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

Optimal. Leaf size=232 \[ \frac{5 \sqrt{\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}-\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}-\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{4 d (a \cot (c+d x)+i a)^2} \]

[Out]

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

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Rubi [A]  time = 0.330622, antiderivative size = 232, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {3673, 3558, 3595, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{5 \sqrt{\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}-\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (\cot (c+d x)-\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (\cot (c+d x)+\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}-\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\cot (c+d x)}+1\right )}{\sqrt{2} a^2 d}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{4 d (a \cot (c+d x)+i a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((9/16 - (5*I)/16)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) - ((9/16 - (5*I)/16)*ArcTan[1 + Sqr
t[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + (5*Sqrt[Cot[c + d*x]])/(8*a^2*d*(I + Cot[c + d*x])) + Cot[c + d*x]
^(3/2)/(4*d*(I*a + a*Cot[c + d*x])^2) - ((9/32 + (5*I)/32)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])
/(Sqrt[2]*a^2*d) + ((9/32 + (5*I)/32)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(Sqrt[2]*a^2*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 3558

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1))/(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 - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1))
- d*(b*d*(m - n + 1) - a*c*(m + n - 1))*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] && LtQ[m, 0] && GtQ[n, 1] && (IntegerQ[m] || IntegersQ[2*m,
2*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+i a \tan (c+d x))^2} \, dx &=\int \frac{\cot ^{\frac{5}{2}}(c+d x)}{(i a+a \cot (c+d x))^2} \, dx\\ &=\frac{\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 i a}{2}+\frac{7}{2} a \cot (c+d x)\right )}{i a+a \cot (c+d x)} \, dx}{4 a^2}\\ &=\frac{5 \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac{\int \frac{-\frac{5 i a^2}{2}+\frac{9}{2} a^2 \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx}{8 a^4}\\ &=\frac{5 \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+\frac{\operatorname{Subst}\left (\int \frac{\frac{5 i a^2}{2}-\frac{9 a^2 x^2}{2}}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{4 a^4 d}\\ &=\frac{5 \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+-\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\cot (c+d x)}\right )}{a^2 d}+\frac{\left (\frac{9}{16}+\frac{5 i}{16}\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 \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}+-\frac{\left (\frac{9}{32}-\frac{5 i}{32}\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 (\frac{9}{32}-\frac{5 i}{32}\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 (\frac{9}{32}+\frac{5 i}{32}\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^2 d}+-\frac{\left (\frac{9}{32}+\frac{5 i}{32}\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^2 d}\\ &=\frac{5 \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a^2 d}+-\frac{\left (\frac{9}{16}-\frac{5 i}{16}\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{\left (\frac{9}{16}-\frac{5 i}{16}\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{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}-\frac{\left (\frac{9}{16}-\frac{5 i}{16}\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\cot (c+d x)}\right )}{\sqrt{2} a^2 d}+\frac{5 \sqrt{\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac{\cot ^{\frac{3}{2}}(c+d x)}{4 d (i a+a \cot (c+d x))^2}-\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (1-\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a^2 d}+\frac{\left (\frac{9}{32}+\frac{5 i}{32}\right ) \log \left (1+\sqrt{2} \sqrt{\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt{2} a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.879167, size = 232, normalized size = 1. \[ \frac{\cot ^{\frac{3}{2}}(c+d x) \csc (c+d x) \sec ^2(c+d x) \left (7 \sin (c+d x)+7 \sin (3 (c+d x))+5 i \cos (c+d x)-5 i \cos (3 (c+d x))-(5+9 i) \sqrt{\sin (2 (c+d x))} \sin ^{-1}(\cos (c+d x)-\sin (c+d x)) (\sin (2 (c+d x))-i \cos (2 (c+d x)))+(-5+9 i) \sin ^{\frac{3}{2}}(2 (c+d x)) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )+(9+5 i) \sqrt{\sin (2 (c+d x))} \cos (2 (c+d x)) \log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )\right )}{32 a^2 d (\cot (c+d x)+i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[c + d*x]^(3/2)*Csc[c + d*x]*Sec[c + d*x]^2*((5*I)*Cos[c + d*x] - (5*I)*Cos[3*(c + d*x)] + 7*Sin[c + d*x]
+ (9 + 5*I)*Cos[2*(c + d*x)]*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]]
- (5 - 9*I)*Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sin[2*(c + d*x)]^(3/2) - (5 + 9*I)*ArcSi
n[Cos[c + d*x] - Sin[c + d*x]]*Sqrt[Sin[2*(c + d*x)]]*((-I)*Cos[2*(c + d*x)] + Sin[2*(c + d*x)]) + 7*Sin[3*(c
+ d*x)]))/(32*a^2*d*(I + Cot[c + d*x])^2)

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Maple [C]  time = 0.376, size = 778, normalized size = 3.4 \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))^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)*(4*I*2^(1/2)*cos(d*x+c)^4*si
n(d*x+c)-4*I*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)-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)+7*I*sin(d*x+c)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+s
in(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))-4*2^(1/2)*cos(d*x+c)^5+5*I*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+4*cos(
d*x+c)^4*2^(1/2)+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)/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)+7*((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)-9*(-(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)*((co
s(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)-
5*I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-3*2^(1/2)*cos(d*x+c)^3+3*2^(1/2)*cos(d*x+c)^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+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [B]  time = 1.43151, size = 1442, normalized size = 6.22 \begin{align*} \frac{{\left (4 \, a^{2} d \sqrt{-\frac{i}{16 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left ({\left (8 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, 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}{16 \, a^{4} 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 ) - 4 \, a^{2} d \sqrt{-\frac{i}{16 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left ({\left (-8 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, 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}{16 \, a^{4} 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 ) - 4 \, a^{2} d \sqrt{\frac{49 i}{64 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-\frac{{\left (8 \,{\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{49 i}{64 \, a^{4} d^{2}}} + 7\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 4 \, a^{2} d \sqrt{\frac{49 i}{64 \, a^{4} d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{{\left (8 \,{\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{49 i}{64 \, a^{4} d^{2}}} - 7\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, 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}}{\left (-6 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{16 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

1/16*(4*a^2*d*sqrt(-1/16*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(((8*I*a^2*d*e^(2*I*d*x + 2*I*c) - 8*I*a^2*d)*sqr
t((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/16*I/(a^4*d^2)) + 2*I*e^(2*I*d*x + 2*I*c))*e^
(-2*I*d*x - 2*I*c)) - 4*a^2*d*sqrt(-1/16*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(((-8*I*a^2*d*e^(2*I*d*x + 2*I*c)
+ 8*I*a^2*d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/16*I/(a^4*d^2)) + 2*I*e^(2*I
*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) - 4*a^2*d*sqrt(49/64*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(-1/8*(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(49/64*I/(a^4*d^2
)) + 7)*e^(-2*I*d*x - 2*I*c)/(a^2*d)) + 4*a^2*d*sqrt(49/64*I/(a^4*d^2))*e^(4*I*d*x + 4*I*c)*log(1/8*(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(49/64*I/(a^4*d^2
)) - 7)*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))*(-6*I*e^(4
*I*d*x + 4*I*c) + 5*I*e^(2*I*d*x + 2*I*c) + I))*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+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{\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+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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