\(\int \frac {\sin (x)}{i+\tan (x)} \, dx\) [4]

Optimal result

Integrand size = 11, antiderivative size = 19 \[ \int \frac {\sin (x)}{i+\tan (x)} \, dx=\frac {1}{3} i \cos ^3(x)+\frac {\sin ^3(x)}{3} \]

[Out]

1/3*I*cos(x)^3+1/3*sin(x)^3

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3599, 3187, 3186, 2645, 30, 2644} \[ \int \frac {\sin (x)}{i+\tan (x)} \, dx=\frac {\sin ^3(x)}{3}+\frac {1}{3} i \cos ^3(x) \]

[In]

Int[Sin[x]/(I + Tan[x]),x]

[Out]

(I/3)*Cos[x]^3 + Sin[x]^3/3

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 3186

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 3187

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 3599

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]))

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos (x) \sin (x)}{i \cos (x)+\sin (x)} \, dx \\ & = -(i \int \cos (x) (\cos (x)+i \sin (x)) \sin (x) \, dx) \\ & = -\left (i \int \left (\cos ^2(x) \sin (x)+i \cos (x) \sin ^2(x)\right ) \, dx\right ) \\ & = -\left (i \int \cos ^2(x) \sin (x) \, dx\right )+\int \cos (x) \sin ^2(x) \, dx \\ & = i \text {Subst}\left (\int x^2 \, dx,x,\cos (x)\right )+\text {Subst}\left (\int x^2 \, dx,x,\sin (x)\right ) \\ & = \frac {1}{3} i \cos ^3(x)+\frac {\sin ^3(x)}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {\sin (x)}{i+\tan (x)} \, dx=\frac {1}{4} i \cos (x)+\frac {1}{12} i \cos (3 x)+\frac {\sin (x)}{4}-\frac {1}{12} \sin (3 x) \]

[In]

Integrate[Sin[x]/(I + Tan[x]),x]

[Out]

(I/4)*Cos[x] + (I/12)*Cos[3*x] + Sin[x]/4 - Sin[3*x]/12

Maple [A] (verified)

Time = 1.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {i {\mathrm e}^{3 i x}}{12}+\frac {i {\mathrm e}^{-i x}}{4}\)	\(18\)
parallelrisch	\(\frac {2 i}{3}+\frac {i \cos \left (3 x \right )}{12}+\frac {i \cos \left (x \right )}{4}-\frac {\sin \left (3 x \right )}{12}+\frac {\sin \left (x \right )}{4}\)	\(26\)
default	\(\frac {1}{2 \tan \left (\frac {x}{2}\right )-2 i}+\frac {i}{\left (\tan \left (\frac {x}{2}\right )+i\right )^{2}}+\frac {2}{3 \left (\tan \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {1}{2 \left (\tan \left (\frac {x}{2}\right )+i\right )}\)	\(47\)
norman	\(\frac {\frac {i \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{3}+\frac {4 i \left (\tan ^{2}\left (x \right )\right )}{3}+\frac {2 \left (\tan ^{2}\left (x \right )\right ) \tan \left (\frac {x}{2}\right )}{3}-\frac {\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) \tan \left (x \right )}{3}-\frac {4 i \tan \left (x \right ) \tan \left (\frac {x}{2}\right )}{3}+\frac {\tan \left (x \right )}{3}-\frac {2 \tan \left (\frac {x}{2}\right )}{3}+i}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right ) \left (\tan ^{2}\left (x \right )+1\right )}\)	\(78\)

[In]

int(sin(x)/(I+tan(x)),x,method=_RETURNVERBOSE)

[Out]

1/12*I*exp(3*I*x)+1/4*I*exp(-I*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int \frac {\sin (x)}{i+\tan (x)} \, dx=\frac {1}{12} \, {\left (i \, e^{\left (4 i \, x\right )} + 3 i\right )} e^{\left (-i \, x\right )} \]

[In]

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

[Out]

1/12*(I*e^(4*I*x) + 3*I)*e^(-I*x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (x)}{i+\tan (x)} \, dx=\frac {i e^{3 i x}}{12} + \frac {i e^{- i x}}{4} \]

[In]

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

[Out]

I*exp(3*I*x)/12 + I*exp(-I*x)/4

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin (x)}{i+\tan (x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (13) = 26\).

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {\sin (x)}{i+\tan (x)} \, dx=-\frac {i}{2 \, {\left (-i \, \tan \left (\frac {1}{2} \, x\right ) - 1\right )}} - \frac {3 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 1}{6 \, {\left (\tan \left (\frac {1}{2} \, x\right ) + i\right )}^{3}} \]

[In]

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

[Out]

-1/2*I/(-I*tan(1/2*x) - 1) - 1/6*(3*tan(1/2*x)^2 - 1)/(tan(1/2*x) + I)^3

Mupad [B] (verification not implemented)

Time = 4.91 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.05 \[ \int \frac {\sin (x)}{i+\tan (x)} \, dx=-\frac {2\,\left (3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\mathrm {tan}\left (\frac {x}{2}\right )\,2{}\mathrm {i}-1\right )}{3\,\left (1+\mathrm {tan}\left (\frac {x}{2}\right )\,1{}\mathrm {i}\right )\,{\left (\mathrm {tan}\left (\frac {x}{2}\right )+1{}\mathrm {i}\right )}^3} \]

[In]

int(sin(x)/(tan(x) + 1i),x)

[Out]

-(2*(tan(x/2)*2i + 3*tan(x/2)^2 - 1))/(3*(tan(x/2)*1i + 1)*(tan(x/2) + 1i)^3)