3.21 \(\int \frac{\cot ^2(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=61 \[ \frac{2 \sqrt{a^2-b^2} \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a b}-\frac{x}{a}-\frac{\tanh ^{-1}(\cos (x))}{b} \]

[Out]

-(x/a) - ArcTanh[Cos[x]]/b + (2*Sqrt[a^2 - b^2]*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a*b)

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Rubi [A]  time = 0.176036, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {3894, 4051, 3770, 3919, 3831, 2660, 618, 206} \[ \frac{2 \sqrt{a^2-b^2} \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a b}-\frac{x}{a}-\frac{\tanh ^{-1}(\cos (x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^2/(a + b*Csc[x]),x]

[Out]

-(x/a) - ArcTanh[Cos[x]]/b + (2*Sqrt[a^2 - b^2]*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a*b)

Rule 3894

Int[cot[(c_.) + (d_.)*(x_)]^2*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(-1 + Csc[c + d*x]
^2)*(a + b*Csc[c + d*x])^n, x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0]

Rule 4051

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[C/b, I
nt[Csc[e + f*x], x], x] + Dist[1/b, Int[(A*b - a*C*Csc[e + f*x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, A, C}, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[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 206

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

Rubi steps

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

Mathematica [A]  time = 0.0676879, size = 71, normalized size = 1.16 \[ \frac{2 \sqrt{b^2-a^2} \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )+a \log \left (\sin \left (\frac{x}{2}\right )\right )-a \log \left (\cos \left (\frac{x}{2}\right )\right )-b x}{a b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^2/(a + b*Csc[x]),x]

[Out]

(-(b*x) + 2*Sqrt[-a^2 + b^2]*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]] - a*Log[Cos[x/2]] + a*Log[Sin[x/2]])/(a
*b)

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Maple [A]  time = 0.064, size = 105, normalized size = 1.7 \begin{align*} -2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{a}}+{\frac{1}{b}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }-2\,{\frac{a}{b\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) }+2\,{\frac{b}{a\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^2/(a+b*csc(x)),x)

[Out]

-2/a*arctan(tan(1/2*x))+1/b*ln(tan(1/2*x))-2*a/b/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(
1/2))+2*b/a/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^2/(a+b*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.590871, size = 539, normalized size = 8.84 \begin{align*} \left [-\frac{2 \, b x + a \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) - a \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) - \sqrt{a^{2} - b^{2}} \log \left (\frac{{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} + 2 \,{\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt{a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right )}{2 \, a b}, -\frac{2 \, b x + a \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) - a \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) - 2 \, \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right )}{2 \, a b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^2/(a+b*csc(x)),x, algorithm="fricas")

[Out]

[-1/2*(2*b*x + a*log(1/2*cos(x) + 1/2) - a*log(-1/2*cos(x) + 1/2) - sqrt(a^2 - b^2)*log(((a^2 - 2*b^2)*cos(x)^
2 + 2*a*b*sin(x) + a^2 + b^2 + 2*(b*cos(x)*sin(x) + a*cos(x))*sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) -
a^2 - b^2)))/(a*b), -1/2*(2*b*x + a*log(1/2*cos(x) + 1/2) - a*log(-1/2*cos(x) + 1/2) - 2*sqrt(-a^2 + b^2)*arct
an(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x))))/(a*b)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{2}{\left (x \right )}}{a + b \csc{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**2/(a+b*csc(x)),x)

[Out]

Integral(cot(x)**2/(a + b*csc(x)), x)

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Giac [A]  time = 1.30169, size = 108, normalized size = 1.77 \begin{align*} -\frac{x}{a} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{b} - \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (\frac{1}{2} \, x\right ) + a}{\sqrt{-a^{2} + b^{2}}}\right )\right )}{\left (a^{2} - b^{2}\right )}}{\sqrt{-a^{2} + b^{2}} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^2/(a+b*csc(x)),x, algorithm="giac")

[Out]

-x/a + log(abs(tan(1/2*x)))/b - 2*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2
)))*(a^2 - b^2)/(sqrt(-a^2 + b^2)*a*b)