Integrand size = 12, antiderivative size = 46 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{5/2}} \, dx=-\frac {\cot (x)}{4 \sin ^2(x)^{3/2}}-\frac {3 \cot (x)}{8 \sqrt {\sin ^2(x)}}-\frac {3 \text {arctanh}(\cos (x)) \sin (x)}{8 \sqrt {\sin ^2(x)}} \] Output:
-1/4*cot(x)/(sin(x)^2)^(3/2)-3/8*cot(x)/(sin(x)^2)^(1/2)-3/8*arctanh(cos(x ))*sin(x)/(sin(x)^2)^(1/2)
Time = 0.18 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{5/2}} \, dx=-\frac {\left (6 \csc ^2\left (\frac {x}{2}\right )+\csc ^4\left (\frac {x}{2}\right )+24 \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )-6 \sec ^2\left (\frac {x}{2}\right )-\sec ^4\left (\frac {x}{2}\right )\right ) \sin (x)}{64 \sqrt {\sin ^2(x)}} \] Input:
Integrate[(1 - Cos[x]^2)^(-5/2),x]
Output:
-1/64*((6*Csc[x/2]^2 + Csc[x/2]^4 + 24*(Log[Cos[x/2]] - Log[Sin[x/2]]) - 6 *Sec[x/2]^2 - Sec[x/2]^4)*Sin[x])/Sqrt[Sin[x]^2]
Time = 0.38 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3042, 3655, 3042, 3683, 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.
\(\displaystyle \int \frac {1}{\left (1-\cos ^2(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (1-\sin \left (x+\frac {\pi }{2}\right )^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \frac {1}{\sin ^2(x)^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (\sin (x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3683 |
\(\displaystyle \frac {3}{4} \int \frac {1}{\sin ^2(x)^{3/2}}dx-\frac {\cot (x)}{4 \sin ^2(x)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} \int \frac {1}{\left (\sin (x)^2\right )^{3/2}}dx-\frac {\cot (x)}{4 \sin ^2(x)^{3/2}}\) |
\(\Big \downarrow \) 3683 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {\sin ^2(x)}}dx-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\right )-\frac {\cot (x)}{4 \sin ^2(x)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {\sin (x)^2}}dx-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\right )-\frac {\cot (x)}{4 \sin ^2(x)^{3/2}}\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {3}{4} \left (\frac {\sin (x) \int \csc (x)dx}{2 \sqrt {\sin ^2(x)}}-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\right )-\frac {\cot (x)}{4 \sin ^2(x)^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{4} \left (\frac {\sin (x) \int \csc (x)dx}{2 \sqrt {\sin ^2(x)}}-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\right )-\frac {\cot (x)}{4 \sin ^2(x)^{3/2}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {3}{4} \left (-\frac {\sin (x) \text {arctanh}(\cos (x))}{2 \sqrt {\sin ^2(x)}}-\frac {\cot (x)}{2 \sqrt {\sin ^2(x)}}\right )-\frac {\cot (x)}{4 \sin ^2(x)^{3/2}}\) |
Input:
Int[(1 - Cos[x]^2)^(-5/2),x]
Output:
-1/4*Cot[x]/(Sin[x]^2)^(3/2) + (3*(-1/2*Cot[x]/Sqrt[Sin[x]^2] - (ArcTanh[C os[x]]*Sin[x])/(2*Sqrt[Sin[x]^2])))/4
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 = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26
method | result | size |
default | \(\frac {6 \sin \left (x \right )^{2} \cos \left (x \right )+4 \cos \left (x \right )+\left (3 \ln \left (1+\cos \left (x \right )\right )-3 \ln \left (-1+\cos \left (x \right )\right )\right ) \sin \left (x \right )^{4}}{8 \left (1+\cos \left (x \right )\right ) \left (-1+\cos \left (x \right )\right ) \sin \left (x \right ) \sqrt {2-2 \cos \left (2 x \right )}}\) | \(58\) |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{6 i x}-11 \,{\mathrm e}^{4 i x}-11 \,{\mathrm e}^{2 i x}+3\right )}{4 \left ({\mathrm e}^{2 i x}-1\right )^{3} \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {3 \ln \left ({\mathrm e}^{i x}+1\right ) \sin \left (x \right )}{4 \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {3 \ln \left ({\mathrm e}^{i x}-1\right ) \sin \left (x \right )}{4 \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}\) | \(115\) |
Input:
int(1/(1-cos(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/16*(6*sin(x)^2*cos(x)+4*cos(x)+(3*ln(1+cos(x))-3*ln(-1+cos(x)))*sin(x)^4 )/(1+cos(x))/(-1+cos(x))/sin(x)/(sin(x)^2)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{5/2}} \, dx=\frac {6 \, \cos \left (x\right )^{3} - 3 \, {\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) - 10 \, \cos \left (x\right )}{16 \, {\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )}} \] Input:
integrate(1/(1-cos(x)^2)^(5/2),x, algorithm="fricas")
Output:
1/16*(6*cos(x)^3 - 3*(cos(x)^4 - 2*cos(x)^2 + 1)*log(1/2*cos(x) + 1/2) + 3 *(cos(x)^4 - 2*cos(x)^2 + 1)*log(-1/2*cos(x) + 1/2) - 10*cos(x))/(cos(x)^4 - 2*cos(x)^2 + 1)
\[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (1 - \cos ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(1-cos(x)**2)**(5/2),x)
Output:
Integral((1 - cos(x)**2)**(-5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 869 vs. \(2 (34) = 68\).
Time = 0.19 (sec) , antiderivative size = 869, normalized size of antiderivative = 18.89 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(1-cos(x)^2)^(5/2),x, algorithm="maxima")
Output:
1/16*(4*(3*cos(7*x) - 11*cos(5*x) - 11*cos(3*x) + 3*cos(x))*cos(8*x) - 12* (4*cos(6*x) - 6*cos(4*x) + 4*cos(2*x) - 1)*cos(7*x) + 16*(11*cos(5*x) + 11 *cos(3*x) - 3*cos(x))*cos(6*x) - 44*(6*cos(4*x) - 4*cos(2*x) + 1)*cos(5*x) - 24*(11*cos(3*x) - 3*cos(x))*cos(4*x) + 44*(4*cos(2*x) - 1)*cos(3*x) - 4 8*cos(2*x)*cos(x) + 3*(2*(4*cos(6*x) - 6*cos(4*x) + 4*cos(2*x) - 1)*cos(8* x) - cos(8*x)^2 + 8*(6*cos(4*x) - 4*cos(2*x) + 1)*cos(6*x) - 16*cos(6*x)^2 + 12*(4*cos(2*x) - 1)*cos(4*x) - 36*cos(4*x)^2 - 16*cos(2*x)^2 + 4*(2*sin (6*x) - 3*sin(4*x) + 2*sin(2*x))*sin(8*x) - sin(8*x)^2 + 16*(3*sin(4*x) - 2*sin(2*x))*sin(6*x) - 16*sin(6*x)^2 - 36*sin(4*x)^2 + 48*sin(4*x)*sin(2*x ) - 16*sin(2*x)^2 + 8*cos(2*x) - 1)*log(cos(x)^2 + sin(x)^2 + 2*cos(x) + 1 ) - 3*(2*(4*cos(6*x) - 6*cos(4*x) + 4*cos(2*x) - 1)*cos(8*x) - cos(8*x)^2 + 8*(6*cos(4*x) - 4*cos(2*x) + 1)*cos(6*x) - 16*cos(6*x)^2 + 12*(4*cos(2*x ) - 1)*cos(4*x) - 36*cos(4*x)^2 - 16*cos(2*x)^2 + 4*(2*sin(6*x) - 3*sin(4* x) + 2*sin(2*x))*sin(8*x) - sin(8*x)^2 + 16*(3*sin(4*x) - 2*sin(2*x))*sin( 6*x) - 16*sin(6*x)^2 - 36*sin(4*x)^2 + 48*sin(4*x)*sin(2*x) - 16*sin(2*x)^ 2 + 8*cos(2*x) - 1)*log(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1) + 4*(3*sin(7*x ) - 11*sin(5*x) - 11*sin(3*x) + 3*sin(x))*sin(8*x) - 24*(2*sin(6*x) - 3*si n(4*x) + 2*sin(2*x))*sin(7*x) + 16*(11*sin(5*x) + 11*sin(3*x) - 3*sin(x))* sin(6*x) - 88*(3*sin(4*x) - 2*sin(2*x))*sin(5*x) - 24*(11*sin(3*x) - 3*sin (x))*sin(4*x) + 176*sin(3*x)*sin(2*x) - 48*sin(2*x)*sin(x) + 12*cos(x))...
Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (34) = 68\).
Time = 0.17 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.26 \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{5/2}} \, dx=\frac {1}{64} \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{4} + \frac {1}{8} \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2} + \frac {3 \, \log \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right )}{16 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )} - \frac {18 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 8 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 1}{64 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{4}} \] Input:
integrate(1/(1-cos(x)^2)^(5/2),x, algorithm="giac")
Output:
1/64*sgn(tan(1/2*x)^3 + tan(1/2*x))*tan(1/2*x)^4 + 1/8*sgn(tan(1/2*x)^3 + tan(1/2*x))*tan(1/2*x)^2 + 3/16*log(tan(1/2*x)^2)/sgn(tan(1/2*x)^3 + tan(1 /2*x)) - 1/64*(18*tan(1/2*x)^4 + 8*tan(1/2*x)^2 + 1)/(sgn(tan(1/2*x)^3 + t an(1/2*x))*tan(1/2*x)^4)
Timed out. \[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (1-{\cos \left (x\right )}^2\right )}^{5/2}} \,d x \] Input:
int(1/(1 - cos(x)^2)^(5/2),x)
Output:
int(1/(1 - cos(x)^2)^(5/2), x)
\[ \int \frac {1}{\left (1-\cos ^2(x)\right )^{5/2}} \, dx=-\left (\int \frac {\sqrt {-\cos \left (x \right )^{2}+1}}{\cos \left (x \right )^{6}-3 \cos \left (x \right )^{4}+3 \cos \left (x \right )^{2}-1}d x \right ) \] Input:
int(1/(1-cos(x)^2)^(5/2),x)
Output:
- int(sqrt( - cos(x)**2 + 1)/(cos(x)**6 - 3*cos(x)**4 + 3*cos(x)**2 - 1), x)