3.8.42 \(\int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx\) [742]

Optimal. Leaf size=234 \[ \frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {3 i \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {\sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{32}+\frac {3 i}{32}\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{\sqrt {2} a^2 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.22, antiderivative size = 234, 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, 3677, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {3 i \sqrt {\cot (c+d x)}}{8 a^2 d (\cot (c+d x)+i)}+\frac {\left (\frac {1}{32}+\frac {3 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 {1}{32}+\frac {3 i}{32}\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} a^2 d}+\frac {\sqrt {\cot (c+d x)}}{4 d (a \cot (c+d x)+i a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^2),x]

[Out]

((1/16 - (3*I)/16)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) - ((1/16 - (3*I)/16)*ArcTan[1 + Sqr
t[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*a^2*d) + (((3*I)/8)*Sqrt[Cot[c + d*x]])/(a^2*d*(I + Cot[c + d*x])) + Sqrt[C
ot[c + d*x]]/(4*d*(I*a + a*Cot[c + d*x])^2) + ((1/32 + (3*I)/32)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c +
d*x]])/(Sqrt[2]*a^2*d) - ((1/32 + (3*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 3677

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*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*
f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^2} \, dx &=\int \frac {\cot ^{\frac {3}{2}}(c+d x)}{(i a+a \cot (c+d x))^2} \, dx\\ &=\frac {\sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {-\frac {i a}{2}+\frac {5}{2} a \cot (c+d x)}{\sqrt {\cot (c+d x)} (i a+a \cot (c+d x))} \, dx}{4 a^2}\\ &=\frac {3 i \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {\sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\int \frac {\frac {a^2}{2}-\frac {3}{2} i a^2 \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{8 a^4}\\ &=\frac {3 i \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {\sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\text {Subst}\left (\int \frac {-\frac {a^2}{2}+\frac {3}{2} i a^2 x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a^4 d}\\ &=\frac {3 i \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {\sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+-\frac {\left (\frac {1}{16}+\frac {3 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 {1}{16}-\frac {3 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 {3 i \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {\sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+-\frac {\left (\frac {1}{32}-\frac {3 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 {1}{32}-\frac {3 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 {1}{32}+\frac {3 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 {1}{32}+\frac {3 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 {3 i \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {\sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\left (\frac {1}{32}+\frac {3 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 {1}{32}+\frac {3 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 {1}{16}-\frac {3 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 {1}{16}-\frac {3 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 {1}{16}-\frac {3 i}{16}\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}-\frac {\left (\frac {1}{16}-\frac {3 i}{16}\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} a^2 d}+\frac {3 i \sqrt {\cot (c+d x)}}{8 a^2 d (i+\cot (c+d x))}+\frac {\sqrt {\cot (c+d x)}}{4 d (i a+a \cot (c+d x))^2}+\frac {\left (\frac {1}{32}+\frac {3 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 {1}{32}+\frac {3 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 = 0.92, size = 224, normalized size = 0.96 \begin {gather*} \frac {\csc ^3(c+d x) \left (-\cos (c+d x)+\cos (3 (c+d x))+3 i \sin (c+d x)-(1+3 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))}+(3-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))-(3+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)))+3 i \sin (3 (c+d x))\right )}{32 a^2 d \sqrt {\cot (c+d x)} (i+\cot (c+d x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^2),x]

[Out]

(Csc[c + d*x]^3*(-Cos[c + d*x] + Cos[3*(c + d*x)] + (3*I)*Sin[c + d*x] - (1 + 3*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)]] + (3 - I)*Log[Cos[c + d*x] + Sin[c + d*
x] + Sqrt[Sin[2*(c + d*x)]]]*Sin[2*(c + d*x)]^(3/2) - (3 + I)*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)]) + (3*I)*Sin[3*(c + d*x)]))/(32*a^2*d*Sqrt[Cot[c + d*x]]*
(I + Cot[c + d*x])^2)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 13.38, size = 6915, normalized size = 29.55

method	result	size
default	\(\text {Expression too large to display}\)	\(6915\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

result too large to display

________________________________________________________________________________________

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(1/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.

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 509 vs. \(2 (171) = 342\).
time = 0.45, size = 509, normalized size = 2.18 \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 (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 {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 (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 {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}{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 {i}{64 \, a^{4} d^{2}}} + i\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a^{2} d}\right ) + 4 \, a^{2} d \sqrt {-\frac {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 {i}{64 \, a^{4} d^{2}}} - i\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 (2 \, e^{\left (4 i \, d x + 4 i \, c\right )} - e^{\left (2 i \, d x + 2 i \, c\right )} - 1\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(1/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*(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(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*(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(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(-1/64*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(-1/64*I/(a^4*d^2)) + I)*e^(-2*I*
d*x - 2*I*c)/(a^2*d)) + 4*a^2*d*sqrt(-1/64*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(-1/64*I/(a^4*d^2)) - I)*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))*(2*e^(4*I*d*x + 4*I*c) -
e^(2*I*d*x + 2*I*c) - 1))*e^(-4*I*d*x - 4*I*c)/(a^2*d)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {1}{\tan ^{2}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}} - 2 i \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}} - \sqrt {\cot {\left (c + d x \right )}}}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{\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(1/(cot(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)^2),x)

[Out]

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

________________________________________________________________________________________