3.21 \(\int \cot ^4(c+d x) (a+i a \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=74 \[ -\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{i a^2 \cot ^2(c+d x)}{d}+\frac{2 a^2 \cot (c+d x)}{d}-\frac{2 i a^2 \log (\sin (c+d x))}{d}+2 a^2 x \]

[Out]

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

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Rubi [A]  time = 0.113194, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3542, 3529, 3531, 3475} \[ -\frac{a^2 \cot ^3(c+d x)}{3 d}-\frac{i a^2 \cot ^2(c+d x)}{d}+\frac{2 a^2 \cot (c+d x)}{d}-\frac{2 i a^2 \log (\sin (c+d x))}{d}+2 a^2 x \]

Antiderivative was successfully verified.

[In]

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

Mathematica [C]  time = 0.4579, size = 105, normalized size = 1.42 \[ -\frac{a^2 \cot ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\tan ^2(c+d x)\right )}{3 d}+\frac{a^2 \cot (c+d x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\tan ^2(c+d x)\right )}{d}-\frac{i a^2 \left (\cot ^2(c+d x)+2 \log (\tan (c+d x))+2 \log (\cos (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^4*(a + I*a*Tan[c + d*x])^2,x]

[Out]

-(a^2*Cot[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[c + d*x]^2])/(3*d) + (a^2*Cot[c + d*x]*Hypergeometr
ic2F1[-1/2, 1, 1/2, -Tan[c + d*x]^2])/d - (I*a^2*(Cot[c + d*x]^2 + 2*Log[Cos[c + d*x]] + 2*Log[Tan[c + d*x]]))
/d

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Maple [A]  time = 0.046, size = 80, normalized size = 1.1 \begin{align*} 2\,{a}^{2}x+2\,{\frac{{a}^{2}\cot \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}c}{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 ) ^{3}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 2.36703, size = 112, normalized size = 1.51 \begin{align*} \frac{6 \,{\left (d x + c\right )} a^{2} + 3 i \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 6 i \, a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac{6 \, a^{2} \tan \left (d x + c\right )^{2} - 3 i \, a^{2} \tan \left (d x + c\right ) - a^{2}}{\tan \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.45229, size = 394, normalized size = 5.32 \begin{align*} \frac{30 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 36 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 14 i \, a^{2} +{\left (-6 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + 18 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 18 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 i \, a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{3 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(30*I*a^2*e^(4*I*d*x + 4*I*c) - 36*I*a^2*e^(2*I*d*x + 2*I*c) + 14*I*a^2 + (-6*I*a^2*e^(6*I*d*x + 6*I*c) +
18*I*a^2*e^(4*I*d*x + 4*I*c) - 18*I*a^2*e^(2*I*d*x + 2*I*c) + 6*I*a^2)*log(e^(2*I*d*x + 2*I*c) - 1))/(d*e^(6*I
*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)

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Sympy [B]  time = 3.32621, size = 141, normalized size = 1.91 \begin{align*} - \frac{2 i a^{2} \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{\frac{10 i a^{2} e^{- 2 i c} e^{4 i d x}}{d} - \frac{12 i a^{2} e^{- 4 i c} e^{2 i d x}}{d} + \frac{14 i a^{2} e^{- 6 i c}}{3 d}}{e^{6 i d x} - 3 e^{- 2 i c} e^{4 i d x} + 3 e^{- 4 i c} e^{2 i d x} - e^{- 6 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(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 + (10*I*a**2*exp(-2*I*c)*exp(4*I*d*x)/d - 12*I*a**2*exp(-4*I*c)*ex
p(2*I*d*x)/d + 14*I*a**2*exp(-6*I*c)/(3*d))/(exp(6*I*d*x) - 3*exp(-2*I*c)*exp(4*I*d*x) + 3*exp(-4*I*c)*exp(2*I
*d*x) - exp(-6*I*c))

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Giac [B]  time = 1.36316, size = 198, normalized size = 2.68 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 96 i \, a^{2} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 48 i \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 27 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{-88 i \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 27 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 6 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 )^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*d*x + 1/2*c)^3 - 6*I*a^2*tan(1/2*d*x + 1/2*c)^2 + 96*I*a^2*log(tan(1/2*d*x + 1/2*c) + I) - 4
8*I*a^2*log(abs(tan(1/2*d*x + 1/2*c))) - 27*a^2*tan(1/2*d*x + 1/2*c) - (-88*I*a^2*tan(1/2*d*x + 1/2*c)^3 - 27*
a^2*tan(1/2*d*x + 1/2*c)^2 + 6*I*a^2*tan(1/2*d*x + 1/2*c) + a^2)/tan(1/2*d*x + 1/2*c)^3)/d