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

Optimal. Leaf size=29 \[ \frac{\sin ^5(x)}{5}-\frac{1}{5} i \cos ^5(x)+\frac{1}{3} i \cos ^3(x) \]

[Out]

(I/3)*Cos[x]^3 - (I/5)*Cos[x]^5 + Sin[x]^5/5

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Rubi [A]  time = 0.136883, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3518, 3108, 3107, 2565, 14, 2564, 30} \[ \frac{\sin ^5(x)}{5}-\frac{1}{5} i \cos ^5(x)+\frac{1}{3} i \cos ^3(x) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(I + Tan[x]),x]

[Out]

(I/3)*Cos[x]^3 - (I/5)*Cos[x]^5 + Sin[x]^5/5

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 2565

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2564

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 30

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

Rubi steps

\begin{align*} \int \frac{\sin ^3(x)}{i+\tan (x)} \, dx &=\int \frac{\cos (x) \sin ^3(x)}{i \cos (x)+\sin (x)} \, dx\\ &=-\left (i \int \cos (x) (\cos (x)+i \sin (x)) \sin ^3(x) \, dx\right )\\ &=-\left (i \int \left (\cos ^2(x) \sin ^3(x)+i \cos (x) \sin ^4(x)\right ) \, dx\right )\\ &=-\left (i \int \cos ^2(x) \sin ^3(x) \, dx\right )+\int \cos (x) \sin ^4(x) \, dx\\ &=i \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (x)\right )+\operatorname{Subst}\left (\int x^4 \, dx,x,\sin (x)\right )\\ &=\frac{\sin ^5(x)}{5}+i \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (x)\right )\\ &=\frac{1}{3} i \cos ^3(x)-\frac{1}{5} i \cos ^5(x)+\frac{\sin ^5(x)}{5}\\ \end{align*}

Mathematica [A]  time = 0.019013, size = 51, normalized size = 1.76 \[ \frac{\sin (x)}{8}-\frac{1}{16} \sin (3 x)+\frac{1}{80} \sin (5 x)+\frac{1}{8} i \cos (x)+\frac{1}{48} i \cos (3 x)-\frac{1}{80} i \cos (5 x) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^3/(I + Tan[x]),x]

[Out]

(I/8)*Cos[x] + (I/48)*Cos[3*x] - (I/80)*Cos[5*x] + Sin[x]/8 - Sin[3*x]/16 + Sin[5*x]/80

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Maple [B]  time = 0.051, size = 81, normalized size = 2.8 \begin{align*}{-{\frac{i}{4}} \left ( \tan \left ({\frac{x}{2}} \right ) -i \right ) ^{-2}}+{\frac{1}{6} \left ( \tan \left ({\frac{x}{2}} \right ) -i \right ) ^{-3}}+{\frac{1}{8} \left ( \tan \left ({\frac{x}{2}} \right ) -i \right ) ^{-1}}+{i \left ( \tan \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}+{\frac{2}{5} \left ( \tan \left ({\frac{x}{2}} \right ) +i \right ) ^{-5}}-{\frac{2}{3} \left ( \tan \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{\frac{1}{8} \left ( \tan \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(I+tan(x)),x)

[Out]

-1/4*I/(tan(1/2*x)-I)^2+1/6/(tan(1/2*x)-I)^3+1/8/(tan(1/2*x)-I)+I/(tan(1/2*x)+I)^4+2/5/(tan(1/2*x)+I)^5-2/3/(t
an(1/2*x)+I)^3-1/8/(tan(1/2*x)+I)

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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)^3/(I+tan(x)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.06113, size = 101, normalized size = 3.48 \begin{align*} \frac{1}{240} \,{\left (-3 i \, e^{\left (8 i \, x\right )} + 10 i \, e^{\left (6 i \, x\right )} + 30 i \, e^{\left (2 i \, x\right )} - 5 i\right )} e^{\left (-3 i \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(I+tan(x)),x, algorithm="fricas")

[Out]

1/240*(-3*I*e^(8*I*x) + 10*I*e^(6*I*x) + 30*I*e^(2*I*x) - 5*I)*e^(-3*I*x)

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Sympy [A]  time = 0.289148, size = 37, normalized size = 1.28 \begin{align*} - \frac{i e^{5 i x}}{80} + \frac{i e^{3 i x}}{24} + \frac{i e^{- i x}}{8} - \frac{i e^{- 3 i x}}{48} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**3/(I+tan(x)),x)

[Out]

-I*exp(5*I*x)/80 + I*exp(3*I*x)/24 + I*exp(-I*x)/8 - I*exp(-3*I*x)/48

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Giac [B]  time = 1.29783, size = 96, normalized size = 3.31 \begin{align*} -\frac{-3 i \, \tan \left (\frac{1}{2} \, x\right )^{2} - 12 \, \tan \left (\frac{1}{2} \, x\right ) + 5 i}{24 \,{\left (-i \, \tan \left (\frac{1}{2} \, x\right ) - 1\right )}^{3}} - \frac{15 \, \tan \left (\frac{1}{2} \, x\right )^{4} + 60 i \, \tan \left (\frac{1}{2} \, x\right )^{3} - 10 \, \tan \left (\frac{1}{2} \, x\right )^{2} - 20 i \, \tan \left (\frac{1}{2} \, x\right ) + 7}{120 \,{\left (\tan \left (\frac{1}{2} \, x\right ) + i\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(I+tan(x)),x, algorithm="giac")

[Out]

-1/24*(-3*I*tan(1/2*x)^2 - 12*tan(1/2*x) + 5*I)/(-I*tan(1/2*x) - 1)^3 - 1/120*(15*tan(1/2*x)^4 + 60*I*tan(1/2*
x)^3 - 10*tan(1/2*x)^2 - 20*I*tan(1/2*x) + 7)/(tan(1/2*x) + I)^5