Integrand size = 26, antiderivative size = 68 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\frac {\sqrt [4]{-1} \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{2 a d}+\frac {i \sqrt {\cot (c+d x)}}{2 d (i a+a \cot (c+d x))} \] Output:
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))
Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (i \sqrt {2} \arctan \left (\frac {(1+i) \sqrt {\cot (c+d x)}}{\sqrt {2}}\right )+\frac {(1+i) \sqrt {\cot (c+d x)}}{i+\cot (c+d x)}\right )}{a d} \] Input:
Integrate[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])),x]
Output:
((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)
Time = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 4156, 3042, 4032, 27, 3042, 4016, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))}dx\) |
\(\Big \downarrow \) 4156 |
\(\displaystyle \int \frac {\sqrt {\cot (c+d x)}}{a \cot (c+d x)+i a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}{-a \tan \left (c+d x+\frac {\pi }{2}\right )+i a}dx\) |
\(\Big \downarrow \) 4032 |
\(\displaystyle \frac {i \sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}-\frac {\int -\frac {a-i a \cot (c+d x)}{2 \sqrt {\cot (c+d x)}}dx}{2 a^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a-i a \cot (c+d x)}{\sqrt {\cot (c+d x)}}dx}{4 a^2}+\frac {i \sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {i \tan \left (c+d x+\frac {\pi }{2}\right ) a+a}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx}{4 a^2}+\frac {i \sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}\) |
\(\Big \downarrow \) 4016 |
\(\displaystyle \frac {\int \frac {1}{-i \cot (c+d x) a-a}d\sqrt {\cot (c+d x)}}{2 d}+\frac {i \sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\sqrt [4]{-1} \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{2 a d}+\frac {i \sqrt {\cot (c+d x)}}{2 d (a \cot (c+d x)+i a)}\) |
Input:
Int[1/(Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])),x]
Output:
((-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]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2*(c^2/f) Subst[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]
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] + Simp[1/(2*a*(b*c - a*d)) Int[(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] && Ne Q[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[0, n, 1]
Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x _)]^(n_.))^(p_.), x_Symbol] :> Simp[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]
Time = 0.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01
method | result | size |
derivativedivides | \(\frac {\frac {i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}+\frac {i \sqrt {\cot \left (d x +c \right )}}{2 i+2 \cot \left (d x +c \right )}}{d a}\) | \(69\) |
default | \(\frac {\frac {i \arctan \left (\frac {2 \sqrt {\cot \left (d x +c \right )}}{\sqrt {2}-i \sqrt {2}}\right )}{\sqrt {2}-i \sqrt {2}}+\frac {i \sqrt {\cot \left (d x +c \right )}}{2 i+2 \cot \left (d x +c \right )}}{d a}\) | \(69\) |
Input:
int(1/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(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)))
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 = 0.08 (sec) , antiderivative size = 269, normalized size of antiderivative = 3.96 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=-\frac {{\left (a d \sqrt {\frac {i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (2 \, {\left (2 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{4 \, a^{2} d^{2}}} + i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - a d \sqrt {\frac {i}{4 \, a^{2} d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-2 \, {\left (2 \, {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{4 \, a^{2} d^{2}}} - i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \] Input:
integrate(1/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
-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)) - s qrt((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)
\[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=- \frac {i \int \frac {1}{\tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}} - i \sqrt {\cot {\left (c + d x \right )}}}\, dx}{a} \] Input:
integrate(1/cot(d*x+c)**(1/2)/(a+I*a*tan(d*x+c)),x)
Output:
-I*Integral(1/(tan(c + d*x)*sqrt(cot(c + d*x)) - I*sqrt(cot(c + d*x))), x) /a
Exception generated. \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(1/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.17 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\frac {i \, {\left (-\left (i + 1\right ) \, \sqrt {2} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right ) - \frac {2 i \, \sqrt {\tan \left (d x + c\right )}}{\tan \left (d x + c\right ) - i}\right )}}{4 \, a d} \] Input:
integrate(1/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
1/4*I*(-(I + 1)*sqrt(2)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c))) - 2*I*sqrt(tan(d*x + c))/(tan(d*x + c) - I))/(a*d)
Timed out. \[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\int \frac {1}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )} \,d x \] Input:
int(1/(cot(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)),x)
Output:
int(1/(cot(c + d*x)^(1/2)*(a + a*tan(c + d*x)*1i)), x)
\[ \int \frac {1}{\sqrt {\cot (c+d x)} (a+i a \tan (c+d x))} \, dx=\frac {\int \frac {\sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right ) \tan \left (d x +c \right ) i +\cot \left (d x +c \right )}d x}{a} \] Input:
int(1/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c)),x)
Output:
int(sqrt(cot(c + d*x))/(cot(c + d*x)*tan(c + d*x)*i + cot(c + d*x)),x)/a