3.748 \(\int \frac{1}{\cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx\)
Optimal. Leaf size=141 \[ \frac{\sqrt{\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}-\frac{(-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{8 a^3 d}+\frac{i \sqrt{\cot (c+d x)}}{6 a d (a \cot (c+d x)+i a)^2}+\frac{\sqrt{\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3} \]
[Out]
-((-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/(8*a^3*d) + Sqrt[Cot[c + d*x]]/(6*d*(I*a + a*Cot[c + d*x]
)^3) + ((I/6)*Sqrt[Cot[c + d*x]])/(a*d*(I*a + a*Cot[c + d*x])^2) + Sqrt[Cot[c + d*x]]/(8*d*(I*a^3 + a^3*Cot[c
+ d*x]))
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Rubi [A] time = 0.303264, antiderivative size = 141, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used =
{3673, 3558, 3596, 12, 3549, 3533, 208} \[ \frac{\sqrt{\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}-\frac{(-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{8 a^3 d}+\frac{i \sqrt{\cot (c+d x)}}{6 a d (a \cot (c+d x)+i a)^2}+\frac{\sqrt{\cot (c+d x)}}{6 d (a \cot (c+d x)+i a)^3} \]
Antiderivative was successfully verified.
[In]
Int[1/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^3),x]
[Out]
-((-1)^(3/4)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/(8*a^3*d) + Sqrt[Cot[c + d*x]]/(6*d*(I*a + a*Cot[c + d*x]
)^3) + ((I/6)*Sqrt[Cot[c + d*x]])/(a*d*(I*a + a*Cot[c + d*x])^2) + Sqrt[Cot[c + d*x]]/(8*d*(I*a^3 + a^3*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 3558
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*m), x] + Dist[1/(2*a^2*m), Int[(a
+ b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Tan[e + f*x], x], x], x] /; FreeQ[{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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*n])
Rule 3596
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n,
0]
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 3549
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
a*c + b*d)*(c + d*Tan[e + f*x])^n)/(2*(b*c - a*d)*f*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a*(b*c - a*d)), Int[
(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*d*(n - 1) + b*c^2 + b*d^2*n - d*(b*c - a*d)*(n - 1)*Tan[e + f*x], x], x]
, x] /; FreeQ[{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] && LtQ[0,
n, 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 \frac{1}{\cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^3} \, dx &=\int \frac{\cot ^{\frac{3}{2}}(c+d x)}{(i a+a \cot (c+d x))^3} \, dx\\ &=\frac{\sqrt{\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac{\int \frac{-\frac{i a}{2}+\frac{7}{2} a \cot (c+d x)}{\sqrt{\cot (c+d x)} (i a+a \cot (c+d x))^2} \, dx}{6 a^2}\\ &=\frac{\sqrt{\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac{i \sqrt{\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac{\int -\frac{6 i a^2 \sqrt{\cot (c+d x)}}{i a+a \cot (c+d x)} \, dx}{24 a^4}\\ &=\frac{\sqrt{\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac{i \sqrt{\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}-\frac{i \int \frac{\sqrt{\cot (c+d x)}}{i a+a \cot (c+d x)} \, dx}{4 a^2}\\ &=\frac{\sqrt{\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac{i \sqrt{\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac{\sqrt{\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac{i \int \frac{-\frac{a}{2}+\frac{1}{2} i a \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx}{8 a^4}\\ &=\frac{\sqrt{\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac{i \sqrt{\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac{\sqrt{\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac{i \operatorname{Subst}\left (\int \frac{1}{\frac{a}{2}+\frac{1}{2} i a x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{16 a^2 d}\\ &=-\frac{(-1)^{3/4} \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{8 a^3 d}+\frac{\sqrt{\cot (c+d x)}}{6 d (i a+a \cot (c+d x))^3}+\frac{i \sqrt{\cot (c+d x)}}{6 a d (i a+a \cot (c+d x))^2}+\frac{\sqrt{\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 1.52611, size = 154, normalized size = 1.09 \[ \frac{i \sqrt{\cot (c+d x)} \csc ^3(c+d x) \left (3 i \sin (c+d x)-3 i \sin (3 (c+d x))+5 \cos (c+d x)-5 \cos (3 (c+d x))+6 \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))\right )}{48 a^3 d (\cot (c+d x)+i)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^3),x]
[Out]
((I/48)*Sqrt[Cot[c + d*x]]*Csc[c + d*x]^3*(5*Cos[c + d*x] - 5*Cos[3*(c + d*x)] + (3*I)*Sin[c + d*x] - (3*I)*Si
n[3*(c + d*x)] + 6*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*(Cos[3*(c + d*x)] + I*S
in[3*(c + d*x)])*Sqrt[I*Tan[c + d*x]]))/(a^3*d*(I + Cot[c + d*x])^3)
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Maple [C] time = 0.306, size = 1983, normalized size = 14.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x)
[Out]
1/48/a^3/d*2^(1/2)*(cos(d*x+c)-1)*(15*I*((cos(d*x+c)-1)/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))/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)^2-3*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))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/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))-15*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)^2-12*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)-1)/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))/sin(d*x+c))
^(1/2)*cos(d*x+c)^4-12*sin(d*x+c)*((cos(d*x+c)-1)/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))/sin(d*x+c))^(1/2)*cos(d*x+c)^3*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c)
)^(1/2),1/2*2^(1/2))+12*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)-1)/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))/sin(d*x+c
))^(1/2)*cos(d*x+c)^3*sin(d*x+c)+3*I*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((co
s(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)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x
+c))^(1/2)+12*I*((cos(d*x+c)-1)/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))/sin(d*x+c))^(1/2)*cos(d*x+c)^4*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2
))+10*I*sin(d*x+c)*2^(1/2)*cos(d*x+c)^3-12*I*((cos(d*x+c)-1)/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))/sin(d*x+c))^(1/2)*cos(d*x+c)^4*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))+15*(-(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)^2+9*(-(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)-9*(-(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)-10*I*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-12*I*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)-1)/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))/sin(d*x+c))^(1/2)*cos(d*x+c)^3*sin(d*x+c)+9*I*Elli
pticPi((-(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)*((cos(d*x+c)
-1)/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))/sin(d*x+c))^(1/
2)+6*cos(d*x+c)^4*2^(1/2)-6*2^(1/2)*cos(d*x+c)^3-3*(-(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))-3*2^(1/2)*cos(d*x+c)^2+3*2^(1/2)*cos(d*x+c))*cos(d*x+c)*(cos(d*x+c)+1)^2/(
4*I*sin(d*x+c)*cos(d*x+c)^2+4*cos(d*x+c)^3-I*sin(d*x+c)-3*cos(d*x+c))/(cos(d*x+c)/sin(d*x+c))^(3/2)/sin(d*x+c)
^5
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
[Out]
Exception raised: RuntimeError
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Fricas [B] time = 1.40116, size = 888, normalized size = 6.3 \begin{align*} -\frac{{\left (12 \, a^{3} d \sqrt{-\frac{i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left ({\left ({\left (16 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 16 i \, a^{3} 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}} \sqrt{-\frac{i}{64 \, a^{6} d^{2}}} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - 12 \, a^{3} d \sqrt{-\frac{i}{64 \, a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left ({\left ({\left (-16 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, a^{3} 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}} \sqrt{-\frac{i}{64 \, a^{6} d^{2}}} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\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}}{\left (-4 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 4 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{48 \, a^{3} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
[Out]
-1/48*(12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(((16*I*a^3*d*e^(2*I*d*x + 2*I*c) - 16*I*a^3*d)
*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/64*I/(a^6*d^2)) + 2*I*e^(2*I*d*x + 2*I*c)
)*e^(-2*I*d*x - 2*I*c)) - 12*a^3*d*sqrt(-1/64*I/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(((-16*I*a^3*d*e^(2*I*d*x +
2*I*c) + 16*I*a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/64*I/(a^6*d^2)) + 2*I
*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) - sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(-4*
I*e^(6*I*d*x + 6*I*c) + 4*I*e^(4*I*d*x + 4*I*c) + I*e^(2*I*d*x + 2*I*c) - I))*e^(-6*I*d*x - 6*I*c)/(a^3*d)
________________________________________________________________________________________
Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**3,x)
[Out]
Exception raised: AttributeError
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\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(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
[Out]
integrate(1/((I*a*tan(d*x + c) + a)^3*cot(d*x + c)^(3/2)), x)