3.17 \(\int \frac{\cot ^4(x)}{a+b \cos (x)} \, dx\)

Optimal. Leaf size=138 \[ -\frac{a \cot ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a^3 \cot (x)}{\left (a^2-b^2\right )^2}+\frac{b \csc ^3(x)}{3 \left (a^2-b^2\right )}-\frac{a^2 b \csc (x)}{\left (a^2-b^2\right )^2}-\frac{b \csc (x)}{a^2-b^2}+\frac{2 a^4 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \]

[Out]

(2*a^4*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)) + (a^3*Cot[x])/(a^2 - b^2)^2
- (a*Cot[x]^3)/(3*(a^2 - b^2)) - (a^2*b*Csc[x])/(a^2 - b^2)^2 - (b*Csc[x])/(a^2 - b^2) + (b*Csc[x]^3)/(3*(a^2
- b^2))

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Rubi [A]  time = 0.206036, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {2727, 2607, 30, 2606, 3767, 8, 2659, 205} \[ -\frac{a \cot ^3(x)}{3 \left (a^2-b^2\right )}+\frac{a^3 \cot (x)}{\left (a^2-b^2\right )^2}+\frac{b \csc ^3(x)}{3 \left (a^2-b^2\right )}-\frac{a^2 b \csc (x)}{\left (a^2-b^2\right )^2}-\frac{b \csc (x)}{a^2-b^2}+\frac{2 a^4 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^4/(a + b*Cos[x]),x]

[Out]

(2*a^4*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)) + (a^3*Cot[x])/(a^2 - b^2)^2
- (a*Cot[x]^3)/(3*(a^2 - b^2)) - (a^2*b*Csc[x])/(a^2 - b^2)^2 - (b*Csc[x])/(a^2 - b^2) + (b*Csc[x]^3)/(3*(a^2
- b^2))

Rule 2727

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[a/(a^2 - b^
2), Int[(g*Tan[e + f*x])^p/Sin[e + f*x]^2, x], x] + (-Dist[(b*g)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 1)/Cos
[e + f*x], x], x] - Dist[(a^2*g^2)/(a^2 - b^2), Int[(g*Tan[e + f*x])^(p - 2)/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*p] && GtQ[p, 1]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.671035, size = 112, normalized size = 0.81 \[ -\frac{\csc ^3(x) \left (6 b \left (b^2-2 a^2\right ) \cos (2 x)+\left (4 a^2-b^2\right ) (a \cos (3 x)+2 b)-3 a b^2 \cos (x)\right )}{12 (a-b)^2 (a+b)^2}-\frac{2 a^4 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^4/(a + b*Cos[x]),x]

[Out]

(-2*a^4*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) - ((-3*a*b^2*Cos[x] + 6*b*(-2*a^2 + b
^2)*Cos[2*x] + (4*a^2 - b^2)*(2*b + a*Cos[3*x]))*Csc[x]^3)/(12*(a - b)^2*(a + b)^2)

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Maple [A]  time = 0.093, size = 153, normalized size = 1.1 \begin{align*}{\frac{a}{24\, \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{b}{24\, \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{5\,a}{8\, \left ( a-b \right ) ^{2}}\tan \left ({\frac{x}{2}} \right ) }+{\frac{3\,b}{8\, \left ( a-b \right ) ^{2}}\tan \left ({\frac{x}{2}} \right ) }-{\frac{1}{24\,a+24\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-3}}+{\frac{5\,a}{8\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{3\,b}{8\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+2\,{\frac{{a}^{4}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^4/(a+b*cos(x)),x)

[Out]

1/24/(a-b)^2*tan(1/2*x)^3*a-1/24/(a-b)^2*tan(1/2*x)^3*b-5/8/(a-b)^2*a*tan(1/2*x)+3/8/(a-b)^2*tan(1/2*x)*b-1/24
/(a+b)/tan(1/2*x)^3+5/8/(a+b)^2/tan(1/2*x)*a+3/8/(a+b)^2/tan(1/2*x)*b+2*a^4/(a+b)^2/(a-b)^2/((a-b)*(a+b))^(1/2
)*arctan(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(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)^4/(a+b*cos(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60391, size = 1029, normalized size = 7.46 \begin{align*} \left [-\frac{10 \, a^{4} b - 14 \, a^{2} b^{3} + 4 \, b^{5} + 2 \,{\left (4 \, a^{5} - 5 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{3} - 3 \,{\left (a^{4} \cos \left (x\right )^{2} - a^{4}\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) \sin \left (x\right ) - 6 \,{\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{2} - 6 \,{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (x\right )}{6 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6} -{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}, -\frac{5 \, a^{4} b - 7 \, a^{2} b^{3} + 2 \, b^{5} +{\left (4 \, a^{5} - 5 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{3} + 3 \,{\left (a^{4} \cos \left (x\right )^{2} - a^{4}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right ) \sin \left (x\right ) - 3 \,{\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{2} - 3 \,{\left (a^{5} - a^{3} b^{2}\right )} \cos \left (x\right )}{3 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6} -{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^4/(a+b*cos(x)),x, algorithm="fricas")

[Out]

[-1/6*(10*a^4*b - 14*a^2*b^3 + 4*b^5 + 2*(4*a^5 - 5*a^3*b^2 + a*b^4)*cos(x)^3 - 3*(a^4*cos(x)^2 - a^4)*sqrt(-a
^2 + b^2)*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2)
/(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2))*sin(x) - 6*(2*a^4*b - 3*a^2*b^3 + b^5)*cos(x)^2 - 6*(a^5 - a^3*b^2)*cos(
x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(x)^2)*sin(x)), -1/3*(5*a^4*b
- 7*a^2*b^3 + 2*b^5 + (4*a^5 - 5*a^3*b^2 + a*b^4)*cos(x)^3 + 3*(a^4*cos(x)^2 - a^4)*sqrt(a^2 - b^2)*arctan(-(
a*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x)))*sin(x) - 3*(2*a^4*b - 3*a^2*b^3 + b^5)*cos(x)^2 - 3*(a^5 - a^3*b^2)*co
s(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 - (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(x)^2)*sin(x))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{4}{\left (x \right )}}{a + b \cos{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**4/(a+b*cos(x)),x)

[Out]

Integral(cot(x)**4/(a + b*cos(x)), x)

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Giac [A]  time = 1.61148, size = 284, normalized size = 2.06 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )} a^{4}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{a^{2} \tan \left (\frac{1}{2} \, x\right )^{3} - 2 \, a b \tan \left (\frac{1}{2} \, x\right )^{3} + b^{2} \tan \left (\frac{1}{2} \, x\right )^{3} - 15 \, a^{2} \tan \left (\frac{1}{2} \, x\right ) + 24 \, a b \tan \left (\frac{1}{2} \, x\right ) - 9 \, b^{2} \tan \left (\frac{1}{2} \, x\right )}{24 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac{15 \, a \tan \left (\frac{1}{2} \, x\right )^{2} + 9 \, b \tan \left (\frac{1}{2} \, x\right )^{2} - a - b}{24 \,{\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (\frac{1}{2} \, x\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^4/(a+b*cos(x)),x, algorithm="giac")

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))*a^4/((a
^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + 1/24*(a^2*tan(1/2*x)^3 - 2*a*b*tan(1/2*x)^3 + b^2*tan(1/2*x)^3 - 15*a
^2*tan(1/2*x) + 24*a*b*tan(1/2*x) - 9*b^2*tan(1/2*x))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 1/24*(15*a*tan(1/2*x)^
2 + 9*b*tan(1/2*x)^2 - a - b)/((a^2 + 2*a*b + b^2)*tan(1/2*x)^3)