3.9 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=50 \[ -\frac{a \cot ^2(c+d x)}{2 d}-\frac{i a \cot (c+d x)}{d}-\frac{a \log (\sin (c+d x))}{d}-i a x \]

[Out]

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

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Rubi [A]  time = 0.0670244, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {3529, 3531, 3475} \[ -\frac{a \cot ^2(c+d x)}{2 d}-\frac{i a \cot (c+d x)}{d}-\frac{a \log (\sin (c+d x))}{d}-i a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 ^3(c+d x) (a+i a \tan (c+d x)) \, dx &=-\frac{a \cot ^2(c+d x)}{2 d}+\int \cot ^2(c+d x) (i a-a \tan (c+d x)) \, dx\\ &=-\frac{i a \cot (c+d x)}{d}-\frac{a \cot ^2(c+d x)}{2 d}+\int \cot (c+d x) (-a-i a \tan (c+d x)) \, dx\\ &=-i a x-\frac{i a \cot (c+d x)}{d}-\frac{a \cot ^2(c+d x)}{2 d}-a \int \cot (c+d x) \, dx\\ &=-i a x-\frac{i a \cot (c+d x)}{d}-\frac{a \cot ^2(c+d x)}{2 d}-\frac{a \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.16178, size = 68, normalized size = 1.36 \[ -\frac{a \left (\cot ^2(c+d x)+2 \log (\tan (c+d x))+2 \log (\cos (c+d x))\right )}{2 d}-\frac{i a \cot (c+d x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\tan ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 55, normalized size = 1.1 \begin{align*} -iax-{\frac{ia\cot \left ( dx+c \right ) }{d}}-{\frac{iac}{d}}-{\frac{a \left ( \cot \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{a\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 2.36444, size = 78, normalized size = 1.56 \begin{align*} -\frac{2 i \,{\left (d x + c\right )} a - a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, a \log \left (\tan \left (d x + c\right )\right ) + \frac{2 i \, a \tan \left (d x + c\right ) + a}{\tan \left (d x + c\right )^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.14454, size = 232, normalized size = 4.64 \begin{align*} \frac{4 \, a e^{\left (2 i \, d x + 2 i \, c\right )} -{\left (a e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 2 \, a}{d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, 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)^3*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 1.70003, size = 87, normalized size = 1.74 \begin{align*} - \frac{a \log{\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac{\frac{4 a e^{- 2 i c} e^{2 i d x}}{d} - \frac{2 a e^{- 4 i c}}{d}}{e^{4 i d x} - 2 e^{- 2 i c} e^{2 i d x} + e^{- 4 i c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c)),x)

[Out]

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

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Giac [B]  time = 1.27432, size = 139, normalized size = 2.78 \begin{align*} -\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 16 \, a \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) + 8 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - 4 i \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{12 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 i \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

-1/8*(a*tan(1/2*d*x + 1/2*c)^2 - 16*a*log(tan(1/2*d*x + 1/2*c) + I) + 8*a*log(abs(tan(1/2*d*x + 1/2*c))) - 4*I
*a*tan(1/2*d*x + 1/2*c) - (12*a*tan(1/2*d*x + 1/2*c)^2 - 4*I*a*tan(1/2*d*x + 1/2*c) - a)/tan(1/2*d*x + 1/2*c)^
2)/d