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

Optimal. Leaf size=60 \[ -\frac{a^3+i a^3 \tan (c+d x)}{d}+\frac{a^3 \log (\sin (c+d x))}{d}+\frac{3 a^3 \log (\cos (c+d x))}{d}+4 i a^3 x \]

[Out]

(4*I)*a^3*x + (3*a^3*Log[Cos[c + d*x]])/d + (a^3*Log[Sin[c + d*x]])/d - (a^3 + I*a^3*Tan[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.101905, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3556, 3589, 3475, 3531} \[ -\frac{a^3+i a^3 \tan (c+d x)}{d}+\frac{a^3 \log (\sin (c+d x))}{d}+\frac{3 a^3 \log (\cos (c+d x))}{d}+4 i a^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*I)*a^3*x + (3*a^3*Log[Cos[c + d*x]])/d + (a^3*Log[Sin[c + d*x]])/d - (a^3 + I*a^3*Tan[c + d*x])/d

Rule 3556

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[a/(d*(m + n - 1
)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + a*d*(m + 2*n) + (a*c*(m - 2) +
b*d*(3*m + 2*n - 4))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a
^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] && (IntegerQ[m] || Intege
rsQ[2*m, 2*n])

Rule 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*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]

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]

Rubi steps

\begin{align*} \int \cot (c+d x) (a+i a \tan (c+d x))^3 \, dx &=-\frac{a^3+i a^3 \tan (c+d x)}{d}+a \int \cot (c+d x) (a+i a \tan (c+d x)) (a+3 i a \tan (c+d x)) \, dx\\ &=-\frac{a^3+i a^3 \tan (c+d x)}{d}+a \int \cot (c+d x) \left (a^2+4 i a^2 \tan (c+d x)\right ) \, dx-\left (3 a^3\right ) \int \tan (c+d x) \, dx\\ &=4 i a^3 x+\frac{3 a^3 \log (\cos (c+d x))}{d}-\frac{a^3+i a^3 \tan (c+d x)}{d}+a^3 \int \cot (c+d x) \, dx\\ &=4 i a^3 x+\frac{3 a^3 \log (\cos (c+d x))}{d}+\frac{a^3 \log (\sin (c+d x))}{d}-\frac{a^3+i a^3 \tan (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 1.22954, size = 95, normalized size = 1.58 \[ \frac{a^3 \sec (c) \sec (c+d x) \left (\cos (d x) \left (\log \left (\sin ^2(c+d x)\right )+3 \log \left (\cos ^2(c+d x)\right )+8 i d x\right )+\cos (2 c+d x) \left (\log \left (\sin ^2(c+d x)\right )+3 \log \left (\cos ^2(c+d x)\right )+8 i d x\right )-4 i \sin (d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sec[c]*Sec[c + d*x]*(Cos[d*x]*((8*I)*d*x + 3*Log[Cos[c + d*x]^2] + Log[Sin[c + d*x]^2]) + Cos[2*c + d*x]*
((8*I)*d*x + 3*Log[Cos[c + d*x]^2] + Log[Sin[c + d*x]^2]) - (4*I)*Sin[d*x]))/(4*d)

________________________________________________________________________________________

Maple [A]  time = 0.047, size = 63, normalized size = 1.1 \begin{align*} 4\,i{a}^{3}x-{\frac{i{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{4\,i{a}^{3}c}{d}}+3\,{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{3}\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)*(a+I*a*tan(d*x+c))^3,x)

[Out]

4*I*a^3*x-I/d*tan(d*x+c)*a^3+4*I/d*a^3*c+3*a^3*ln(cos(d*x+c))/d+a^3*ln(sin(d*x+c))/d

________________________________________________________________________________________

Maxima [A]  time = 2.36846, size = 72, normalized size = 1.2 \begin{align*} \frac{4 i \,{\left (d x + c\right )} a^{3} - 2 \, a^{3} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + a^{3} \log \left (\tan \left (d x + c\right )\right ) - i \, a^{3} \tan \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.21336, size = 223, normalized size = 3.72 \begin{align*} \frac{2 \, a^{3} + 3 \,{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) +{\left (a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}{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)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.7647, size = 73, normalized size = 1.22 \begin{align*} \frac{a^{3} \left (\log{\left (e^{2 i d x} - e^{- 2 i c} \right )} + 3 \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}\right )}{d} + \frac{2 a^{3} e^{- 2 i c}}{d \left (e^{2 i d x} + e^{- 2 i c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*(log(exp(2*I*d*x) - exp(-2*I*c)) + 3*log(exp(2*I*d*x) + exp(-2*I*c)))/d + 2*a**3*exp(-2*I*c)/(d*(exp(2*I*
d*x) + exp(-2*I*c)))

________________________________________________________________________________________

Giac [B]  time = 1.32291, size = 171, normalized size = 2.85 \begin{align*} -\frac{8 \, a^{3} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 3 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) + \frac{3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 i \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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