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]

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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Rubi [A]  time = 0.05, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.09, size = 26, normalized size = 0.70 \[ \frac {1}{60} \sec ^4(x) \left (15-4 i \sin ^2(x) (\cos (2 x)+4) \tan (x)\right ) \]

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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fricas [A]  time = 0.80, size = 48, normalized size = 1.30 \[ \frac {4 \, {\left (20 \, e^{\left (4 i \, x\right )} - 5 \, e^{\left (2 i \, x\right )} - 1\right )}}{15 \, {\left (e^{\left (10 i \, x\right )} + 5 \, e^{\left (8 i \, x\right )} + 10 \, e^{\left (6 i \, x\right )} + 10 \, e^{\left (4 i \, x\right )} + 5 \, e^{\left (2 i \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(20*e^(4*I*x) - 5*e^(2*I*x) - 1)/(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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giac [A]  time = 0.62, size = 25, normalized size = 0.68 \[ -\frac {1}{5} i \, \tan \relax (x)^{5} + \frac {1}{4} \, \tan \relax (x)^{4} - \frac {1}{3} i \, \tan \relax (x)^{3} + \frac {1}{2} \, \tan \relax (x)^{2} \]

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

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maple [A]  time = 0.35, size = 28, normalized size = 0.76 \[ \frac {\left (\tan ^{2}\relax (x )\right )}{2}-\frac {i \left (\tan ^{3}\relax (x )\right )}{3}+\frac {\left (\tan ^{4}\relax (x )\right )}{4}-\frac {i \left (\tan ^{5}\relax (x )\right )}{5} \]

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 = 0.86, size = 25, normalized size = 0.68 \[ -\frac {1}{5} i \, \tan \relax (x)^{5} + \frac {1}{4} \, \tan \relax (x)^{4} - \frac {1}{3} i \, \tan \relax (x)^{3} + \frac {1}{2} \, \tan \relax (x)^{2} \]

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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mupad [B]  time = 0.21, size = 27, normalized size = 0.73 \[ -\frac {{\mathrm {tan}\relax (x)}^5\,1{}\mathrm {i}}{5}+\frac {{\mathrm {tan}\relax (x)}^4}{4}-\frac {{\mathrm {tan}\relax (x)}^3\,1{}\mathrm {i}}{3}+\frac {{\mathrm {tan}\relax (x)}^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(x)^6*(cot(x) + 1i)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{6}{\relax (x )}}{\cot {\relax (x )} + i}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(x)**6/(cot(x) + I), x)

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