3.2235 \(\int \frac {2 ((\frac {a}{b})^{\frac {1}{n}}-x \cos (\frac {(-1+2 k) \pi }{n}))}{(\frac {a}{b})^{2/n}+x^2-2 (\frac {a}{b})^{\frac {1}{n}} x \cos (\frac {(-1+2 k) \pi }{n})} \, dx\)

Optimal. Leaf size=114 \[ 2 \sin \left (\frac {\pi -2 \pi k}{n}\right ) \tan ^{-1}\left (\left (\frac {a}{b}\right )^{-1/n} \csc \left (\frac {\pi -2 \pi k}{n}\right ) \left (x-\left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi -2 \pi k}{n}\right )\right )\right )-\cos \left (\frac {\pi -2 \pi k}{n}\right ) \log \left (-2 x \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi -2 \pi k}{n}\right )+\left (\frac {a}{b}\right )^{2/n}+x^2\right ) \]

[Out]

-cos((-2*Pi*k+Pi)/n)*ln((a/b)^(2/n)+x^2-2*(a/b)^(1/n)*x*cos((-2*Pi*k+Pi)/n))+2*arctan((x-(a/b)^(1/n)*cos((-2*P
i*k+Pi)/n))*csc((-2*Pi*k+Pi)/n)/((a/b)^(1/n)))*sin((-2*Pi*k+Pi)/n)

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Rubi [A]  time = 0.26, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {12, 634, 618, 204, 628} \[ 2 \sin \left (\frac {\pi -2 \pi k}{n}\right ) \tan ^{-1}\left (\left (\frac {a}{b}\right )^{-1/n} \csc \left (\frac {\pi -2 \pi k}{n}\right ) \left (x-\left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi -2 \pi k}{n}\right )\right )\right )-\cos \left (\frac {\pi -2 \pi k}{n}\right ) \log \left (-2 x \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi -2 \pi k}{n}\right )+\left (\frac {a}{b}\right )^{2/n}+x^2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[(Pi - 2*k*Pi)/n]*Log[(a/b)^(2/n) + x^2 - 2*(a/b)^n^(-1)*x*Cos[(Pi - 2*k*Pi)/n]]) + 2*ArcTan[((x - (a/b)^
n^(-1)*Cos[(Pi - 2*k*Pi)/n])*Csc[(Pi - 2*k*Pi)/n])/(a/b)^n^(-1)]*Sin[(Pi - 2*k*Pi)/n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

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 628

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

Rule 634

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

Rubi steps

\begin {align*} \int \frac {2 \left (\left (\frac {a}{b}\right )^{\frac {1}{n}}-x \cos \left (\frac {(-1+2 k) \pi }{n}\right )\right )}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {(-1+2 k) \pi }{n}\right )} \, dx &=2 \int \frac {\left (\frac {a}{b}\right )^{\frac {1}{n}}-x \cos \left (\frac {(-1+2 k) \pi }{n}\right )}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {(-1+2 k) \pi }{n}\right )} \, dx\\ &=-\left (\cos \left (\frac {(-1+2 k) \pi }{n}\right ) \int \frac {2 x-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {(-1+2 k) \pi }{n}\right )}{\left (\frac {a}{b}\right )^{2/n}+x^2-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} x \cos \left (\frac {(-1+2 k) \pi }{n}\right )} \, dx\right )+\left (2 \left (\frac {a}{b}\right )^{\frac {1}{n}}-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos ^2\left (\frac {(-1+2 k) \pi }{n}\right )\right ) \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 {(-1+2 k) \pi }{n}\right )} \, dx\\ &=-\cos \left (\frac {(1-2 k) \pi }{n}\right ) \log \left (\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 )\right )+\left (2 \left (-2 \left (\frac {a}{b}\right )^{\frac {1}{n}}+2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos ^2\left (\frac {(-1+2 k) \pi }{n}\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-x^2-4 \left (\frac {a}{b}\right )^{2/n} \sin ^2\left (\frac {(1-2 k) \pi }{n}\right )} \, dx,x,2 x-2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {(-1+2 k) \pi }{n}\right )\right )\\ &=-\cos \left (\frac {(1-2 k) \pi }{n}\right ) \log \left (\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 )\right )+2 \tan ^{-1}\left (\left (\frac {a}{b}\right )^{-1/n} \left (x-\left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {(1-2 k) \pi }{n}\right )\right ) \csc \left (\frac {\pi -2 k \pi }{n}\right )\right ) \csc \left (\frac {\pi -2 k \pi }{n}\right ) \sin ^2\left (\frac {(1-2 k) \pi }{n}\right )\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 111, normalized size = 0.97 \[ 2 \left (\sin \left (\frac {\pi (2 k-1)}{n}\right ) \tan ^{-1}\left (\frac {\tan \left (\frac {\pi (2 k-1)}{2 n}\right ) \left (\left (\frac {a}{b}\right )^{\frac {1}{n}}+x\right )}{\left (\frac {a}{b}\right )^{\frac {1}{n}}-x}\right )-\frac {1}{2} \cos \left (\frac {\pi (2 k-1)}{n}\right ) \log \left (-2 x \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi (2 k-1)}{n}\right )+\left (\frac {a}{b}\right )^{2/n}+x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*(-1/2*(Cos[((-1 + 2*k)*Pi)/n]*Log[(a/b)^(2/n) + x^2 - 2*(a/b)^n^(-1)*x*Cos[((-1 + 2*k)*Pi)/n]]) + ArcTan[(((
a/b)^n^(-1) + x)*Tan[((-1 + 2*k)*Pi)/(2*n)])/((a/b)^n^(-1) - x)]*Sin[((-1 + 2*k)*Pi)/n])

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fricas [A]  time = 0.68, size = 162, normalized size = 1.42 \[ -\cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) \log \left (-\frac {2 \, {\left (2 \, x \left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) - x^{2} - \left (\frac {a}{b}\right )^{\frac {2}{n}}\right )}}{\cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) + 1}\right ) - 2 \, \arctan \left (\frac {\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) - x}{\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \sin \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )}\right ) \sin \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 202, normalized size = 1.77 \[ -\cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) \log \left (-2 \, x \left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) + x^{2} + \left (\frac {a}{b}\right )^{\frac {2}{n}}\right ) - \frac {2 \, {\left (\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )^{2} - \left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )}\right )} \arctan \left (-\frac {\left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )} \cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right ) - x}{\sqrt {-\cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )^{2} + 1} \left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )}}\right )}{\sqrt {-\cos \left (\frac {2 \, \pi k}{n} - \frac {\pi }{n}\right )^{2} + 1} \left (\frac {a}{b}\right )^{\left (\frac {1}{n}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.53, size = 311, normalized size = 2.73 \[ \frac {2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \arctan \left (\frac {2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi \left (2 k -1\right )}{n}\right )-2 x}{2 \sqrt {-\left (\frac {a}{b}\right )^{\frac {2}{n}} \left (\cos ^{2}\left (\frac {\pi \left (2 k -1\right )}{n}\right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{n}}}}\right ) \left (\cos ^{2}\left (\frac {\pi \left (2 k -1\right )}{n}\right )\right )}{\sqrt {-\left (\frac {a}{b}\right )^{\frac {2}{n}} \left (\cos ^{2}\left (\frac {\pi \left (2 k -1\right )}{n}\right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{n}}}}-\frac {2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \arctan \left (\frac {2 \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi \left (2 k -1\right )}{n}\right )-2 x}{2 \sqrt {-\left (\frac {a}{b}\right )^{\frac {2}{n}} \left (\cos ^{2}\left (\frac {\pi \left (2 k -1\right )}{n}\right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{n}}}}\right )}{\sqrt {-\left (\frac {a}{b}\right )^{\frac {2}{n}} \left (\cos ^{2}\left (\frac {\pi \left (2 k -1\right )}{n}\right )\right )+\left (\frac {a}{b}\right )^{\frac {2}{n}}}}-\cos \left (\frac {\pi \left (2 k -1\right )}{n}\right ) \ln \left (2 x \left (\frac {a}{b}\right )^{\frac {1}{n}} \cos \left (\frac {\pi \left (2 k -1\right )}{n}\right )-x^{2}-\left (\frac {a}{b}\right )^{\frac {2}{n}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*((a/b)^(1/n)-x*cos((-1+2*k)*pi/n))/((a/b)^(2/n)+x^2-2*(a/b)^(1/n)*x*cos((-1+2*k)*pi/n)),x, algorit
hm="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 zero or nonzero?

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mupad [B]  time = 1.03, size = 178, normalized size = 1.56 \[ -2\,\mathrm {atan}\left (\frac {2\,x\,\sqrt {1-{\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )}^2}-2\,\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )\,{\left (\frac {a}{b}\right )}^{1/n}\,\sqrt {1-{\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )}^2}}{2\,{\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )}^2\,{\left (\frac {a}{b}\right )}^{1/n}-2\,{\left (\frac {a}{b}\right )}^{1/n}}\right )\,\sqrt {1-{\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )}^2}-\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )\,\ln \left ({\left (\frac {a}{b}\right )}^{2/n}+x^2-2\,x\,\cos \left (\frac {\Pi \,\left (2\,k-1\right )}{n}\right )\,{\left (\frac {a}{b}\right )}^{1/n}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.87, size = 177, normalized size = 1.55 \[ - \left (- \sqrt {\left (\cos {\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1\right ) \left (\cos {\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} + 1\right )} + \cos {\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )}\right ) \log {\left (x - \left (\frac {a}{b}\right )^{\frac {1}{n}} \left (- \sqrt {\left (\cos {\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1\right ) \left (\cos {\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} + 1\right )} + \cos {\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )}\right ) \right )} - \left (\sqrt {\left (\cos {\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1\right ) \left (\cos {\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} + 1\right )} + \cos {\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )}\right ) \log {\left (x - \left (\frac {a}{b}\right )^{\frac {1}{n}} \left (\sqrt {\left (\cos {\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} - 1\right ) \left (\cos {\left (\frac {\pi \left (2 k - 1\right )}{n} \right )} + 1\right )} + \cos {\left (\frac {2 \pi k}{n} - \frac {\pi }{n} \right )}\right ) \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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