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

Optimal. Leaf size=219 \[ -\frac{317 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}+\frac{151 \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{17 \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[Out]

((1/8 + I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[
Tan[c + d*x]])/(a^(5/2)*d) + Sqrt[Cot[c + d*x]]/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (17*Sqrt[Cot[c + d*x]])/(
30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (151*Sqrt[Cot[c + d*x]])/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) - (317*S
qrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d)

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Rubi [A]  time = 0.667005, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4241, 3559, 3596, 3598, 12, 3544, 205} \[ -\frac{317 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}+\frac{151 \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{17 \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/8 + I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[
Tan[c + d*x]])/(a^(5/2)*d) + Sqrt[Cot[c + d*x]]/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (17*Sqrt[Cot[c + d*x]])/(
30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (151*Sqrt[Cot[c + d*x]])/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) - (317*S
qrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d)

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 3559

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(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*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*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] && LtQ[m, 0] && (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 3598

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*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 12

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

Rule 3544

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{\cot ^{\frac{3}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac{\sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{11 a}{2}-3 i a \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{\sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\frac{83 a^2}{4}-17 i a^2 \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{\sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{151 \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)} \left (\frac{317 a^3}{8}-\frac{151}{4} i a^3 \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac{\sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{151 \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{317 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}+\frac{\left (2 \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{15 i a^4 \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{15 a^7}\\ &=\frac{\sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{151 \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{317 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}+\frac{\left (i \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac{\sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{151 \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{317 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{4 a d}\\ &=\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right ) \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}}{a^{5/2} d}+\frac{\sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{17 \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{151 \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{317 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}\\ \end{align*}

Mathematica [A]  time = 2.66995, size = 165, normalized size = 0.75 \[ \frac{\cot ^{\frac{3}{2}}(c+d x) \sec (c+d x) \left ((340-460 \cos (2 (c+d x))) \csc (c+d x)-i (466 \cos (2 (c+d x))+149) \sec (c+d x)+15 e^{2 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \csc (2 (c+d x)) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{60 a^2 d (\cot (c+d x)+i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[c + d*x]^(3/2)*Sec[c + d*x]*((340 - 460*Cos[2*(c + d*x)])*Csc[c + d*x] + 15*E^((2*I)*(c + d*x))*Sqrt[-1 +
E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]]*Csc[2*(c + d*x)] - I*(149 + 466*
Cos[2*(c + d*x)])*Sec[c + d*x]))/(60*a^2*d*(I + Cot[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.4, size = 398, normalized size = 1.8 \begin{align*}{\frac{ \left ( -{\frac{1}{120}}-{\frac{i}{120}} \right ) \sin \left ( dx+c \right ) }{d{a}^{3}\cos \left ( dx+c \right ) } \left ({\frac{\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( -15\,i\arctan \left ( \left ({\frac{1}{2}}+{\frac{i}{2}} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\sin \left ( dx+c \right ) +48\,i \left ( \cos \left ( dx+c \right ) \right ) ^{6}-48\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}+48\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +32\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+56\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +151\,i\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) -32\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}+56\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +15\,\arctan \left ( \left ( 1/2+i/2 \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2} \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\cos \left ( dx+c \right ) +117\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+48\,i \left ( \cos \left ( dx+c \right ) \right ) ^{5}\sin \left ( dx+c \right ) +15\,\sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2}\arctan \left ( \left ( 1/2+i/2 \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) -1}{\sin \left ( dx+c \right ) }}}\sqrt{2} \right ) -117\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+151\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +317-317\,i \right ) } \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))^(5/2),x)

[Out]

(-1/120-1/120*I)/d/a^3*(cos(d*x+c)/sin(d*x+c))^(3/2)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*sin(d*x+c)
*(-15*I*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2
)*sin(d*x+c)+48*I*cos(d*x+c)^6-48*cos(d*x+c)^6+48*cos(d*x+c)^5*sin(d*x+c)+32*I*cos(d*x+c)^4+56*I*cos(d*x+c)^3*
sin(d*x+c)+151*I*sin(d*x+c)*cos(d*x+c)-32*cos(d*x+c)^4+56*cos(d*x+c)^3*sin(d*x+c)+15*arctan((1/2+1/2*I)*((cos(
d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*cos(d*x+c)+117*I*cos(d*x+c)^2+4
8*I*cos(d*x+c)^5*sin(d*x+c)+15*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/si
n(d*x+c))^(1/2)*2^(1/2))-117*cos(d*x+c)^2+151*cos(d*x+c)*sin(d*x+c)+317-317*I)/cos(d*x+c)

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

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.66099, size = 1069, normalized size = 4.88 \begin{align*} \frac{{\left (30 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{4} \,{\left (4 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 30 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac{1}{4} \,{\left (4 \, a^{3} d \sqrt{\frac{i}{8 \, a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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 (463 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 194 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 26 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{120 \, a^{3} 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))^(5/2),x, algorithm="fricas")

[Out]

1/120*(30*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(1/4*(4*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(2*I*d*x +
2*I*c) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*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) - 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 30*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(6*I*d*x + 6*
I*c)*log(-1/4*(4*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(2*I*d*x + 2*I*c) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*s
qrt((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) - 1)*e^(I*d*x + I*c))*e^(-I*d*
x - I*c)) - sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) -
1))*(463*e^(6*I*d*x + 6*I*c) - 194*e^(4*I*d*x + 4*I*c) - 26*e^(2*I*d*x + 2*I*c) - 3)*e^(I*d*x + I*c))*e^(-6*I*
d*x - 6*I*c)/(a^3*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))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (d x + c\right )^{\frac{3}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{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))^(5/2),x, algorithm="giac")

[Out]

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