3.58 \(\int \frac{\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=104 \[ \frac{i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac{9 \tan (c+d x)}{4 a^2 d}+\frac{2 i \log (\cos (c+d x))}{a^2 d}+\frac{9 x}{4 a^2}-\frac{\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

(9*x)/(4*a^2) + ((2*I)*Log[Cos[c + d*x]])/(a^2*d) - (9*Tan[c + d*x])/(4*a^2*d) + (I*Tan[c + d*x]^2)/(a^2*d*(1
+ I*Tan[c + d*x])) - Tan[c + d*x]^3/(4*d*(a + I*a*Tan[c + d*x])^2)

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Rubi [A]  time = 0.164888, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3558, 3595, 3525, 3475} \[ \frac{i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac{9 \tan (c+d x)}{4 a^2 d}+\frac{2 i \log (\cos (c+d x))}{a^2 d}+\frac{9 x}{4 a^2}-\frac{\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^2,x]

[Out]

(9*x)/(4*a^2) + ((2*I)*Log[Cos[c + d*x]])/(a^2*d) - (9*Tan[c + d*x])/(4*a^2*d) + (I*Tan[c + d*x]^2)/(a^2*d*(1
+ I*Tan[c + d*x])) - Tan[c + d*x]^3/(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 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 ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac{\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{\int \frac{\tan ^2(c+d x) (-3 a+5 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac{i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac{\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac{\int \tan (c+d x) \left (-16 i a^2-18 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac{9 x}{4 a^2}-\frac{9 \tan (c+d x)}{4 a^2 d}+\frac{i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac{\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac{(2 i) \int \tan (c+d x) \, dx}{a^2}\\ &=\frac{9 x}{4 a^2}+\frac{2 i \log (\cos (c+d x))}{a^2 d}-\frac{9 \tan (c+d x)}{4 a^2 d}+\frac{i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac{\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end{align*}

Mathematica [B]  time = 1.44543, size = 273, normalized size = 2.62 \[ -\frac{\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (64 d x \sin ^2(c)+36 i d x \sin (2 c)-i \sin (2 c) \sin (4 d x)-32 i d x \tan (c)+\sin (2 c) \cos (4 d x)-8 i \sec (c) \cos (2 c-d x) \sec (c+d x)+8 i \sec (c) \cos (2 c+d x) \sec (c+d x)+8 \sec (c) \sin (2 c-d x) \sec (c+d x)-8 \sec (c) \sin (2 c+d x) \sec (c+d x)-16 \sin (2 c) \log \left (\cos ^2(c+d x)\right )+32 (\cos (2 c)+i \sin (2 c)) \tan ^{-1}(\tan (d x))+\cos (2 c) \left (-32 i d x \tan (c)+16 i \log \left (\cos ^2(c+d x)\right )+36 d x+\sin (4 d x)+i \cos (4 d x)\right )-32 d x-12 \sin (2 d x)-12 i \cos (2 d x)\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^2,x]

[Out]

-(Sec[c + d*x]^2*(Cos[d*x] + I*Sin[d*x])^2*(-32*d*x - (12*I)*Cos[2*d*x] - (8*I)*Cos[2*c - d*x]*Sec[c]*Sec[c +
d*x] + (8*I)*Cos[2*c + d*x]*Sec[c]*Sec[c + d*x] + 64*d*x*Sin[c]^2 + 32*ArcTan[Tan[d*x]]*(Cos[2*c] + I*Sin[2*c]
) + (36*I)*d*x*Sin[2*c] + Cos[4*d*x]*Sin[2*c] - 16*Log[Cos[c + d*x]^2]*Sin[2*c] - 12*Sin[2*d*x] - I*Sin[2*c]*S
in[4*d*x] + 8*Sec[c]*Sec[c + d*x]*Sin[2*c - d*x] - 8*Sec[c]*Sec[c + d*x]*Sin[2*c + d*x] - (32*I)*d*x*Tan[c] +
Cos[2*c]*(36*d*x + I*Cos[4*d*x] + (16*I)*Log[Cos[c + d*x]^2] + Sin[4*d*x] - (32*I)*d*x*Tan[c])))/(16*a^2*d*(-I
+ Tan[c + d*x])^2)

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Maple [A]  time = 0.024, size = 93, normalized size = 0.9 \begin{align*} -{\frac{\tan \left ( dx+c \right ) }{{a}^{2}d}}-{\frac{{\frac{17\,i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{{a}^{2}d}}-{\frac{{\frac{i}{4}}}{{a}^{2}d \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{7}{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)^4/(a+I*a*tan(d*x+c))^2,x)

[Out]

-tan(d*x+c)/a^2/d-17/8*I/d/a^2*ln(tan(d*x+c)-I)-1/4*I/d/a^2/(tan(d*x+c)-I)^2-7/4/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)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.24476, size = 331, normalized size = 3.18 \begin{align*} \frac{68 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (68 \, d x - 44 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (32 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 32 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 ) - 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{16 \,{\left (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)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/16*(68*d*x*e^(6*I*d*x + 6*I*c) + (68*d*x - 44*I)*e^(4*I*d*x + 4*I*c) + (32*I*e^(6*I*d*x + 6*I*c) + 32*I*e^(4
*I*d*x + 4*I*c))*log(e^(2*I*d*x + 2*I*c) + 1) - 11*I*e^(2*I*d*x + 2*I*c) + I)/(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 = 4.56554, size = 134, normalized size = 1.29 \begin{align*} \frac{\left (\begin{cases} 17 x e^{4 i c} - \frac{3 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 (17 e^{4 i c} - 6 e^{2 i c} + 1\right ) & \text{otherwise} \end{cases}\right ) e^{- 4 i c}}{4 a^{2}} + \frac{2 i \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} - \frac{2 i e^{- 2 i c}}{a^{2} 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)**4/(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise((17*x*exp(4*I*c) - 3*I*exp(2*I*c)*exp(-2*I*d*x)/d + I*exp(-4*I*d*x)/(4*d), Ne(d, 0)), (x*(17*exp(4*I
*c) - 6*exp(2*I*c) + 1), True))*exp(-4*I*c)/(4*a**2) + 2*I*log(exp(2*I*d*x) + exp(-2*I*c))/(a**2*d) - 2*I*exp(
-2*I*c)/(a**2*d*(exp(2*I*d*x) + exp(-2*I*c)))

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Giac [A]  time = 2.4558, size = 107, normalized size = 1.03 \begin{align*} -\frac{-\frac{2 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac{34 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac{16 \, \tan \left (d x + c\right )}{a^{2}} + \frac{-51 i \, \tan \left (d x + c\right )^{2} - 74 \, \tan \left (d x + c\right ) + 27 i}{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)^4/(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/16*(-2*I*log(tan(d*x + c) + I)/a^2 + 34*I*log(tan(d*x + c) - I)/a^2 + 16*tan(d*x + c)/a^2 + (-51*I*tan(d*x
+ c)^2 - 74*tan(d*x + c) + 27*I)/(a^2*(tan(d*x + c) - I)^2))/d