + c\right ) - 61}{a^{2}{\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(2*log(tan(d*x + c) + I)/a^2 + 62*log(tan(d*x + c) - I)/a^2 - 8*(a^2*tan(d*x + c)^2 + 4*I*a^2*tan(d*x + c
))/a^4 - (93*tan(d*x + c)^2 - 150*I*tan(d*x + c) - 61)/(a^2*(tan(d*x + c) - I)^2))/d