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

Optimal. Leaf size=71 \[ \frac{4 i a^4 \log (\sin (c+d x))}{d}+\frac{4 i a^4 \log (\cos (c+d x))}{d}-\frac{\cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-8 a^4 x \]

[Out]

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

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Rubi [A]  time = 0.0874361, antiderivative size = 71, 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 = {3553, 12, 3541, 3475} \[ \frac{4 i a^4 \log (\sin (c+d x))}{d}+\frac{4 i a^4 \log (\cos (c+d x))}{d}-\frac{\cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-8 a^4 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3553

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3541

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d*(2
*b*c - a*d)*x)/b^2, x] + (Dist[d^2/b, Int[Tan[e + f*x], x], x] + Dist[(b*c - a*d)^2/b^2, Int[1/(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]

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 ^2(c+d x) (a+i a \tan (c+d x))^4 \, dx &=-\frac{\cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}-\int -4 i a^2 \cot (c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=-\frac{\cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\left (4 i a^2\right ) \int \cot (c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=-8 a^4 x-\frac{\cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\left (4 i a^4\right ) \int \cot (c+d x) \, dx-\left (4 i a^4\right ) \int \tan (c+d x) \, dx\\ &=-8 a^4 x+\frac{4 i a^4 \log (\cos (c+d x))}{d}+\frac{4 i a^4 \log (\sin (c+d x))}{d}-\frac{\cot (c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}\\ \end{align*}

Mathematica [B]  time = 2.28419, size = 151, normalized size = 2.13 \[ \frac{a^4 \csc (c) \sec (c) \csc (c+d x) \sec (c+d x) \left (6 d x \cos (4 c+2 d x)+4 \sin (2 c) \sin (2 (c+d x)) \tan ^{-1}(\tan (5 c+d x))-i \cos (4 c+2 d x) \log \left (\cos ^2(c+d x)\right )+\cos (2 d x) \left (i \log \left (\sin ^2(c+d x)\right )+i \log \left (\cos ^2(c+d x)\right )-6 d x\right )-i \cos (4 c+2 d x) \log \left (\sin ^2(c+d x)\right )+2 \sin (2 d x)\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*Csc[c]*Csc[c + d*x]*Sec[c]*Sec[c + d*x]*(6*d*x*Cos[4*c + 2*d*x] - I*Cos[4*c + 2*d*x]*Log[Cos[c + d*x]^2]
+ Cos[2*d*x]*(-6*d*x + I*Log[Cos[c + d*x]^2] + I*Log[Sin[c + d*x]^2]) - I*Cos[4*c + 2*d*x]*Log[Sin[c + d*x]^2]
+ 2*Sin[2*d*x] + 4*ArcTan[Tan[5*c + d*x]]*Sin[2*c]*Sin[2*(c + d*x)]))/(4*d)

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Maple [A]  time = 0.044, size = 76, normalized size = 1.1 \begin{align*}{\frac{4\,i{a}^{4}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+{\frac{4\,i{a}^{4}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-8\,{a}^{4}x-{\frac{{a}^{4}\cot \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}\tan \left ( dx+c \right ) }{d}}-8\,{\frac{{a}^{4}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*I*a^4*ln(sin(d*x+c))/d+4*I*a^4*ln(cos(d*x+c))/d-8*a^4*x-a^4*cot(d*x+c)/d+a^4*tan(d*x+c)/d-8/d*a^4*c

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Maxima [A]  time = 1.66617, size = 90, normalized size = 1.27 \begin{align*} -\frac{8 \,{\left (d x + c\right )} a^{4} + 4 i \, a^{4} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 4 i \, a^{4} \log \left (\tan \left (d x + c\right )\right ) - a^{4} \tan \left (d x + c\right ) + \frac{a^{4}}{\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)^2*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.20713, size = 150, normalized size = 2.11 \begin{align*} \frac{-4 i \, a^{4} +{\left (4 i \, a^{4} e^{\left (4 i \, d x + 4 i \, c\right )} - 4 i \, a^{4}\right )} \log \left (e^{\left (4 i \, d x + 4 i \, c\right )} - 1\right )}{d e^{\left (4 i \, d x + 4 i \, c\right )} - d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.2065, size = 58, normalized size = 0.82 \begin{align*} \frac{4 i a^{4} \log{\left (e^{4 i d x} - e^{- 4 i c} \right )}}{d} - \frac{4 i a^{4} e^{- 4 i c}}{d \left (e^{4 i d x} - e^{- 4 i c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.67399, size = 224, normalized size = 3.15 \begin{align*} -\frac{32 i \, a^{4} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right ) - 8 i \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 8 i \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - 8 i \, a^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right ) - a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \frac{-8 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 i \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{4}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(32*I*a^4*log(tan(1/2*d*x + 1/2*c) + I) - 8*I*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*I*a^4*log(abs(ta
n(1/2*d*x + 1/2*c) - 1)) - 8*I*a^4*log(abs(tan(1/2*d*x + 1/2*c))) - a^4*tan(1/2*d*x + 1/2*c) - (-8*I*a^4*tan(1
/2*d*x + 1/2*c)^3 - 5*a^4*tan(1/2*d*x + 1/2*c)^2 + 8*I*a^4*tan(1/2*d*x + 1/2*c) + a^4)/(tan(1/2*d*x + 1/2*c)^3
- tan(1/2*d*x + 1/2*c)))/d