Integrand size = 13, antiderivative size = 100 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx=-\frac {1}{4 (a+b) (1-\cos (x))}+\frac {1}{4 (a-b) (1+\cos (x))}+\frac {(a+2 b) \log (1-\cos (x))}{4 (a+b)^2}-\frac {(a-2 b) \log (1+\cos (x))}{4 (a-b)^2}-\frac {b^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2} \] Output:
-1/4/(a+b)/(1-cos(x))+1/4/(a-b)/(1+cos(x))+1/4*(a+2*b)*ln(1-cos(x))/(a+b)^ 2-1/4*(a-2*b)*ln(1+cos(x))/(a-b)^2-b^3*ln(a+b*cos(x))/(a^2-b^2)^2
Time = 0.55 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.99 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx=\frac {1}{8} \left (-\frac {\csc ^2\left (\frac {x}{2}\right )}{a+b}-\frac {4 (a-2 b) \log \left (\cos \left (\frac {x}{2}\right )\right )}{(a-b)^2}-\frac {8 b^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}+\frac {4 (a+2 b) \log \left (\sin \left (\frac {x}{2}\right )\right )}{(a+b)^2}+\frac {\sec ^2\left (\frac {x}{2}\right )}{a-b}\right ) \] Input:
Integrate[Csc[x]^3/(a + b*Cos[x]),x]
Output:
(-(Csc[x/2]^2/(a + b)) - (4*(a - 2*b)*Log[Cos[x/2]])/(a - b)^2 - (8*b^3*Lo g[a + b*Cos[x]])/(a^2 - b^2)^2 + (4*(a + 2*b)*Log[Sin[x/2]])/(a + b)^2 + S ec[x/2]^2/(a - b))/8
Time = 0.38 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3147, 477, 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.
\(\displaystyle \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos \left (x-\frac {\pi }{2}\right )^3 \left (a-b \sin \left (x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle -b^3 \int \frac {1}{(a+b \cos (x)) \left (b^2-b^2 \cos ^2(x)\right )^2}d(b \cos (x))\) |
\(\Big \downarrow \) 477 |
\(\displaystyle -\frac {\int \left (\frac {b^4}{\left (a^2-b^2\right )^2 (a+b \cos (x))}+\frac {b^2}{4 (a+b) (b-b \cos (x))^2}+\frac {b^2}{4 (a-b) (\cos (x) b+b)^2}+\frac {(a+2 b) b}{4 (a+b)^2 (b-b \cos (x))}+\frac {(a-2 b) b}{4 (a-b)^2 (\cos (x) b+b)}\right )d(b \cos (x))}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {b^4 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}+\frac {b^2}{4 (a+b) (b-b \cos (x))}-\frac {b^2}{4 (a-b) (b \cos (x)+b)}-\frac {b (a+2 b) \log (b-b \cos (x))}{4 (a+b)^2}+\frac {b (a-2 b) \log (b \cos (x)+b)}{4 (a-b)^2}}{b}\) |
Input:
Int[Csc[x]^3/(a + b*Cos[x]),x]
Output:
-((b^2/(4*(a + b)*(b - b*Cos[x])) - b^2/(4*(a - b)*(b + b*Cos[x])) - (b*(a + 2*b)*Log[b - b*Cos[x]])/(4*(a + b)^2) + (b^4*Log[a + b*Cos[x]])/(a^2 - b^2)^2 + ((a - 2*b)*b*Log[b + b*Cos[x]])/(4*(a - b)^2))/b)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(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] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.83 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.91
method | result | size |
parallelrisch | \(\frac {-8 b^{3} \ln \left (2 b +\sec \left (\frac {x}{2}\right )^{2} \left (a -b \right )\right )-\left (-4 \left (a +2 b \right ) \left (a -b \right ) \ln \left (\tan \left (\frac {x}{2}\right )\right )+\left (\left (a -b \right ) \cot \left (\frac {x}{2}\right )^{2}-\tan \left (\frac {x}{2}\right )^{2} \left (a +b \right )\right ) \left (a +b \right )\right ) \left (a -b \right )}{8 \left (a -b \right )^{2} \left (a +b \right )^{2}}\) | \(91\) |
default | \(\frac {1}{\left (4 a -4 b \right ) \left (\cos \left (x \right )+1\right )}+\frac {\left (-a +2 b \right ) \ln \left (\cos \left (x \right )+1\right )}{4 \left (a -b \right )^{2}}-\frac {b^{3} \ln \left (a +b \cos \left (x \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (x \right )\right )}+\frac {\left (a +2 b \right ) \ln \left (-1+\cos \left (x \right )\right )}{4 \left (a +b \right )^{2}}\) | \(96\) |
norman | \(\frac {-\frac {1}{8 \left (a +b \right )}+\frac {\tan \left (\frac {x}{2}\right )^{4}}{8 a -8 b}}{\tan \left (\frac {x}{2}\right )^{2}}-\frac {b^{3} \ln \left (a \tan \left (\frac {x}{2}\right )^{2}-b \tan \left (\frac {x}{2}\right )^{2}+a +b \right )}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {\left (a +2 b \right ) \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a^{2}+4 a b +2 b^{2}}\) | \(100\) |
risch | \(-\frac {i x a}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {i x b}{a^{2}+2 a b +b^{2}}+\frac {i x a}{2 a^{2}-4 a b +2 b^{2}}-\frac {i x b}{a^{2}-2 a b +b^{2}}+\frac {2 i x \,b^{3}}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {{\mathrm e}^{3 i x} a -2 \,{\mathrm e}^{2 i x} b +a \,{\mathrm e}^{i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2} \left (-a^{2}+b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}-1\right ) a}{2 a^{2}+4 a b +2 b^{2}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right ) b}{a^{2}+2 a b +b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+1\right ) a}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}+1\right ) b}{a^{2}-2 a b +b^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i x}+\frac {2 a \,{\mathrm e}^{i x}}{b}+1\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\) | \(278\) |
Input:
int(csc(x)^3/(a+b*cos(x)),x,method=_RETURNVERBOSE)
Output:
1/8*(-8*b^3*ln(2*b+sec(1/2*x)^2*(a-b))-(-4*(a+2*b)*(a-b)*ln(tan(1/2*x))+(( a-b)*cot(1/2*x)^2-tan(1/2*x)^2*(a+b))*(a+b))*(a-b))/(a-b)^2/(a+b)^2
Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (90) = 180\).
Time = 0.12 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.81 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx=\frac {2 \, a^{2} b - 2 \, b^{3} - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (x\right ) + 4 \, {\left (b^{3} \cos \left (x\right )^{2} - b^{3}\right )} \log \left (-b \cos \left (x\right ) - a\right ) - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3} - {\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + {\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3} - {\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2}\right )}} \] Input:
integrate(csc(x)^3/(a+b*cos(x)),x, algorithm="fricas")
Output:
1/4*(2*a^2*b - 2*b^3 - 2*(a^3 - a*b^2)*cos(x) + 4*(b^3*cos(x)^2 - b^3)*log (-b*cos(x) - a) - (a^3 - 3*a*b^2 - 2*b^3 - (a^3 - 3*a*b^2 - 2*b^3)*cos(x)^ 2)*log(1/2*cos(x) + 1/2) + (a^3 - 3*a*b^2 + 2*b^3 - (a^3 - 3*a*b^2 + 2*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)
\[ \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx=\int \frac {\csc ^{3}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \] Input:
integrate(csc(x)**3/(a+b*cos(x)),x)
Output:
Integral(csc(x)**3/(a + b*cos(x)), x)
Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.15 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx=-\frac {b^{3} \log \left (b \cos \left (x\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {{\left (a + 2 \, b\right )} \log \left (\cos \left (x\right ) - 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {a \cos \left (x\right ) - b}{2 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{2}\right )}} \] Input:
integrate(csc(x)^3/(a+b*cos(x)),x, algorithm="maxima")
Output:
-b^3*log(b*cos(x) + a)/(a^4 - 2*a^2*b^2 + b^4) - 1/4*(a - 2*b)*log(cos(x) + 1)/(a^2 - 2*a*b + b^2) + 1/4*(a + 2*b)*log(cos(x) - 1)/(a^2 + 2*a*b + b^ 2) + 1/2*(a*cos(x) - b)/((a^2 - b^2)*cos(x)^2 - a^2 + b^2)
Time = 0.12 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.36 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx=-\frac {b^{4} \log \left ({\left | b \cos \left (x\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {{\left (a - 2 \, b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {{\left (a + 2 \, b\right )} \log \left (-\cos \left (x\right ) + 1\right )}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac {a^{2} b - b^{3} - {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \, {\left (a + b\right )}^{2} {\left (a - b\right )}^{2} {\left (\cos \left (x\right ) + 1\right )} {\left (\cos \left (x\right ) - 1\right )}} \] Input:
integrate(csc(x)^3/(a+b*cos(x)),x, algorithm="giac")
Output:
-b^4*log(abs(b*cos(x) + a))/(a^4*b - 2*a^2*b^3 + b^5) - 1/4*(a - 2*b)*log( cos(x) + 1)/(a^2 - 2*a*b + b^2) + 1/4*(a + 2*b)*log(-cos(x) + 1)/(a^2 + 2* a*b + b^2) - 1/2*(a^2*b - b^3 - (a^3 - a*b^2)*cos(x))/((a + b)^2*(a - b)^2 *(cos(x) + 1)*(cos(x) - 1))
Time = 43.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx=\ln \left (\cos \left (x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{4\,\left (a+b\right )}\right )+\frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\cos \left (x\right )}{2\,\left (a^2-b^2\right )}}{{\sin \left (x\right )}^2}-\frac {b^3\,\ln \left (a+b\,\cos \left (x\right )\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {\ln \left (\cos \left (x\right )+1\right )\,\left (a-2\,b\right )}{4\,{\left (a-b\right )}^2} \] Input:
int(1/(sin(x)^3*(a + b*cos(x))),x)
Output:
log(cos(x) - 1)*(b/(4*(a + b)^2) + 1/(4*(a + b))) + (b/(2*(a^2 - b^2)) - ( a*cos(x))/(2*(a^2 - b^2)))/sin(x)^2 - (b^3*log(a + b*cos(x)))/(a^4 + b^4 - 2*a^2*b^2) - (log(cos(x) + 1)*(a - 2*b))/(4*(a - b)^2)
Time = 0.18 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.41 \[ \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx=\frac {-2 \cos \left (x \right ) a^{3}+2 \cos \left (x \right ) a \,b^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )^{2} a -\tan \left (\frac {x}{2}\right )^{2} b +a +b \right ) \sin \left (x \right )^{2} b^{3}+2 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2} a^{3}-6 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2} a \,b^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2} b^{3}-\sin \left (x \right )^{2} a^{2} b +\sin \left (x \right )^{2} b^{3}+2 a^{2} b -2 b^{3}}{4 \sin \left (x \right )^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )} \] Input:
int(csc(x)^3/(a+b*cos(x)),x)
Output:
( - 2*cos(x)*a**3 + 2*cos(x)*a*b**2 - 4*log(tan(x/2)**2*a - tan(x/2)**2*b + a + b)*sin(x)**2*b**3 + 2*log(tan(x/2))*sin(x)**2*a**3 - 6*log(tan(x/2)) *sin(x)**2*a*b**2 + 4*log(tan(x/2))*sin(x)**2*b**3 - sin(x)**2*a**2*b + si n(x)**2*b**3 + 2*a**2*b - 2*b**3)/(4*sin(x)**2*(a**4 - 2*a**2*b**2 + b**4) )