3.8.36 \(\int \frac {\sqrt {\cot (c+d x)}}{a+i a \tan (c+d x)} \, dx\) [736]

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

[Out]

(-3/8+1/8*I)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)+(-3/8+1/8*I)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/a
/d*2^(1/2)-(3/16+1/16*I)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)+(3/16+1/16*I)*ln(1+cot(d*x+c)+2
^(1/2)*cot(d*x+c)^(1/2))/a/d*2^(1/2)+1/2*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))

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Rubi [A]
time = 0.14, antiderivative size = 200, normalized size of antiderivative = 1.00, 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 = {3754, 3631, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a d}-\frac {\left (\frac {3}{4}-\frac {i}{4}\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a d}+\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} \end {gather*}

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 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 3631

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 3754

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]

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 {\text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 1.01, size = 174, normalized size = 0.87 \begin {gather*} \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \sqrt {\cot (c+d x)} \left ((2-i) \csc (c+d x) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right )-(1-2 i) \text {ArcSin}(\cos (c+d x)-\sin (c+d x)) (\csc (c+d x)+i \sec (c+d x))+(1+2 i) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sec (c+d x)+\frac {2-2 i}{\sqrt {\sin (2 (c+d x))}}\right ) \sqrt {\sin (2 (c+d x))}}{a d (i+\cot (c+d x))} \end {gather*}

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]))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 14.07, size = 710, normalized size = 3.55

method	result	size
default	\(-\frac {\left (-i \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c \right )+2 i \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}+\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c \right )+2 \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right ) \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticF \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (d x +c \right )-i \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-\left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+\sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\cos \left (d x +c \right )+1\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}}{4 a d \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}}\)	\(710\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 471 vs. \(2 (145) = 290\).
time = 0.49, size = 471, normalized size = 2.36 \begin {gather*} -\frac {{\left (a d \sqrt {-\frac {i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-2 \, {\left (2 \, {\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}{4 \, a^{2} d^{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 (-2 \, {\left (2 \, {\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}{4 \, a^{2} d^{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 {gather*}

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(-2*(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(-1/4*I/(a^2*d^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(-2*(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(-1/4*I/(a^2*d^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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {i \int \frac {\sqrt {\cot {\left (c + d x \right )}}}{\tan {\left (c + d x \right )} - i}\, dx}{a} \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}}{a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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