Integrand size = 10, antiderivative size = 36 \[ \int \frac {1}{\left (-1+\cos ^2(x)\right )^{3/2}} \, dx=\frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}+\frac {\text {arctanh}(\cos (x)) \sin (x)}{2 \sqrt {-\sin ^2(x)}} \]
Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\left (-1+\cos ^2(x)\right )^{3/2}} \, dx=\frac {\left (\csc ^2\left (\frac {x}{2}\right )+4 \log \left (\cos \left (\frac {x}{2}\right )\right )-4 \log \left (\sin \left (\frac {x}{2}\right )\right )-\sec ^2\left (\frac {x}{2}\right )\right ) \sin (x)}{8 \sqrt {-\sin ^2(x)}} \]
Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {3042, 3655, 3042, 3683, 3042, 3686, 3042, 4257}
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.
\(\displaystyle \int \frac {1}{\left (\cos ^2(x)-1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\sin \left (x+\frac {\pi }{2}\right )^2-1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \frac {1}{\left (-\sin ^2(x)\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (-\sin (x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3683 |
\(\displaystyle \frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}-\frac {1}{2} \int \frac {1}{\sqrt {-\sin ^2(x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}-\frac {1}{2} \int \frac {1}{\sqrt {-\sin (x)^2}}dx\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}-\frac {\sin (x) \int \csc (x)dx}{2 \sqrt {-\sin ^2(x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}-\frac {\sin (x) \int \csc (x)dx}{2 \sqrt {-\sin ^2(x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\sin (x) \text {arctanh}(\cos (x))}{2 \sqrt {-\sin ^2(x)}}+\frac {\cot (x)}{2 \sqrt {-\sin ^2(x)}}\) |
3.1.54.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*cos[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a + b, 0]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[Cot[e + f*x]* ((b*Sin[e + f*x]^2)^(p + 1)/(b*f*(2*p + 1))), x] + Simp[2*((p + 1)/(b*(2*p + 1))) Int[(b*Sin[e + f*x]^2)^(p + 1), x], x] /; FreeQ[{b, e, f}, x] && !IntegerQ[p] && LtQ[p, -1]
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.64 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08
method | result | size |
default | \(-\frac {-\frac {\cos \left (x \right )}{2}+\frac {\left (\ln \left (\cos \left (x \right )-1\right )-\ln \left (1+\cos \left (x \right )\right )\right ) \left (\sin ^{2}\left (x \right )\right )}{4}}{\sin \left (x \right ) \sqrt {-\left (\sin ^{2}\left (x \right )\right )}}\) | \(39\) |
risch | \(\frac {i \left ({\mathrm e}^{2 i x}+1\right )}{\left ({\mathrm e}^{2 i x}-1\right ) \sqrt {\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {\ln \left ({\mathrm e}^{i x}+1\right ) \sin \left (x \right )}{\sqrt {\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {\ln \left ({\mathrm e}^{i x}-1\right ) \sin \left (x \right )}{\sqrt {\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}\) | \(95\) |
Exception generated. \[ \int \frac {1}{\left (-1+\cos ^2(x)\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]
\[ \int \frac {1}{\left (-1+\cos ^2(x)\right )^{3/2}} \, dx=\int \frac {1}{\left (\cos ^{2}{\left (x \right )} - 1\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (28) = 56\).
Time = 0.33 (sec) , antiderivative size = 284, normalized size of antiderivative = 7.89 \[ \int \frac {1}{\left (-1+\cos ^2(x)\right )^{3/2}} \, dx=\frac {{\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )} \arctan \left (\sin \left (x\right ), \cos \left (x\right ) + 1\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )} \arctan \left (\sin \left (x\right ), \cos \left (x\right ) - 1\right ) + 2 \, {\left (\sin \left (3 \, x\right ) + \sin \left (x\right )\right )} \cos \left (4 \, x\right ) - 2 \, {\left (\cos \left (3 \, x\right ) + \cos \left (x\right )\right )} \sin \left (4 \, x\right ) - 2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \sin \left (3 \, x\right ) + 4 \, \cos \left (3 \, x\right ) \sin \left (2 \, x\right ) + 4 \, \cos \left (x\right ) \sin \left (2 \, x\right ) - 4 \, \cos \left (2 \, x\right ) \sin \left (x\right ) + 2 \, \sin \left (x\right )}{2 \, {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )}} \]
1/2*((2*(2*cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 - 4*cos(2*x)^2 - sin(4*x)^2 + 4*sin(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)*arctan2(sin(x), co s(x) + 1) - (2*(2*cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 - 4*cos(2*x)^2 - sin (4*x)^2 + 4*sin(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)*arctan2(sin (x), cos(x) - 1) + 2*(sin(3*x) + sin(x))*cos(4*x) - 2*(cos(3*x) + cos(x))* sin(4*x) - 2*(2*cos(2*x) - 1)*sin(3*x) + 4*cos(3*x)*sin(2*x) + 4*cos(x)*si n(2*x) - 4*cos(2*x)*sin(x) + 2*sin(x))/(2*(2*cos(2*x) - 1)*cos(4*x) - cos( 4*x)^2 - 4*cos(2*x)^2 - sin(4*x)^2 + 4*sin(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)
Result contains complex when optimal does not.
Time = 0.38 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.50 \[ \int \frac {1}{\left (-1+\cos ^2(x)\right )^{3/2}} \, dx=-\frac {i \, \tan \left (\frac {1}{2} \, x\right )^{2}}{8 \, \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, x\right )^{3} - \tan \left (\frac {1}{2} \, x\right )\right )} - \frac {i \, \log \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right )}{4 \, \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, x\right )^{3} - \tan \left (\frac {1}{2} \, x\right )\right )} + \frac {2 i \, \tan \left (\frac {1}{2} \, x\right )^{2} + i}{8 \, \mathrm {sgn}\left (-\tan \left (\frac {1}{2} \, x\right )^{3} - \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2}} \]
-1/8*I*tan(1/2*x)^2/sgn(-tan(1/2*x)^3 - tan(1/2*x)) - 1/4*I*log(tan(1/2*x) ^2)/sgn(-tan(1/2*x)^3 - tan(1/2*x)) + 1/8*(2*I*tan(1/2*x)^2 + I)/(sgn(-tan (1/2*x)^3 - tan(1/2*x))*tan(1/2*x)^2)
Timed out. \[ \int \frac {1}{\left (-1+\cos ^2(x)\right )^{3/2}} \, dx=\int \frac {1}{{\left ({\cos \left (x\right )}^2-1\right )}^{3/2}} \,d x \]