Integrand size = 12, antiderivative size = 32 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{3/2}} \, dx=-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}-\frac {\text {arctanh}(\cos (x)) \sin (x)}{2 \sqrt {\sin ^2(x)}} \] Output:
-1/2*cot(x)/(sin(x)^2)^(1/2)-1/2*arctanh(cos(x))*sin(x)/(sin(x)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.59 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{3/2}} \, dx=-\frac {\left (\csc ^2\left (\frac {x}{2}\right )+4 \log \left (\cos \left (\frac {x}{2}\right )\right )-4 \log \left (\sin \left (\frac {x}{2}\right )\right )-\sec ^2\left (\frac {x}{2}\right )\right ) \sin (x)}{8 \sqrt {\sin ^2(x)}} \] Input:
Integrate[(1 - Cos[x]^2)^(-3/2),x]
Output:
-1/8*((Csc[x/2]^2 + 4*Log[Cos[x/2]] - 4*Log[Sin[x/2]] - Sec[x/2]^2)*Sin[x] )/Sqrt[Sin[x]^2]
Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, 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.
\(\displaystyle \int \frac {1}{\left (1-\cos ^2(x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (1-\sin \left (x+\frac {\pi }{2}\right )^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \frac {1}{\sin ^2(x)^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\sin (x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3683 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt {\sin ^2(x)}}dx-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt {\sin (x)^2}}dx-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {\sin (x) \int \csc (x)dx}{2 \sqrt {\sin ^2(x)}}-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sin (x) \int \csc (x)dx}{2 \sqrt {\sin ^2(x)}}-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -\frac {\sin (x) \text {arctanh}(\cos (x))}{2 \sqrt {\sin ^2(x)}}-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\) |
Input:
Int[(1 - Cos[x]^2)^(-3/2),x]
Output:
-1/2*Cot[x]/Sqrt[Sin[x]^2] - (ArcTanh[Cos[x]]*Sin[x])/(2*Sqrt[Sin[x]^2])
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 = 0.08 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16
method | result | size |
default | \(-\frac {2 \left (\frac {\cos \left (x \right )}{2}+\frac {\left (\ln \left (1+\cos \left (x \right )\right )-\ln \left (-1+\cos \left (x \right )\right )\right ) \sin \left (x \right )^{2}}{4}\right )}{\sin \left (x \right ) \sqrt {2-2 \cos \left (2 x \right )}}\) | \(37\) |
risch | \(-\frac {i \left ({\mathrm e}^{2 i x}+1\right )}{\left ({\mathrm e}^{2 i x}-1\right ) \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {\ln \left ({\mathrm e}^{i x}+1\right ) \sin \left (x \right )}{\sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right ) \sin \left (x \right )}{\sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}\) | \(98\) |
Input:
int(1/(1-cos(x)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-(1/2*cos(x)+1/4*(ln(1+cos(x))-ln(-1+cos(x)))*sin(x)^2)/sin(x)/(sin(x)^2)^ (1/2)
Time = 0.11 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{3/2}} \, dx=-\frac {{\left (\cos \left (x\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - 2 \, \cos \left (x\right )}{4 \, {\left (\cos \left (x\right )^{2} - 1\right )}} \] Input:
integrate(1/(1-cos(x)^2)^(3/2),x, algorithm="fricas")
Output:
-1/4*((cos(x)^2 - 1)*log(1/2*cos(x) + 1/2) - (cos(x)^2 - 1)*log(-1/2*cos(x ) + 1/2) - 2*cos(x))/(cos(x)^2 - 1)
\[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{3/2}} \, dx=\int \frac {1}{\left (1 - \cos ^{2}{\left (x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(1-cos(x)**2)**(3/2),x)
Output:
Integral((1 - cos(x)**2)**(-3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (24) = 48\).
Time = 0.16 (sec) , antiderivative size = 300, normalized size of antiderivative = 9.38 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{3/2}} \, dx=\frac {4 \, {\left (\cos \left (3 \, x\right ) + \cos \left (x\right )\right )} \cos \left (4 \, x\right ) - 4 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (3 \, x\right ) - 8 \, \cos \left (2 \, x\right ) \cos \left (x\right ) + {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \cos \left (x\right ) + 1\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )} \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right ) + 4 \, {\left (\sin \left (3 \, x\right ) + \sin \left (x\right )\right )} \sin \left (4 \, x\right ) - 8 \, \sin \left (3 \, x\right ) \sin \left (2 \, x\right ) - 8 \, \sin \left (2 \, x\right ) \sin \left (x\right ) + 4 \, \cos \left (x\right )}{4 \, {\left (2 \, {\left (2 \, \cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - 4 \, \cos \left (2 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 4 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - 4 \, \sin \left (2 \, x\right )^{2} + 4 \, \cos \left (2 \, x\right ) - 1\right )}} \] Input:
integrate(1/(1-cos(x)^2)^(3/2),x, algorithm="maxima")
Output:
1/4*(4*(cos(3*x) + cos(x))*cos(4*x) - 4*(2*cos(2*x) - 1)*cos(3*x) - 8*cos( 2*x)*cos(x) + (2*(2*cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 - 4*cos(2*x)^2 - s in(4*x)^2 + 4*sin(4*x)*sin(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)*log(cos(x )^2 + sin(x)^2 + 2*cos(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)*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) + 4*(sin(3*x) + sin(x))* sin(4*x) - 8*sin(3*x)*sin(2*x) - 8*sin(2*x)*sin(x) + 4*cos(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)*si n(2*x) - 4*sin(2*x)^2 + 4*cos(2*x) - 1)
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (24) = 48\).
Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.44 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{3/2}} \, dx=\frac {\tan \left (\frac {1}{2} \, x\right )^{2}}{8 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )} + \frac {\log \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right )}{4 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )} - \frac {2 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 1}{8 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2}} \] Input:
integrate(1/(1-cos(x)^2)^(3/2),x, algorithm="giac")
Output:
1/8*tan(1/2*x)^2/sgn(tan(1/2*x)^3 + tan(1/2*x)) + 1/4*log(tan(1/2*x)^2)/sg n(tan(1/2*x)^3 + tan(1/2*x)) - 1/8*(2*tan(1/2*x)^2 + 1)/(sgn(tan(1/2*x)^3 + tan(1/2*x))*tan(1/2*x)^2)
Timed out. \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{3/2}} \, dx=\int \frac {1}{{\left (1-{\cos \left (x\right )}^2\right )}^{3/2}} \,d x \] Input:
int(1/(1 - cos(x)^2)^(3/2),x)
Output:
int(1/(1 - cos(x)^2)^(3/2), x)
\[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{3/2}} \, dx=\int \frac {\sqrt {-\cos \left (x \right )^{2}+1}}{\cos \left (x \right )^{4}-2 \cos \left (x \right )^{2}+1}d x \] Input:
int(1/(1-cos(x)^2)^(3/2),x)
Output:
int(sqrt( - cos(x)**2 + 1)/(cos(x)**4 - 2*cos(x)**2 + 1),x)