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

Optimal. Leaf size=91 \[ -\frac{\left (2 a^2-3 b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac{2 \left (a^2-b^2\right )^{3/2} \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a b^3}+\frac{a \cot (x)}{b^2}+\frac{x}{a}-\frac{\cot (x) \csc (x)}{2 b} \]

[Out]

x/a - ((2*a^2 - 3*b^2)*ArcTanh[Cos[x]])/(2*b^3) + (2*(a^2 - b^2)^(3/2)*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2
]])/(a*b^3) + (a*Cot[x])/b^2 - (Cot[x]*Csc[x])/(2*b)

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Rubi [A]  time = 0.288846, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3898, 2893, 3057, 2660, 618, 206, 3770} \[ -\frac{\left (2 a^2-3 b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac{2 \left (a^2-b^2\right )^{3/2} \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a b^3}+\frac{a \cot (x)}{b^2}+\frac{x}{a}-\frac{\cot (x) \csc (x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^4/(a + b*Csc[x]),x]

[Out]

x/a - ((2*a^2 - 3*b^2)*ArcTanh[Cos[x]])/(2*b^3) + (2*(a^2 - b^2)^(3/2)*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2
]])/(a*b^3) + (a*Cot[x])/b^2 - (Cot[x]*Csc[x])/(2*b)

Rule 3898

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

Rule 2893

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(-Dist[1/(a^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

Rule 3057

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(C*x)/(b*d), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^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])

Rule 3770

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

Rubi steps

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

Mathematica [A]  time = 0.410146, size = 158, normalized size = 1.74 \[ \frac{-16 \left (b^2-a^2\right )^{3/2} \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )-4 a^2 b \tan \left (\frac{x}{2}\right )+4 a^2 b \cot \left (\frac{x}{2}\right )+8 a^3 \log \left (\sin \left (\frac{x}{2}\right )\right )-8 a^3 \log \left (\cos \left (\frac{x}{2}\right )\right )-a b^2 \csc ^2\left (\frac{x}{2}\right )+a b^2 \sec ^2\left (\frac{x}{2}\right )-12 a b^2 \log \left (\sin \left (\frac{x}{2}\right )\right )+12 a b^2 \log \left (\cos \left (\frac{x}{2}\right )\right )+8 b^3 x}{8 a b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^4/(a + b*Csc[x]),x]

[Out]

(8*b^3*x - 16*(-a^2 + b^2)^(3/2)*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]] + 4*a^2*b*Cot[x/2] - a*b^2*Csc[x/2]
^2 - 8*a^3*Log[Cos[x/2]] + 12*a*b^2*Log[Cos[x/2]] + 8*a^3*Log[Sin[x/2]] - 12*a*b^2*Log[Sin[x/2]] + a*b^2*Sec[x
/2]^2 - 4*a^2*b*Tan[x/2])/(8*a*b^3)

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Maple [B]  time = 0.067, size = 206, normalized size = 2.3 \begin{align*}{\frac{1}{8\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}}-{\frac{a}{2\,{b}^{2}}\tan \left ({\frac{x}{2}} \right ) }+2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{a}}-{\frac{1}{8\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{3}{2\,b}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }+{\frac{a}{2\,{b}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-2\,{\frac{{a}^{3}}{{b}^{3}\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 ) }+4\,{\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)^4/(a+b*csc(x)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.934268, size = 1096, normalized size = 12.04 \begin{align*} \left [\frac{4 \, b^{3} x \cos \left (x\right )^{2} - 4 \, a^{2} b \cos \left (x\right ) \sin \left (x\right ) - 4 \, b^{3} x + 2 \, a b^{2} \cos \left (x\right ) - 2 \,{\left ({\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{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 ) +{\left (2 \, a^{3} - 3 \, a b^{2} -{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) -{\left (2 \, a^{3} - 3 \, a b^{2} -{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{4 \,{\left (a b^{3} \cos \left (x\right )^{2} - a b^{3}\right )}}, \frac{4 \, b^{3} x \cos \left (x\right )^{2} - 4 \, a^{2} b \cos \left (x\right ) \sin \left (x\right ) - 4 \, b^{3} x + 2 \, a b^{2} \cos \left (x\right ) + 4 \,{\left ({\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{2}\right )} \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 ) +{\left (2 \, a^{3} - 3 \, a b^{2} -{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) -{\left (2 \, a^{3} - 3 \, a b^{2} -{\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{4 \,{\left (a b^{3} \cos \left (x\right )^{2} - a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(4*b^3*x*cos(x)^2 - 4*a^2*b*cos(x)*sin(x) - 4*b^3*x + 2*a*b^2*cos(x) - 2*((a^2 - b^2)*cos(x)^2 - a^2 + b^
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))*sq
rt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + (2*a^3 - 3*a*b^2 - (2*a^3 - 3*a*b^2)*cos(x)^2)*log
(1/2*cos(x) + 1/2) - (2*a^3 - 3*a*b^2 - (2*a^3 - 3*a*b^2)*cos(x)^2)*log(-1/2*cos(x) + 1/2))/(a*b^3*cos(x)^2 -
a*b^3), 1/4*(4*b^3*x*cos(x)^2 - 4*a^2*b*cos(x)*sin(x) - 4*b^3*x + 2*a*b^2*cos(x) + 4*((a^2 - b^2)*cos(x)^2 - a
^2 + b^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x))) + (2*a^3 - 3*a*b^2 -
(2*a^3 - 3*a*b^2)*cos(x)^2)*log(1/2*cos(x) + 1/2) - (2*a^3 - 3*a*b^2 - (2*a^3 - 3*a*b^2)*cos(x)^2)*log(-1/2*co
s(x) + 1/2))/(a*b^3*cos(x)^2 - a*b^3)]

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

[Out]

Integral(cot(x)**4/(a + b*csc(x)), x)

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Giac [B]  time = 1.40509, size = 220, normalized size = 2.42 \begin{align*} \frac{x}{a} + \frac{b \tan \left (\frac{1}{2} \, x\right )^{2} - 4 \, a \tan \left (\frac{1}{2} \, x\right )}{8 \, b^{2}} + \frac{{\left (2 \, a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{2 \, b^{3}} - \frac{2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\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 )}}{\sqrt{-a^{2} + b^{2}} a b^{3}} - \frac{12 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 18 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 4 \, a b \tan \left (\frac{1}{2} \, x\right ) + b^{2}}{8 \, b^{3} \tan \left (\frac{1}{2} \, x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a + 1/8*(b*tan(1/2*x)^2 - 4*a*tan(1/2*x))/b^2 + 1/2*(2*a^2 - 3*b^2)*log(abs(tan(1/2*x)))/b^3 - 2*(a^4 - 2*a^
2*b^2 + b^4)*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))/(sqrt(-a^2 + b^2)
*a*b^3) - 1/8*(12*a^2*tan(1/2*x)^2 - 18*b^2*tan(1/2*x)^2 - 4*a*b*tan(1/2*x) + b^2)/(b^3*tan(1/2*x)^2)