\(\int \frac {1}{(\frac {a}{b})^{2/n}+x^2-2 (\frac {a}{b})^{\frac {1}{n}} x \cos (\frac {\pi -2 k \pi }{n})} \, dx\) [98]

Optimal result

Integrand size = 40, antiderivative size = 62 \[ \int \frac {1}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {\pi -2 k \pi }{n}\right )} \, dx=-\left (\frac {a}{b}\right )^{-1/n} \arctan \left (\cot \left (\frac {\pi -2 k \pi }{n}\right )-\left (\frac {a}{b}\right )^{-1/n} x \csc \left (\frac {\pi -2 k \pi }{n}\right )\right ) \csc \left (\frac {\pi -2 k \pi }{n}\right ) \]

[Out]

arctan(-cot((-2*Pi*k+Pi)/n)+x*csc((-2*Pi*k+Pi)/n)/((a/b)^(1/n)))*csc((-2*Pi*k+Pi)/n)/((a/b)^(1/n))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {632, 210} \[ \int \frac {1}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {\pi -2 k \pi }{n}\right )} \, dx=-\left (\frac {a}{b}\right )^{-1/n} \csc \left (\frac {\pi -2 \pi k}{n}\right ) \arctan \left (\cot \left (\frac {\pi -2 \pi k}{n}\right )-x \left (\frac {a}{b}\right )^{-1/n} \csc \left (\frac {\pi -2 \pi k}{n}\right )\right ) \]

[In]

Int[((a/b)^(2/n) + x^2 - 2*(a/b)^n^(-1)*x*Cos[(Pi - 2*k*Pi)/n])^(-1),x]

[Out]

-((ArcTan[Cot[(Pi - 2*k*Pi)/n] - (x*Csc[(Pi - 2*k*Pi)/n])/(a/b)^n^(-1)]*Csc[(Pi - 2*k*Pi)/n])/(a/b)^n^(-1))

Rule 210

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

Rule 632

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]

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {1}{-x^2-4 \left (\frac {a}{b}\right )^{2/n} \left (1-\cos ^2\left (\frac {\pi -2 k \pi }{n}\right )\right )} \, dx,x,2 x-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi -2 k \pi }{n}\right )\right )\right ) \\ & = -\left (\frac {a}{b}\right )^{-1/n} \tan ^{-1}\left (\cot \left (\frac {\pi -2 k \pi }{n}\right )-\left (\frac {a}{b}\right )^{-1/n} x \csc \left (\frac {\pi -2 k \pi }{n}\right )\right ) \csc \left (\frac {\pi -2 k \pi }{n}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {\pi -2 k \pi }{n}\right )} \, dx=-\left (\frac {a}{b}\right )^{-1/n} \arctan \left (\cot \left (\frac {\pi -2 k \pi }{n}\right )-\left (\frac {a}{b}\right )^{-1/n} x \csc \left (\frac {\pi -2 k \pi }{n}\right )\right ) \csc \left (\frac {\pi -2 k \pi }{n}\right ) \]

[In]

Integrate[((a/b)^(2/n) + x^2 - 2*(a/b)^n^(-1)*x*Cos[(Pi - 2*k*Pi)/n])^(-1),x]

[Out]

-((ArcTan[Cot[(Pi - 2*k*Pi)/n] - (x*Csc[(Pi - 2*k*Pi)/n])/(a/b)^n^(-1)]*Csc[(Pi - 2*k*Pi)/n])/(a/b)^n^(-1))

Maple [A] (verified)

Time = 3.14 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.79

method	result	size
default	\(\frac {\arctan \left (\frac {2 x -2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi \left (2 k -1\right )}{n}\right )}{2 \sqrt {-\left (\frac {a}{b}\right )^{\frac {2}{n}} \left (\cos ^{2}\left (\frac {\pi \left (2 k -1\right )}{n}\right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{n}}}}\right )}{\sqrt {-\left (\frac {a}{b}\right )^{\frac {2}{n}} \left (\cos ^{2}\left (\frac {\pi \left (2 k -1\right )}{n}\right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{n}}}}\)	\(111\)
risch	\(\text {Expression too large to display}\)	\(1343\)

[In]

int(1/((a/b)^(2/n)+x^2-2*(a/b)^(1/n)*x*cos((-2*Pi*k+Pi)/n)),x,method=_RETURNVERBOSE)

[Out]

1/(-((a/b)^(1/n))^2*cos(Pi*(2*k-1)/n)^2+(a/b)^(2/n))^(1/2)*arctan(1/2*(2*x-2*(a/b)^(1/n)*cos(Pi*(2*k-1)/n))/(-
((a/b)^(1/n))^2*cos(Pi*(2*k-1)/n)^2+(a/b)^(2/n))^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {\pi -2 k \pi }{n}\right )} \, dx=-\frac {\arctan \left (\frac {\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) - x}{\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \sin \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )}\right )}{\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \sin \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )} \]

[In]

integrate(1/((a/b)^(2/n)+x^2-2*(a/b)^(1/n)*x*cos((-2*pi*k+pi)/n)),x, algorithm="fricas")

[Out]

-arctan(((a/b)^(1/n)*cos(2*pi*k/n - pi/n) - x)/((a/b)^(1/n)*sin(2*pi*k/n - pi/n)))/((a/b)^(1/n)*sin(2*pi*k/n -
pi/n))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (46) = 92\).

Time = 0.49 (sec) , antiderivative size = 212, normalized size of antiderivative = 3.42 \[ \int \frac {1}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {\pi -2 k \pi }{n}\right )} \, dx=- \frac {\sqrt {\frac {\left (\frac {a}{b}\right )^{- \frac {2}{n}}}{\cos ^{2}{\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1}} \log {\left (x - \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos {\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )} - \frac {\sqrt {\frac {\left (\frac {a}{b}\right )^{- \frac {2}{n}}}{\cos ^{2}{\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1}} \left (- 2 \left (\frac {a}{b}\right )^{\frac {2}{n}} \cos ^{2}{\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )} + 2 \left (\frac {a}{b}\right )^{\frac {2}{n}}\right )}{2} \right )}}{2} + \frac {\sqrt {\frac {\left (\frac {a}{b}\right )^{- \frac {2}{n}}}{\cos ^{2}{\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1}} \log {\left (x - \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos {\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )} + \frac {\sqrt {\frac {\left (\frac {a}{b}\right )^{- \frac {2}{n}}}{\cos ^{2}{\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1}} \left (- 2 \left (\frac {a}{b}\right )^{\frac {2}{n}} \cos ^{2}{\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )} + 2 \left (\frac {a}{b}\right )^{\frac {2}{n}}\right )}{2} \right )}}{2} \]

[In]

integrate(1/((a/b)**(2/n)+x**2-2*(a/b)**(1/n)*x*cos((-2*pi*k+pi)/n)),x)

[Out]

-sqrt(1/((a/b)**(2/n)*(cos(pi*(2*k - 1)/n)**2 - 1)))*log(x - (a/b)**(1/n)*cos(2*pi*k/n - pi/n) - sqrt(1/((a/b)
**(2/n)*(cos(pi*(2*k - 1)/n)**2 - 1)))*(-2*(a/b)**(2/n)*cos(2*pi*k/n - pi/n)**2 + 2*(a/b)**(2/n))/2)/2 + sqrt(
1/((a/b)**(2/n)*(cos(pi*(2*k - 1)/n)**2 - 1)))*log(x - (a/b)**(1/n)*cos(2*pi*k/n - pi/n) + sqrt(1/((a/b)**(2/n
)*(cos(pi*(2*k - 1)/n)**2 - 1)))*(-2*(a/b)**(2/n)*cos(2*pi*k/n - pi/n)**2 + 2*(a/b)**(2/n))/2)/2

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {\pi -2 k \pi }{n}\right )} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/((a/b)^(2/n)+x^2-2*(a/b)^(1/n)*x*cos((-2*pi*k+pi)/n)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(1>0)', see `assume?` for more
details)Is 1

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.61 \[ \int \frac {1}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {\pi -2 k \pi }{n}\right )} \, dx=\frac {\arctan \left (-\frac {\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \cos \left (-\frac {2 \, \pi k}{n} + \frac {\pi }{n}\right ) - x}{\sqrt {-\cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )^{2} + 1} \left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )}}\right )}{\sqrt {-\cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )^{2} + 1} \left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )}} \]

[In]

integrate(1/((a/b)^(2/n)+x^2-2*(a/b)^(1/n)*x*cos((-2*pi*k+pi)/n)),x, algorithm="giac")

[Out]

arctan(-((a/b)^(1/n)*cos(-2*pi*k/n + pi/n) - x)/(sqrt(-cos(2*pi*k/n - pi/n)^2 + 1)*(a/b)^(1/n)))/(sqrt(-cos(2*
pi*k/n - pi/n)^2 + 1)*(a/b)^(1/n))

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.77 \[ \int \frac {1}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {\pi -2 k \pi }{n}\right )} \, dx=-\frac {\mathrm {atanh}\left (\frac {x-\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )\,{\left (\frac {a}{b}\right )}^{1/n}}{\sqrt {\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )-1}\,\sqrt {\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )+1}\,{\left (\frac {a}{b}\right )}^{1/n}}\right )}{\sqrt {\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )-1}\,\sqrt {\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )+1}\,{\left (\frac {a}{b}\right )}^{1/n}} \]

[In]

int(1/((a/b)^(2/n) + x^2 - 2*x*cos((Pi - 2*Pi*k)/n)*(a/b)^(1/n)),x)

[Out]

-atanh((x - cos((Pi*(2*k - 1))/n)*(a/b)^(1/n))/((cos((Pi*(2*k - 1))/n) - 1)^(1/2)*(cos((Pi*(2*k - 1))/n) + 1)^
(1/2)*(a/b)^(1/n)))/((cos((Pi*(2*k - 1))/n) - 1)^(1/2)*(cos((Pi*(2*k - 1))/n) + 1)^(1/2)*(a/b)^(1/n))