3.11 \(\int \frac {\csc ^7(x)}{i+\cot (x)} \, dx\)

Optimal. Leaf size=40 \[ -\frac {1}{5} \csc ^5(x)+\frac {3}{8} i \tanh ^{-1}(\cos (x))+\frac {1}{4} i \cot (x) \csc ^3(x)+\frac {3}{8} i \cot (x) \csc (x) \]

[Out]

3/8*I*arctanh(cos(x))+3/8*I*cot(x)*csc(x)+1/4*I*cot(x)*csc(x)^3-1/5*csc(x)^5

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Rubi [A]  time = 0.05, antiderivative size = 40, 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 = {3501, 3768, 3770} \[ -\frac {1}{5} \csc ^5(x)+\frac {3}{8} i \tanh ^{-1}(\cos (x))+\frac {1}{4} i \cot (x) \csc ^3(x)+\frac {3}{8} i \cot (x) \csc (x) \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^7/(I + Cot[x]),x]

[Out]

((3*I)/8)*ArcTanh[Cos[x]] + ((3*I)/8)*Cot[x]*Csc[x] + (I/4)*Cot[x]*Csc[x]^3 - Csc[x]^5/5

Rule 3501

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

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 ^7(x)}{i+\cot (x)} \, dx &=-\frac {1}{5} \csc ^5(x)-i \int \csc ^5(x) \, dx\\ &=\frac {1}{4} i \cot (x) \csc ^3(x)-\frac {\csc ^5(x)}{5}-\frac {3}{4} i \int \csc ^3(x) \, dx\\ &=\frac {3}{8} i \cot (x) \csc (x)+\frac {1}{4} i \cot (x) \csc ^3(x)-\frac {\csc ^5(x)}{5}-\frac {3}{8} i \int \csc (x) \, dx\\ &=\frac {3}{8} i \tanh ^{-1}(\cos (x))+\frac {3}{8} i \cot (x) \csc (x)+\frac {1}{4} i \cot (x) \csc ^3(x)-\frac {\csc ^5(x)}{5}\\ \end {align*}

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Mathematica [B]  time = 0.15, size = 99, normalized size = 2.48 \[ \frac {1}{640} i \csc ^5(x) \left (140 \sin (2 x)-30 \sin (4 x)+75 \sin (3 x) \log \left (\sin \left (\frac {x}{2}\right )\right )-15 \sin (5 x) \log \left (\sin \left (\frac {x}{2}\right )\right )+150 \sin (x) \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )-75 \sin (3 x) \log \left (\cos \left (\frac {x}{2}\right )\right )+15 \sin (5 x) \log \left (\cos \left (\frac {x}{2}\right )\right )+128 i\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^7/(I + Cot[x]),x]

[Out]

(I/640)*Csc[x]^5*(128*I + 150*(Log[Cos[x/2]] - Log[Sin[x/2]])*Sin[x] + 140*Sin[2*x] - 75*Log[Cos[x/2]]*Sin[3*x
] + 75*Log[Sin[x/2]]*Sin[3*x] - 30*Sin[4*x] + 15*Log[Cos[x/2]]*Sin[5*x] - 15*Log[Sin[x/2]]*Sin[5*x])

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fricas [B]  time = 0.76, size = 145, normalized size = 3.62 \[ \frac {{\left (15 i \, e^{\left (10 i \, x\right )} - 75 i \, e^{\left (8 i \, x\right )} + 150 i \, e^{\left (6 i \, x\right )} - 150 i \, e^{\left (4 i \, x\right )} + 75 i \, e^{\left (2 i \, x\right )} - 15 i\right )} \log \left (e^{\left (i \, x\right )} + 1\right ) + {\left (-15 i \, e^{\left (10 i \, x\right )} + 75 i \, e^{\left (8 i \, x\right )} - 150 i \, e^{\left (6 i \, x\right )} + 150 i \, e^{\left (4 i \, x\right )} - 75 i \, e^{\left (2 i \, x\right )} + 15 i\right )} \log \left (e^{\left (i \, x\right )} - 1\right ) - 30 i \, e^{\left (9 i \, x\right )} + 140 i \, e^{\left (7 i \, x\right )} - 256 i \, e^{\left (5 i \, x\right )} - 140 i \, e^{\left (3 i \, x\right )} + 30 i \, e^{\left (i \, x\right )}}{40 \, {\left (e^{\left (10 i \, x\right )} - 5 \, e^{\left (8 i \, x\right )} + 10 \, e^{\left (6 i \, x\right )} - 10 \, e^{\left (4 i \, x\right )} + 5 \, e^{\left (2 i \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^7/(I+cot(x)),x, algorithm="fricas")

[Out]

1/40*((15*I*e^(10*I*x) - 75*I*e^(8*I*x) + 150*I*e^(6*I*x) - 150*I*e^(4*I*x) + 75*I*e^(2*I*x) - 15*I)*log(e^(I*
x) + 1) + (-15*I*e^(10*I*x) + 75*I*e^(8*I*x) - 150*I*e^(6*I*x) + 150*I*e^(4*I*x) - 75*I*e^(2*I*x) + 15*I)*log(
e^(I*x) - 1) - 30*I*e^(9*I*x) + 140*I*e^(7*I*x) - 256*I*e^(5*I*x) - 140*I*e^(3*I*x) + 30*I*e^(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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giac [B]  time = 0.37, size = 94, normalized size = 2.35 \[ -\frac {1}{160} \, \tan \left (\frac {1}{2} \, x\right )^{5} - \frac {1}{64} i \, \tan \left (\frac {1}{2} \, x\right )^{4} - \frac {1}{32} \, \tan \left (\frac {1}{2} \, x\right )^{3} - \frac {1}{8} i \, \tan \left (\frac {1}{2} \, x\right )^{2} - \frac {-274 i \, \tan \left (\frac {1}{2} \, x\right )^{5} + 20 \, \tan \left (\frac {1}{2} \, x\right )^{4} - 40 i \, \tan \left (\frac {1}{2} \, x\right )^{3} + 10 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 5 i \, \tan \left (\frac {1}{2} \, x\right ) + 2}{320 \, \tan \left (\frac {1}{2} \, x\right )^{5}} - \frac {3}{8} i \, \log \left (\tan \left (\frac {1}{2} \, x\right )\right ) - \frac {1}{16} \, \tan \left (\frac {1}{2} \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^7/(I+cot(x)),x, algorithm="giac")

[Out]

-1/160*tan(1/2*x)^5 - 1/64*I*tan(1/2*x)^4 - 1/32*tan(1/2*x)^3 - 1/8*I*tan(1/2*x)^2 - 1/320*(-274*I*tan(1/2*x)^
5 + 20*tan(1/2*x)^4 - 40*I*tan(1/2*x)^3 + 10*tan(1/2*x)^2 - 5*I*tan(1/2*x) + 2)/tan(1/2*x)^5 - 3/8*I*log(tan(1
/2*x)) - 1/16*tan(1/2*x)

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maple [B]  time = 0.33, size = 92, normalized size = 2.30 \[ -\frac {\tan \left (\frac {x}{2}\right )}{16}-\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{160}-\frac {i \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{64}-\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{32}-\frac {i \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{8}+\frac {i}{8 \tan \left (\frac {x}{2}\right )^{2}}-\frac {1}{32 \tan \left (\frac {x}{2}\right )^{3}}+\frac {i}{64 \tan \left (\frac {x}{2}\right )^{4}}-\frac {3 i \ln \left (\tan \left (\frac {x}{2}\right )\right )}{8}-\frac {1}{160 \tan \left (\frac {x}{2}\right )^{5}}-\frac {1}{16 \tan \left (\frac {x}{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^7/(I+cot(x)),x)

[Out]

-1/16*tan(1/2*x)-1/160*tan(1/2*x)^5-1/64*I*tan(1/2*x)^4-1/32*tan(1/2*x)^3-1/8*I*tan(1/2*x)^2+1/8*I/tan(1/2*x)^
2-1/32/tan(1/2*x)^3+1/64*I/tan(1/2*x)^4-3/8*I*ln(tan(1/2*x))-1/160/tan(1/2*x)^5-1/16/tan(1/2*x)

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maxima [B]  time = 0.42, size = 131, normalized size = 3.28 \[ -\frac {{\left (-\frac {15 i \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {30 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {120 i \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {60 \, \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + 6\right )} {\left (\cos \relax (x) + 1\right )}^{5}}{960 \, \sin \relax (x)^{5}} - \frac {\sin \relax (x)}{16 \, {\left (\cos \relax (x) + 1\right )}} - \frac {i \, \sin \relax (x)^{2}}{8 \, {\left (\cos \relax (x) + 1\right )}^{2}} - \frac {\sin \relax (x)^{3}}{32 \, {\left (\cos \relax (x) + 1\right )}^{3}} - \frac {i \, \sin \relax (x)^{4}}{64 \, {\left (\cos \relax (x) + 1\right )}^{4}} - \frac {\sin \relax (x)^{5}}{160 \, {\left (\cos \relax (x) + 1\right )}^{5}} - \frac {3}{8} i \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^7/(I+cot(x)),x, algorithm="maxima")

[Out]

-1/960*(-15*I*sin(x)/(cos(x) + 1) + 30*sin(x)^2/(cos(x) + 1)^2 - 120*I*sin(x)^3/(cos(x) + 1)^3 + 60*sin(x)^4/(
cos(x) + 1)^4 + 6)*(cos(x) + 1)^5/sin(x)^5 - 1/16*sin(x)/(cos(x) + 1) - 1/8*I*sin(x)^2/(cos(x) + 1)^2 - 1/32*s
in(x)^3/(cos(x) + 1)^3 - 1/64*I*sin(x)^4/(cos(x) + 1)^4 - 1/160*sin(x)^5/(cos(x) + 1)^5 - 3/8*I*log(sin(x)/(co
s(x) + 1))

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mupad [B]  time = 0.30, size = 89, normalized size = 2.22 \[ -\frac {\mathrm {cot}\left (\frac {x}{2}\right )}{16}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{16}-\frac {{\mathrm {cot}\left (\frac {x}{2}\right )}^3}{32}-\frac {{\mathrm {cot}\left (\frac {x}{2}\right )}^5}{160}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{32}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{160}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,3{}\mathrm {i}}{8}+\frac {{\mathrm {cot}\left (\frac {x}{2}\right )}^2\,1{}\mathrm {i}}{8}+\frac {{\mathrm {cot}\left (\frac {x}{2}\right )}^4\,1{}\mathrm {i}}{64}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,1{}\mathrm {i}}{8}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,1{}\mathrm {i}}{64} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^7*(cot(x) + 1i)),x)

[Out]

(cot(x/2)^2*1i)/8 - tan(x/2)/16 - (log(tan(x/2))*3i)/8 - cot(x/2)/16 - cot(x/2)^3/32 + (cot(x/2)^4*1i)/64 - co
t(x/2)^5/160 - (tan(x/2)^2*1i)/8 - tan(x/2)^3/32 - (tan(x/2)^4*1i)/64 - tan(x/2)^5/160

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**7/(I+cot(x)),x)

[Out]

Timed out

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