3.731 \(\int \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=88 \[ -\frac{16 i a^3 \sqrt{\cot (c+d x)}}{3 d}-\frac{2 \sqrt{\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}-\frac{8 (-1)^{3/4} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

[Out]

(-8*(-1)^(3/4)*a^3*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (((16*I)/3)*a^3*Sqrt[Cot[c + d*x]])/d - (2*Sqrt
[Cot[c + d*x]]*(I*a^3 + a^3*Cot[c + d*x]))/(3*d)

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Rubi [A]  time = 0.16995, antiderivative size = 88, 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, 3556, 3592, 3533, 208} \[ -\frac{16 i a^3 \sqrt{\cot (c+d x)}}{3 d}-\frac{2 \sqrt{\cot (c+d x)} \left (a^3 \cot (c+d x)+i a^3\right )}{3 d}-\frac{8 (-1)^{3/4} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(-8*(-1)^(3/4)*a^3*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (((16*I)/3)*a^3*Sqrt[Cot[c + d*x]])/d - (2*Sqrt
[Cot[c + d*x]]*(I*a^3 + a^3*Cot[c + d*x]))/(3*d)

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 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 3592

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

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{5}{2}}(c+d x) (a+i a \tan (c+d x))^3 \, dx &=\int \frac{(i a+a \cot (c+d x))^3}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 \sqrt{\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{3 d}-\frac{1}{3} (2 i a) \int \frac{(-2 i a-4 a \cot (c+d x)) (i a+a \cot (c+d x))}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{16 i a^3 \sqrt{\cot (c+d x)}}{3 d}-\frac{2 \sqrt{\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{3 d}-\frac{1}{3} (2 i a) \int \frac{6 a^2-6 i a^2 \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{16 i a^3 \sqrt{\cot (c+d x)}}{3 d}-\frac{2 \sqrt{\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{3 d}-\frac{\left (48 i a^5\right ) \operatorname{Subst}\left (\int \frac{1}{-6 a^2-6 i a^2 x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{8 (-1)^{3/4} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{16 i a^3 \sqrt{\cot (c+d x)}}{3 d}-\frac{2 \sqrt{\cot (c+d x)} \left (i a^3+a^3 \cot (c+d x)\right )}{3 d}\\ \end{align*}

Mathematica [A]  time = 2.50784, size = 125, normalized size = 1.42 \[ -\frac{2 a^3 e^{-3 i c} \sqrt{\cot (c+d x)} (\cos (3 (c+d x))+i \sin (3 (c+d x))) \left (\cot (c+d x)-12 i \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+9 i\right )}{3 d (\cos (d x)+i \sin (d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(-2*a^3*Sqrt[Cot[c + d*x]]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)])*(9*I + Cot[c + d*x] - (12*I)*ArcTanh[Sqrt[(
-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Sqrt[I*Tan[c + d*x]]))/(3*d*E^((3*I)*c)*(Cos[d*x] + I*Si
n[d*x])^3)

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Maple [C]  time = 0.285, size = 791, normalized size = 9. \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)^(5/2)*(a+I*a*tan(d*x+c))^3,x)

[Out]

1/3*a^3/d*2^(1/2)*(12*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)*sin(d*x+c)-12*(-(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)*sin(d*x+c)+12*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))*sin(d*x+c)+12*(-(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)*sin(d*x+c)-12*(-(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))*sin(d*x+c)+12*(-(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))*sin(d*x+c)-9*I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-2^(1/2)*cos(d*x+c)^2)*(cos(d*x+c)/sin(d*x+c))^(5
/2)*sin(d*x+c)/cos(d*x+c)^3

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Maxima [B]  time = 1.66982, size = 196, normalized size = 2.23 \begin{align*} \frac{3 \,{\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} - \frac{18 i \, a^{3}}{\sqrt{\tan \left (d x + c\right )}} - \frac{2 \, a^{3}}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/3*(3*((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 - 18*I*a^3/sqrt(tan(d*x + c))
- 2*a^3/tan(d*x + c)^(3/2))/d

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Fricas [B]  time = 1.42049, size = 811, normalized size = 9.22 \begin{align*} -\frac{3 \, \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 (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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 ) - 3 \, \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 (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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 ) -{\left (-80 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 64 i \, a^{3}\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}}}{12 \,{\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)^(5/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/12*(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)*(I*d*e^(2*I*d*x + 2*I*c) - I*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) - 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)*(-I*d*e^(2*I*d*x + 2*I*c) + I*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) - (-80*I*a^3*e^(2*I*d*x + 2*I*c) + 64*I*a^3)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*
I*d*x + 2*I*c) - 1)))/(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)**(5/2)*(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{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)*(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)^(5/2), x)