\(\int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx\) [524]

Optimal result

Integrand size = 36, antiderivative size = 307 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (i A-B)}{2 a d \sqrt {\cot (c+d x)}}+\frac {i A-B}{2 d \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))}+\frac {((3-5 i) A+(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((1+4 i) A-(6+i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a d} \]

[Out]

1/6*(-3*A-7*I*B)/a/d/cot(d*x+c)^(3/2)+1/2*(I*A-B)/d/cot(d*x+c)^(3/2)/(I*a+a*cot(d*x+c))-(1/8+1/8*I)*((4+I)*A+(
1+6*I)*B)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)-(1/8+1/8*I)*((4+I)*A+(1+6*I)*B)*arctan(1+2^(1/2)*cot
(d*x+c)^(1/2))/a/d*2^(1/2)+1/16*((3-5*I)*A+(5+7*I)*B)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)+(1
/16+1/16*I)*((1+4*I)*A-(6+I)*B)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)-5/2*(I*A-B)/a/d/cot(d*x+
c)^(1/2)

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3662, 3677, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) ((4+i) A+(1+6 i) B) \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d}+\frac {-B+i A}{2 d \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+i a)}-\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (-B+i A)}{2 a d \sqrt {\cot (c+d x)}}+\frac {((3-5 i) A+(5+7 i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{8 \sqrt {2} a d}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) ((1+4 i) A-(6+i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d} \]

[In]

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

[Out]

((1/4 + I/4)*((4 + I)*A + (1 + 6*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a*d) - ((1/4 + I/4)*((
4 + I)*A + (1 + 6*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a*d) - (3*A + (7*I)*B)/(6*a*d*Cot[c +
d*x]^(3/2)) - (5*(I*A - B))/(2*a*d*Sqrt[Cot[c + d*x]]) + (I*A - B)/(2*d*Cot[c + d*x]^(3/2)*(I*a + a*Cot[c + d
*x])) + (((3 - 5*I)*A + (5 + 7*I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(8*Sqrt[2]*a*d) + ((1
/8 + I/8)*((1 + 4*I)*A - (6 + I)*B)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a*d)

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1182

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]

Rule 3610

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b
*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3615

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3662

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3677

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {B+A \cot (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (i a+a \cot (c+d x))} \, dx \\ & = \frac {i A-B}{2 d \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))}+\frac {\int \frac {-\frac {1}{2} a (3 A+7 i B)-\frac {5}{2} a (i A-B) \cot (c+d x)}{\cot ^{\frac {5}{2}}(c+d x)} \, dx}{2 a^2} \\ & = -\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}+\frac {i A-B}{2 d \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))}+\frac {\int \frac {-\frac {5}{2} a (i A-B)+\frac {1}{2} a (3 A+7 i B) \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx}{2 a^2} \\ & = -\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (i A-B)}{2 a d \sqrt {\cot (c+d x)}}+\frac {i A-B}{2 d \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))}+\frac {\int \frac {\frac {1}{2} a (3 A+7 i B)+\frac {5}{2} a (i A-B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (i A-B)}{2 a d \sqrt {\cot (c+d x)}}+\frac {i A-B}{2 d \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))}+\frac {\text {Subst}\left (\int \frac {-\frac {1}{2} a (3 A+7 i B)-\frac {5}{2} a (i A-B) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d} \\ & = -\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (i A-B)}{2 a d \sqrt {\cot (c+d x)}}+\frac {i A-B}{2 d \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))}-\frac {((3+5 i) A-(5-7 i) B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a d}-\frac {((3-5 i) A+(5+7 i) B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a d} \\ & = -\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (i A-B)}{2 a d \sqrt {\cot (c+d x)}}+\frac {i A-B}{2 d \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))}-\frac {((3+5 i) A-(5-7 i) B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{8 a d}-\frac {((3+5 i) A-(5-7 i) B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{8 a d}+\frac {((3-5 i) A+(5+7 i) B) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a d}+\frac {((3-5 i) A+(5+7 i) B) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{8 \sqrt {2} a d} \\ & = -\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (i A-B)}{2 a d \sqrt {\cot (c+d x)}}+\frac {i A-B}{2 d \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))}+\frac {((3-5 i) A+(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d}-\frac {((3-5 i) A+(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d}-\frac {((3+5 i) A-(5-7 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{4 \sqrt {2} a d}+\frac {((3+5 i) A-(5-7 i) B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{4 \sqrt {2} a d} \\ & = \frac {((3+5 i) A-(5-7 i) B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {((3+5 i) A-(5-7 i) B) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{4 \sqrt {2} a d}-\frac {3 A+7 i B}{6 a d \cot ^{\frac {3}{2}}(c+d x)}-\frac {5 (i A-B)}{2 a d \sqrt {\cot (c+d x)}}+\frac {i A-B}{2 d \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))}+\frac {((3-5 i) A+(5+7 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d}-\frac {((3-5 i) A+(5+7 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{8 \sqrt {2} a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.57 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.58 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-3 (-1)^{3/4} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (-i+\tan (c+d x))+6 \sqrt [4]{-1} (-2 i A+3 B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (-i+\tan (c+d x))+\sqrt {\tan (c+d x)} \left (-15 (A+i B)+4 (-3 i A+2 B) \tan (c+d x)-4 i B \tan ^2(c+d x)\right )\right )}{6 a d (-i+\tan (c+d x))} \]

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-3*(-1)^(3/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(-I + Ta
n[c + d*x]) + 6*(-1)^(1/4)*((-2*I)*A + 3*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(-I + Tan[c + d*x]) + Sqrt[
Tan[c + d*x]]*(-15*(A + I*B) + 4*((-3*I)*A + 2*B)*Tan[c + d*x] - (4*I)*B*Tan[c + d*x]^2)))/(6*a*d*(-I + Tan[c
+ d*x]))

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.52

method	result	size
derivativedivides	\(\frac {\frac {4 \left (-\frac {i A}{4}-\frac {B}{4}\right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}+\frac {-2 i A +2 B}{\sqrt {\cot \left (d x +c \right )}}-\frac {2 i B}{3 \cot \left (d x +c \right )^{\frac {3}{2}}}-\frac {i \left (-\frac {i \left (i A -B \right ) \sqrt {\cot \left (d x +c \right )}}{i+\cot \left (d x +c \right )}+\frac {4 \left (3 i B +2 A \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2}}{a d}\)	\(161\)
default	\(\frac {\frac {4 \left (-\frac {i A}{4}-\frac {B}{4}\right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}+\frac {-2 i A +2 B}{\sqrt {\cot \left (d x +c \right )}}-\frac {2 i B}{3 \cot \left (d x +c \right )^{\frac {3}{2}}}-\frac {i \left (-\frac {i \left (i A -B \right ) \sqrt {\cot \left (d x +c \right )}}{i+\cot \left (d x +c \right )}+\frac {4 \left (3 i B +2 A \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}\right )}{2}}{a d}\)	\(161\)

[In]

int((A+B*tan(d*x+c))/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/a/d*(4*(-1/4*I*A-1/4*B)/(2^(1/2)-I*2^(1/2))*arctan(2*cot(d*x+c)^(1/2)/(2^(1/2)-I*2^(1/2)))+2*(-I*A+B)/cot(d*
x+c)^(1/2)-2/3*I*B/cot(d*x+c)^(3/2)-1/2*I*(-I*(I*A-B)*cot(d*x+c)^(1/2)/(I+cot(d*x+c))+4*(2*A+3*I*B)/(2^(1/2)+I
*2^(1/2))*arctan(2*cot(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 794 vs. \(2 (222) = 444\).

Time = 0.27 (sec) , antiderivative size = 794, normalized size of antiderivative = 2.59 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=-\frac {3 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a 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 \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) - 3 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {2 \, {\left ({\left (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a 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 \, A^{2} + 2 \, A B - i \, B^{2}}{a^{2} d^{2}}} + {\left (A - i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{i \, A + B}\right ) + 6 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} \log \left (-\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a 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 {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} + 2 i \, A - 3 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) - 6 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {\frac {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} \log \left (\frac {{\left ({\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a 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 {-4 i \, A^{2} + 12 \, A B + 9 i \, B^{2}}{a^{2} d^{2}}} - 2 i \, A + 3 \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + 2 \, {\left ({\left (27 \, A + 19 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (3 \, A + 19 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (27 \, A + 35 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\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}}}{24 \, {\left (a d e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \]

[In]

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

[Out]

-1/24*(3*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2 + 2*A*B -
I*B^2)/(a^2*d^2))*log(-2*((I*a*d*e^(2*I*d*x + 2*I*c) - I*a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x +
2*I*c) - 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2)) + (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*
A + B)) - 3*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((I*A^2 + 2*A*
B - I*B^2)/(a^2*d^2))*log(-2*((-I*a*d*e^(2*I*d*x + 2*I*c) + I*a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*
x + 2*I*c) - 1))*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^2*d^2)) + (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)
/(I*A + B)) + 6*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((-4*I*A^2
+ 12*A*B + 9*I*B^2)/(a^2*d^2))*log(-((a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I
*d*x + 2*I*c) - 1))*sqrt((-4*I*A^2 + 12*A*B + 9*I*B^2)/(a^2*d^2)) + 2*I*A - 3*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) -
6*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*sqrt((-4*I*A^2 + 12*A*B + 9
*I*B^2)/(a^2*d^2))*log(((a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c)
- 1))*sqrt((-4*I*A^2 + 12*A*B + 9*I*B^2)/(a^2*d^2)) - 2*I*A + 3*B)*e^(-2*I*d*x - 2*I*c)/(a*d)) + 2*((27*A + 19
*I*B)*e^(6*I*d*x + 6*I*c) + (3*A + 19*I*B)*e^(4*I*d*x + 4*I*c) - (27*A + 35*I*B)*e^(2*I*d*x + 2*I*c) - 3*A - 3
*I*B)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x
+ 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))

Sympy [F]

\[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=- \frac {i \left (\int \frac {A}{\tan {\left (c + d x \right )} \cot ^{\frac {5}{2}}{\left (c + d x \right )} - i \cot ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \tan {\left (c + d x \right )}}{\tan {\left (c + d x \right )} \cot ^{\frac {5}{2}}{\left (c + d x \right )} - i \cot ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx\right )}{a} \]

[In]

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

[Out]

-I*(Integral(A/(tan(c + d*x)*cot(c + d*x)**(5/2) - I*cot(c + d*x)**(5/2)), x) + Integral(B*tan(c + d*x)/(tan(c
+ d*x)*cot(c + d*x)**(5/2) - I*cot(c + d*x)**(5/2)), x))/a

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F]

\[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\int { \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )} \cot \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]

[In]

int((A + B*tan(c + d*x))/(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)),x)

[Out]

int((A + B*tan(c + d*x))/(cot(c + d*x)^(5/2)*(a + a*tan(c + d*x)*1i)), x)