3.83 \(\int \frac{1}{1-\cos ^5(x)} \, dx\)

Optimal. Leaf size=205 \[ \frac{2 \tan ^{-1}\left (\sqrt{\frac{1-\sqrt [5]{-1}}{1+\sqrt [5]{-1}}} \tan \left (\frac{x}{2}\right )\right )}{5 \sqrt{1-(-1)^{2/5}}}+\frac{2 \tan ^{-1}\left (\sqrt{\frac{1-(-1)^{3/5}}{1+(-1)^{3/5}}} \tan \left (\frac{x}{2}\right )\right )}{5 \sqrt{1+\sqrt [5]{-1}}}-\frac{\sin (x)}{5 (1-\cos (x))}-\frac{2 \tanh ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right )}{\sqrt{-\frac{1-(-1)^{2/5}}{1+(-1)^{2/5}}}}\right )}{5 \sqrt{(-1)^{4/5}-1}}+\frac{2 \tanh ^{-1}\left (\sqrt{-\frac{1+(-1)^{4/5}}{1-(-1)^{4/5}}} \tan \left (\frac{x}{2}\right )\right )}{5 \sqrt{-1-(-1)^{3/5}}} \]

[Out]

(2*ArcTan[Sqrt[(1 - (-1)^(1/5))/(1 + (-1)^(1/5))]*Tan[x/2]])/(5*Sqrt[1 - (-1)^(2/5)]) + (2*ArcTan[Sqrt[(1 - (-
1)^(3/5))/(1 + (-1)^(3/5))]*Tan[x/2]])/(5*Sqrt[1 + (-1)^(1/5)]) - (2*ArcTanh[Tan[x/2]/Sqrt[-((1 - (-1)^(2/5))/
(1 + (-1)^(2/5)))]])/(5*Sqrt[-1 + (-1)^(4/5)]) + (2*ArcTanh[Sqrt[-((1 + (-1)^(4/5))/(1 - (-1)^(4/5)))]*Tan[x/2
]])/(5*Sqrt[-1 - (-1)^(3/5)]) - Sin[x]/(5*(1 - Cos[x]))

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Rubi [A]  time = 0.473475, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3213, 2648, 2659, 205, 208} \[ \frac{2 \tan ^{-1}\left (\sqrt{\frac{1-\sqrt [5]{-1}}{1+\sqrt [5]{-1}}} \tan \left (\frac{x}{2}\right )\right )}{5 \sqrt{1-(-1)^{2/5}}}+\frac{2 \tan ^{-1}\left (\sqrt{\frac{1-(-1)^{3/5}}{1+(-1)^{3/5}}} \tan \left (\frac{x}{2}\right )\right )}{5 \sqrt{1+\sqrt [5]{-1}}}-\frac{\sin (x)}{5 (1-\cos (x))}-\frac{2 \tanh ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right )}{\sqrt{-\frac{1-(-1)^{2/5}}{1+(-1)^{2/5}}}}\right )}{5 \sqrt{(-1)^{4/5}-1}}+\frac{2 \tanh ^{-1}\left (\sqrt{-\frac{1+(-1)^{4/5}}{1-(-1)^{4/5}}} \tan \left (\frac{x}{2}\right )\right )}{5 \sqrt{-1-(-1)^{3/5}}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cos[x]^5)^(-1),x]

[Out]

(2*ArcTan[Sqrt[(1 - (-1)^(1/5))/(1 + (-1)^(1/5))]*Tan[x/2]])/(5*Sqrt[1 - (-1)^(2/5)]) + (2*ArcTan[Sqrt[(1 - (-
1)^(3/5))/(1 + (-1)^(3/5))]*Tan[x/2]])/(5*Sqrt[1 + (-1)^(1/5)]) - (2*ArcTanh[Tan[x/2]/Sqrt[-((1 - (-1)^(2/5))/
(1 + (-1)^(2/5)))]])/(5*Sqrt[-1 + (-1)^(4/5)]) + (2*ArcTanh[Sqrt[-((1 + (-1)^(4/5))/(1 - (-1)^(4/5)))]*Tan[x/2
]])/(5*Sqrt[-1 - (-1)^(3/5)]) - Sin[x]/(5*(1 - Cos[x]))

Rule 3213

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> Int[ExpandTrig[(a + b*(c*sin[e + f*
x])^n)^p, x], x] /; FreeQ[{a, b, c, e, f, n}, x] && (IGtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

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

Rule 208

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

Rubi steps

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

Mathematica [C]  time = 0.119013, size = 378, normalized size = 1.84 \[ -\frac{1}{5} \cot \left (\frac{x}{2}\right )+\frac{1}{10} \text{RootSum}\left [\text{$\#$1}^8+2 \text{$\#$1}^7+8 \text{$\#$1}^6+14 \text{$\#$1}^5+30 \text{$\#$1}^4+14 \text{$\#$1}^3+8 \text{$\#$1}^2+2 \text{$\#$1}+1\& ,\frac{-i \text{$\#$1}^6 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )-4 i \text{$\#$1}^5 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )-15 i \text{$\#$1}^4 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )-40 i \text{$\#$1}^3 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )-15 i \text{$\#$1}^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )-4 i \text{$\#$1} \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )-i \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )+2 \text{$\#$1}^6 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )+8 \text{$\#$1}^5 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )+30 \text{$\#$1}^4 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )+80 \text{$\#$1}^3 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )+30 \text{$\#$1}^2 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )+8 \text{$\#$1} \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )+2 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )}{4 \text{$\#$1}^7+7 \text{$\#$1}^6+24 \text{$\#$1}^5+35 \text{$\#$1}^4+60 \text{$\#$1}^3+21 \text{$\#$1}^2+8 \text{$\#$1}+1}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cos[x]^5)^(-1),x]

[Out]

-Cot[x/2]/5 + RootSum[1 + 2*#1 + 8*#1^2 + 14*#1^3 + 30*#1^4 + 14*#1^5 + 8*#1^6 + 2*#1^7 + #1^8 & , (2*ArcTan[S
in[x]/(Cos[x] - #1)] - I*Log[1 - 2*Cos[x]*#1 + #1^2] + 8*ArcTan[Sin[x]/(Cos[x] - #1)]*#1 - (4*I)*Log[1 - 2*Cos
[x]*#1 + #1^2]*#1 + 30*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^2 - (15*I)*Log[1 - 2*Cos[x]*#1 + #1^2]*#1^2 + 80*ArcTan
[Sin[x]/(Cos[x] - #1)]*#1^3 - (40*I)*Log[1 - 2*Cos[x]*#1 + #1^2]*#1^3 + 30*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^4 -
(15*I)*Log[1 - 2*Cos[x]*#1 + #1^2]*#1^4 + 8*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^5 - (4*I)*Log[1 - 2*Cos[x]*#1 + #
1^2]*#1^5 + 2*ArcTan[Sin[x]/(Cos[x] - #1)]*#1^6 - I*Log[1 - 2*Cos[x]*#1 + #1^2]*#1^6)/(1 + 8*#1 + 21*#1^2 + 60
*#1^3 + 35*#1^4 + 24*#1^5 + 7*#1^6 + 4*#1^7) & ]/10

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Maple [C]  time = 0.02, size = 62, normalized size = 0.3 \begin{align*} -{\frac{1}{5} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{1}{10}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}+10\,{{\it \_Z}}^{4}+5 \right ) }{\frac{{{\it \_R}}^{6}+5\,{{\it \_R}}^{4}+5\,{{\it \_R}}^{2}+5}{{{\it \_R}}^{7}+5\,{{\it \_R}}^{3}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cos(x)^5),x)

[Out]

-1/5/tan(1/2*x)+1/10*sum((_R^6+5*_R^4+5*_R^2+5)/(_R^7+5*_R^3)*ln(tan(1/2*x)-_R),_R=RootOf(_Z^8+10*_Z^4+5))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^5),x, algorithm="maxima")

[Out]

1/5*(5*(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1)*integrate(2/5*((cos(7*x) + 4*cos(6*x) + 15*cos(5*x) + 40*cos(4*x)
+ 15*cos(3*x) + 4*cos(2*x) + cos(x))*cos(8*x) + (16*cos(6*x) + 44*cos(5*x) + 110*cos(4*x) + 44*cos(3*x) + 16*c
os(2*x) + 4*cos(x) + 1)*cos(7*x) + 2*cos(7*x)^2 + 4*(44*cos(5*x) + 110*cos(4*x) + 44*cos(3*x) + 16*cos(2*x) +
4*cos(x) + 1)*cos(6*x) + 32*cos(6*x)^2 + (1010*cos(4*x) + 420*cos(3*x) + 176*cos(2*x) + 44*cos(x) + 15)*cos(5*
x) + 210*cos(5*x)^2 + 10*(101*cos(3*x) + 44*cos(2*x) + 11*cos(x) + 4)*cos(4*x) + 1200*cos(4*x)^2 + (176*cos(2*
x) + 44*cos(x) + 15)*cos(3*x) + 210*cos(3*x)^2 + 4*(4*cos(x) + 1)*cos(2*x) + 32*cos(2*x)^2 + 2*cos(x)^2 + (sin
(7*x) + 4*sin(6*x) + 15*sin(5*x) + 40*sin(4*x) + 15*sin(3*x) + 4*sin(2*x) + sin(x))*sin(8*x) + 2*(8*sin(6*x) +
22*sin(5*x) + 55*sin(4*x) + 22*sin(3*x) + 8*sin(2*x) + 2*sin(x))*sin(7*x) + 2*sin(7*x)^2 + 8*(22*sin(5*x) + 5
5*sin(4*x) + 22*sin(3*x) + 8*sin(2*x) + 2*sin(x))*sin(6*x) + 32*sin(6*x)^2 + 2*(505*sin(4*x) + 210*sin(3*x) +
88*sin(2*x) + 22*sin(x))*sin(5*x) + 210*sin(5*x)^2 + 10*(101*sin(3*x) + 44*sin(2*x) + 11*sin(x))*sin(4*x) + 12
00*sin(4*x)^2 + 44*(4*sin(2*x) + sin(x))*sin(3*x) + 210*sin(3*x)^2 + 32*sin(2*x)^2 + 16*sin(2*x)*sin(x) + 2*si
n(x)^2 + cos(x))/(2*(2*cos(7*x) + 8*cos(6*x) + 14*cos(5*x) + 30*cos(4*x) + 14*cos(3*x) + 8*cos(2*x) + 2*cos(x)
+ 1)*cos(8*x) + cos(8*x)^2 + 4*(8*cos(6*x) + 14*cos(5*x) + 30*cos(4*x) + 14*cos(3*x) + 8*cos(2*x) + 2*cos(x)
+ 1)*cos(7*x) + 4*cos(7*x)^2 + 16*(14*cos(5*x) + 30*cos(4*x) + 14*cos(3*x) + 8*cos(2*x) + 2*cos(x) + 1)*cos(6*
x) + 64*cos(6*x)^2 + 28*(30*cos(4*x) + 14*cos(3*x) + 8*cos(2*x) + 2*cos(x) + 1)*cos(5*x) + 196*cos(5*x)^2 + 60
*(14*cos(3*x) + 8*cos(2*x) + 2*cos(x) + 1)*cos(4*x) + 900*cos(4*x)^2 + 28*(8*cos(2*x) + 2*cos(x) + 1)*cos(3*x)
+ 196*cos(3*x)^2 + 16*(2*cos(x) + 1)*cos(2*x) + 64*cos(2*x)^2 + 4*cos(x)^2 + 4*(sin(7*x) + 4*sin(6*x) + 7*sin
(5*x) + 15*sin(4*x) + 7*sin(3*x) + 4*sin(2*x) + sin(x))*sin(8*x) + sin(8*x)^2 + 8*(4*sin(6*x) + 7*sin(5*x) + 1
5*sin(4*x) + 7*sin(3*x) + 4*sin(2*x) + sin(x))*sin(7*x) + 4*sin(7*x)^2 + 32*(7*sin(5*x) + 15*sin(4*x) + 7*sin(
3*x) + 4*sin(2*x) + sin(x))*sin(6*x) + 64*sin(6*x)^2 + 56*(15*sin(4*x) + 7*sin(3*x) + 4*sin(2*x) + sin(x))*sin
(5*x) + 196*sin(5*x)^2 + 120*(7*sin(3*x) + 4*sin(2*x) + sin(x))*sin(4*x) + 900*sin(4*x)^2 + 56*(4*sin(2*x) + s
in(x))*sin(3*x) + 196*sin(3*x)^2 + 64*sin(2*x)^2 + 32*sin(2*x)*sin(x) + 4*sin(x)^2 + 4*cos(x) + 1), x) - 2*sin
(x))/(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1)

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Fricas [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(1/(1-cos(x)^5),x, algorithm="fricas")

[Out]

Timed out

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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(1/(1-cos(x)**5),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{1}{\cos \left (x\right )^{5} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^5),x, algorithm="giac")

[Out]

integrate(-1/(cos(x)^5 - 1), x)