\(\int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx\) [528]

Optimal result

Integrand size = 36, antiderivative size = 284 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((2+i) A+(7-2 i) B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {((1+3 i) A+(9+5 i) B) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {(A+5 i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(i A-B) \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {((1-3 i) A-(9-5 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+2 i) A+(2-7 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^2 d} \]

[Out]

(1/32+1/32*I)*((2+I)*A+(7-2*I)*B)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a^2/d*2^(1/2)+1/32*((1+3*I)*A+(9+5*I)*B)
*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/a^2/d*2^(1/2)+1/64*((1-3*I)*A+(-9+5*I)*B)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+
c)^(1/2))/a^2/d*2^(1/2)+(1/64+1/64*I)*((1+2*I)*A+(2-7*I)*B)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/a^2/d*2^
(1/2)+1/8*(A+5*I*B)*cot(d*x+c)^(1/2)/a^2/d/(I+cot(d*x+c))+1/4*(I*A-B)*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))^2

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3662, 3677, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {\left (\frac {1}{16}+\frac {i}{16}\right ) ((2+i) A+(7-2 i) B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {((1+3 i) A+(9+5 i) B) \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{16 \sqrt {2} a^2 d}+\frac {(A+5 i B) \sqrt {\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac {((1-3 i) A-(9-5 i) B) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{32 \sqrt {2} a^2 d}+\frac {\left (\frac {1}{32}+\frac {i}{32}\right ) ((1+2 i) A+(2-7 i) B) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {(-B+i A) \sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2} \]

[In]

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

[Out]

((-1/16 - I/16)*((2 + I)*A + (7 - 2*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + (((1 + 3*I
)*A + (9 + 5*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(16*Sqrt[2]*a^2*d) + ((A + (5*I)*B)*Sqrt[Cot[c + d*
x]])/(8*a^2*d*(I + Cot[c + d*x])) + ((I*A - B)*Sqrt[Cot[c + d*x]])/(4*d*(I*a + a*Cot[c + d*x])^2) + (((1 - 3*I
)*A - (9 - 5*I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(32*Sqrt[2]*a^2*d) + ((1/32 + I/32)*((1
+ 2*I)*A + (2 - 7*I)*B)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^2*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 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)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2} \, dx \\ & = \frac {(i A-B) \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {\frac {1}{2} a (A-7 i B)-\frac {3}{2} a (i A-B) \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \, dx}{4 a^2} \\ & = \frac {(A+5 i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(i A-B) \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-\frac {3}{2} a^2 (i A+3 B)-\frac {1}{2} a^2 (A+5 i B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{8 a^4} \\ & = \frac {(A+5 i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(i A-B) \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\text {Subst}\left (\int \frac {\frac {3}{2} a^2 (i A+3 B)+\frac {1}{2} a^2 (A+5 i B) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a^4 d} \\ & = \frac {(A+5 i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(i A-B) \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\left (\left (\frac {1}{16}+\frac {i}{16}\right ) ((1+2 i) A+(2-7 i) B)\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 d}+\frac {((1+3 i) A+(9+5 i) B) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{16 a^2 d} \\ & = \frac {(A+5 i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(i A-B) \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {((1-3 i) A-(9-5 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^2 d}+\frac {((1-3 i) A-(9-5 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^2 d}+\frac {((1+3 i) A+(9+5 i) B) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{32 a^2 d}+\frac {((1+3 i) A+(9+5 i) B) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{32 a^2 d} \\ & = \frac {(A+5 i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(i A-B) \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {((1-3 i) A-(9-5 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1-3 i) A-(9-5 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}+\frac {((1+3 i) A+(9+5 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^2 d}-\frac {((1+3 i) A+(9+5 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^2 d} \\ & = -\frac {((1+3 i) A+(9+5 i) B) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {((1+3 i) A+(9+5 i) B) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{16 \sqrt {2} a^2 d}+\frac {(A+5 i B) \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {(i A-B) \sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {((1-3 i) A-(9-5 i) B) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d}-\frac {((1-3 i) A-(9-5 i) B) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt {2} a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.38 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.69 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-2 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))+\sqrt [4]{-1} (A-7 i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right ) \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-\sqrt {\tan (c+d x)} (A+5 i B+(3 i A-7 B) \tan (c+d x))\right )}{8 a^2 d (-i+\tan (c+d x))^2} \]

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-2*(-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]*Sec[c +
d*x]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) + (-1)^(1/4)*(A - (7*I)*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]
]]*Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]) - Sqrt[Tan[c + d*x]]*(A + (5*I)*B + ((3*I)*A - 7*B)*
Tan[c + d*x])))/(8*a^2*d*(-I + Tan[c + d*x])^2)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.51

method	result	size
derivativedivides	\(\frac {-\frac {\left (-\frac {5 i B}{2}-\frac {A}{2}\right ) \cot \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {7 B}{2}-\frac {3 i A}{2}\right ) \sqrt {\cot \left (d x +c \right )}}{4 \left (i+\cot \left (d x +c \right )\right )^{2}}-\frac {\left (-7 i B +A \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{4 \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{2 \left (\sqrt {2}-i \sqrt {2}\right )}}{a^{2} d}\)	\(144\)
default	\(\frac {-\frac {\left (-\frac {5 i B}{2}-\frac {A}{2}\right ) \cot \left (d x +c \right )^{\frac {3}{2}}+\left (\frac {7 B}{2}-\frac {3 i A}{2}\right ) \sqrt {\cot \left (d x +c \right )}}{4 \left (i+\cot \left (d x +c \right )\right )^{2}}-\frac {\left (-7 i B +A \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}+i \sqrt {2}}\right )}{4 \left (\sqrt {2}+i \sqrt {2}\right )}-\frac {i \left (i A +B \right ) \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{2 \left (\sqrt {2}-i \sqrt {2}\right )}}{a^{2} d}\)	\(144\)

[In]

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

[Out]

1/a^2/d*(-1/4*((-5/2*I*B-1/2*A)*cot(d*x+c)^(3/2)+(7/2*B-3/2*I*A)*cot(d*x+c)^(1/2))/(I+cot(d*x+c))^2-1/4*(-7*I*
B+A)/(2^(1/2)+I*2^(1/2))*arctan(2*cot(d*x+c)^(1/2)/(2^(1/2)+I*2^(1/2)))-1/2*I*(I*A+B)/(2^(1/2)-I*2^(1/2))*arct
an(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. 666 vs. \(2 (209) = 418\).

Time = 0.26 (sec) , antiderivative size = 666, normalized size of antiderivative = 2.35 \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=-\frac {{\left (2 \, a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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^{4} 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 ) - 2 \, a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {2 \, {\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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^{4} 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 ) + a^{2} d \sqrt {\frac {i \, A^{2} + 14 \, A B - 49 i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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} + 14 \, A B - 49 i \, B^{2}}{a^{4} d^{2}}} + A - 7 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) - a^{2} d \sqrt {\frac {i \, A^{2} + 14 \, A B - 49 i \, B^{2}}{a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac {{\left ({\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2} 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} + 14 \, A B - 49 i \, B^{2}}{a^{4} d^{2}}} - A + 7 i \, B\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 2 \, {\left (2 \, {\left (i \, A - 3 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (3 i \, A - 7 \, 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 (-4 i \, d x - 4 i \, c\right )}}{32 \, a^{2} d} \]

[In]

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

[Out]

-1/32*(2*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-2*((a^2*d*e^(2*I*d*x + 2*I*c)
- a^2*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^4*d^2))
+ (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/(I*A + B)) - 2*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^4
*d^2))*e^(4*I*d*x + 4*I*c)*log(2*((a^2*d*e^(2*I*d*x + 2*I*c) - a^2*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^4*d^2)) - (A - I*B)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*
I*c)/(I*A + B)) + a^2*d*sqrt((I*A^2 + 14*A*B - 49*I*B^2)/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-1/8*((a^2*d*e^(2*
I*d*x + 2*I*c) - a^2*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 + 14*A*B - 49*
I*B^2)/(a^4*d^2)) + A - 7*I*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) - a^2*d*sqrt((I*A^2 + 14*A*B - 49*I*B^2)/(a^4*d^2
))*e^(4*I*d*x + 4*I*c)*log(1/8*((a^2*d*e^(2*I*d*x + 2*I*c) - a^2*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 + 14*A*B - 49*I*B^2)/(a^4*d^2)) - A + 7*I*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + 2*(
2*(I*A - 3*B)*e^(4*I*d*x + 4*I*c) - (3*I*A - 7*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^(-4*I*d*x - 4*I*c)/(a^2*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^2} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]

[In]

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

[Out]

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