3.1 \(\int \frac{\sin ^4(x)}{i+\tan (x)} \, dx\)

Optimal. Leaf size=78 \[ -\frac{i x}{16}-\frac{i}{8 (-\tan (x)+i)}-\frac{3 i}{16 (\tan (x)+i)}-\frac{1}{32 (-\tan (x)+i)^2}-\frac{5}{32 (\tan (x)+i)^2}+\frac{i}{24 (\tan (x)+i)^3} \]

[Out]

(-I/16)*x - 1/(32*(I - Tan[x])^2) - (I/8)/(I - Tan[x]) + (I/24)/(I + Tan[x])^3 - 5/(32*(I + Tan[x])^2) - ((3*I
)/16)/(I + Tan[x])

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Rubi [A]  time = 0.0659966, antiderivative size = 78, 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} \[ -\frac{i x}{16}-\frac{i}{8 (-\tan (x)+i)}-\frac{3 i}{16 (\tan (x)+i)}-\frac{1}{32 (-\tan (x)+i)^2}-\frac{5}{32 (\tan (x)+i)^2}+\frac{i}{24 (\tan (x)+i)^3} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^4/(I + Tan[x]),x]

[Out]

(-I/16)*x - 1/(32*(I - Tan[x])^2) - (I/8)/(I - Tan[x]) + (I/24)/(I + Tan[x])^3 - 5/(32*(I + Tan[x])^2) - ((3*I
)/16)/(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 ^4(x)}{i+\tan (x)} \, dx &=\operatorname{Subst}\left (\int \frac{x^4}{(i+x) \left (1+x^2\right )^3} \, dx,x,\tan (x)\right )\\ &=\operatorname{Subst}\left (\int \frac{x^4}{(-i+x)^3 (i+x)^4} \, dx,x,\tan (x)\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{1}{16 (-i+x)^3}-\frac{i}{8 (-i+x)^2}-\frac{i}{8 (i+x)^4}+\frac{5}{16 (i+x)^3}+\frac{3 i}{16 (i+x)^2}-\frac{i}{16 \left (1+x^2\right )}\right ) \, dx,x,\tan (x)\right )\\ &=-\frac{1}{32 (i-\tan (x))^2}-\frac{i}{8 (i-\tan (x))}+\frac{i}{24 (i+\tan (x))^3}-\frac{5}{32 (i+\tan (x))^2}-\frac{3 i}{16 (i+\tan (x))}-\frac{1}{16} i \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (x)\right )\\ &=-\frac{i x}{16}-\frac{1}{32 (i-\tan (x))^2}-\frac{i}{8 (i-\tan (x))}+\frac{i}{24 (i+\tan (x))^3}-\frac{5}{32 (i+\tan (x))^2}-\frac{3 i}{16 (i+\tan (x))}\\ \end{align*}

Mathematica [A]  time = 0.0800587, size = 67, normalized size = 0.86 \[ \frac{\sec (x) \left (-32 \sin (x)-27 \sin (3 x)+5 \sin (5 x)-56 i \cos (x)-9 i \cos (3 x)+i \cos (5 x)+24 \tan ^{-1}(\tan (x)) (\cos (x)-i \sin (x))\right )}{384 (\tan (x)+i)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^4/(I + Tan[x]),x]

[Out]

(Sec[x]*((-56*I)*Cos[x] - (9*I)*Cos[3*x] + I*Cos[5*x] + 24*ArcTan[Tan[x]]*(Cos[x] - I*Sin[x]) - 32*Sin[x] - 27
*Sin[3*x] + 5*Sin[5*x]))/(384*(I + Tan[x]))

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Maple [A]  time = 0.075, size = 66, normalized size = 0.9 \begin{align*}{\frac{{\frac{i}{8}}}{\tan \left ( x \right ) -i}}-{\frac{1}{32\, \left ( \tan \left ( x \right ) -i \right ) ^{2}}}-{\frac{\ln \left ( \tan \left ( x \right ) -i \right ) }{32}}+{\frac{{\frac{i}{24}}}{ \left ( i+\tan \left ( x \right ) \right ) ^{3}}}-{\frac{{\frac{3\,i}{16}}}{i+\tan \left ( x \right ) }}-{\frac{5}{32\, \left ( i+\tan \left ( x \right ) \right ) ^{2}}}+{\frac{\ln \left ( i+\tan \left ( x \right ) \right ) }{32}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^4/(I+tan(x)),x)

[Out]

1/8*I/(tan(x)-I)-1/32/(tan(x)-I)^2-1/32*ln(tan(x)-I)+1/24*I/(I+tan(x))^3-3/16*I/(I+tan(x))-5/32/(I+tan(x))^2+1
/32*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)^4/(I+tan(x)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.06533, size = 136, normalized size = 1.74 \begin{align*} \frac{1}{384} \,{\left (-24 i \, x e^{\left (4 i \, x\right )} - 2 \, e^{\left (10 i \, x\right )} + 9 \, e^{\left (8 i \, x\right )} - 12 \, e^{\left (6 i \, x\right )} - 18 \, e^{\left (2 i \, x\right )} + 3\right )} e^{\left (-4 i \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(I+tan(x)),x, algorithm="fricas")

[Out]

1/384*(-24*I*x*e^(4*I*x) - 2*e^(10*I*x) + 9*e^(8*I*x) - 12*e^(6*I*x) - 18*e^(2*I*x) + 3)*e^(-4*I*x)

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Sympy [A]  time = 0.395444, size = 51, normalized size = 0.65 \begin{align*} - \frac{i x}{16} - \frac{e^{6 i x}}{192} + \frac{3 e^{4 i x}}{128} - \frac{e^{2 i x}}{32} - \frac{3 e^{- 2 i x}}{64} + \frac{e^{- 4 i x}}{128} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**4/(I+tan(x)),x)

[Out]

-I*x/16 - exp(6*I*x)/192 + 3*exp(4*I*x)/128 - exp(2*I*x)/32 - 3*exp(-2*I*x)/64 + exp(-4*I*x)/128

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Giac [A]  time = 1.3062, size = 72, normalized size = 0.92 \begin{align*} -\frac{3 i \, \tan \left (x\right )^{4} + 21 \, \tan \left (x\right )^{3} + 13 i \, \tan \left (x\right )^{2} + 11 \, \tan \left (x\right ) + 8 i}{48 \,{\left (\tan \left (x\right ) + i\right )}^{3}{\left (\tan \left (x\right ) - i\right )}^{2}} + \frac{1}{32} \, \log \left (\tan \left (x\right ) + i\right ) - \frac{1}{32} \, \log \left (\tan \left (x\right ) - i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^4/(I+tan(x)),x, algorithm="giac")

[Out]

-1/48*(3*I*tan(x)^4 + 21*tan(x)^3 + 13*I*tan(x)^2 + 11*tan(x) + 8*I)/((tan(x) + I)^3*(tan(x) - I)^2) + 1/32*lo
g(tan(x) + I) - 1/32*log(tan(x) - I)