3.8 \(\int \frac{\sec ^4(x)}{i+\cot (x)} \, dx\)

Optimal. Leaf size=19 \[ \frac{\tan ^2(x)}{2}-\frac{1}{3} i \tan ^3(x) \]

[Out]

Tan[x]^2/2 - (I/3)*Tan[x]^3

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Rubi [A]  time = 0.0447802, antiderivative size = 19, 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, 43} \[ \frac{\tan ^2(x)}{2}-\frac{1}{3} i \tan ^3(x) \]

Antiderivative was successfully verified.

[In]

Int[Sec[x]^4/(I + Cot[x]),x]

[Out]

Tan[x]^2/2 - (I/3)*Tan[x]^3

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0542781, size = 24, normalized size = 1.26 \[ \frac{1}{6} \left (2 i \tan (x)+(3-2 i \tan (x)) \sec ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[x]^4/(I + Cot[x]),x]

[Out]

(Sec[x]^2*(3 - (2*I)*Tan[x]) + (2*I)*Tan[x])/6

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Maple [A]  time = 0.034, size = 15, 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(x)^4/(I+cot(x)),x)

[Out]

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

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Maxima [A]  time = 1.20208, size = 18, normalized size = 0.95 \begin{align*} -\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)^4/(I+cot(x)),x, algorithm="maxima")

[Out]

-1/3*I*tan(x)^3 + 1/2*tan(x)^2

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Fricas [B]  time = 1.85279, size = 140, normalized size = 7.37 \begin{align*} \frac{2 \,{\left ({\left (3 \, e^{\left (2 i \, x\right )} + 1\right )} e^{\left (2 i \, x\right )} - 2 \, e^{\left (2 i \, x\right )}\right )} e^{\left (-2 i \, x\right )}}{3 \,{\left (e^{\left (6 i \, x\right )} + 3 \, e^{\left (4 i \, x\right )} + 3 \, e^{\left (2 i \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(x)^4/(I+cot(x)),x, algorithm="fricas")

[Out]

2/3*((3*e^(2*I*x) + 1)*e^(2*I*x) - 2*e^(2*I*x))*e^(-2*I*x)/(e^(6*I*x) + 3*e^(4*I*x) + 3*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)**4/(I+cot(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.27745, size = 18, normalized size = 0.95 \begin{align*} -\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)^4/(I+cot(x)),x, algorithm="giac")

[Out]

-1/3*I*tan(x)^3 + 1/2*tan(x)^2