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

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Rubi [A]  time = 0.09, antiderivative size = 58, normalized size of antiderivative = 1.00, 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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 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]

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*}

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Mathematica [C]  time = 0.20, size = 64, normalized size = 1.10 \[ -\frac {a^2 \left (4 i \cot (c+d x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-\tan ^2(c+d x)\right )+\cot ^2(c+d x)+4 (\log (\tan (c+d x))+\log (\cos (c+d x)))\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(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]])))/d

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fricas [A]  time = 0.42, size = 94, normalized size = 1.62 \[ \frac {2 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, a^{2} - {\left (a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )\right )}}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d} \]

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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giac [B]  time = 1.88, size = 116, normalized size = 2.00 \[ -\frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 32 \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 16 \, a^{2} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 8 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {24 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 i \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{8 \, d} \]

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(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

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maple [A]  time = 0.37, size = 65, normalized size = 1.12 \[ -\frac {2 a^{2} \ln \left (\sin \left (d x +c \right )\right )}{d}-2 i a^{2} x -\frac {2 i a^{2} \cot \left (d x +c \right )}{d}-\frac {2 i a^{2} c}{d}-\frac {a^{2} \left (\cot ^{2}\left (d x +c \right )\right )}{2 d} \]

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 = 0.81, size = 68, normalized size = 1.17 \[ -\frac {4 i \, {\left (d x + c\right )} a^{2} - 2 \, a^{2} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 4 \, a^{2} \log \left (\tan \left (d x + c\right )\right ) + \frac {4 i \, a^{2} \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2}}}{2 \, d} \]

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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mupad [B]  time = 3.84, size = 53, normalized size = 0.91 \[ -\frac {\frac {a^2}{2}+a^2\,\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^2}-\frac {a^2\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,4{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^3*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

- (a^2*tan(c + d*x)*2i + a^2/2)/(d*tan(c + d*x)^2) - (a^2*atan(2*tan(c + d*x) + 1i)*4i)/d

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sympy [A]  time = 0.77, size = 95, normalized size = 1.64 \[ - \frac {2 a^{2} \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {6 i a^{2} e^{2 i c} e^{2 i d x} - 4 i a^{2}}{i d e^{4 i c} e^{4 i d x} - 2 i d e^{2 i c} e^{2 i d x} + i d} \]

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*I*a**2*exp(2*I*c)*exp(2*I*d*x) - 4*I*a**2)/(I*d*exp(4*I*c)*exp(
4*I*d*x) - 2*I*d*exp(2*I*c)*exp(2*I*d*x) + I*d)

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