\(\int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx\) [737]

Optimal result

Integrand size = 26, antiderivative size = 68 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\frac {\sqrt [4]{-1} \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{2 a d}+\frac {i \sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))} \]

[Out]

1/2*(-1)^(1/4)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/a/d+1/2*I*cot(d*x+c)^(1/2)/d/(I*a+a*cot(d*x+c))

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3754, 3630, 3614, 214} \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\frac {\sqrt [4]{-1} \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{2 a d}+\frac {i \sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)} \]

[In]

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

[Out]

((-1)^(1/4)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/(2*a*d) + ((I/2)*Sqrt[Cot[c + d*x]])/(d*(I*a + a*Cot[c + d
*x]))

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3614

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2*(c^2/f), S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3630

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
a*c + b*d))*((c + d*Tan[e + f*x])^n/(2*(b*c - a*d)*f*(a + b*Tan[e + f*x]))), x] + Dist[1/(2*a*(b*c - a*d)), In
t[(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c*d*(n - 1) + b*c^2 + b*d^2*n - d*(b*c - a*d)*(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[0
, 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*} \text {integral}& = \int \frac {\sqrt {\cot (c+d x)}}{i a+a \cot (c+d x)} \, dx \\ & = \frac {i \sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {\int \frac {-\frac {a}{2}+\frac {1}{2} i a \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{2 a^2} \\ & = \frac {i \sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))}-\frac {\text {Subst}\left (\int \frac {1}{\frac {a}{2}+\frac {1}{2} i a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 d} \\ & = \frac {\sqrt [4]{-1} \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{2 a d}+\frac {i \sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (i \sqrt {2} \arctan \left (\frac {(1+i) \sqrt {\cot (c+d x)}}{\sqrt {2}}\right )+\frac {(1+i) \sqrt {\cot (c+d x)}}{i+\cot (c+d x)}\right )}{a d} \]

[In]

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

[Out]

((1/4 + I/4)*(I*Sqrt[2]*ArcTan[((1 + I)*Sqrt[Cot[c + d*x]])/Sqrt[2]] + ((1 + I)*Sqrt[Cot[c + d*x]])/(I + Cot[c
+ d*x])))/(a*d)

Maple [A] (verified)

Time = 1.62 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(-\frac {-\frac {i \arctan \left (\frac {2 \left (\sqrt {\cot }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}-\frac {i \left (\sqrt {\cot }\left (d x +c \right )\right )}{2 \left (i+\cot \left (d x +c \right )\right )}}{a d}\)	\(70\)
default	\(-\frac {-\frac {i \arctan \left (\frac {2 \left (\sqrt {\cot }\left (d x +c \right )\right )}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}-\frac {i \left (\sqrt {\cot }\left (d x +c \right )\right )}{2 \left (i+\cot \left (d x +c \right )\right )}}{a d}\)	\(70\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (52) = 104\).

Time = 0.26 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.96 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=-\frac {{\left (a d \sqrt {\frac {i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (2 \, {\left (2 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{4 \, a^{2} d^{2}}} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - a d \sqrt {\frac {i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-2 \, {\left (2 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{4 \, a^{2} d^{2}}} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \]

[In]

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

Sympy [F]

\[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=- \frac {i \int \frac {1}{\tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}} - i \sqrt {\cot {\left (c + d x \right )}}}\, dx}{a} \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F]

\[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )} \sqrt {\cot \left (d x + c\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\int \frac {1}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \]

[In]

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

[Out]

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