3.5 \(\int \frac{\sec (x)}{i+\cot (x)} \, dx\)

Optimal. Leaf size=18 \[ i \sin (x)-\cos (x)-i \tanh ^{-1}(\sin (x)) \]

[Out]

(-I)*ArcTanh[Sin[x]] - Cos[x] + I*Sin[x]

________________________________________________________________________________________

Rubi [A]  time = 0.101674, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {3518, 3108, 3107, 2638, 2592, 321, 206} \[ i \sin (x)-\cos (x)-i \tanh ^{-1}(\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]/(I + Cot[x]),x]

[Out]

(-I)*ArcTanh[Sin[x]] - Cos[x] + I*Sin[x]

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3108

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[a^p*b^p, Int[(Cos[c + d*x]^m*Sin[c + d*x]^n)/(b*Cos[c + d*x] + a*Sin
[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[a^2 + b^2, 0] && ILtQ[p, 0]

Rule 3107

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*sin[c + d*x]^n*(a*cos[c + d*x] + b*sin[c +
d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[p, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec (x)}{i+\cot (x)} \, dx &=-\int \frac{\tan (x)}{-\cos (x)-i \sin (x)} \, dx\\ &=i \int (-i \cos (x)-\sin (x)) \tan (x) \, dx\\ &=i \int (-i \sin (x)-\sin (x) \tan (x)) \, dx\\ &=-(i \int \sin (x) \tan (x) \, dx)+\int \sin (x) \, dx\\ &=-\cos (x)-i \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\sin (x)\right )\\ &=-\cos (x)+i \sin (x)-i \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (x)\right )\\ &=-i \tanh ^{-1}(\sin (x))-\cos (x)+i \sin (x)\\ \end{align*}

Mathematica [B]  time = 0.033853, size = 44, normalized size = 2.44 \[ -\cos (x)+i \left (\sin (x)+\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]/(I + Cot[x]),x]

[Out]

-Cos[x] + I*(Log[Cos[x/2] - Sin[x/2]] - Log[Cos[x/2] + Sin[x/2]] + Sin[x])

________________________________________________________________________________________

Maple [B]  time = 0.041, size = 34, normalized size = 1.9 \begin{align*}{2\,i \left ( \tan \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}-i\ln \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) +i\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)/(I+cot(x)),x)

[Out]

2*I/(tan(1/2*x)-I)-I*ln(tan(1/2*x)+1)+I*ln(tan(1/2*x)-1)

________________________________________________________________________________________

Maxima [B]  time = 1.26699, size = 61, normalized size = 3.39 \begin{align*} -\frac{2}{\frac{i \, \sin \left (x\right )}{\cos \left (x\right ) + 1} + 1} - i \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right ) + i \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(I+cot(x)),x, algorithm="maxima")

[Out]

-2/(I*sin(x)/(cos(x) + 1) + 1) - I*log(sin(x)/(cos(x) + 1) + 1) + I*log(sin(x)/(cos(x) + 1) - 1)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (-i \, e^{\left (3 i \, x\right )} + i \, e^{\left (i \, x\right )}\right )} e^{\left (-2 i \, x\right )}}{e^{\left (2 i \, x\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(I+cot(x)),x, algorithm="fricas")

[Out]

integral((-I*e^(3*I*x) + I*e^(I*x))*e^(-2*I*x)/(e^(2*I*x) + 1), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec{\left (x \right )}}{\cot{\left (x \right )} + i}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(I+cot(x)),x)

[Out]

Integral(sec(x)/(cot(x) + I), x)

________________________________________________________________________________________

Giac [B]  time = 1.31099, size = 42, normalized size = 2.33 \begin{align*} \frac{2 i}{\tan \left (\frac{1}{2} \, x\right ) - i} - i \, \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right ) + i \, \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)/(I+cot(x)),x, algorithm="giac")

[Out]

2*I/(tan(1/2*x) - I) - I*log(abs(tan(1/2*x) + 1)) + I*log(abs(tan(1/2*x) - 1))