3.5 \(\int \frac{\csc (x)}{i+\tan (x)} \, dx\)

Optimal. Leaf size=16 \[ \sin (x)-i \cos (x)+i \tanh ^{-1}(\cos (x)) \]

[Out]

I*ArcTanh[Cos[x]] - I*Cos[x] + Sin[x]

________________________________________________________________________________________

Rubi [A]  time = 0.0864813, antiderivative size = 16, 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, 2637, 2592, 321, 206} \[ \sin (x)-i \cos (x)+i \tanh ^{-1}(\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(I + Tan[x]),x]

[Out]

I*ArcTanh[Cos[x]] - I*Cos[x] + 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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{\csc (x)}{i+\tan (x)} \, dx &=\int \frac{\cot (x)}{i \cos (x)+\sin (x)} \, dx\\ &=-(i \int \cot (x) (\cos (x)+i \sin (x)) \, dx)\\ &=-(i \int (i \cos (x)+\cos (x) \cot (x)) \, dx)\\ &=-(i \int \cos (x) \cot (x) \, dx)+\int \cos (x) \, dx\\ &=\sin (x)+i \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (x)\right )\\ &=-i \cos (x)+\sin (x)+i \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (x)\right )\\ &=i \tanh ^{-1}(\cos (x))-i \cos (x)+\sin (x)\\ \end{align*}

Mathematica [A]  time = 0.015746, size = 31, normalized size = 1.94 \[ \sin (x)-i \cos (x)-i \log \left (\sin \left (\frac{x}{2}\right )\right )+i \log \left (\cos \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(I + Tan[x]),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.033, size = 21, normalized size = 1.3 \begin{align*} 2\, \left ( \tan \left ( x/2 \right ) +i \right ) ^{-1}-i\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)/(I+tan(x)),x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.47837, size = 38, normalized size = 2.38 \begin{align*} \frac{2}{\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + i} - i \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(I+tan(x)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.20494, size = 73, normalized size = 4.56 \begin{align*} -i \, e^{\left (i \, x\right )} + i \, \log \left (e^{\left (i \, x\right )} + 1\right ) - i \, \log \left (e^{\left (i \, x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(I+tan(x)),x, algorithm="fricas")

[Out]

-I*e^(I*x) + I*log(e^(I*x) + 1) - I*log(e^(I*x) - 1)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(I+tan(x)),x)

[Out]

Integral(csc(x)/(tan(x) + I), x)

________________________________________________________________________________________

Giac [A]  time = 1.33305, size = 30, normalized size = 1.88 \begin{align*} -\frac{2 i}{-i \, \tan \left (\frac{1}{2} \, x\right ) + 1} - i \, \log \left (-i \, \tan \left (\frac{1}{2} \, x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(I+tan(x)),x, algorithm="giac")

[Out]

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