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

Optimal. Leaf size=108 \[ -\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{i \cot ^2(c+d x)}{a d}+\frac{5 \cot (c+d x)}{2 a d}+\frac{2 i \log (\sin (c+d x))}{a d}+\frac{\cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{5 x}{2 a} \]

[Out]

(5*x)/(2*a) + (5*Cot[c + d*x])/(2*a*d) + (I*Cot[c + d*x]^2)/(a*d) - (5*Cot[c + d*x]^3)/(6*a*d) + ((2*I)*Log[Si
n[c + d*x]])/(a*d) + Cot[c + d*x]^3/(2*d*(a + I*a*Tan[c + d*x]))

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Rubi [A]  time = 0.145177, antiderivative size = 108, 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 = {3552, 3529, 3531, 3475} \[ -\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{i \cot ^2(c+d x)}{a d}+\frac{5 \cot (c+d x)}{2 a d}+\frac{2 i \log (\sin (c+d x))}{a d}+\frac{\cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{5 x}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x)/(2*a) + (5*Cot[c + d*x])/(2*a*d) + (I*Cot[c + d*x]^2)/(a*d) - (5*Cot[c + d*x]^3)/(6*a*d) + ((2*I)*Log[Si
n[c + d*x]])/(a*d) + Cot[c + d*x]^3/(2*d*(a + I*a*Tan[c + d*x]))

Rule 3552

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

Rule 3529

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

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*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] && NeQ[a*c + b*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{\cot ^4(c+d x)}{a+i a \tan (c+d x)} \, dx &=\frac{\cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{\int \cot ^4(c+d x) (-5 a+4 i a \tan (c+d x)) \, dx}{2 a^2}\\ &=-\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{\cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{\int \cot ^3(c+d x) (4 i a+5 a \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{i \cot ^2(c+d x)}{a d}-\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{\cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{\int \cot ^2(c+d x) (5 a-4 i a \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{5 \cot (c+d x)}{2 a d}+\frac{i \cot ^2(c+d x)}{a d}-\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{\cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{\int \cot (c+d x) (-4 i a-5 a \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{5 x}{2 a}+\frac{5 \cot (c+d x)}{2 a d}+\frac{i \cot ^2(c+d x)}{a d}-\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{\cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{(2 i) \int \cot (c+d x) \, dx}{a}\\ &=\frac{5 x}{2 a}+\frac{5 \cot (c+d x)}{2 a d}+\frac{i \cot ^2(c+d x)}{a d}-\frac{5 \cot ^3(c+d x)}{6 a d}+\frac{2 i \log (\sin (c+d x))}{a d}+\frac{\cot ^3(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end{align*}

Mathematica [B]  time = 3.09389, size = 365, normalized size = 3.38 \[ \frac{\csc (c) (\cos (d x)+i \sin (d x)) \left (14 \csc (c+d x)-24 d x \sec (c+d x)+14 i \sec (c+d x)+24 d x \cos ^2(c) \sec (c+d x)-28 \cos (c-d x) \csc (2 (c+d x))-28 i \sin (c-d x) \csc (2 (c+d x))+30 d x \sin ^2(c) \sec (c+d x)-3 \sin ^2(c) \sin (2 d x) \sec (c+d x)-3 i d x \sin (2 c) \sec (c+d x)+12 i \sin ^2(c) \sec (c+d x) \log \left (\sin ^2(c+d x)\right )+6 \sin (2 c) \sec (c+d x) \log \left (\sin ^2(c+d x)\right )-3 i \sin ^2(c) \cos (2 d x) \sec (c+d x)+3 \sin (c) \cos (c) \cos (2 d x) \sec (c+d x)-3 i \sin (c) \cos (c) \sin (2 d x) \sec (c+d x)+2 \csc ^3(c+d x) (\cos (c-d x) \sec (c+d x)+i \sin (c-d x) \sec (c+d x)-1)+2 (\cos (c)+i \sin (c)) (2 \sin (c)+i \cos (c)) \csc ^2(c+d x) \sec (c+d x)+24 \sin (c) (\sin (c)-i \cos (c)) \tan ^{-1}(\tan (d x)) \sec (c+d x)\right )}{12 a d (\tan (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c]*(Cos[d*x] + I*Sin[d*x])*(14*Csc[c + d*x] - 28*Cos[c - d*x]*Csc[2*(c + d*x)] + (14*I)*Sec[c + d*x] - 24
*d*x*Sec[c + d*x] + 24*d*x*Cos[c]^2*Sec[c + d*x] + 3*Cos[c]*Cos[2*d*x]*Sec[c + d*x]*Sin[c] + 30*d*x*Sec[c + d*
x]*Sin[c]^2 - (3*I)*Cos[2*d*x]*Sec[c + d*x]*Sin[c]^2 + (12*I)*Log[Sin[c + d*x]^2]*Sec[c + d*x]*Sin[c]^2 + 24*A
rcTan[Tan[d*x]]*Sec[c + d*x]*Sin[c]*((-I)*Cos[c] + Sin[c]) + 2*Csc[c + d*x]^2*Sec[c + d*x]*(Cos[c] + I*Sin[c])
*(I*Cos[c] + 2*Sin[c]) - (3*I)*d*x*Sec[c + d*x]*Sin[2*c] + 6*Log[Sin[c + d*x]^2]*Sec[c + d*x]*Sin[2*c] - (3*I)
*Cos[c]*Sec[c + d*x]*Sin[c]*Sin[2*d*x] - 3*Sec[c + d*x]*Sin[c]^2*Sin[2*d*x] - (28*I)*Csc[2*(c + d*x)]*Sin[c -
d*x] + 2*Csc[c + d*x]^3*(-1 + Cos[c - d*x]*Sec[c + d*x] + I*Sec[c + d*x]*Sin[c - d*x])))/(12*a*d*(-I + Tan[c +
d*x]))

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Maple [A]  time = 0.07, size = 124, normalized size = 1.2 \begin{align*}{\frac{-{\frac{9\,i}{4}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{ad}}+{\frac{1}{2\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{i}{4}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{ad}}-{\frac{1}{3\,ad \left ( \tan \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{\frac{i}{2}}}{ad \left ( \tan \left ( dx+c \right ) \right ) ^{2}}}+{\frac{2\,i\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}}+2\,{\frac{1}{ad\tan \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4/(a+I*a*tan(d*x+c)),x)

[Out]

-9/4*I/d/a*ln(tan(d*x+c)-I)+1/2/a/d/(tan(d*x+c)-I)+1/4*I/d/a*ln(tan(d*x+c)+I)-1/3/d/a/tan(d*x+c)^3+1/2*I/d/a/t
an(d*x+c)^2+2*I/d/a*ln(tan(d*x+c))+2/d/a/tan(d*x+c)

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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(cot(d*x+c)^4/(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.23132, size = 545, normalized size = 5.05 \begin{align*} \frac{54 \, d x e^{\left (8 i \, d x + 8 i \, c\right )} -{\left (162 \, d x - 51 i\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (162 \, d x - 81 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} -{\left (54 \, d x - 65 i\right )} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (24 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 72 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 72 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 24 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 3 i}{12 \,{\left (a d e^{\left (8 i \, d x + 8 i \, c\right )} - 3 \, a d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} - a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(54*d*x*e^(8*I*d*x + 8*I*c) - (162*d*x - 51*I)*e^(6*I*d*x + 6*I*c) + (162*d*x - 81*I)*e^(4*I*d*x + 4*I*c)
- (54*d*x - 65*I)*e^(2*I*d*x + 2*I*c) + (24*I*e^(8*I*d*x + 8*I*c) - 72*I*e^(6*I*d*x + 6*I*c) + 72*I*e^(4*I*d*
x + 4*I*c) - 24*I*e^(2*I*d*x + 2*I*c))*log(e^(2*I*d*x + 2*I*c) - 1) - 3*I)/(a*d*e^(8*I*d*x + 8*I*c) - 3*a*d*e^
(6*I*d*x + 6*I*c) + 3*a*d*e^(4*I*d*x + 4*I*c) - a*d*e^(2*I*d*x + 2*I*c))

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Sympy [A]  time = 2.97439, size = 180, normalized size = 1.67 \begin{align*} \frac{\frac{4 i e^{- 2 i c} e^{4 i d x}}{a d} - \frac{6 i e^{- 4 i c} e^{2 i d x}}{a d} + \frac{14 i e^{- 6 i c}}{3 a 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} 9 x e^{2 i c} + \frac{i e^{- 2 i d x}}{2 d} & \text{for}\: d \neq 0 \\x \left (9 e^{2 i c} + 1\right ) & \text{otherwise} \end{cases}\right ) e^{- 2 i c}}{2 a} + \frac{2 i \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4/(a+I*a*tan(d*x+c)),x)

[Out]

(4*I*exp(-2*I*c)*exp(4*I*d*x)/(a*d) - 6*I*exp(-4*I*c)*exp(2*I*d*x)/(a*d) + 14*I*exp(-6*I*c)/(3*a*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((9*x*exp(2*I*c) + I*
exp(-2*I*d*x)/(2*d), Ne(d, 0)), (x*(9*exp(2*I*c) + 1), True))*exp(-2*I*c)/(2*a) + 2*I*log(exp(2*I*d*x) - exp(-
2*I*c))/(a*d)

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Giac [A]  time = 1.42347, size = 158, normalized size = 1.46 \begin{align*} -\frac{\frac{27 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} - \frac{3 i \, \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a} - \frac{24 i \, \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a} + \frac{3 \,{\left (-9 i \, \tan \left (d x + c\right ) - 11\right )}}{a{\left (\tan \left (d x + c\right ) - i\right )}} + \frac{2 i \,{\left (22 \, \tan \left (d x + c\right )^{3} + 12 i \, \tan \left (d x + c\right )^{2} - 3 \, \tan \left (d x + c\right ) - 2 i\right )}}{a \tan \left (d x + c\right )^{3}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4/(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/12*(27*I*log(tan(d*x + c) - I)/a - 3*I*log(-I*tan(d*x + c) + 1)/a - 24*I*log(abs(tan(d*x + c)))/a + 3*(-9*I
*tan(d*x + c) - 11)/(a*(tan(d*x + c) - I)) + 2*I*(22*tan(d*x + c)^3 + 12*I*tan(d*x + c)^2 - 3*tan(d*x + c) - 2
*I)/(a*tan(d*x + c)^3))/d