3.23 \(\int \cot ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=112 \[ -\frac{a^2 \cot ^5(c+d x)}{5 d}-\frac{i a^2 \cot ^4(c+d x)}{2 d}+\frac{2 a^2 \cot ^3(c+d x)}{3 d}+\frac{i a^2 \cot ^2(c+d x)}{d}-\frac{2 a^2 \cot (c+d x)}{d}+\frac{2 i a^2 \log (\sin (c+d x))}{d}-2 a^2 x \]

[Out]

-2*a^2*x - (2*a^2*Cot[c + d*x])/d + (I*a^2*Cot[c + d*x]^2)/d + (2*a^2*Cot[c + d*x]^3)/(3*d) - ((I/2)*a^2*Cot[c
+ d*x]^4)/d - (a^2*Cot[c + d*x]^5)/(5*d) + ((2*I)*a^2*Log[Sin[c + d*x]])/d

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Rubi [A]  time = 0.167012, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3542, 3529, 3531, 3475} \[ -\frac{a^2 \cot ^5(c+d x)}{5 d}-\frac{i a^2 \cot ^4(c+d x)}{2 d}+\frac{2 a^2 \cot ^3(c+d x)}{3 d}+\frac{i a^2 \cot ^2(c+d x)}{d}-\frac{2 a^2 \cot (c+d x)}{d}+\frac{2 i a^2 \log (\sin (c+d x))}{d}-2 a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]

[Out]

-2*a^2*x - (2*a^2*Cot[c + d*x])/d + (I*a^2*Cot[c + d*x]^2)/d + (2*a^2*Cot[c + d*x]^3)/(3*d) - ((I/2)*a^2*Cot[c
+ d*x]^4)/d - (a^2*Cot[c + d*x]^5)/(5*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 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 ^6(c+d x) (a+i a \tan (c+d x))^2 \, dx &=-\frac{a^2 \cot ^5(c+d x)}{5 d}+\int \cot ^5(c+d x) \left (2 i a^2-2 a^2 \tan (c+d x)\right ) \, dx\\ &=-\frac{i a^2 \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 d}+\int \cot ^4(c+d x) \left (-2 a^2-2 i a^2 \tan (c+d x)\right ) \, dx\\ &=\frac{2 a^2 \cot ^3(c+d x)}{3 d}-\frac{i a^2 \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 d}+\int \cot ^3(c+d x) \left (-2 i a^2+2 a^2 \tan (c+d x)\right ) \, dx\\ &=\frac{i a^2 \cot ^2(c+d x)}{d}+\frac{2 a^2 \cot ^3(c+d x)}{3 d}-\frac{i a^2 \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 d}+\int \cot ^2(c+d x) \left (2 a^2+2 i a^2 \tan (c+d x)\right ) \, dx\\ &=-\frac{2 a^2 \cot (c+d x)}{d}+\frac{i a^2 \cot ^2(c+d x)}{d}+\frac{2 a^2 \cot ^3(c+d x)}{3 d}-\frac{i a^2 \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 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{2 a^2 \cot (c+d x)}{d}+\frac{i a^2 \cot ^2(c+d x)}{d}+\frac{2 a^2 \cot ^3(c+d x)}{3 d}-\frac{i a^2 \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 d}+\left (2 i a^2\right ) \int \cot (c+d x) \, dx\\ &=-2 a^2 x-\frac{2 a^2 \cot (c+d x)}{d}+\frac{i a^2 \cot ^2(c+d x)}{d}+\frac{2 a^2 \cot ^3(c+d x)}{3 d}-\frac{i a^2 \cot ^4(c+d x)}{2 d}-\frac{a^2 \cot ^5(c+d x)}{5 d}+\frac{2 i a^2 \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.799427, size = 124, normalized size = 1.11 \[ -\frac{a^2 \cot ^5(c+d x) \, _2F_1\left (-\frac{5}{2},1;-\frac{3}{2};-\tan ^2(c+d x)\right )}{5 d}+\frac{a^2 \cot ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\tan ^2(c+d x)\right )}{3 d}+\frac{i a^2 \left (-\cot ^4(c+d x)+2 \cot ^2(c+d x)+4 \log (\tan (c+d x))+4 \log (\cos (c+d x))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x])^2,x]

[Out]

-(a^2*Cot[c + d*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[c + d*x]^2])/(5*d) + (a^2*Cot[c + d*x]^3*Hypergeome
tric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/(3*d) + ((I/2)*a^2*(2*Cot[c + d*x]^2 - Cot[c + d*x]^4 + 4*Log[Cos[c +
d*x]] + 4*Log[Tan[c + d*x]]))/d

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Maple [A]  time = 0.049, size = 113, normalized size = 1. \begin{align*}{\frac{2\,{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-2\,{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}-2\,{a}^{2}x-2\,{\frac{{a}^{2}c}{d}}-{\frac{{\frac{i}{2}}{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{4}}{d}}+{\frac{i{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{2\,i{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2} \left ( \cot \left ( dx+c \right ) \right ) ^{5}}{5\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x)

[Out]

2/3*a^2*cot(d*x+c)^3/d-2*a^2*cot(d*x+c)/d-2*a^2*x-2/d*a^2*c-1/2*I*a^2*cot(d*x+c)^4/d+I*a^2*cot(d*x+c)^2/d+2*I*
a^2*ln(sin(d*x+c))/d-1/5*a^2*cot(d*x+c)^5/d

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Maxima [A]  time = 2.01789, size = 147, normalized size = 1.31 \begin{align*} -\frac{60 \,{\left (d x + c\right )} a^{2} + 30 i \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 i \, a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac{60 \, a^{2} \tan \left (d x + c\right )^{4} - 30 i \, a^{2} \tan \left (d x + c\right )^{3} - 20 \, a^{2} \tan \left (d x + c\right )^{2} + 15 i \, a^{2} \tan \left (d x + c\right ) + 6 \, a^{2}}{\tan \left (d x + c\right )^{5}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/30*(60*(d*x + c)*a^2 + 30*I*a^2*log(tan(d*x + c)^2 + 1) - 60*I*a^2*log(tan(d*x + c)) + (60*a^2*tan(d*x + c)
^4 - 30*I*a^2*tan(d*x + c)^3 - 20*a^2*tan(d*x + c)^2 + 15*I*a^2*tan(d*x + c) + 6*a^2)/tan(d*x + c)^5)/d

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Fricas [B]  time = 2.55803, size = 655, normalized size = 5.85 \begin{align*} \frac{-270 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 600 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 740 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 400 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 86 i \, a^{2} +{\left (30 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} - 150 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} + 300 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 300 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 150 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 30 i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{15 \,{\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, 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)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/15*(-270*I*a^2*e^(8*I*d*x + 8*I*c) + 600*I*a^2*e^(6*I*d*x + 6*I*c) - 740*I*a^2*e^(4*I*d*x + 4*I*c) + 400*I*a
^2*e^(2*I*d*x + 2*I*c) - 86*I*a^2 + (30*I*a^2*e^(10*I*d*x + 10*I*c) - 150*I*a^2*e^(8*I*d*x + 8*I*c) + 300*I*a^
2*e^(6*I*d*x + 6*I*c) - 300*I*a^2*e^(4*I*d*x + 4*I*c) + 150*I*a^2*e^(2*I*d*x + 2*I*c) - 30*I*a^2)*log(e^(2*I*d
*x + 2*I*c) - 1))/(d*e^(10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*
d*x + 4*I*c) + 5*d*e^(2*I*d*x + 2*I*c) - d)

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Sympy [B]  time = 10.1137, size = 226, normalized size = 2.02 \begin{align*} \frac{2 i a^{2} \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{- \frac{18 i a^{2} e^{- 2 i c} e^{8 i d x}}{d} + \frac{40 i a^{2} e^{- 4 i c} e^{6 i d x}}{d} - \frac{148 i a^{2} e^{- 6 i c} e^{4 i d x}}{3 d} + \frac{80 i a^{2} e^{- 8 i c} e^{2 i d x}}{3 d} - \frac{86 i a^{2} e^{- 10 i c}}{15 d}}{e^{10 i d x} - 5 e^{- 2 i c} e^{8 i d x} + 10 e^{- 4 i c} e^{6 i d x} - 10 e^{- 6 i c} e^{4 i d x} + 5 e^{- 8 i c} e^{2 i d x} - e^{- 10 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(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 + (-18*I*a**2*exp(-2*I*c)*exp(8*I*d*x)/d + 40*I*a**2*exp(-4*I*c)*ex
p(6*I*d*x)/d - 148*I*a**2*exp(-6*I*c)*exp(4*I*d*x)/(3*d) + 80*I*a**2*exp(-8*I*c)*exp(2*I*d*x)/(3*d) - 86*I*a**
2*exp(-10*I*c)/(15*d))/(exp(10*I*d*x) - 5*exp(-2*I*c)*exp(8*I*d*x) + 10*exp(-4*I*c)*exp(6*I*d*x) - 10*exp(-6*I
*c)*exp(4*I*d*x) + 5*exp(-8*I*c)*exp(2*I*d*x) - exp(-10*I*c))

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Giac [B]  time = 1.38177, size = 288, normalized size = 2.57 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 15 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 55 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 180 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1920 i \, a^{2} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 960 i \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + 630 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \frac{-2192 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 630 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 180 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 55 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 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 )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/480*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 15*I*a^2*tan(1/2*d*x + 1/2*c)^4 - 55*a^2*tan(1/2*d*x + 1/2*c)^3 + 180*I*
a^2*tan(1/2*d*x + 1/2*c)^2 - 1920*I*a^2*log(tan(1/2*d*x + 1/2*c) + I) + 960*I*a^2*log(abs(tan(1/2*d*x + 1/2*c)
)) + 630*a^2*tan(1/2*d*x + 1/2*c) + (-2192*I*a^2*tan(1/2*d*x + 1/2*c)^5 - 630*a^2*tan(1/2*d*x + 1/2*c)^4 + 180
*I*a^2*tan(1/2*d*x + 1/2*c)^3 + 55*a^2*tan(1/2*d*x + 1/2*c)^2 - 15*I*a^2*tan(1/2*d*x + 1/2*c) - 3*a^2)/tan(1/2
*d*x + 1/2*c)^5)/d