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

Optimal. Leaf size=40 \[ -\frac {\cot ^5(x)}{5 a}+\frac {\csc ^5(x)}{5 a}-\frac {2 \csc ^3(x)}{3 a}+\frac {\csc (x)}{a} \]

[Out]

-1/5*cot(x)^5/a+csc(x)/a-2/3*csc(x)^3/a+1/5*csc(x)^5/a

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Rubi [A]  time = 0.08, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2706, 2607, 30, 2606, 194} \[ -\frac {\cot ^5(x)}{5 a}+\frac {\csc ^5(x)}{5 a}-\frac {2 \csc ^3(x)}{3 a}+\frac {\csc (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^4/(a + a*Cos[x]),x]

[Out]

-Cot[x]^5/(5*a) + Csc[x]/a - (2*Csc[x]^3)/(3*a) + Csc[x]^5/(5*a)

Rule 30

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

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

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 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 2706

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 41, normalized size = 1.02 \[ -\frac {(8 \cos (x)+36 \cos (2 x)+24 \cos (3 x)-3 \cos (4 x)-25) \csc ^3(x)}{120 a (\cos (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^4/(a + a*Cos[x]),x]

[Out]

-1/120*((-25 + 8*Cos[x] + 36*Cos[2*x] + 24*Cos[3*x] - 3*Cos[4*x])*Csc[x]^3)/(a*(1 + Cos[x]))

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fricas [A]  time = 2.40, size = 53, normalized size = 1.32 \[ -\frac {3 \, \cos \relax (x)^{4} - 12 \, \cos \relax (x)^{3} - 12 \, \cos \relax (x)^{2} + 8 \, \cos \relax (x) + 8}{15 \, {\left (a \cos \relax (x)^{3} + a \cos \relax (x)^{2} - a \cos \relax (x) - a\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*cos(x)^4 - 12*cos(x)^3 - 12*cos(x)^2 + 8*cos(x) + 8)/((a*cos(x)^3 + a*cos(x)^2 - a*cos(x) - a)*sin(x)
)

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giac [A]  time = 0.37, size = 59, normalized size = 1.48 \[ \frac {12 \, \tan \left (\frac {1}{2} \, x\right )^{2} - 1}{48 \, a \tan \left (\frac {1}{2} \, x\right )^{3}} + \frac {3 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{5} - 20 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{3} + 90 \, a^{4} \tan \left (\frac {1}{2} \, x\right )}{240 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(12*tan(1/2*x)^2 - 1)/(a*tan(1/2*x)^3) + 1/240*(3*a^4*tan(1/2*x)^5 - 20*a^4*tan(1/2*x)^3 + 90*a^4*tan(1/2
*x))/a^5

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maple [A]  time = 0.06, size = 45, normalized size = 1.12 \[ \frac {\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{5}-\frac {4 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3}+6 \tan \left (\frac {x}{2}\right )-\frac {1}{3 \tan \left (\frac {x}{2}\right )^{3}}+\frac {4}{\tan \left (\frac {x}{2}\right )}}{16 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/a*(1/5*tan(1/2*x)^5-4/3*tan(1/2*x)^3+6*tan(1/2*x)-1/3/tan(1/2*x)^3+4/tan(1/2*x))

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maxima [B]  time = 1.72, size = 70, normalized size = 1.75 \[ \frac {\frac {90 \, \sin \relax (x)}{\cos \relax (x) + 1} - \frac {20 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {3 \, \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}}}{240 \, a} + \frac {{\left (\frac {12 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - 1\right )} {\left (\cos \relax (x) + 1\right )}^{3}}{48 \, a \sin \relax (x)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(90*sin(x)/(cos(x) + 1) - 20*sin(x)^3/(cos(x) + 1)^3 + 3*sin(x)^5/(cos(x) + 1)^5)/a + 1/48*(12*sin(x)^2/
(cos(x) + 1)^2 - 1)*(cos(x) + 1)^3/(a*sin(x)^3)

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mupad [B]  time = 0.45, size = 45, normalized size = 1.12 \[ \frac {3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {x}{2}\right )}^6+90\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+60\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-5}{240\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^4/(a + a*cos(x)),x)

[Out]

(60*tan(x/2)^2 + 90*tan(x/2)^4 - 20*tan(x/2)^6 + 3*tan(x/2)^8 - 5)/(240*a*tan(x/2)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cot ^{4}{\relax (x )}}{\cos {\relax (x )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(x)**4/(cos(x) + 1), x)/a

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