3.20 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=58 \[ -\frac{a^2 \cot ^2(c+d x)}{2 d}-\frac{2 i a^2 \cot (c+d x)}{d}-\frac{2 a^2 \log (\sin (c+d x))}{d}-2 i a^2 x \]

[Out]

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

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Rubi [A]  time = 0.0926823, antiderivative size = 58, 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 = {3542, 3529, 3531, 3475} \[ -\frac{a^2 \cot ^2(c+d x)}{2 d}-\frac{2 i a^2 \cot (c+d x)}{d}-\frac{2 a^2 \log (\sin (c+d x))}{d}-2 i a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

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

Mathematica [C]  time = 0.200672, size = 64, normalized size = 1.1 \[ -\frac{a^2 \left (4 i \cot (c+d x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\tan ^2(c+d x)\right )+\cot ^2(c+d x)+4 (\log (\tan (c+d x))+\log (\cos (c+d x)))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]

[Out]

-(a^2*(Cot[c + d*x]^2 + (4*I)*Cot[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2] + 4*(Log[Cos[c + d
*x]] + Log[Tan[c + d*x]])))/(2*d)

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Maple [A]  time = 0.049, size = 65, normalized size = 1.1 \begin{align*} -2\,{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,i{a}^{2}x-{\frac{2\,i{a}^{2}\cot \left ( dx+c \right ) }{d}}-{\frac{2\,i{a}^{2}c}{d}}-{\frac{{a}^{2} \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))^2,x)

[Out]

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

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Maxima [A]  time = 2.20466, size = 92, normalized size = 1.59 \begin{align*} -\frac{4 i \,{\left (d x + c\right )} a^{2} - 2 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 4 \, a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac{4 i \, a^{2} \tan \left (d x + c\right ) + a^{2}}{\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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.18332, size = 248, normalized size = 4.28 \begin{align*} \frac{2 \,{\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, a^{2} -{\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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))^2,x, algorithm="fricas")

[Out]

2*(3*a^2*e^(2*I*d*x + 2*I*c) - 2*a^2 - (a^2*e^(4*I*d*x + 4*I*c) - 2*a^2*e^(2*I*d*x + 2*I*c) + a^2)*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 = 2.21715, size = 94, normalized size = 1.62 \begin{align*} - \frac{2 a^{2} \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{\frac{6 a^{2} e^{- 2 i c} e^{2 i d x}}{d} - \frac{4 a^{2} 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))**2,x)

[Out]

-2*a**2*log(exp(2*I*d*x) - exp(-2*I*c))/d + (6*a**2*exp(-2*I*c)*exp(2*I*d*x)/d - 4*a**2*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 [B]  time = 1.40915, size = 158, normalized size = 2.72 \begin{align*} -\frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 32 \, a^{2} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 16 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 8 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{24 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 8 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2}}{\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))^2,x, algorithm="giac")

[Out]

-1/8*(a^2*tan(1/2*d*x + 1/2*c)^2 - 32*a^2*log(tan(1/2*d*x + 1/2*c) + I) + 16*a^2*log(abs(tan(1/2*d*x + 1/2*c))
) - 8*I*a^2*tan(1/2*d*x + 1/2*c) - (24*a^2*tan(1/2*d*x + 1/2*c)^2 - 8*I*a^2*tan(1/2*d*x + 1/2*c) - a^2)/tan(1/
2*d*x + 1/2*c)^2)/d