Integrand size = 13, antiderivative size = 93 \[ \int \frac {\cot ^3(x)}{a+b \cos (x)} \, dx=-\frac {(a-b \cos (x)) \csc ^2(x)}{2 \left (a^2-b^2\right )}-\frac {(2 a+b) \log (1-\cos (x))}{4 (a+b)^2}-\frac {(2 a-b) \log (1+\cos (x))}{4 (a-b)^2}+\frac {a^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2} \]
-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(1+cos(x))/(a-b)^2+a^3*ln(a+b*cos(x))/(a^2-b^2)^2
Time = 0.38 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.08 \[ \int \frac {\cot ^3(x)}{a+b \cos (x)} \, dx=\frac {1}{8} \left (-\frac {\csc ^2\left (\frac {x}{2}\right )}{a+b}+\frac {4 (-2 a+b) \log \left (\cos \left (\frac {x}{2}\right )\right )}{(a-b)^2}+\frac {8 a^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}-\frac {4 (2 a+b) \log \left (\sin \left (\frac {x}{2}\right )\right )}{(a+b)^2}-\frac {\sec ^2\left (\frac {x}{2}\right )}{a-b}\right ) \]
(-(Csc[x/2]^2/(a + b)) + (4*(-2*a + b)*Log[Cos[x/2]])/(a - b)^2 + (8*a^3*L og[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
Time = 0.39 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.45, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 25, 3200, 601, 25, 27, 657, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\cot ^3(x)}{a+b \cos (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\tan \left (x-\frac {\pi }{2}\right )^3}{a-b \sin \left (x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\tan \left (x-\frac {\pi }{2}\right )^3}{a-b \sin \left (x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle -\int \frac {b^3 \cos ^3(x)}{(a+b \cos (x)) \left (b^2-b^2 \cos ^2(x)\right )^2}d(b \cos (x))\) |
\(\Big \downarrow \) 601 |
\(\displaystyle \frac {\int -\frac {b^2 \left (\frac {a b^2}{a^2-b^2}-\frac {b \left (2 a^2-b^2\right ) \cos (x)}{a^2-b^2}\right )}{(a+b \cos (x)) \left (b^2-b^2 \cos ^2(x)\right )}d(b \cos (x))}{2 b^2}-\frac {b^2 (a-b \cos (x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \cos ^2(x)\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {b^2 \left (a b^2-b \left (2 a^2-b^2\right ) \cos (x)\right )}{\left (a^2-b^2\right ) (a+b \cos (x)) \left (b^2-b^2 \cos ^2(x)\right )}d(b \cos (x))}{2 b^2}-\frac {b^2 (a-b \cos (x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \cos ^2(x)\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {a b^2-b \left (2 a^2-b^2\right ) \cos (x)}{(a+b \cos (x)) \left (b^2-b^2 \cos ^2(x)\right )}d(b \cos (x))}{2 \left (a^2-b^2\right )}-\frac {b^2 (a-b \cos (x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \cos ^2(x)\right )}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle -\frac {\int \left (-\frac {2 a^3}{(a-b) (a+b) (a+b \cos (x))}+\frac {-2 a^2+b a+b^2}{2 (a+b) (b-b \cos (x))}+\frac {(2 a-b) (a+b)}{2 (a-b) (\cos (x) b+b)}\right )d(b \cos (x))}{2 \left (a^2-b^2\right )}-\frac {b^2 (a-b \cos (x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \cos ^2(x)\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {b^2 (a-b \cos (x))}{2 \left (a^2-b^2\right ) \left (b^2-b^2 \cos ^2(x)\right )}-\frac {-\frac {2 a^3 \log (a+b \cos (x))}{a^2-b^2}+\frac {(a-b) (2 a+b) \log (b-b \cos (x))}{2 (a+b)}+\frac {(2 a-b) (a+b) \log (b \cos (x)+b)}{2 (a-b)}}{2 \left (a^2-b^2\right )}\) |
-1/2*(b^2*(a - b*Cos[x]))/((a^2 - b^2)*(b^2 - b^2*Cos[x]^2)) - (((a - b)*( 2*a + b)*Log[b - b*Cos[x]])/(2*(a + b)) - (2*a^3*Log[a + b*Cos[x]])/(a^2 - b^2) + ((2*a - b)*(a + b)*Log[b + b*Cos[x]])/(2*(a - b)))/(2*(a^2 - b^2))
3.1.16.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) *((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* (2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(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]
Time = 0.90 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03
method | result | size |
default | \(\frac {a^{3} \ln \left (a +\cos \left (x \right ) b \right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (\cos \left (x \right )+1\right )}+\frac {\left (-2 a +b \right ) \ln \left (\cos \left (x \right )+1\right )}{4 \left (a -b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (\cos \left (x \right )-1\right )}+\frac {\left (-2 a -b \right ) \ln \left (\cos \left (x \right )-1\right )}{4 \left (a +b \right )^{2}}\) | \(96\) |
risch | \(\frac {i x a}{a^{2}-2 a b +b^{2}}-\frac {i x b}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {i x a}{a^{2}+2 a b +b^{2}}+\frac {i x b}{2 a^{2}+4 a b +2 b^{2}}-\frac {2 i x \,a^{3}}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {{\mathrm e}^{3 i x} b -2 a \,{\mathrm e}^{2 i x}+{\mathrm e}^{i x} b}{\left ({\mathrm e}^{2 i x}-1\right )^{2} \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i x}+1\right ) a}{a^{2}-2 a b +b^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+1\right ) b}{2 a^{2}-4 a b +2 b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-1\right ) a}{a^{2}+2 a b +b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}-1\right ) b}{2 \left (a^{2}+2 a b +b^{2}\right )}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i x}+\frac {2 a \,{\mathrm e}^{i x}}{b}+1\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\) | \(279\) |
a^3/(a+b)^2/(a-b)^2*ln(a+cos(x)*b)-1/(4*a-4*b)/(cos(x)+1)+1/4/(a-b)^2*(-2* a+b)*ln(cos(x)+1)+1/(4*a+4*b)/(cos(x)-1)+1/4/(a+b)^2*(-2*a-b)*ln(cos(x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (88) = 176\).
Time = 0.32 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.99 \[ \int \frac {\cot ^3(x)}{a+b \cos (x)} \, dx=-\frac {2 \, a^{3} - 2 \, a b^{2} - 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (x\right ) + 4 \, {\left (a^{3} \cos \left (x\right )^{2} - a^{3}\right )} \log \left (-b \cos \left (x\right ) - a\right ) + {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3} - {\left (2 \, a^{3} + 3 \, a^{2} b - b^{3}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (2 \, a^{3} - 3 \, a^{2} b + b^{3} - {\left (2 \, a^{3} - 3 \, a^{2} b + b^{3}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2}\right )}} \]
-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)*lo g(-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)
\[ \int \frac {\cot ^3(x)}{a+b \cos (x)} \, dx=\int \frac {\cot ^{3}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.25 \[ \int \frac {\cot ^3(x)}{a+b \cos (x)} \, dx=\frac {a^{3} \log \left (b \cos \left (x\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (2 \, a - b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (2 \, a + b\right )} \log \left (\cos \left (x\right ) - 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {b \cos \left (x\right ) - a}{2 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{2}\right )}} \]
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)
Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.48 \[ \int \frac {\cot ^3(x)}{a+b \cos (x)} \, dx=\frac {a^{3} b \log \left ({\left | b \cos \left (x\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {{\left (2 \, a - b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} - \frac {{\left (2 \, a + b\right )} \log \left (-\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {a^{3} - a b^{2} - {\left (a^{2} b - b^{3}\right )} \cos \left (x\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} {\left (\cos \left (x\right ) + 1\right )} {\left (\cos \left (x\right ) - 1\right )}} \]
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))
Time = 14.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.25 \[ \int \frac {\cot ^3(x)}{a+b \cos (x)} \, dx=\frac {a^3\,\ln \left (a+b+a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{2\,\left (4\,a-4\,b\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (2\,a+b\right )}{2\,a^2+4\,a\,b+2\,b^2}-\frac {a-b}{2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a+b\right )\,\left (4\,a-4\,b\right )} \]