\(\int \frac {\sin ^2(x)}{i+\tan (x)} \, dx\) [3]

Optimal result

Integrand size = 13, antiderivative size = 50 \[ \int \frac {\sin ^2(x)}{i+\tan (x)} \, dx=-\frac {i x}{8}-\frac {i}{8 (i-\tan (x))}-\frac {1}{8 (i+\tan (x))^2}-\frac {i}{4 (i+\tan (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3597, 862, 90, 209} \[ \int \frac {\sin ^2(x)}{i+\tan (x)} \, dx=-\frac {i x}{8}-\frac {i}{8 (-\tan (x)+i)}-\frac {i}{4 (\tan (x)+i)}-\frac {1}{8 (\tan (x)+i)^2} \]

[In]

Int[Sin[x]^2/(I + Tan[x]),x]

[Out]

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

Rule 90

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 209

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

Rule 862

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/e)*x)^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 3597

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]

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

Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^2(x)}{i+\tan (x)} \, dx=-\frac {i (3+\cos (2 x)-3 i \sin (2 x)+2 \arctan (\tan (x)) (i+\tan (x)))}{16 (i+\tan (x))} \]

[In]

Integrate[Sin[x]^2/(I + Tan[x]),x]

[Out]

((-1/16*I)*(3 + Cos[2*x] - (3*I)*Sin[2*x] + 2*ArcTan[Tan[x]]*(I + Tan[x])))/(I + Tan[x])

Maple [A] (verified)

Time = 2.68 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.38

method	result	size
risch	\(-\frac {i x}{8}+\frac {{\mathrm e}^{4 i x}}{32}-\frac {\cos \left (2 x \right )}{8}\)	\(19\)
parallelrisch	\(-\frac {i x}{4}-\frac {7}{96}+\ln \left (\frac {1}{\left (i+\tan \left (x \right )\right )^{\frac {1}{8}}}\right )+\ln \left (\left (\sec ^{2}\left (x \right )\right )^{\frac {1}{16}}\right )+\frac {i \sin \left (4 x \right )}{32}+\frac {\cos \left (4 x \right )}{32}-\frac {\cos \left (2 x \right )}{8}\)	\(41\)
default	\(\frac {i}{8 \tan \left (x \right )-8 i}-\frac {\ln \left (\tan \left (x \right )-i\right )}{16}-\frac {i}{4 \left (i+\tan \left (x \right )\right )}-\frac {1}{8 \left (i+\tan \left (x \right )\right )^{2}}+\frac {\ln \left (i+\tan \left (x \right )\right )}{16}\)	\(47\)
norman	\(\frac {-\frac {1}{4}-\frac {\left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4}-\frac {\left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}-\frac {i x}{8}+i x \tan \left (x \right ) \tan \left (\frac {x}{2}\right )-\frac {3 i x \left (\tan ^{2}\left (x \right )\right ) \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{8}-i x \tan \left (x \right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+\frac {i x \left (\tan ^{2}\left (x \right )\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{4}-\frac {x \left (\tan ^{2}\left (x \right )\right ) \tan \left (\frac {x}{2}\right )}{2}+\frac {x \left (\tan ^{2}\left (x \right )\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}-i \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+\frac {\tan \left (x \right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}+\frac {x \tan \left (\frac {x}{2}\right )}{2}-\frac {5 i x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{4}-\frac {3 i \tan \left (x \right ) \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{8}+\frac {3 x \tan \left (x \right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}-\frac {3 i x \left (\tan ^{2}\left (x \right )\right )}{8}-\frac {i x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{8}+\frac {i \tan \left (x \right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{4}-\frac {x \tan \left (x \right ) \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4}-\frac {x \tan \left (x \right )}{4}-\frac {x \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}-\frac {3 i \tan \left (x \right )}{8}+i \tan \left (\frac {x}{2}\right )-\frac {\tan \left (x \right ) \tan \left (\frac {x}{2}\right )}{2}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2} \left (\tan ^{2}\left (x \right )+1\right )}\)	\(248\)

[In]

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

[Out]

-1/8*I*x+1/32*exp(4*I*x)-1/8*cos(2*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.50 \[ \int \frac {\sin ^2(x)}{i+\tan (x)} \, dx=\frac {1}{32} \, {\left (-4 i \, x e^{\left (2 i \, x\right )} + e^{\left (6 i \, x\right )} - 2 \, e^{\left (4 i \, x\right )} - 2\right )} e^{\left (-2 i \, x\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.62 \[ \int \frac {\sin ^2(x)}{i+\tan (x)} \, dx=- \frac {i x}{8} + \frac {e^{4 i x}}{32} - \frac {e^{2 i x}}{16} - \frac {e^{- 2 i x}}{16} \]

[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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sin ^2(x)}{i+\tan (x)} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.82 \[ \int \frac {\sin ^2(x)}{i+\tan (x)} \, dx=-\frac {i \, \tan \left (x\right )^{2} + 3 \, \tan \left (x\right ) + 2 i}{8 \, {\left (\tan \left (x\right ) + i\right )}^{2} {\left (\tan \left (x\right ) - i\right )}} + \frac {1}{16} \, \log \left (\tan \left (x\right ) + i\right ) - \frac {1}{16} \, \log \left (\tan \left (x\right ) - i\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 4.17 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.70 \[ \int \frac {\sin ^2(x)}{i+\tan (x)} \, dx=-\frac {x\,1{}\mathrm {i}}{8}+\frac {\frac {{\mathrm {tan}\left (x\right )}^2}{8}-\frac {\mathrm {tan}\left (x\right )\,3{}\mathrm {i}}{8}+\frac {1}{4}}{{\left (\mathrm {tan}\left (x\right )+1{}\mathrm {i}\right )}^2\,\left (1+\mathrm {tan}\left (x\right )\,1{}\mathrm {i}\right )} \]

[In]

int(sin(x)^2/(tan(x) + 1i),x)

[Out]

(tan(x)^2/8 - (tan(x)*3i)/8 + 1/4)/((tan(x) + 1i)^2*(tan(x)*1i + 1)) - (x*1i)/8