\(\int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\) [531]

Optimal result

Integrand size = 36, antiderivative size = 318 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {((-7+5 i) A+2 i B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((-7+5 i) A+2 i B) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {(4 A+i B) \cot ^{\frac {3}{2}}(c+d x)}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 A \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}-\frac {((7+5 i) A-2 i B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d} \]

[Out]

1/6*(A+I*B)*cot(d*x+c)^(5/2)/d/(I*a+a*cot(d*x+c))^3+1/12*(4*A+I*B)*cot(d*x+c)^(3/2)/a/d/(I*a+a*cot(d*x+c))^2+1
/32*((-7+5*I)*A+2*I*B)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+1/32*((-7+5*I)*A+2*I*B)*arctan(1+2^(1
/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)-1/64*((7+5*I)*A-2*I*B)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(
1/2)+1/64*((7+5*I)*A-2*I*B)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a^3/d*2^(1/2)+5/8*A*cot(d*x+c)^(1/2)/d/(
I*a^3+a^3*cot(d*x+c))

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3662, 3676, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {(2 i B-(7-5 i) A) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {(2 i B-(7-5 i) A) \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{16 \sqrt {2} a^3 d}-\frac {((7+5 i) A-2 i B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^3 d}+\frac {5 A \sqrt {\cot (c+d x)}}{8 d \left (a^3 \cot (c+d x)+i a^3\right )}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (a \cot (c+d x)+i a)^3}+\frac {(4 A+i B) \cot ^{\frac {3}{2}}(c+d x)}{12 a d (a \cot (c+d x)+i a)^2} \]

[In]

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

[Out]

-1/16*(((-7 + 5*I)*A + (2*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^3*d) + (((-7 + 5*I)*A + (2*
I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(16*Sqrt[2]*a^3*d) + ((A + I*B)*Cot[c + d*x]^(5/2))/(6*d*(I*a +
a*Cot[c + d*x])^3) + ((4*A + I*B)*Cot[c + d*x]^(3/2))/(12*a*d*(I*a + a*Cot[c + d*x])^2) + (5*A*Sqrt[Cot[c + d*
x]])/(8*d*(I*a^3 + a^3*Cot[c + d*x])) - (((7 + 5*I)*A - (2*I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c +
d*x]])/(32*Sqrt[2]*a^3*d) + (((7 + 5*I)*A - (2*I)*B)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(32*S
qrt[2]*a^3*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 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 3676

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cot ^{\frac {5}{2}}(c+d x) (B+A \cot (c+d x))}{(i a+a \cot (c+d x))^3} \, dx \\ & = \frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {\int \frac {\cot ^{\frac {3}{2}}(c+d x) \left (-\frac {5}{2} a (i A-B)+\frac {1}{2} a (11 A-i B) \cot (c+d x)\right )}{(i a+a \cot (c+d x))^2} \, dx}{6 a^2} \\ & = \frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {(4 A+i B) \cot ^{\frac {3}{2}}(c+d x)}{12 a d (i a+a \cot (c+d x))^2}+\frac {\int \frac {\sqrt {\cot (c+d x)} \left (-3 a^2 (4 i A-B)+3 a^2 (6 A-i B) \cot (c+d x)\right )}{i a+a \cot (c+d x)} \, dx}{24 a^4} \\ & = \frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {(4 A+i B) \cot ^{\frac {3}{2}}(c+d x)}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 A \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\int \frac {-15 i a^3 A+3 a^3 (7 A-2 i B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{48 a^6} \\ & = \frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {(4 A+i B) \cot ^{\frac {3}{2}}(c+d x)}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 A \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {\text {Subst}\left (\int \frac {15 i a^3 A-3 a^3 (7 A-2 i B) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{24 a^6 d} \\ & = \frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {(4 A+i B) \cot ^{\frac {3}{2}}(c+d x)}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 A \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}+\frac {((7+5 i) A-2 i B) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^3 d}+\frac {((-7+5 i) A+2 i B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^3 d} \\ & = \frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {(4 A+i B) \cot ^{\frac {3}{2}}(c+d x)}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 A \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}-\frac {((7+5 i) A-2 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 )}{32 \sqrt {2} a^3 d}-\frac {((7+5 i) A-2 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 )}{32 \sqrt {2} a^3 d}+\frac {((-7+5 i) A+2 i B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{32 a^3 d}+\frac {((-7+5 i) A+2 i B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{32 a^3 d} \\ & = \frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {(4 A+i B) \cot ^{\frac {3}{2}}(c+d x)}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 A \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}-\frac {((7+5 i) A-2 i B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((-7+5 i) A+2 i B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}-\frac {((-7+5 i) A+2 i B) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d} \\ & = -\frac {((-7+5 i) A+2 i B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {((-7+5 i) A+2 i B) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^3 d}+\frac {(A+i B) \cot ^{\frac {5}{2}}(c+d x)}{6 d (i a+a \cot (c+d x))^3}+\frac {(4 A+i B) \cot ^{\frac {3}{2}}(c+d x)}{12 a d (i a+a \cot (c+d x))^2}+\frac {5 A \sqrt {\cot (c+d x)}}{8 d \left (i a^3+a^3 \cot (c+d x)\right )}-\frac {((7+5 i) A-2 i B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d}+\frac {((7+5 i) A-2 i B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.59 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=-\frac {i \sqrt {\cot (c+d x)} \sec ^3(c+d x) \left (12 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))+12 \sqrt [4]{-1} (6 A-i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) (\cos (3 (c+d x))+i \sin (3 (c+d x)))-4 \cos (c+d x) (6 A+3 i B+3 (7 A+i B) \cos (2 (c+d x))+(19 i A-B) \sin (2 (c+d x))) \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)}}{96 a^3 d (-i+\tan (c+d x))^3} \]

[In]

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

[Out]

((-1/96*I)*Sqrt[Cot[c + d*x]]*Sec[c + d*x]^3*(12*(-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(C
os[3*(c + d*x)] + I*Sin[3*(c + d*x)]) + 12*(-1)^(1/4)*(6*A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*(Cos[
3*(c + d*x)] + I*Sin[3*(c + d*x)]) - 4*Cos[c + d*x]*(6*A + (3*I)*B + 3*(7*A + I*B)*Cos[2*(c + d*x)] + ((19*I)*
A - B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])*Sqrt[Tan[c + d*x]])/(a^3*d*(-I + Tan[c + d*x])^3)

Maple [A] (verified)

Time = 0.45 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.51

method	result	size
derivativedivides	\(\frac {\frac {i \left (\frac {-i \left (2 i B +9 A \right ) \cot \left (d x +c \right )^{\frac {5}{2}}+\left (\frac {2 i B}{3}+\frac {38 A}{3}\right ) \cot \left (d x +c \right )^{\frac {3}{2}}+5 i A \sqrt {\cot \left (d x +c \right )}}{\left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {2 \left (6 i A +B \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}\right )}{8}+\frac {4 \left (-\frac {A}{16}+\frac {i B}{16}\right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}}{a^{3} d}\)	\(162\)
default	\(\frac {\frac {i \left (\frac {-i \left (2 i B +9 A \right ) \cot \left (d x +c \right )^{\frac {5}{2}}+\left (\frac {2 i B}{3}+\frac {38 A}{3}\right ) \cot \left (d x +c \right )^{\frac {3}{2}}+5 i A \sqrt {\cot \left (d x +c \right )}}{\left (i+\cot \left (d x +c \right )\right )^{3}}+\frac {2 \left (6 i A +B \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{\sqrt {2}+i \sqrt {2}}\right )}{8}+\frac {4 \left (-\frac {A}{16}+\frac {i B}{16}\right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}}{a^{3} d}\)	\(162\)

[In]

int(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/a^3/d*(1/8*I*((-I*(2*I*B+9*A)*cot(d*x+c)^(5/2)+(2/3*I*B+38/3*A)*cot(d*x+c)^(3/2)+5*I*A*cot(d*x+c)^(1/2))/(I+
cot(d*x+c))^3+2*(6*I*A+B)/(2^(1/2)+I*2^(1/2))*arctan(2*cot(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2))))+4*(-1/16*A+1/16*
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. 683 vs. \(2 (247) = 494\).

Time = 0.27 (sec) , antiderivative size = 683, normalized size of antiderivative = 2.15 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {{\left (3 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 \, A^{2} - 2 \, A B + i \, B^{2}}{a^{6} 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 \, a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 \, A^{2} - 2 \, A B + i \, B^{2}}{a^{6} 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 \, a^{3} d \sqrt {\frac {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} + 6 \, A - i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) + 3 \, a^{3} d \sqrt {\frac {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac {{\left ({\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 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 {36 i \, A^{2} + 12 \, A B - i \, B^{2}}{a^{6} d^{2}}} - 6 \, A + i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{3} d}\right ) - 2 \, {\left (2 \, {\left (10 i \, A - B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - {\left (14 i \, A + B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (5 i \, A - 2 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, A + 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}}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]

[In]

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

1/96*(3*a^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-2*((a^3*d*e^(2*I*d*x + 2*I*c)
- a^3*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^6*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^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^6*
d^2))*e^(6*I*d*x + 6*I*c)*log(2*((a^3*d*e^(2*I*d*x + 2*I*c) - a^3*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^6*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^3*d*sqrt((36*I*A^2 + 12*A*B - I*B^2)/(a^6*d^2))*e^(6*I*d*x + 6*I*c)*log(-1/8*((a^3*d*e^(2
*I*d*x + 2*I*c) - a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt((36*I*A^2 + 12*A*B -
I*B^2)/(a^6*d^2)) + 6*A - I*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) + 3*a^3*d*sqrt((36*I*A^2 + 12*A*B - I*B^2)/(a^6*
d^2))*e^(6*I*d*x + 6*I*c)*log(1/8*((a^3*d*e^(2*I*d*x + 2*I*c) - a^3*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*
I*d*x + 2*I*c) - 1))*sqrt((36*I*A^2 + 12*A*B - I*B^2)/(a^6*d^2)) - 6*A + I*B)*e^(-2*I*d*x - 2*I*c)/(a^3*d)) -
2*(2*(10*I*A - B)*e^(6*I*d*x + 6*I*c) - (14*I*A + B)*e^(4*I*d*x + 4*I*c) - (5*I*A - 2*B)*e^(2*I*d*x + 2*I*c) -
I*A + B)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-6*I*d*x - 6*I*c)/(a^3*d)

Sympy [F]

\[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \left (\int \frac {A \sqrt {\cot {\left (c + d x \right )}}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx + \int \frac {B \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx\right )}{a^{3}} \]

[In]

integrate(cot(d*x+c)**(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**3,x)

[Out]

I*(Integral(A*sqrt(cot(c + d*x))/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan(c + d*x) + I), x) + Integral(B
*tan(c + d*x)*sqrt(cot(c + d*x))/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan(c + d*x) + I), x))/a**3

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Not invertible Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx=\int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]

[In]

int((cot(c + d*x)^(1/2)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int((cot(c + d*x)^(1/2)*(A + B*tan(c + d*x)))/(a + a*tan(c + d*x)*1i)^3, x)