3.1.16 \(\int \frac {\cot ^3(x)}{a+b \cos (x)} \, dx\) [16]

Optimal. Leaf size=93 \[ -\frac {(a-b \cos (x)) \csc ^2(x)}{2 \left (a^2-b^2\right )}-\frac {(2 a+b) \log (1-\cos (x))}{4 (a+b)^2}-\frac {(2 a-b) \log (1+\cos (x))}{4 (a-b)^2}+\frac {a^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2} \]

[Out]

-1/2*(a-b*cos(x))*csc(x)^2/(a^2-b^2)-1/4*(2*a+b)*ln(1-cos(x))/(a+b)^2-1/4*(2*a-b)*ln(cos(x)+1)/(a-b)^2+a^3*ln(
a+b*cos(x))/(a^2-b^2)^2

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Rubi [A]
time = 0.13, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2800, 1661, 815} \begin {gather*} -\frac {\csc ^2(x) (a-b \cos (x))}{2 \left (a^2-b^2\right )}+\frac {a^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}-\frac {(2 a+b) \log (1-\cos (x))}{4 (a+b)^2}-\frac {(2 a-b) \log (\cos (x)+1)}{4 (a-b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[x]^3/(a + b*Cos[x]),x]

[Out]

-1/2*((a - b*Cos[x])*Csc[x]^2)/(a^2 - b^2) - ((2*a + b)*Log[1 - Cos[x]])/(4*(a + b)^2) - ((2*a - b)*Log[1 + Co
s[x]])/(4*(a - b)^2) + (a^3*Log[a + b*Cos[x]])/(a^2 - b^2)^2

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 1661

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[(a*g - c*f*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

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

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Mathematica [A]
time = 0.59, size = 100, normalized size = 1.08 \begin {gather*} \frac {1}{8} \left (-\frac {\csc ^2\left (\frac {x}{2}\right )}{a+b}+\frac {4 (-2 a+b) \log \left (\cos \left (\frac {x}{2}\right )\right )}{(a-b)^2}+\frac {8 a^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}-\frac {4 (2 a+b) \log \left (\sin \left (\frac {x}{2}\right )\right )}{(a+b)^2}-\frac {\sec ^2\left (\frac {x}{2}\right )}{a-b}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^3/(a + b*Cos[x]),x]

[Out]

(-(Csc[x/2]^2/(a + b)) + (4*(-2*a + b)*Log[Cos[x/2]])/(a - b)^2 + (8*a^3*Log[a + b*Cos[x]])/(a^2 - b^2)^2 - (4
*(2*a + b)*Log[Sin[x/2]])/(a + b)^2 - Sec[x/2]^2/(a - b))/8

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Maple [A]
time = 0.18, size = 96, normalized size = 1.03

method	result	size
default	\(\frac {a^{3} \ln \left (a +b \cos \left (x \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (x \right )\right )}+\frac {\left (-2 a -b \right ) \ln \left (-1+\cos \left (x \right )\right )}{4 \left (a +b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (\cos \left (x \right )+1\right )}+\frac {\left (-2 a +b \right ) \ln \left (\cos \left (x \right )+1\right )}{4 \left (a -b \right )^{2}}\)	\(96\)
risch	\(\frac {i x a}{a^{2}-2 a b +b^{2}}-\frac {i x b}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {i x a}{a^{2}+2 a b +b^{2}}+\frac {i b x}{2 a^{2}+4 a b +2 b^{2}}-\frac {2 i x \,a^{3}}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {b \,{\mathrm e}^{3 i x}-2 a \,{\mathrm e}^{2 i x}+b \,{\mathrm e}^{i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2} \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i x}+1\right ) a}{a^{2}-2 a b +b^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+1\right ) b}{2 a^{2}-4 a b +2 b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-1\right ) a}{a^{2}+2 a b +b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-1\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i x}+\frac {2 a \,{\mathrm e}^{i x}}{b}+1\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\)	\(279\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^3/(a+b*cos(x)),x,method=_RETURNVERBOSE)

[Out]

a^3/(a+b)^2/(a-b)^2*ln(a+b*cos(x))+1/(4*a+4*b)/(-1+cos(x))+1/4/(a+b)^2*(-2*a-b)*ln(-1+cos(x))-1/(4*a-4*b)/(cos
(x)+1)+1/4/(a-b)^2*(-2*a+b)*ln(cos(x)+1)

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Maxima [A]
time = 0.28, size = 116, normalized size = 1.25 \begin {gather*} \frac {a^{3} \log \left (b \cos \left (x\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (2 \, a - b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (2 \, a + b\right )} \log \left (\cos \left (x\right ) - 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {b \cos \left (x\right ) - a}{2 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^3/(a+b*cos(x)),x, algorithm="maxima")

[Out]

a^3*log(b*cos(x) + a)/(a^4 - 2*a^2*b^2 + b^4) - 1/4*(2*a - b)*log(cos(x) + 1)/(a^2 - 2*a*b + b^2) - 1/4*(2*a +
b)*log(cos(x) - 1)/(a^2 + 2*a*b + b^2) - 1/2*(b*cos(x) - a)/((a^2 - b^2)*cos(x)^2 - a^2 + b^2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (88) = 176\).
time = 0.44, size = 185, normalized size = 1.99 \begin {gather*} -\frac {2 \, a^{3} - 2 \, a b^{2} - 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (x\right ) + 4 \, {\left (a^{3} \cos \left (x\right )^{2} - a^{3}\right )} \log \left (-b \cos \left (x\right ) - a\right ) + {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3} - {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\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^{2} b + b^{3} - {\left (2 \, a^{3} - 3 \, a^{2} b + b^{3}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^3/(a+b*cos(x)),x, algorithm="fricas")

[Out]

-1/4*(2*a^3 - 2*a*b^2 - 2*(a^2*b - b^3)*cos(x) + 4*(a^3*cos(x)^2 - a^3)*log(-b*cos(x) - a) + (2*a^3 + 3*a^2*b
- b^3 - (2*a^3 + 3*a^2*b - b^3)*cos(x)^2)*log(1/2*cos(x) + 1/2) + (2*a^3 - 3*a^2*b + b^3 - (2*a^3 - 3*a^2*b +
b^3)*cos(x)^2)*log(-1/2*cos(x) + 1/2))/(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(x)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{3}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**3/(a+b*cos(x)),x)

[Out]

Integral(cot(x)**3/(a + b*cos(x)), x)

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Giac [A]
time = 0.49, size = 138, normalized size = 1.48 \begin {gather*} \frac {a^{3} b \log \left ({\left | b \cos \left (x\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {{\left (2 \, a - b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (2 \, a + b\right )} \log \left (-\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {a^{3} - a b^{2} - {\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} {\left (\cos \left (x\right ) + 1\right )} {\left (\cos \left (x\right ) - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^3/(a+b*cos(x)),x, algorithm="giac")

[Out]

a^3*b*log(abs(b*cos(x) + a))/(a^4*b - 2*a^2*b^3 + b^5) - 1/4*(2*a - b)*log(cos(x) + 1)/(a^2 - 2*a*b + b^2) - 1
/4*(2*a + b)*log(-cos(x) + 1)/(a^2 + 2*a*b + b^2) + 1/2*(a^3 - a*b^2 - (a^2*b - b^3)*cos(x))/((a + b)^2*(a - b
)^2*(cos(x) + 1)*(cos(x) - 1))

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Mupad [B]
time = 0.58, size = 116, normalized size = 1.25 \begin {gather*} \frac {a^3\,\ln \left (a+b+a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{2\,\left (4\,a-4\,b\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (2\,a+b\right )}{2\,a^2+4\,a\,b+2\,b^2}-\frac {a-b}{2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a+b\right )\,\left (4\,a-4\,b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^3/(a + b*cos(x)),x)

[Out]

(a^3*log(a + b + a*tan(x/2)^2 - b*tan(x/2)^2))/(a^4 + b^4 - 2*a^2*b^2) - tan(x/2)^2/(2*(4*a - 4*b)) - (log(tan
(x/2))*(2*a + b))/(4*a*b + 2*a^2 + 2*b^2) - (a - b)/(2*tan(x/2)^2*(a + b)*(4*a - 4*b))

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