3.8.41 \(\int \frac {\sqrt {\cot (c+d x)}}{(a+i a \tan (c+d x))^2} \, dx\) [741]

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

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 232, normalized size of antiderivative = 1.00, 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 = {3754, 3639, 3676, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (\frac {9}{16}-\frac {5 i}{16}\right ) \text {ArcTan}\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 ) \text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\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 {\cot ^{\frac {3}{2}}(c+d x)}{4 d (a \cot (c+d x)+i a)^2} \end {gather*}

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 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 3639

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-(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 3676

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 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))^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 {\text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \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^2 d}+-\frac {\left (\frac {9}{32}+\frac {5 i}{32}\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^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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.94, size = 232, normalized size = 1.00 \begin {gather*} \frac {\cot ^{\frac {3}{2}}(c+d x) \csc (c+d x) \sec ^2(c+d x) \left (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 \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sqrt {\sin (2 (c+d x))}-(5-9 i) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sin ^{\frac {3}{2}}(2 (c+d x))-(5+9 i) \text {ArcSin}(\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))\right )}{32 a^2 d (i+\cot (c+d x))^2} \end {gather*}

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] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 12.46, size = 778, normalized size = 3.35

method	result	size
default	\(\frac {\left (-4 i \sqrt {2}\, \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+2 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 )-7 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 )+4 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}+4 \sqrt {2}\, \left (\cos ^{5}\left (d x +c \right )\right )-5 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {2}-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 )-7 \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 )+9 \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 )-4 \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}+5 i \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+3 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}-3 \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}}{16 a^{2} d \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}}\)	\(778\)

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,method=_RETURNVERBOSE)

[Out]

1/16/a^2/d*(-4*I*cos(d*x+c)^4*sin(d*x+c)*2^(1/2)+2*I*sin(d*x+c)*((-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)*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*sin(d*x+c)*((-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)*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))+4*I*cos(d*x+c)^3*sin(d*x+c)*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)-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)*((-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)-7*((-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))+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))/s
in(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)-4*cos(d*x+c)^4*2^(1/2)+5*I*cos(d*x+c)*sin(d*x+c)*2^(1/2)+3*cos(d*x+c)^3*2^(1/2)-3*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))^2,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. 511 vs. \(2 (171) = 342\).
time = 0.53, size = 511, normalized size = 2.20 \begin {gather*} -\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 (-2 \, {\left (4 \, {\left (i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - 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}}} - 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 (-2 \, {\left (4 \, {\left (-i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + 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}}} - 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 {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))^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(-2*(4*(I*a^2*d*e^(2*I*d*x + 2*I*c) - I*a^2*d)*s
qrt((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)) - 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(-2*(4*(-I*a^2*d*e^(2*I*d*x + 2*I
*c) + 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)) - 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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\sqrt {\cot {\left (c + d x \right )}}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \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))**2,x)

[Out]

-Integral(sqrt(cot(c + d*x))/(tan(c + d*x)**2 - 2*I*tan(c + d*x) - 1), x)/a**2

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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))^2,x, algorithm="giac")

[Out]

integrate(sqrt(cot(d*x + c))/(I*a*tan(d*x + c) + a)^2, 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 )}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,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)^2,x)

[Out]

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

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