3.1.3 \(\int \frac {\cos ^2(x)}{i+\cot (x)} \, dx\) [3]

Optimal. Leaf size=50 \[ -\frac {i x}{8}+\frac {i}{8 (i-\cot (x))}+\frac {1}{8 (i+\cot (x))^2}+\frac {i}{4 (i+\cot (x))} \]

[Out]

-1/8*I*x+1/8*I/(I-cot(x))+1/8/(I+cot(x))^2+1/4*I/(I+cot(x))

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Rubi [A]
time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, 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 = {3597, 862, 90, 209} \begin {gather*} -\frac {i x}{8}+\frac {i}{8 (-\cot (x)+i)}+\frac {i}{4 (\cot (x)+i)}+\frac {1}{8 (\cot (x)+i)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2/(I + Cot[x]),x]

[Out]

(-1/8*I)*x + (I/8)/(I - Cot[x]) + 1/(8*(I + Cot[x])^2) + (I/4)/(I + Cot[x])

Rule 90

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 862

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/e)*x)^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 3597

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 {\cos ^2(x)}{i+\cot (x)} \, dx &=-\text {Subst}\left (\int \frac {x^2}{(i+x) \left (1+x^2\right )^2} \, dx,x,\cot (x)\right )\\ &=-\text {Subst}\left (\int \frac {x^2}{(-i+x)^2 (i+x)^3} \, dx,x,\cot (x)\right )\\ &=-\text {Subst}\left (\int \left (-\frac {i}{8 (-i+x)^2}+\frac {1}{4 (i+x)^3}+\frac {i}{4 (i+x)^2}-\frac {i}{8 \left (1+x^2\right )}\right ) \, dx,x,\cot (x)\right )\\ &=\frac {i}{8 (i-\cot (x))}+\frac {1}{8 (i+\cot (x))^2}+\frac {i}{4 (i+\cot (x))}+\frac {1}{8} i \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (x)\right )\\ &=-\frac {i x}{8}+\frac {i}{8 (i-\cot (x))}+\frac {1}{8 (i+\cot (x))^2}+\frac {i}{4 (i+\cot (x))}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 32, normalized size = 0.64 \begin {gather*} -\frac {1}{32} i (4 x-4 i \cos (2 x)-i \cos (4 x)-\sin (4 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^2/(I + Cot[x]),x]

[Out]

(-1/32*I)*(4*x - (4*I)*Cos[2*x] - I*Cos[4*x] - Sin[4*x])

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Maple [A]
time = 0.26, size = 37, normalized size = 0.74

method	result	size
risch	\(-\frac {i x}{8}-\frac {{\mathrm e}^{-4 i x}}{32}-\frac {\cos \left (2 x \right )}{8}\)	\(19\)
default	\(\frac {1}{8 \left (\tan \left (x \right )-i\right )^{2}}-\frac {\ln \left (\tan \left (x \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (x \right )+i\right )}+\frac {\ln \left (\tan \left (x \right )+i\right )}{16}\)	\(37\)
norman	\(\frac {\frac {x \tan \left (x \right ) \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4}-\frac {x \left (\tan ^{2}\left (x \right )\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}-\frac {i x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{8}-\frac {3 i x \left (\tan ^{2}\left (x \right )\right )}{8}-\frac {1}{4}+\frac {5 i \tan \left (x \right ) \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{8}+\frac {5 i \tan \left (x \right )}{8}+i x \tan \left (x \right ) \tan \left (\frac {x}{2}\right )-\frac {5 i x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{4}-\frac {3 x \tan \left (x \right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}+\frac {x \left (\tan ^{2}\left (x \right )\right ) \tan \left (\frac {x}{2}\right )}{2}-\frac {7 i \tan \left (x \right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{4}-\frac {i x}{8}-i \tan \left (\frac {x}{2}\right )-\frac {\left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}+\frac {x \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}-i x \tan \left (x \right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )-\frac {3 i x \left (\tan ^{2}\left (x \right )\right ) \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{8}-\frac {x \tan \left (\frac {x}{2}\right )}{2}+\frac {x \tan \left (x \right )}{4}+\frac {\tan \left (x \right ) \tan \left (\frac {x}{2}\right )}{2}+\frac {i x \left (\tan ^{2}\left (x \right )\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{4}-\frac {\tan \left (x \right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}+i \left (\tan ^{3}\left (\frac {x}{2}\right )\right )-\frac {\left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2} \left (1+\tan ^{2}\left (x \right )\right )}\)	\(248\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/(I+cot(x)),x,method=_RETURNVERBOSE)

[Out]

1/8/(tan(x)-I)^2-1/16*ln(tan(x)-I)-1/8*I/(tan(x)+I)+1/16*ln(tan(x)+I)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^2/(I+cot(x)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [A]
time = 2.37, size = 27, normalized size = 0.54 \begin {gather*} \frac {1}{32} \, {\left (-4 i \, x e^{\left (4 i \, x\right )} - 2 \, e^{\left (6 i \, x\right )} - 2 \, e^{\left (2 i \, x\right )} - 1\right )} e^{\left (-4 i \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^2/(I+cot(x)),x, algorithm="fricas")

[Out]

1/32*(-4*I*x*e^(4*I*x) - 2*e^(6*I*x) - 2*e^(2*I*x) - 1)*e^(-4*I*x)

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Sympy [A]
time = 0.06, size = 34, normalized size = 0.68 \begin {gather*} - \frac {i x}{8} - \frac {e^{2 i x}}{16} - \frac {e^{- 2 i x}}{16} - \frac {e^{- 4 i x}}{32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**2/(I+cot(x)),x)

[Out]

-I*x/8 - exp(2*I*x)/16 - exp(-2*I*x)/16 - exp(-4*I*x)/32

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Giac [A]
time = 0.41, size = 51, normalized size = 1.02 \begin {gather*} -\frac {-i \, \tan \left (x\right ) + 3}{16 \, {\left (-i \, \tan \left (x\right ) + 1\right )}} + \frac {3 \, \tan \left (x\right )^{2} - 6 i \, \tan \left (x\right ) + 1}{32 \, {\left (\tan \left (x\right ) - i\right )}^{2}} + \frac {1}{16} \, \log \left (\tan \left (x\right ) + i\right ) - \frac {1}{16} \, \log \left (\tan \left (x\right ) - i\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^2/(I+cot(x)),x, algorithm="giac")

[Out]

-1/16*(-I*tan(x) + 3)/(-I*tan(x) + 1) + 1/32*(3*tan(x)^2 - 6*I*tan(x) + 1)/(tan(x) - I)^2 + 1/16*log(tan(x) +
I) - 1/16*log(tan(x) - I)

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Mupad [B]
time = 0.25, size = 36, normalized size = 0.72 \begin {gather*} -\frac {x\,1{}\mathrm {i}}{8}+\frac {\frac {{\mathrm {tan}\left (x\right )}^2\,1{}\mathrm {i}}{8}+\frac {\mathrm {tan}\left (x\right )}{8}-\frac {1}{4}{}\mathrm {i}}{\left (\mathrm {tan}\left (x\right )+1{}\mathrm {i}\right )\,{\left (1+\mathrm {tan}\left (x\right )\,1{}\mathrm {i}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/(cot(x) + 1i),x)

[Out]

(tan(x)/8 + (tan(x)^2*1i)/8 - 1i/4)/((tan(x) + 1i)*(tan(x)*1i + 1)^2) - (x*1i)/8

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