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

Optimal. Leaf size=214 \[ \frac{13 A-37 i B}{60 a^2 d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{-B+i A}{5 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{A+11 i B}{30 a d \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]

[Out]

((1/8 + I/8)*(I*A + 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) + (I*A - B)/(5*d*Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) + (A +
(11*I)*B)/(30*a*d*Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + (13*A - (37*I)*B)/(60*a^2*d*Sqrt[Cot[c +
d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.757366, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4241, 3595, 3596, 12, 3544, 205} \[ \frac{13 A-37 i B}{60 a^2 d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) (B+i A) \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac{-B+i A}{5 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{A+11 i B}{30 a d \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/8 + I/8)*(I*A + 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) + (I*A - B)/(5*d*Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) + (A +
(11*I)*B)/(30*a*d*Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + (13*A - (37*I)*B)/(60*a^2*d*Sqrt[Cot[c +
d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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 3595

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

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 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{A+B \tan (c+d x)}{\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{\tan ^{\frac{3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac{i A-B}{5 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}-\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{\tan (c+d x)} \left (\frac{3}{2} a (i A-B)-a (A-4 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{i A-B}{5 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{A+11 i B}{30 a d \sqrt{\cot (c+d x)} (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 (A+11 i B)-\frac{1}{2} a^2 (7 i A+13 B) \tan (c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{i A-B}{5 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{A+11 i B}{30 a d \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{13 A-37 i B}{60 a^2 d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int -\frac{15 a^3 (A-i B) \sqrt{a+i a \tan (c+d x)}}{8 \sqrt{\tan (c+d x)}} \, dx}{15 a^6}\\ &=\frac{i A-B}{5 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{A+11 i B}{30 a d \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{13 A-37 i B}{60 a^2 d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}-\frac{\left ((A-i 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{i A-B}{5 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{A+11 i B}{30 a d \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{13 A-37 i B}{60 a^2 d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}+\frac{\left (i (A-i 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{i A-B}{5 d \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac{A+11 i B}{30 a d \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac{13 A-37 i B}{60 a^2 d \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 7.57418, size = 169, normalized size = 0.79 \[ \frac{\cot ^{\frac{3}{2}}(c+d x) \sec ^2(c+d x) \left (40 (B+i A) \sin (2 (c+d x))+4 (7 A-13 i B) \cos (2 (c+d x))-\frac{30 (A-i B) e^{3 i (c+d x)} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )}{\sqrt{-1+e^{2 i (c+d x)}}}+2 A+22 i B\right )}{120 a^2 d (\cot (c+d x)+i)^2 \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(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]^2*(2*A + (22*I)*B - (30*(A - I*B)*E^((3*I)*(c + d*x))*ArcTanh[E^(I*(c + d*x))
/Sqrt[-1 + E^((2*I)*(c + d*x))]])/Sqrt[-1 + E^((2*I)*(c + d*x))] + 4*(7*A - (13*I)*B)*Cos[2*(c + d*x)] + 40*(I
*A + B)*Sin[2*(c + d*x)]))/(120*a^2*d*(I + Cot[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.535, size = 1212, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/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*(60*I*A*cos(d*x+c)^2*sin(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))-13*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+15*A*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)*2^(1/2))*2^(1/2)+37*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-40*B*cos(d*x+c)^3*((cos(d*x+c)-
1)/sin(d*x+c))^(1/2)+40*B*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-30*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)*arct
an((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))-30*I*A*cos(d*x+c)*sin(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))-28*I*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^
2-52*I*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2-40*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(
d*x+c)^3+40*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)-60*I*B*cos(d*x+c)^3*2^(1/2)*arctan((1/2+1/2*I)*((co
s(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))+30*I*B*cos(d*x+c)^2*2^(1/2)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+
c))^(1/2)*2^(1/2))-15*I*A*sin(d*x+c)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+45*
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))+60*B*cos(d*x+c)^2*sin(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))-45*A*cos(d*x+c)*arctan((1/2+1/2*I)*(
(cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)-15*B*sin(d*x+c)*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c
))^(1/2)*2^(1/2))*2^(1/2)-30*A*2^(1/2)*cos(d*x+c)^2*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/
2))+28*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-52*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(
d*x+c)^2*sin(d*x+c)-40*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^3+40*I*A*cos(d*x+c)*((cos(d*x+c)-1)/si
n(d*x+c))^(1/2)+13*I*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+37*I*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+
c))^(1/2)+40*I*B*cos(d*x+c)^3*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-40*I*B*cos(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^
(1/2)-15*I*B*arctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+60*A*cos(d*x+c)^3*2^(1/2)*a
rctan((1/2+1/2*I)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)))*cos(d*x+c)^2*(a*(I*sin(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))/sin(d*x+c)^2/(cos(d*x+c)/si
n(d*x+c))^(3/2)/((cos(d*x+c)-1)/sin(d*x+c))^(1/2)

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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((A+B*tan(d*x+c))/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 = 2.0327, size = 1359, normalized size = 6.35 \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 \, \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 \, \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 (-17 i \, A - 23 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (16 i \, A + 34 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (4 i \, A - 14 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, A + 3 \, 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((A+B*tan(d*x+c))/cot(d*x+c)^(3/2)/(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*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*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)*((-17*I*A - 23*B)*e^(6*I*d*x + 6
*I*c) + (16*I*A + 34*B)*e^(4*I*d*x + 4*I*c) + (4*I*A - 14*B)*e^(2*I*d*x + 2*I*c) - 3*I*A + 3*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((A+B*tan(d*x+c))/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{B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \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((A+B*tan(d*x+c))/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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