Integrand size = 13, antiderivative size = 55 \[ \int \frac {\csc ^3(x)}{a+a \cos (x)} \, dx=-\frac {3 \text {arctanh}(\cos (x))}{8 a}-\frac {1}{8 (a-a \cos (x))}+\frac {1}{4 (a+a \cos (x))}+\frac {a^3}{8 \left (a^2+a^2 \cos (x)\right )^2} \] Output:
-3/8*arctanh(cos(x))/a-1/(8*a-8*a*cos(x))+1/(4*a+4*a*cos(x))+1/8*a^3/(a^2+ a^2*cos(x))^2
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.09 \[ \int \frac {\csc ^3(x)}{a+a \cos (x)} \, dx=\frac {4-2 \cot ^2\left (\frac {x}{2}\right )-12 \cos ^2\left (\frac {x}{2}\right ) \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )+\sec ^2\left (\frac {x}{2}\right )}{16 a (1+\cos (x))} \] Input:
Integrate[Csc[x]^3/(a + a*Cos[x]),x]
Output:
(4 - 2*Cot[x/2]^2 - 12*Cos[x/2]^2*(Log[Cos[x/2]] - Log[Sin[x/2]]) + Sec[x/ 2]^2)/(16*a*(1 + Cos[x]))
Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3146, 54, 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.
\(\displaystyle \int \frac {\csc ^3(x)}{a \cos (x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos \left (x-\frac {\pi }{2}\right )^3 \left (a-a \sin \left (x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle -a^3 \int \frac {1}{(a-a \cos (x))^2 (\cos (x) a+a)^3}d(a \cos (x))\) |
\(\Big \downarrow \) 54 |
\(\displaystyle -a^3 \int \left (\frac {1}{8 a^3 (a-a \cos (x))^2}+\frac {1}{4 a^3 (\cos (x) a+a)^2}+\frac {1}{4 a^2 (\cos (x) a+a)^3}+\frac {3}{8 a^3 \left (a^2-a^2 \cos ^2(x)\right )}\right )d(a \cos (x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -a^3 \left (\frac {3 \text {arctanh}(\cos (x))}{8 a^4}+\frac {1}{8 a^3 (a-a \cos (x))}-\frac {1}{4 a^3 (a \cos (x)+a)}-\frac {1}{8 a^2 (a \cos (x)+a)^2}\right )\) |
Input:
Int[Csc[x]^3/(a + a*Cos[x]),x]
Output:
-(a^3*((3*ArcTanh[Cos[x]])/(8*a^4) + 1/(8*a^3*(a - a*Cos[x])) - 1/(8*a^2*( a + a*Cos[x])^2) - 1/(4*a^3*(a + a*Cos[x]))))
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
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 + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 0.56 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65
method | result | size |
parallelrisch | \(\frac {\tan \left (\frac {x}{2}\right )^{4}-2 \cot \left (\frac {x}{2}\right )^{2}+6 \tan \left (\frac {x}{2}\right )^{2}+12 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{32 a}\) | \(36\) |
default | \(\frac {\frac {1}{8 \left (\cos \left (x \right )+1\right )^{2}}+\frac {1}{4 \cos \left (x \right )+4}-\frac {3 \ln \left (\cos \left (x \right )+1\right )}{16}+\frac {1}{-8+8 \cos \left (x \right )}+\frac {3 \ln \left (-1+\cos \left (x \right )\right )}{16}}{a}\) | \(44\) |
norman | \(\frac {-\frac {1}{16 a}+\frac {3 \tan \left (\frac {x}{2}\right )^{4}}{16 a}+\frac {\tan \left (\frac {x}{2}\right )^{6}}{32 a}}{\tan \left (\frac {x}{2}\right )^{2}}+\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{8 a}\) | \(47\) |
risch | \(\frac {3 \,{\mathrm e}^{5 i x}+6 \,{\mathrm e}^{4 i x}-2 \,{\mathrm e}^{3 i x}+6 \,{\mathrm e}^{2 i x}+3 \,{\mathrm e}^{i x}}{4 \left ({\mathrm e}^{i x}+1\right )^{4} a \left ({\mathrm e}^{i x}-1\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i x}-1\right )}{8 a}-\frac {3 \ln \left ({\mathrm e}^{i x}+1\right )}{8 a}\) | \(87\) |
Input:
int(csc(x)^3/(a+a*cos(x)),x,method=_RETURNVERBOSE)
Output:
1/32*(tan(1/2*x)^4-2*cot(1/2*x)^2+6*tan(1/2*x)^2+12*ln(tan(1/2*x)))/a
Time = 0.08 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.51 \[ \int \frac {\csc ^3(x)}{a+a \cos (x)} \, dx=\frac {6 \, \cos \left (x\right )^{2} - 3 \, {\left (\cos \left (x\right )^{3} + \cos \left (x\right )^{2} - \cos \left (x\right ) - 1\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (x\right )^{3} + \cos \left (x\right )^{2} - \cos \left (x\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + 6 \, \cos \left (x\right ) - 4}{16 \, {\left (a \cos \left (x\right )^{3} + a \cos \left (x\right )^{2} - a \cos \left (x\right ) - a\right )}} \] Input:
integrate(csc(x)^3/(a+a*cos(x)),x, algorithm="fricas")
Output:
1/16*(6*cos(x)^2 - 3*(cos(x)^3 + cos(x)^2 - cos(x) - 1)*log(1/2*cos(x) + 1 /2) + 3*(cos(x)^3 + cos(x)^2 - cos(x) - 1)*log(-1/2*cos(x) + 1/2) + 6*cos( x) - 4)/(a*cos(x)^3 + a*cos(x)^2 - a*cos(x) - a)
\[ \int \frac {\csc ^3(x)}{a+a \cos (x)} \, dx=\frac {\int \frac {\csc ^{3}{\left (x \right )}}{\cos {\left (x \right )} + 1}\, dx}{a} \] Input:
integrate(csc(x)**3/(a+a*cos(x)),x)
Output:
Integral(csc(x)**3/(cos(x) + 1), x)/a
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05 \[ \int \frac {\csc ^3(x)}{a+a \cos (x)} \, dx=\frac {3 \, \cos \left (x\right )^{2} + 3 \, \cos \left (x\right ) - 2}{8 \, {\left (a \cos \left (x\right )^{3} + a \cos \left (x\right )^{2} - a \cos \left (x\right ) - a\right )}} - \frac {3 \, \log \left (\cos \left (x\right ) + 1\right )}{16 \, a} + \frac {3 \, \log \left (\cos \left (x\right ) - 1\right )}{16 \, a} \] Input:
integrate(csc(x)^3/(a+a*cos(x)),x, algorithm="maxima")
Output:
1/8*(3*cos(x)^2 + 3*cos(x) - 2)/(a*cos(x)^3 + a*cos(x)^2 - a*cos(x) - a) - 3/16*log(cos(x) + 1)/a + 3/16*log(cos(x) - 1)/a
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95 \[ \int \frac {\csc ^3(x)}{a+a \cos (x)} \, dx=-\frac {3 \, \log \left (\cos \left (x\right ) + 1\right )}{16 \, a} + \frac {3 \, \log \left (-\cos \left (x\right ) + 1\right )}{16 \, a} + \frac {3 \, \cos \left (x\right )^{2} + 3 \, \cos \left (x\right ) - 2}{8 \, a {\left (\cos \left (x\right ) + 1\right )}^{2} {\left (\cos \left (x\right ) - 1\right )}} \] Input:
integrate(csc(x)^3/(a+a*cos(x)),x, algorithm="giac")
Output:
-3/16*log(cos(x) + 1)/a + 3/16*log(-cos(x) + 1)/a + 1/8*(3*cos(x)^2 + 3*co s(x) - 2)/(a*(cos(x) + 1)^2*(cos(x) - 1))
Time = 41.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int \frac {\csc ^3(x)}{a+a \cos (x)} \, dx=-\frac {\frac {3\,{\cos \left (x\right )}^2}{8}+\frac {3\,\cos \left (x\right )}{8}-\frac {1}{4}}{-a\,{\cos \left (x\right )}^3-a\,{\cos \left (x\right )}^2+a\,\cos \left (x\right )+a}-\frac {3\,\mathrm {atanh}\left (\cos \left (x\right )\right )}{8\,a} \] Input:
int(1/(sin(x)^3*(a + a*cos(x))),x)
Output:
- ((3*cos(x))/8 + (3*cos(x)^2)/8 - 1/4)/(a + a*cos(x) - a*cos(x)^2 - a*cos (x)^3) - (3*atanh(cos(x)))/(8*a)
Time = 0.19 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {\csc ^3(x)}{a+a \cos (x)} \, dx=\frac {12 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )\right ) \tan \left (\frac {x}{2}\right )^{2}+\tan \left (\frac {x}{2}\right )^{6}+6 \tan \left (\frac {x}{2}\right )^{4}-2}{32 \tan \left (\frac {x}{2}\right )^{2} a} \] Input:
int(csc(x)^3/(a+a*cos(x)),x)
Output:
(12*log(tan(x/2))*tan(x/2)**2 + tan(x/2)**6 + 6*tan(x/2)**4 - 2)/(32*tan(x /2)**2*a)