3.30 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=71 \[ -\frac{2 i a^3 \cot (c+d x)}{d}-\frac{4 a^3 \log (\sin (c+d x))}{d}-4 i a^3 x-\frac{a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \]

[Out]

(-4*I)*a^3*x - ((2*I)*a^3*Cot[c + d*x])/d - (4*a^3*Log[Sin[c + d*x]])/d - (a*Cot[c + d*x]^2*(a + I*a*Tan[c + d
*x])^2)/(2*d)

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Rubi [A]  time = 0.10625, antiderivative size = 71, 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 = {3545, 3542, 3531, 3475} \[ -\frac{2 i a^3 \cot (c+d x)}{d}-\frac{4 a^3 \log (\sin (c+d x))}{d}-4 i a^3 x-\frac{a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*I)*a^3*x - ((2*I)*a^3*Cot[c + d*x])/d - (4*a^3*Log[Sin[c + d*x]])/d - (a*Cot[c + d*x]^2*(a + I*a*Tan[c + d
*x])^2)/(2*d)

Rule 3545

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), 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] && EqQ[m + n, 0] && GtQ[m, 1/2]

Rule 3542

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

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 \cot ^3(c+d x) (a+i a \tan (c+d x))^3 \, dx &=-\frac{a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}+(2 i a) \int \cot ^2(c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=-\frac{2 i a^3 \cot (c+d x)}{d}-\frac{a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}+(2 i a) \int \cot (c+d x) \left (2 i a^2-2 a^2 \tan (c+d x)\right ) \, dx\\ &=-4 i a^3 x-\frac{2 i a^3 \cot (c+d x)}{d}-\frac{a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}-\left (4 a^3\right ) \int \cot (c+d x) \, dx\\ &=-4 i a^3 x-\frac{2 i a^3 \cot (c+d x)}{d}-\frac{4 a^3 \log (\sin (c+d x))}{d}-\frac{a \cot ^2(c+d x) (a+i a \tan (c+d x))^2}{2 d}\\ \end{align*}

Mathematica [A]  time = 1.0422, size = 126, normalized size = 1.77 \[ \frac{a^3 \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \csc ^2(c+d x) (\cos (3 d x)+i \sin (3 d x)) \left (-3 i \cos (c+2 d x)+\sin (c) \left (-2 \log \left (\sin ^2(c+d x)\right )+2 \cos (2 (c+d x)) \left (\log \left (\sin ^2(c+d x)\right )+2 i d x\right )-4 i d x-1\right )+3 i \cos (c)\right )}{4 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.055, size = 65, normalized size = 0.9 \begin{align*} -4\,i{a}^{3}x-{\frac{4\,i{a}^{3}c}{d}}-4\,{\frac{{a}^{3}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{3\,i\cot \left ( dx+c \right ){a}^{3}}{d}}-{\frac{{a}^{3} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x)

[Out]

-4*I*a^3*x-4*I/d*a^3*c-4*a^3*ln(sin(d*x+c))/d-3*I/d*cot(d*x+c)*a^3-1/2*a^3*cot(d*x+c)^2/d

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Maxima [A]  time = 2.52493, size = 92, normalized size = 1.3 \begin{align*} -\frac{8 i \,{\left (d x + c\right )} a^{3} - 4 \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 8 \, a^{3} \log \left (\tan \left (d x + c\right )\right ) + \frac{6 i \, a^{3} \tan \left (d x + c\right ) + a^{3}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2*(8*I*(d*x + c)*a^3 - 4*a^3*log(tan(d*x + c)^2 + 1) + 8*a^3*log(tan(d*x + c)) + (6*I*a^3*tan(d*x + c) + a^
3)/tan(d*x + c)^2)/d

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Fricas [A]  time = 2.19002, size = 251, normalized size = 3.54 \begin{align*} \frac{2 \,{\left (4 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, a^{3} - 2 \,{\left (a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

2*(4*a^3*e^(2*I*d*x + 2*I*c) - 3*a^3 - 2*(a^3*e^(4*I*d*x + 4*I*c) - 2*a^3*e^(2*I*d*x + 2*I*c) + a^3)*log(e^(2*
I*d*x + 2*I*c) - 1))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [A]  time = 2.87831, size = 94, normalized size = 1.32 \begin{align*} - \frac{4 a^{3} \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{\frac{8 a^{3} e^{- 2 i c} e^{2 i d x}}{d} - \frac{6 a^{3} e^{- 4 i c}}{d}}{e^{4 i d x} - 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**3,x)

[Out]

-4*a**3*log(exp(2*I*d*x) - exp(-2*I*c))/d + (8*a**3*exp(-2*I*c)*exp(2*I*d*x)/d - 6*a**3*exp(-4*I*c)/d)/(exp(4*
I*d*x) - 2*exp(-2*I*c)*exp(2*I*d*x) + exp(-4*I*c))

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Giac [A]  time = 1.4501, size = 158, normalized size = 2.23 \begin{align*} -\frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 64 \, a^{3} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 32 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 12 i \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{48 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 i \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/8*(a^3*tan(1/2*d*x + 1/2*c)^2 - 64*a^3*log(tan(1/2*d*x + 1/2*c) + I) + 32*a^3*log(abs(tan(1/2*d*x + 1/2*c))
) - 12*I*a^3*tan(1/2*d*x + 1/2*c) - (48*a^3*tan(1/2*d*x + 1/2*c)^2 - 12*I*a^3*tan(1/2*d*x + 1/2*c) - a^3)/tan(
1/2*d*x + 1/2*c)^2)/d