3.3 \(\int \frac{\sin ^2(x)}{i+\tan (x)} \, dx\)

Optimal. Leaf size=50 \[ -\frac{i x}{8}-\frac{i}{8 (-\tan (x)+i)}-\frac{i}{4 (\tan (x)+i)}-\frac{1}{8 (\tan (x)+i)^2} \]

[Out]

(-I/8)*x - (I/8)/(I - Tan[x]) - 1/(8*(I + Tan[x])^2) - (I/4)/(I + Tan[x])

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Rubi [A]  time = 0.0539709, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3516, 848, 88, 203} \[ -\frac{i x}{8}-\frac{i}{8 (-\tan (x)+i)}-\frac{i}{4 (\tan (x)+i)}-\frac{1}{8 (\tan (x)+i)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(I + Tan[x]),x]

[Out]

(-I/8)*x - (I/8)/(I - Tan[x]) - 1/(8*(I + Tan[x])^2) - (I/4)/(I + Tan[x])

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\sin ^2(x)}{i+\tan (x)} \, dx &=\operatorname{Subst}\left (\int \frac{x^2}{(i+x) \left (1+x^2\right )^2} \, dx,x,\tan (x)\right )\\ &=\operatorname{Subst}\left (\int \frac{x^2}{(-i+x)^2 (i+x)^3} \, dx,x,\tan (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{i}{8 (-i+x)^2}+\frac{1}{4 (i+x)^3}+\frac{i}{4 (i+x)^2}-\frac{i}{8 \left (1+x^2\right )}\right ) \, dx,x,\tan (x)\right )\\ &=-\frac{i}{8 (i-\tan (x))}-\frac{1}{8 (i+\tan (x))^2}-\frac{i}{4 (i+\tan (x))}-\frac{1}{8} i \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )\\ &=-\frac{i x}{8}-\frac{i}{8 (i-\tan (x))}-\frac{1}{8 (i+\tan (x))^2}-\frac{i}{4 (i+\tan (x))}\\ \end{align*}

Mathematica [A]  time = 0.091249, size = 39, normalized size = 0.78 \[ -\frac{i \left (-3 i \sin (2 x)+\cos (2 x)+2 \tan ^{-1}(\tan (x)) (\tan (x)+i)+3\right )}{16 (\tan (x)+i)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(I + Tan[x]),x]

[Out]

((-I/16)*(3 + Cos[2*x] - (3*I)*Sin[2*x] + 2*ArcTan[Tan[x]]*(I + Tan[x])))/(I + Tan[x])

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Maple [A]  time = 0.041, size = 47, normalized size = 0.9 \begin{align*}{\frac{{\frac{i}{8}}}{\tan \left ( x \right ) -i}}-{\frac{\ln \left ( \tan \left ( x \right ) -i \right ) }{16}}-{\frac{{\frac{i}{4}}}{i+\tan \left ( x \right ) }}-{\frac{1}{8\, \left ( i+\tan \left ( x \right ) \right ) ^{2}}}+{\frac{\ln \left ( i+\tan \left ( x \right ) \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(I+tan(x)),x)

[Out]

1/8*I/(tan(x)-I)-1/16*ln(tan(x)-I)-1/4*I/(I+tan(x))-1/8/(I+tan(x))^2+1/16*ln(I+tan(x))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(I+tan(x)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.12543, size = 89, normalized size = 1.78 \begin{align*} \frac{1}{32} \,{\left (-4 i \, x e^{\left (2 i \, x\right )} + e^{\left (6 i \, x\right )} - 2 \, e^{\left (4 i \, x\right )} - 2\right )} e^{\left (-2 i \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(I+tan(x)),x, algorithm="fricas")

[Out]

1/32*(-4*I*x*e^(2*I*x) + e^(6*I*x) - 2*e^(4*I*x) - 2)*e^(-2*I*x)

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Sympy [A]  time = 0.425476, size = 31, normalized size = 0.62 \begin{align*} - \frac{i x}{8} + \frac{e^{4 i x}}{32} - \frac{e^{2 i x}}{16} - \frac{e^{- 2 i x}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**2/(I+tan(x)),x)

[Out]

-I*x/8 + exp(4*I*x)/32 - exp(2*I*x)/16 - exp(-2*I*x)/16

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Giac [A]  time = 1.3412, size = 55, normalized size = 1.1 \begin{align*} -\frac{i \, \tan \left (x\right )^{2} + 3 \, \tan \left (x\right ) + 2 i}{8 \,{\left (\tan \left (x\right ) + i\right )}^{2}{\left (\tan \left (x\right ) - i\right )}} + \frac{1}{16} \, \log \left (\tan \left (x\right ) + i\right ) - \frac{1}{16} \, \log \left (\tan \left (x\right ) - i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(I+tan(x)),x, algorithm="giac")

[Out]

-1/8*(I*tan(x)^2 + 3*tan(x) + 2*I)/((tan(x) + I)^2*(tan(x) - I)) + 1/16*log(tan(x) + I) - 1/16*log(tan(x) - I)