3.736 \(\int \frac{\sqrt{\cot (c+d x)}}{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=200 \[ \frac{\sqrt{\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a d}-\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a d} \]

[Out]

((3/4 - I/4)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a*d) - ((3/4 - I/4)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[
c + d*x]]])/(Sqrt[2]*a*d) + Sqrt[Cot[c + d*x]]/(2*d*(I*a + a*Cot[c + d*x])) - ((3/8 + I/8)*Log[1 - Sqrt[2]*Sqr
t[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a*d) + ((3/8 + I/8)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x
]])/(Sqrt[2]*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.204401, antiderivative size = 200, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {3673, 3550, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{\sqrt{\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a d}+\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a d}-\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3/4 - I/4)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a*d) - ((3/4 - I/4)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[
c + d*x]]])/(Sqrt[2]*a*d) + Sqrt[Cot[c + d*x]]/(2*d*(I*a + a*Cot[c + d*x])) - ((3/8 + I/8)*Log[1 - Sqrt[2]*Sqr
t[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a*d) + ((3/8 + I/8)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x
]])/(Sqrt[2]*a*d)

Rule 3673

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

Rule 3550

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*(c + d*Tan[e + f*x])^(n - 1))/(2*a*f*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a^2), Int[(c + d*Tan[e +
f*x])^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*Tan[e + f*x], x], 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] && GtQ[n, 1]

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+i a \tan (c+d x)} \, dx &=\int \frac{\cot ^{\frac{3}{2}}(c+d x)}{i a+a \cot (c+d x)} \, dx\\ &=\frac{\sqrt{\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac{\int \frac{\frac{i a}{2}-\frac{3}{2} a \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx}{2 a^2}\\ &=\frac{\sqrt{\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{i a}{2}+\frac{3 a x^2}{2}}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 d}\\ &=\frac{\sqrt{\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}--\frac{\left (\frac{3}{4}+\frac{i}{4}\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a d}-\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a d}\\ &=\frac{\sqrt{\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac{\left (\frac{3}{8}-\frac{i}{8}\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a d}-\frac{\left (\frac{3}{8}-\frac{i}{8}\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a d}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{\sqrt{2} a d}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{\sqrt{2} a d}\\ &=\frac{\sqrt{\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a d}+\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a d}--\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a d}-\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a d}\\ &=\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a d}-\frac{\left (\frac{3}{4}-\frac{i}{4}\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a d}+\frac{\sqrt{\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a d}+\frac{\left (\frac{3}{8}+\frac{i}{8}\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a d}\\ \end{align*}

Mathematica [A]  time = 0.923598, size = 174, normalized size = 0.87 \[ \frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \sqrt{\sin (2 (c+d x))} \sqrt{\cot (c+d x)} \left (\frac{2-2 i}{\sqrt{\sin (2 (c+d x))}}+(1+2 i) \sec (c+d x) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )+(2-i) \csc (c+d x) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )-(1-2 i) \sin ^{-1}(\cos (c+d x)-\sin (c+d x)) (\csc (c+d x)+i \sec (c+d x))\right )}{a d (\cot (c+d x)+i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/8 + I/8)*Sqrt[Cot[c + d*x]]*((2 - I)*Csc[c + d*x]*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]
] - (1 - 2*I)*ArcSin[Cos[c + d*x] - Sin[c + d*x]]*(Csc[c + d*x] + I*Sec[c + d*x]) + (1 + 2*I)*Log[Cos[c + d*x]
+ Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sec[c + d*x] + (2 - 2*I)/Sqrt[Sin[2*(c + d*x)]])*Sqrt[Sin[2*(c + d*x
)]])/(a*d*(I + Cot[c + d*x]))

________________________________________________________________________________________

Maple [C]  time = 0.347, size = 711, normalized size = 3.6 \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+I*a*tan(d*x+c)),x)

[Out]

1/4/a/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)*(I*((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))*sin(d*x+c)-2*I*((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))*sin(d*x+c)-I*cos(d*x+c)^2*sin(d*x+c)*2^(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)*((cos(d*x+c)-1)/si
n(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)-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)*(-(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)+3*(-(
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))*sin(d*x+c)+I*sin(d*x+c)*cos(d
*x+c)*2^(1/2)+2^(1/2)*cos(d*x+c)^3-2^(1/2)*cos(d*x+c)^2)/sin(d*x+c)^3/cos(d*x+c)

________________________________________________________________________________________

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+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [B]  time = 1.36098, size = 1297, normalized size = 6.48 \begin{align*} \frac{{\left (a d \sqrt{-\frac{i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left ({\left (4 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - 4 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}{4 \, a^{2} d^{2}}} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - a d \sqrt{-\frac{i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left ({\left ({\left (-4 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 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}{4 \, a^{2} d^{2}}} + 2 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - a d \sqrt{\frac{i}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \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{i}{a^{2} d^{2}}} + 1\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a d}\right ) + a d \sqrt{\frac{i}{a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \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{i}{a^{2} d^{2}}} - 1\right )} 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}}{\left (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a*d*sqrt(-1/4*I/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(((4*I*a*d*e^(2*I*d*x + 2*I*c) - 4*I*a*d)*sqrt((I*e^(2*
I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/4*I/(a^2*d^2)) + 2*I*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x -
2*I*c)) - a*d*sqrt(-1/4*I/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*log(((-4*I*a*d*e^(2*I*d*x + 2*I*c) + 4*I*a*d)*sqrt((
I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/4*I/(a^2*d^2)) + 2*I*e^(2*I*d*x + 2*I*c))*e^(-2*
I*d*x - 2*I*c)) - a*d*sqrt(I/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*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(I/(a^2*d^2)) + 1)*e^(-2*I*d*x - 2*I*c)/(a*d)) + a*d*sqrt(
I/(a^2*d^2))*e^(2*I*d*x + 2*I*c)*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(I/(a^2*d^2)) - 1)*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))*(-I*e^(2*I*d*x + 2*I*c) + I))*e^(-2*I*d*x - 2*I*c)/(a*d)

________________________________________________________________________________________

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+I*a*tan(d*x+c)),x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cot \left (d x + c\right )}}{i \, a \tan \left (d x + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(cot(d*x + c))/(I*a*tan(d*x + c) + a), x)