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

Optimal. Leaf size=260 \[ -\frac{(317 A+67 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}+\frac{(151 A+41 i B) \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(A+i B) \sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(17 A+7 i B) \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

((1/8 + I/8)*(A - I*B)*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) + ((A + I*B)*Sqrt[Cot[c + d*x]])/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((1
7*A + (7*I)*B)*Sqrt[Cot[c + d*x]])/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((151*A + (41*I)*B)*Sqrt[Cot[c + d*
x]])/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) - ((317*A + (67*I)*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]
)/(60*a^3*d)

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Rubi [A]  time = 0.96579, antiderivative size = 260, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4241, 3596, 3598, 12, 3544, 205} \[ -\frac{(317 A+67 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}+\frac{(151 A+41 i B) \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (A-i B) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{(A+i B) \sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(17 A+7 i B) \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/8 + I/8)*(A - I*B)*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) + ((A + I*B)*Sqrt[Cot[c + d*x]])/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((1
7*A + (7*I)*B)*Sqrt[Cot[c + d*x]])/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((151*A + (41*I)*B)*Sqrt[Cot[c + d*
x]])/(60*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) - ((317*A + (67*I)*B)*Sqrt[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 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+B \tan (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{A+B \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac{(A+i B) \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{1}{2} a (11 A+i B)-3 a (i A-B) \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{(A+i B) \sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(17 A+7 i B) \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{1}{4} a^2 (83 A+13 i B)-a^2 (17 i A-7 B) \tan (c+d x)}{\tan ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{(A+i B) \sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(17 A+7 i B) \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(151 A+41 i B) \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{1}{8} a^3 (317 A+67 i B)-\frac{1}{4} a^3 (151 i A-41 B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx}{15 a^6}\\ &=\frac{(A+i B) \sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(17 A+7 i B) \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(151 A+41 i B) \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(317 A+67 i B) \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 a^4 (i A+B) \sqrt{a+i a \tan (c+d x)}}{16 \sqrt{\tan (c+d x)}} \, dx}{15 a^7}\\ &=\frac{(A+i B) \sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(17 A+7 i B) \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(151 A+41 i B) \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(317 A+67 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}+\frac{\left ((i A+B) \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{(A+i B) \sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(17 A+7 i B) \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(151 A+41 i B) \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(317 A+67 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}-\frac{\left (i (i A+B) \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 ) (i A+B) \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{(A+i B) \sqrt{\cot (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{(17 A+7 i B) \sqrt{\cot (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac{(151 A+41 i B) \sqrt{\cot (c+d x)}}{60 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(317 A+67 i B) \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{60 a^3 d}\\ \end{align*}

Mathematica [A]  time = 8.75032, size = 200, normalized size = 0.77 \[ \frac{\cot ^{\frac{3}{2}}(c+d x) \sec (c+d x) \left (-20 \csc (c+d x) ((23 A+4 i B) \cos (2 (c+d x))-17 A-4 i B)+\sec (c+d x) ((86 B-466 i A) \cos (2 (c+d x))-149 i A+19 B)+15 (A-i B) e^{2 i (c+d x)} \sqrt{-1+e^{2 i (c+d x)}} \csc (2 (c+d x)) \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right )}{60 a^2 d (\cot (c+d x)+i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[c + d*x]^(3/2)*Sec[c + d*x]*(-20*(-17*A - (4*I)*B + (23*A + (4*I)*B)*Cos[2*(c + d*x)])*Csc[c + d*x] + 15*
(A - I*B)*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)] + ((-149*I)*A + 19*B + ((-466*I)*A + 86*B)*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 = 0.541, size = 764, normalized size = 2.9 \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)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

(1/120+1/120*I)/d/a^3*sin(d*x+c)*(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)*
(-317*A-67*B+15*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(
1/2))*sin(d*x+c)*2^(1/2)-48*I*A*cos(d*x+c)^6+48*I*B*cos(d*x+c)^6-32*I*A*cos(d*x+c)^4-8*I*B*cos(d*x+c)^4-117*I*
A*cos(d*x+c)^2+27*I*B*cos(d*x+c)^2+15*I*B*2^(1/2)*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)+48*B*cos(d*x+c)^6+15*I*B*cos(d*x+c)*2^(1/2)*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)+117*A*cos(d*x+c)^2+32*A*cos(d*x+c)^4+48*A*co
s(d*x+c)^6-151*A*cos(d*x+c)*sin(d*x+c)+48*B*cos(d*x+c)^5*sin(d*x+c)+41*B*cos(d*x+c)*sin(d*x+c)+16*B*cos(d*x+c)
^3*sin(d*x+c)-48*A*cos(d*x+c)^5*sin(d*x+c)-56*A*cos(d*x+c)^3*sin(d*x+c)-48*I*A*cos(d*x+c)^5*sin(d*x+c)-48*I*B*
cos(d*x+c)^5*sin(d*x+c)-56*I*A*cos(d*x+c)^3*sin(d*x+c)-16*I*B*cos(d*x+c)^3*sin(d*x+c)-151*I*A*cos(d*x+c)*sin(d
*x+c)-41*I*B*sin(d*x+c)*cos(d*x+c)-15*A*cos(d*x+c)*2^(1/2)*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)+15*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan((1/2+1/2
*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+317*I*A-67*I*B-8*cos(d*x+c)^4*B+27*B*cos(d*x+c)^2-15*A*
((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2))/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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 2.14662, size = 1365, normalized size = 5.25 \begin{align*} \frac{{\left (15 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{{\left (2 i \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A - B\right )} \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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{{\left (-2 i \, \sqrt{\frac{1}{2}} a^{3} d \sqrt{\frac{i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{2}{\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A - B\right )} \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}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 i \, A + 4 \, B}\right ) - \sqrt{2}{\left ({\left (463 \, A + 83 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \,{\left (97 \, A + 32 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \,{\left (13 \, A + 8 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\right )} \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}} 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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/120*(15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log((2*I*sqrt(1/2)*a^3*d
*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) - I*A -
B)*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^(I*d*x + I*
c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(6*I*d*x +
6*I*c)*log((-2*I*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A +
B)*e^(2*I*d*x + 2*I*c) - I*A - B)*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^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - sqrt(2)*((463*A + 83*I*B)*e^(6*I*d*x + 6
*I*c) - 2*(97*A + 32*I*B)*e^(4*I*d*x + 4*I*c) - 2*(13*A + 8*I*B)*e^(2*I*d*x + 2*I*c) - 3*A - 3*I*B)*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^(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+B*tan(d*x+c))/(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{{\left (B \tan \left (d x + c\right ) + A\right )} \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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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