3.10 \(\int \frac{\sec ^6(x)}{i+\cot (x)} \, dx\)

Optimal. Leaf size=37 \[ -\frac{1}{5} i \tan ^5(x)+\frac{\tan ^4(x)}{4}-\frac{1}{3} i \tan ^3(x)+\frac{\tan ^2(x)}{2} \]

[Out]

Tan[x]^2/2 - (I/3)*Tan[x]^3 + Tan[x]^4/4 - (I/5)*Tan[x]^5

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Rubi [A]  time = 0.053519, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3516, 848, 75} \[ -\frac{1}{5} i \tan ^5(x)+\frac{\tan ^4(x)}{4}-\frac{1}{3} i \tan ^3(x)+\frac{\tan ^2(x)}{2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^6/(I + Cot[x]),x]

[Out]

Tan[x]^2/2 - (I/3)*Tan[x]^3 + Tan[x]^4/4 - (I/5)*Tan[x]^5

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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^6(x)}{i+\cot (x)} \, dx &=-\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^6 (i+x)} \, dx,x,\cot (x)\right )\\ &=-\operatorname{Subst}\left (\int \frac{(-i+x)^2 (i+x)}{x^6} \, dx,x,\cot (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-\frac{i}{x^6}+\frac{1}{x^5}-\frac{i}{x^4}+\frac{1}{x^3}\right ) \, dx,x,\cot (x)\right )\\ &=\frac{\tan ^2(x)}{2}-\frac{1}{3} i \tan ^3(x)+\frac{\tan ^4(x)}{4}-\frac{1}{5} i \tan ^5(x)\\ \end{align*}

Mathematica [A]  time = 0.0872359, size = 26, normalized size = 0.7 \[ \frac{1}{60} \sec ^4(x) \left (15-4 i \sin ^2(x) (\cos (2 x)+4) \tan (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^6/(I + Cot[x]),x]

[Out]

(Sec[x]^4*(15 - (4*I)*(4 + Cos[2*x])*Sin[x]^2*Tan[x]))/60

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Maple [A]  time = 0.037, size = 28, normalized size = 0.8 \begin{align*}{\frac{ \left ( \tan \left ( x \right ) \right ) ^{2}}{2}}-{\frac{i}{3}} \left ( \tan \left ( x \right ) \right ) ^{3}+{\frac{ \left ( \tan \left ( x \right ) \right ) ^{4}}{4}}-{\frac{i}{5}} \left ( \tan \left ( x \right ) \right ) ^{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^6/(I+cot(x)),x)

[Out]

1/2*tan(x)^2-1/3*I*tan(x)^3+1/4*tan(x)^4-1/5*I*tan(x)^5

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Maxima [A]  time = 1.31848, size = 34, normalized size = 0.92 \begin{align*} -\frac{1}{5} i \, \tan \left (x\right )^{5} + \frac{1}{4} \, \tan \left (x\right )^{4} - \frac{1}{3} i \, \tan \left (x\right )^{3} + \frac{1}{2} \, \tan \left (x\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^6/(I+cot(x)),x, algorithm="maxima")

[Out]

-1/5*I*tan(x)^5 + 1/4*tan(x)^4 - 1/3*I*tan(x)^3 + 1/2*tan(x)^2

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Fricas [B]  time = 1.87751, size = 227, normalized size = 6.14 \begin{align*} \frac{4 \,{\left (2 \,{\left (10 \, e^{\left (4 i \, x\right )} + 5 \, e^{\left (2 i \, x\right )} + 1\right )} e^{\left (2 i \, x\right )} - 15 \, e^{\left (4 i \, x\right )} - 3 \, e^{\left (2 i \, x\right )}\right )} e^{\left (-2 i \, x\right )}}{15 \,{\left (e^{\left (10 i \, x\right )} + 5 \, e^{\left (8 i \, x\right )} + 10 \, e^{\left (6 i \, x\right )} + 10 \, e^{\left (4 i \, x\right )} + 5 \, e^{\left (2 i \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^6/(I+cot(x)),x, algorithm="fricas")

[Out]

4/15*(2*(10*e^(4*I*x) + 5*e^(2*I*x) + 1)*e^(2*I*x) - 15*e^(4*I*x) - 3*e^(2*I*x))*e^(-2*I*x)/(e^(10*I*x) + 5*e^
(8*I*x) + 10*e^(6*I*x) + 10*e^(4*I*x) + 5*e^(2*I*x) + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)**6/(I+cot(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.32192, size = 34, normalized size = 0.92 \begin{align*} -\frac{1}{5} i \, \tan \left (x\right )^{5} + \frac{1}{4} \, \tan \left (x\right )^{4} - \frac{1}{3} i \, \tan \left (x\right )^{3} + \frac{1}{2} \, \tan \left (x\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^6/(I+cot(x)),x, algorithm="giac")

[Out]

-1/5*I*tan(x)^5 + 1/4*tan(x)^4 - 1/3*I*tan(x)^3 + 1/2*tan(x)^2