3.732 \(\int \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=64 \[ -\frac{8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{2 \left (a^3 \cot (c+d x)+i a^3\right )}{d \sqrt{\cot (c+d x)}} \]

[Out]

(-8*(-1)^(1/4)*a^3*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (2*(I*a^3 + a^3*Cot[c + d*x]))/(d*Sqrt[Cot[c +
d*x]])

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Rubi [A]  time = 0.118275, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3673, 3553, 12, 3533, 208} \[ -\frac{8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{2 \left (a^3 \cot (c+d x)+i a^3\right )}{d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*(-1)^(1/4)*a^3*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (2*(I*a^3 + a^3*Cot[c + d*x]))/(d*Sqrt[Cot[c +
d*x]])

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

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 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^3 \, dx &=\int \frac{(i a+a \cot (c+d x))^3}{\cot ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 \left (i a^3+a^3 \cot (c+d x)\right )}{d \sqrt{\cot (c+d x)}}-2 \int -\frac{2 i a^2 (i a+a \cot (c+d x))}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 \left (i a^3+a^3 \cot (c+d x)\right )}{d \sqrt{\cot (c+d x)}}+\left (4 i a^2\right ) \int \frac{i a+a \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 \left (i a^3+a^3 \cot (c+d x)\right )}{d \sqrt{\cot (c+d x)}}-\frac{\left (8 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{-i a+a x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{2 \left (i a^3+a^3 \cot (c+d x)\right )}{d \sqrt{\cot (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.52478, size = 125, normalized size = 1.95 \[ \frac{2 a^3 e^{-3 i c} (\cos (3 (c+d x))+i \sin (3 (c+d x))) \left (-\cot (c+d x)+\frac{4 i \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{i \tan (c+d x)}}-i\right )}{d \sqrt{\cot (c+d x)} (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)])*(-I - Cot[c + d*x] + ((4*I)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*
x)))/(1 + E^((2*I)*(c + d*x)))]])/Sqrt[I*Tan[c + d*x]]))/(d*E^((3*I)*c)*Sqrt[Cot[c + d*x]]*(Cos[d*x] + I*Sin[d
*x])^3)

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Maple [C]  time = 0.32, size = 748, normalized size = 11.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^3,x)

[Out]

-a^3/d*2^(1/2)*(4*I*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)
*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*
x+c)-4*I*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+
c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*
x+c)+4*I*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+
c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-4*I*(-(cos(d*x+c)
-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2
)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-4*(-(cos(d*x+c)-1-sin(d*x+c)
)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi(
(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)-4*(-(cos(d*x+c)-1-sin(d*x+c))/
sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+I*2^(1/2)*sin(d*x+c)+2^(1/2)*cos(d*x+c))*(c
os(d*x+c)/sin(d*x+c))^(3/2)*sin(d*x+c)/cos(d*x+c)^2

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Maxima [B]  time = 1.58386, size = 194, normalized size = 3.03 \begin{align*} -\frac{{\left (\left (2 i - 2\right ) \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + \left (2 i - 2\right ) \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) - \left (i + 1\right ) \, \sqrt{2} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) + \left (i + 1\right ) \, \sqrt{2} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} + 2 i \, a^{3} \sqrt{\tan \left (d x + c\right )} + \frac{2 \, a^{3}}{\sqrt{\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)^(3/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.36589, size = 765, normalized size = 11.95 \begin{align*} -\frac{16 \, a^{3} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{64 i \, a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{64 i \, a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3}}\right ) + \sqrt{\frac{64 i \, a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{64 i \, a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3}}\right )}{4 \,{\left (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)^(3/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/4*(16*a^3*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*e^(2*I*d*x + 2*I*c) - sqrt(64*I*a^6/d
^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(8*I*a^3*e^(2*I*d*x + 2*I*c) + sqrt(64*I*a^6/d^2)*(d*e^(2*I*d*x + 2*I*
c) - d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/a^3) + sqrt(64*I*a^6
/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(8*I*a^3*e^(2*I*d*x + 2*I*c) - sqrt(64*I*a^6/d^2)*(d*e^(2*I*d*x + 2*
I*c) - d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/a^3))/(d*e^(2*I*d*
x + 2*I*c) + d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**(3/2)*(a+I*a*tan(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \cot \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^3*cot(d*x + c)^(3/2), x)