3.258 \(\int \frac{1}{1-\sin ^5(x)} \, dx\)

Optimal. Leaf size=187 \[ -\frac{2 \tan ^{-1}\left (\frac{(-1)^{2/5}-\tan \left (\frac{x}{2}\right )}{\sqrt{1-(-1)^{4/5}}}\right )}{5 \sqrt{1-(-1)^{4/5}}}-\frac{2 \tan ^{-1}\left (\frac{(-1)^{4/5}-\tan \left (\frac{x}{2}\right )}{\sqrt{1+(-1)^{3/5}}}\right )}{5 \sqrt{1+(-1)^{3/5}}}+\frac{2 \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right )+\sqrt [5]{-1}}{\sqrt{1-(-1)^{2/5}}}\right )}{5 \sqrt{1-(-1)^{2/5}}}+\frac{2 \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right )+(-1)^{3/5}}{\sqrt{1+\sqrt [5]{-1}}}\right )}{5 \sqrt{1+\sqrt [5]{-1}}}+\frac{\cos (x)}{5 (1-\sin (x))} \]

[Out]

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

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Rubi [A]  time = 0.268514, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3213, 2648, 2660, 618, 204} \[ -\frac{2 \tan ^{-1}\left (\frac{(-1)^{2/5}-\tan \left (\frac{x}{2}\right )}{\sqrt{1-(-1)^{4/5}}}\right )}{5 \sqrt{1-(-1)^{4/5}}}-\frac{2 \tan ^{-1}\left (\frac{(-1)^{4/5}-\tan \left (\frac{x}{2}\right )}{\sqrt{1+(-1)^{3/5}}}\right )}{5 \sqrt{1+(-1)^{3/5}}}+\frac{2 \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right )+\sqrt [5]{-1}}{\sqrt{1-(-1)^{2/5}}}\right )}{5 \sqrt{1-(-1)^{2/5}}}+\frac{2 \tan ^{-1}\left (\frac{\tan \left (\frac{x}{2}\right )+(-1)^{3/5}}{\sqrt{1+\sqrt [5]{-1}}}\right )}{5 \sqrt{1+\sqrt [5]{-1}}}+\frac{\cos (x)}{5 (1-\sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Sin[x]^5)^(-1),x]

[Out]

(-2*ArcTan[((-1)^(2/5) - Tan[x/2])/Sqrt[1 - (-1)^(4/5)]])/(5*Sqrt[1 - (-1)^(4/5)]) - (2*ArcTan[((-1)^(4/5) - T
an[x/2])/Sqrt[1 + (-1)^(3/5)]])/(5*Sqrt[1 + (-1)^(3/5)]) + (2*ArcTan[((-1)^(1/5) + Tan[x/2])/Sqrt[1 - (-1)^(2/
5)]])/(5*Sqrt[1 - (-1)^(2/5)]) + (2*ArcTan[((-1)^(3/5) + Tan[x/2])/Sqrt[1 + (-1)^(1/5)]])/(5*Sqrt[1 + (-1)^(1/
5)]) + Cos[x]/(5*(1 - Sin[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 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

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

Mathematica [C]  time = 0.134343, size = 413, normalized size = 2.21 \[ \frac{2 \sin \left (\frac{x}{2}\right )}{5 \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )}+\frac{1}{10} i \text{RootSum}\left [\text{$\#$1}^8+2 i \text{$\#$1}^7-8 \text{$\#$1}^6-14 i \text{$\#$1}^5+30 \text{$\#$1}^4+14 i \text{$\#$1}^3-8 \text{$\#$1}^2-2 i \text{$\#$1}+1\& ,\frac{-i \text{$\#$1}^6 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )+4 \text{$\#$1}^5 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )+15 i \text{$\#$1}^4 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )-40 \text{$\#$1}^3 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )-15 i \text{$\#$1}^2 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )+4 \text{$\#$1} \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )+i \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (x)+1\right )+2 \text{$\#$1}^6 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )+8 i \text{$\#$1}^5 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )-30 \text{$\#$1}^4 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )-80 i \text{$\#$1}^3 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )+30 \text{$\#$1}^2 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )+8 i \text{$\#$1} \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )-2 \tan ^{-1}\left (\frac{\sin (x)}{\cos (x)-\text{$\#$1}}\right )}{4 \text{$\#$1}^7+7 i \text{$\#$1}^6-24 \text{$\#$1}^5-35 i \text{$\#$1}^4+60 \text{$\#$1}^3+21 i \text{$\#$1}^2-8 \text{$\#$1}-i}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Sin[x]^5)^(-1),x]

[Out]

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

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Maple [C]  time = 0.082, size = 133, normalized size = 0.7 \begin{align*}{\frac{2}{5}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}+2\,{{\it \_Z}}^{7}+8\,{{\it \_Z}}^{6}+14\,{{\it \_Z}}^{5}+30\,{{\it \_Z}}^{4}+14\,{{\it \_Z}}^{3}+8\,{{\it \_Z}}^{2}+2\,{\it \_Z}+1 \right ) }{\frac{2\,{{\it \_R}}^{6}+3\,{{\it \_R}}^{5}+10\,{{\it \_R}}^{4}+10\,{{\it \_R}}^{3}+10\,{{\it \_R}}^{2}+3\,{\it \_R}+2}{4\,{{\it \_R}}^{7}+7\,{{\it \_R}}^{6}+24\,{{\it \_R}}^{5}+35\,{{\it \_R}}^{4}+60\,{{\it \_R}}^{3}+21\,{{\it \_R}}^{2}+8\,{\it \_R}+1}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -{\it \_R} \right ) }}-{\frac{2}{5} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*sum((2*_R^6+3*_R^5+10*_R^4+10*_R^3+10*_R^2+3*_R+2)/(4*_R^7+7*_R^6+24*_R^5+35*_R^4+60*_R^3+21*_R^2+8*_R+1)*
ln(tan(1/2*x)-_R),_R=RootOf(_Z^8+2*_Z^7+8*_Z^6+14*_Z^5+30*_Z^4+14*_Z^3+8*_Z^2+2*_Z+1))-2/5/(tan(1/2*x)-1)

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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-sin(x)^5),x, algorithm="maxima")

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-1/(sin(x)^5 - 1), x)