3.526 \(\int \frac{\sqrt{\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=284 \[ \frac{(5 A+i B) \sqrt{\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac{\left (\frac{1}{32}+\frac{i}{32}\right ) ((2+i) B-(7-2 i) A) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{((9+5 i) A-(1+3 i) B) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}+\frac{((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{16 \sqrt{2} a^2 d}+\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+2 i) B-(2-7 i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{4 d (a \cot (c+d x)+i a)^2} \]

[Out]

(((9 - 5*I)*A + (1 - 3*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(16*Sqrt[2]*a^2*d) + ((1/16 + I/16)*((-2
+ 7*I)*A + (1 + 2*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + ((5*A + I*B)*Sqrt[Cot[c + d*
x]])/(8*a^2*d*(I + Cot[c + d*x])) + ((A + I*B)*Cot[c + d*x]^(3/2))/(4*d*(I*a + a*Cot[c + d*x])^2) + ((1/32 + I
/32)*((-7 + 2*I)*A + (2 + I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^2*d) + (((9 + 5
*I)*A - (1 + 3*I)*B)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(32*Sqrt[2]*a^2*d)

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Rubi [A]  time = 0.61032, antiderivative size = 284, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3581, 3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{(5 A+i B) \sqrt{\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac{\left (\frac{1}{32}+\frac{i}{32}\right ) ((2+i) B-(7-2 i) A) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{((9+5 i) A-(1+3 i) B) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{32 \sqrt{2} a^2 d}+\frac{((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{16 \sqrt{2} a^2 d}+\frac{\left (\frac{1}{16}+\frac{i}{16}\right ) ((1+2 i) B-(2-7 i) A) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{4 d (a \cot (c+d x)+i a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((9 - 5*I)*A + (1 - 3*I)*B)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(16*Sqrt[2]*a^2*d) + ((1/16 + I/16)*((-2
+ 7*I)*A + (1 + 2*I)*B)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + ((5*A + I*B)*Sqrt[Cot[c + d*
x]])/(8*a^2*d*(I + Cot[c + d*x])) + ((A + I*B)*Cot[c + d*x]^(3/2))/(4*d*(I*a + a*Cot[c + d*x])^2) + ((1/32 + I
/32)*((-7 + 2*I)*A + (2 + I)*B)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^2*d) + (((9 + 5
*I)*A - (1 + 3*I)*B)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(32*Sqrt[2]*a^2*d)

Rule 3581

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 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 3534

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 1168

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 1162

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 617

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 204

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

Rule 1165

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 628

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]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cot (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\int \frac{\cot ^{\frac{3}{2}}(c+d x) (B+A \cot (c+d x))}{(i a+a \cot (c+d x))^2} \, dx\\ &=\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac{\int \frac{\sqrt{\cot (c+d x)} \left (-\frac{3}{2} a (i A-B)+\frac{1}{2} a (7 A-i B) \cot (c+d x)\right )}{i a+a \cot (c+d x)} \, dx}{4 a^2}\\ &=\frac{(5 A+i B) \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac{\int \frac{-\frac{1}{2} a^2 (5 i A-B)+\frac{3}{2} a^2 (3 A-i B) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(5 A+i B) \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} a^2 (5 i A-B)-\frac{3}{2} a^2 (3 A-i B) x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{4 a^4 d}\\ &=\frac{(5 A+i B) \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac{((9+5 i) A-(1+3 i) B) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{16 a^2 d}+\frac{\left (\left (\frac{1}{16}+\frac{i}{16}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 d}\\ &=\frac{(5 A+i B) \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac{((9+5 i) A-(1+3 i) B) \operatorname{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{((9+5 i) A-(1+3 i) B) \operatorname{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{\left (\left (\frac{1}{32}+\frac{i}{32}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 d}+\frac{\left (\left (\frac{1}{32}+\frac{i}{32}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 d}\\ &=\frac{(5 A+i B) \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac{((9+5 i) A-(1+3 i) B) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt{2} a^2 d}+\frac{((9+5 i) A-(1+3 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 (\left (\frac{1}{16}+\frac{i}{16}\right ) ((-2+7 i) A+(1+2 i) B)\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}+\frac{((9-5 i) A+(1-3 i) B) \operatorname{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{((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{16 \sqrt{2} a^2 d}-\frac{((9-5 i) A+(1-3 i) B) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{16 \sqrt{2} a^2 d}+\frac{(5 A+i B) \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{(A+i B) \cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac{((9+5 i) A-(1+3 i) B) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{32 \sqrt{2} a^2 d}+\frac{((9+5 i) A-(1+3 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]  time = 1.99752, size = 243, normalized size = 0.86 \[ \frac{\sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left (4 \cos (c+d x) (\cos (2 d x)-i \sin (2 d x)) ((-B+5 i A) \sin (c+d x)+(7 A+3 i B) \cos (c+d x))+(-\sin (2 c)+i \cos (2 c)) \sqrt{\sin (2 (c+d x))} \csc (c+d x) \left (((5+9 i) A+(3+i) B) \sin ^{-1}(\cos (c+d x)-\sin (c+d x))-(1+i) ((2+7 i) A+(1-2 i) B) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )\right )}{32 d \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*(4*Cos[c + d*x]*(Cos[2*d*x] - I*Sin[2*d*x])*((7*A + (3*I)*B)*Cos[c + d
*x] + ((5*I)*A - B)*Sin[c + d*x]) + Csc[c + d*x]*(((5 + 9*I)*A + (3 + I)*B)*ArcSin[Cos[c + d*x] - Sin[c + d*x]
] - (1 + I)*((2 + 7*I)*A + (1 - 2*I)*B)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*(I*Cos[2*c]
- Sin[2*c])*Sqrt[Sin[2*(c + d*x)]])*(A + B*Tan[c + d*x]))/(32*d*Sqrt[Cot[c + d*x]]*(A*Cos[c + d*x] + B*Sin[c
+ d*x])*(a + I*a*Tan[c + d*x])^2)

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Maple [C]  time = 0.592, size = 1517, normalized size = 5.3 \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)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x)

[Out]

-1/16/a^2/d*2^(1/2)*(cos(d*x+c)/sin(d*x+c))^(1/2)*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)*(B*2^(1/2)*cos(d*x+c)*sin(d*
x+c)+7*A*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*
x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(d*
x+c)-9*A*sin(d*x+c)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)
*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-5*I*A*
cos(d*x+c)*sin(d*x+c)*2^(1/2)+4*B*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)-3*A*cos(d*x+c)^3*2^(1/2)+3*A*2^(1/2)*cos(d*x
+c)^2+2*A*sin(d*x+c)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2
)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(
1/2))+4*A*cos(d*x+c)^4*2^(1/2)+B*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2
)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1
/2*I,1/2*2^(1/2))*sin(d*x+c)+4*I*A*cos(d*x+c)^4*sin(d*x+c)*2^(1/2)-4*I*A*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)-4*A*c
os(d*x+c)^5*2^(1/2)-B*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+3*I*B*sin(d*x+c)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))
^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-
1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-2*I*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1
+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*
x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+7*I*A*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)
-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(
d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))-2*I*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+
c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-si
n(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-I*B*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+
c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-si
n(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))+5*I*A*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)-2*B*sin(d*x+c)*(-(cos
(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c
))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-I*B*2^(1/2)*cos(d*x+c
)^2-4*I*B*cos(d*x+c)^5*2^(1/2)+4*I*B*cos(d*x+c)^4*2^(1/2)-4*B*cos(d*x+c)^4*sin(d*x+c)*2^(1/2)+I*B*cos(d*x+c)^3
*2^(1/2))/sin(d*x+c)^3/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)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.55716, size = 1715, normalized size = 6.04 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(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((49*I*A^2 + 14*A*B - 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((49*I*A^2 + 14*A*B - I
*B^2)/(a^4*d^2)) + 7*A - I*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + a^2*d*sqrt((49*I*A^2 + 14*A*B - 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((49*I*A^2 + 14*A*B - I*B^2)/(a^4*d^2)) - 7*A + I*B)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + 2*((
-6*I*A + 2*B)*e^(4*I*d*x + 4*I*c) + (5*I*A - 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)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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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 )} \sqrt{\cot \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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