3.9 \(\int \frac {\csc ^5(x)}{i+\tan (x)} \, dx\)

Optimal. Leaf size=40 \[ -\frac {\csc ^3(x)}{3}-\frac {1}{8} i \tanh ^{-1}(\cos (x))+\frac {1}{4} i \cot (x) \csc ^3(x)-\frac {1}{8} i \cot (x) \csc (x) \]

[Out]

-1/8*I*arctanh(cos(x))-1/8*I*cot(x)*csc(x)-1/3*csc(x)^3+1/4*I*cot(x)*csc(x)^3

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Rubi [A]  time = 0.15, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3518, 3108, 3107, 2606, 30, 2611, 3768, 3770} \[ -\frac {\csc ^3(x)}{3}-\frac {1}{8} i \tanh ^{-1}(\cos (x))+\frac {1}{4} i \cot (x) \csc ^3(x)-\frac {1}{8} i \cot (x) \csc (x) \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^5/(I + Tan[x]),x]

[Out]

(-I/8)*ArcTanh[Cos[x]] - (I/8)*Cot[x]*Csc[x] - Csc[x]^3/3 + (I/4)*Cot[x]*Csc[x]^3

Rule 30

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

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 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 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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\csc ^5(x)}{i+\tan (x)} \, dx &=\int \frac {\cot (x) \csc ^4(x)}{i \cos (x)+\sin (x)} \, dx\\ &=-\left (i \int \cot (x) \csc ^4(x) (\cos (x)+i \sin (x)) \, dx\right )\\ &=-\left (i \int \left (i \cot (x) \csc ^3(x)+\cot ^2(x) \csc ^3(x)\right ) \, dx\right )\\ &=-\left (i \int \cot ^2(x) \csc ^3(x) \, dx\right )+\int \cot (x) \csc ^3(x) \, dx\\ &=\frac {1}{4} i \cot (x) \csc ^3(x)+\frac {1}{4} i \int \csc ^3(x) \, dx-\operatorname {Subst}\left (\int x^2 \, dx,x,\csc (x)\right )\\ &=-\frac {1}{8} i \cot (x) \csc (x)-\frac {\csc ^3(x)}{3}+\frac {1}{4} i \cot (x) \csc ^3(x)+\frac {1}{8} i \int \csc (x) \, dx\\ &=-\frac {1}{8} i \tanh ^{-1}(\cos (x))-\frac {1}{8} i \cot (x) \csc (x)-\frac {\csc ^3(x)}{3}+\frac {1}{4} i \cot (x) \csc ^3(x)\\ \end {align*}

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Mathematica [B]  time = 0.02, size = 139, normalized size = 3.48 \[ -\frac {1}{12} \tan \left (\frac {x}{2}\right )-\frac {1}{12} \cot \left (\frac {x}{2}\right )+\frac {1}{64} i \csc ^4\left (\frac {x}{2}\right )-\frac {1}{32} i \csc ^2\left (\frac {x}{2}\right )-\frac {1}{64} i \sec ^4\left (\frac {x}{2}\right )+\frac {1}{32} i \sec ^2\left (\frac {x}{2}\right )+\frac {1}{8} i \log \left (\sin \left (\frac {x}{2}\right )\right )-\frac {1}{8} i \log \left (\cos \left (\frac {x}{2}\right )\right )-\frac {1}{24} \cot \left (\frac {x}{2}\right ) \csc ^2\left (\frac {x}{2}\right )-\frac {1}{24} \tan \left (\frac {x}{2}\right ) \sec ^2\left (\frac {x}{2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^5/(I + Tan[x]),x]

[Out]

-1/12*Cot[x/2] - (I/32)*Csc[x/2]^2 - (Cot[x/2]*Csc[x/2]^2)/24 + (I/64)*Csc[x/2]^4 - (I/8)*Log[Cos[x/2]] + (I/8
)*Log[Sin[x/2]] + (I/32)*Sec[x/2]^2 - (I/64)*Sec[x/2]^4 - Tan[x/2]/12 - (Sec[x/2]^2*Tan[x/2])/24

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fricas [B]  time = 0.47, size = 121, normalized size = 3.02 \[ \frac {{\left (-3 i \, e^{\left (8 i \, x\right )} + 12 i \, e^{\left (6 i \, x\right )} - 18 i \, e^{\left (4 i \, x\right )} + 12 i \, e^{\left (2 i \, x\right )} - 3 i\right )} \log \left (e^{\left (i \, x\right )} + 1\right ) + {\left (3 i \, e^{\left (8 i \, x\right )} - 12 i \, e^{\left (6 i \, x\right )} + 18 i \, e^{\left (4 i \, x\right )} - 12 i \, e^{\left (2 i \, x\right )} + 3 i\right )} \log \left (e^{\left (i \, x\right )} - 1\right ) + 6 i \, e^{\left (7 i \, x\right )} + 106 i \, e^{\left (5 i \, x\right )} - 22 i \, e^{\left (3 i \, x\right )} + 6 i \, e^{\left (i \, x\right )}}{24 \, {\left (e^{\left (8 i \, x\right )} - 4 \, e^{\left (6 i \, x\right )} + 6 \, e^{\left (4 i \, x\right )} - 4 \, e^{\left (2 i \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(I+tan(x)),x, algorithm="fricas")

[Out]

1/24*((-3*I*e^(8*I*x) + 12*I*e^(6*I*x) - 18*I*e^(4*I*x) + 12*I*e^(2*I*x) - 3*I)*log(e^(I*x) + 1) + (3*I*e^(8*I
*x) - 12*I*e^(6*I*x) + 18*I*e^(4*I*x) - 12*I*e^(2*I*x) + 3*I)*log(e^(I*x) - 1) + 6*I*e^(7*I*x) + 106*I*e^(5*I*
x) - 22*I*e^(3*I*x) + 6*I*e^(I*x))/(e^(8*I*x) - 4*e^(6*I*x) + 6*e^(4*I*x) - 4*e^(2*I*x) + 1)

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giac [B]  time = 0.20, size = 62, normalized size = 1.55 \[ -\frac {1}{64} i \, \tan \left (\frac {1}{2} \, x\right )^{4} - \frac {1}{24} \, \tan \left (\frac {1}{2} \, x\right )^{3} - \frac {50 i \, \tan \left (\frac {1}{2} \, x\right )^{4} + 24 \, \tan \left (\frac {1}{2} \, x\right )^{3} + 8 \, \tan \left (\frac {1}{2} \, x\right ) - 3 i}{192 \, \tan \left (\frac {1}{2} \, x\right )^{4}} + \frac {1}{8} i \, \log \left (\tan \left (\frac {1}{2} \, x\right )\right ) - \frac {1}{8} \, \tan \left (\frac {1}{2} \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(I+tan(x)),x, algorithm="giac")

[Out]

-1/64*I*tan(1/2*x)^4 - 1/24*tan(1/2*x)^3 - 1/192*(50*I*tan(1/2*x)^4 + 24*tan(1/2*x)^3 + 8*tan(1/2*x) - 3*I)/ta
n(1/2*x)^4 + 1/8*I*log(tan(1/2*x)) - 1/8*tan(1/2*x)

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maple [A]  time = 0.19, size = 58, normalized size = 1.45 \[ -\frac {\tan \left (\frac {x}{2}\right )}{8}-\frac {i \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{64}-\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{24}-\frac {1}{24 \tan \left (\frac {x}{2}\right )^{3}}+\frac {i}{64 \tan \left (\frac {x}{2}\right )^{4}}+\frac {i \ln \left (\tan \left (\frac {x}{2}\right )\right )}{8}-\frac {1}{8 \tan \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^5/(I+tan(x)),x)

[Out]

-1/8*tan(1/2*x)-1/64*I*tan(1/2*x)^4-1/24*tan(1/2*x)^3-1/24/tan(1/2*x)^3+1/64*I/tan(1/2*x)^4+1/8*I*ln(tan(1/2*x
))-1/8/tan(1/2*x)

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maxima [B]  time = 0.51, size = 83, normalized size = 2.08 \[ -\frac {{\left (\frac {8 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {24 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - 3 i\right )} {\left (\cos \relax (x) + 1\right )}^{4}}{192 \, \sin \relax (x)^{4}} - \frac {\sin \relax (x)}{8 \, {\left (\cos \relax (x) + 1\right )}} - \frac {\sin \relax (x)^{3}}{24 \, {\left (\cos \relax (x) + 1\right )}^{3}} - \frac {i \, \sin \relax (x)^{4}}{64 \, {\left (\cos \relax (x) + 1\right )}^{4}} + \frac {1}{8} i \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^5/(I+tan(x)),x, algorithm="maxima")

[Out]

-1/192*(8*sin(x)/(cos(x) + 1) + 24*sin(x)^3/(cos(x) + 1)^3 - 3*I)*(cos(x) + 1)^4/sin(x)^4 - 1/8*sin(x)/(cos(x)
+ 1) - 1/24*sin(x)^3/(cos(x) + 1)^3 - 1/64*I*sin(x)^4/(cos(x) + 1)^4 + 1/8*I*log(sin(x)/(cos(x) + 1))

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mupad [B]  time = 3.70, size = 57, normalized size = 1.42 \[ -\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{8}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,1{}\mathrm {i}}{8}-\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )}{3}-\frac {1}{4}{}\mathrm {i}}{16\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{24}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,1{}\mathrm {i}}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^5*(tan(x) + 1i)),x)

[Out]

(log(tan(x/2))*1i)/8 - tan(x/2)/8 - ((2*tan(x/2))/3 + 2*tan(x/2)^3 - 1i/4)/(16*tan(x/2)^4) - tan(x/2)^3/24 - (
tan(x/2)^4*1i)/64

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{5}{\relax (x )}}{\tan {\relax (x )} + i}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**5/(I+tan(x)),x)

[Out]

Integral(csc(x)**5/(tan(x) + I), x)

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