3.22 \(\int \cot ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=93 \[ -\frac{a^2 \cot ^4(c+d x)}{4 d}-\frac{2 i a^2 \cot ^3(c+d x)}{3 d}+\frac{a^2 \cot ^2(c+d x)}{d}+\frac{2 i a^2 \cot (c+d x)}{d}+\frac{2 a^2 \log (\sin (c+d x))}{d}+2 i a^2 x \]

[Out]

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

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Rubi [A]  time = 0.139981, antiderivative size = 93, 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 = {3542, 3529, 3531, 3475} \[ -\frac{a^2 \cot ^4(c+d x)}{4 d}-\frac{2 i a^2 \cot ^3(c+d x)}{3 d}+\frac{a^2 \cot ^2(c+d x)}{d}+\frac{2 i a^2 \cot (c+d x)}{d}+\frac{2 a^2 \log (\sin (c+d x))}{d}+2 i a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^5*(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)/d - (((2*I)/3)*a^2*Cot[c + d*x]^3)/d - (a^2*Co
t[c + d*x]^4)/(4*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 ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac{a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^4(c+d x) \left (2 i a^2-2 a^2 \tan (c+d x)\right ) \, dx\\ &=-\frac{2 i a^2 \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 d}+\int \cot ^3(c+d x) \left (-2 a^2-2 i a^2 \tan (c+d x)\right ) \, dx\\ &=\frac{a^2 \cot ^2(c+d x)}{d}-\frac{2 i a^2 \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 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)}{d}-\frac{2 i a^2 \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 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)}{d}-\frac{2 i a^2 \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 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)}{d}-\frac{2 i a^2 \cot ^3(c+d x)}{3 d}-\frac{a^2 \cot ^4(c+d x)}{4 d}+\frac{2 a^2 \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.372963, size = 79, normalized size = 0.85 \[ -\frac{a^2 \left (3 \left (\cot ^4(c+d x)-4 \cot ^2(c+d x)-8 (\log (\tan (c+d x))+\log (\cos (c+d x)))\right )+8 i \cot ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\tan ^2(c+d x)\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^5*(a + I*a*Tan[c + d*x])^2,x]

[Out]

-(a^2*((8*I)*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2] + 3*(-4*Cot[c + d*x]^2 + Cot[c +
d*x]^4 - 8*(Log[Cos[c + d*x]] + Log[Tan[c + d*x]]))))/(12*d)

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Maple [A]  time = 0.049, size = 97, normalized size = 1. \begin{align*}{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}+2\,{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{\frac{2\,i}{3}}{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{2\,i{a}^{2}\cot \left ( dx+c \right ) }{d}}+2\,i{a}^{2}x+{\frac{2\,i{a}^{2}c}{d}}-{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 2.01425, size = 131, normalized size = 1.41 \begin{align*} -\frac{-24 i \,{\left (d x + c\right )} a^{2} + 12 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 24 \, a^{2} \log \left (\tan \left (d x + c\right )\right ) - \frac{24 i \, a^{2} \tan \left (d x + c\right )^{3} + 12 \, a^{2} \tan \left (d x + c\right )^{2} - 8 i \, a^{2} \tan \left (d x + c\right ) - 3 \, a^{2}}{\tan \left (d x + c\right )^{4}}}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/12*(-24*I*(d*x + c)*a^2 + 12*a^2*log(tan(d*x + c)^2 + 1) - 24*a^2*log(tan(d*x + c)) - (24*I*a^2*tan(d*x + c
)^3 + 12*a^2*tan(d*x + c)^2 - 8*I*a^2*tan(d*x + c) - 3*a^2)/tan(d*x + c)^4)/d

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Fricas [B]  time = 2.58865, size = 481, normalized size = 5.17 \begin{align*} -\frac{2 \,{\left (21 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 36 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 29 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 8 \, a^{2} - 3 \,{\left (a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, 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 )}}{3 \,{\left (d e^{\left (8 i \, d x + 8 i \, c\right )} - 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} - 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/3*(21*a^2*e^(6*I*d*x + 6*I*c) - 36*a^2*e^(4*I*d*x + 4*I*c) + 29*a^2*e^(2*I*d*x + 2*I*c) - 8*a^2 - 3*(a^2*e^
(8*I*d*x + 8*I*c) - 4*a^2*e^(6*I*d*x + 6*I*c) + 6*a^2*e^(4*I*d*x + 4*I*c) - 4*a^2*e^(2*I*d*x + 2*I*c) + a^2)*l
og(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(8*I*d*x + 8*I*c) - 4*d*e^(6*I*d*x + 6*I*c) + 6*d*e^(4*I*d*x + 4*I*c) - 4*d*
e^(2*I*d*x + 2*I*c) + d)

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Sympy [B]  time = 6.99456, size = 175, normalized size = 1.88 \begin{align*} \frac{2 a^{2} \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{- \frac{14 a^{2} e^{- 2 i c} e^{6 i d x}}{d} + \frac{24 a^{2} e^{- 4 i c} e^{4 i d x}}{d} - \frac{58 a^{2} e^{- 6 i c} e^{2 i d x}}{3 d} + \frac{16 a^{2} e^{- 8 i c}}{3 d}}{e^{8 i d x} - 4 e^{- 2 i c} e^{6 i d x} + 6 e^{- 4 i c} e^{4 i d x} - 4 e^{- 6 i c} e^{2 i d x} + e^{- 8 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**5*(a+I*a*tan(d*x+c))**2,x)

[Out]

2*a**2*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-14*a**2*exp(-2*I*c)*exp(6*I*d*x)/d + 24*a**2*exp(-4*I*c)*exp(4*I*
d*x)/d - 58*a**2*exp(-6*I*c)*exp(2*I*d*x)/(3*d) + 16*a**2*exp(-8*I*c)/(3*d))/(exp(8*I*d*x) - 4*exp(-2*I*c)*exp
(6*I*d*x) + 6*exp(-4*I*c)*exp(4*I*d*x) - 4*exp(-6*I*c)*exp(2*I*d*x) + exp(-8*I*c))

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Giac [B]  time = 1.42618, size = 244, normalized size = 2.62 \begin{align*} -\frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 16 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 60 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 768 \, a^{2} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 384 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 240 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{800 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 240 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 60 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 16 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4}}}{192 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/192*(3*a^2*tan(1/2*d*x + 1/2*c)^4 - 16*I*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*a^2*tan(1/2*d*x + 1/2*c)^2 + 768*a
^2*log(tan(1/2*d*x + 1/2*c) + I) - 384*a^2*log(abs(tan(1/2*d*x + 1/2*c))) + 240*I*a^2*tan(1/2*d*x + 1/2*c) + (
800*a^2*tan(1/2*d*x + 1/2*c)^4 - 240*I*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*a^2*tan(1/2*d*x + 1/2*c)^2 + 16*I*a^2*t
an(1/2*d*x + 1/2*c) + 3*a^2)/tan(1/2*d*x + 1/2*c)^4)/d