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

Optimal. Leaf size=38 \[ -\frac{a^2 \cot (c+d x)}{d}+\frac{2 i a^2 \log (\sin (c+d x))}{d}-2 a^2 x \]

[Out]

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

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Rubi [A]  time = 0.0618372, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3542, 3531, 3475} \[ -\frac{a^2 \cot (c+d x)}{d}+\frac{2 i a^2 \log (\sin (c+d x))}{d}-2 a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 0.766765, size = 100, normalized size = 2.63 \[ \frac{a^2 \csc (c) \csc (c+d x) \left (4 d x \cos (2 c+d x)+4 \sin (c) \sin (c+d x) \tan ^{-1}(\tan (3 c+d x))-i \cos (2 c+d x) \log \left (\sin ^2(c+d x)\right )+\cos (d x) \left (-4 d x+i \log \left (\sin ^2(c+d x)\right )\right )+2 \sin (d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 47, normalized size = 1.2 \begin{align*}{\frac{2\,i{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-2\,{a}^{2}x-{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}-2\,{\frac{{a}^{2}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.14196, size = 76, normalized size = 2. \begin{align*} -\frac{2 \,{\left (d x + c\right )} a^{2} + i \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 i \, a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac{a^{2}}{\tan \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.23629, size = 150, normalized size = 3.95 \begin{align*} \frac{-2 i \, a^{2} +{\left (2 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{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)^2*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.7, size = 58, normalized size = 1.53 \begin{align*} \frac{2 i a^{2} \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} - \frac{2 i a^{2} e^{- 2 i c}}{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(cot(d*x+c)**2*(a+I*a*tan(d*x+c))**2,x)

[Out]

2*I*a**2*log(exp(2*I*d*x) - exp(-2*I*c))/d - 2*I*a**2*exp(-2*I*c)/(d*(exp(2*I*d*x) - exp(-2*I*c)))

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Giac [B]  time = 1.35024, size = 116, normalized size = 3.05 \begin{align*} -\frac{8 i \, a^{2} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 4 i \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{-4 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 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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