Integrand size = 12, antiderivative size = 43 \[ \int \left (1-\cos ^2(x)\right )^{5/2} \, dx=-\frac {8}{15} \cot (x) \sqrt {\sin ^2(x)}-\frac {4}{15} \cot (x) \sin ^2(x)^{3/2}-\frac {1}{5} \cot (x) \sin ^2(x)^{5/2} \] Output:
-8/15*cot(x)*(sin(x)^2)^(1/2)-4/15*cot(x)*(sin(x)^2)^(3/2)-1/5*cot(x)*(sin (x)^2)^(5/2)
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.72 \[ \int \left (1-\cos ^2(x)\right )^{5/2} \, dx=-\frac {1}{240} (150 \cos (x)-25 \cos (3 x)+3 \cos (5 x)) \csc (x) \sqrt {\sin ^2(x)} \] Input:
Integrate[(1 - Cos[x]^2)^(5/2),x]
Output:
-1/240*((150*Cos[x] - 25*Cos[3*x] + 3*Cos[5*x])*Csc[x]*Sqrt[Sin[x]^2])
Time = 0.39 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.12, 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, 3682, 3042, 3682, 3042, 3686, 3042, 3118}
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.
\(\displaystyle \int \left (1-\cos ^2(x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (1-\sin \left (x+\frac {\pi }{2}\right )^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \sin ^2(x)^{5/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (\sin (x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 3682 |
\(\displaystyle \frac {4}{5} \int \sin ^2(x)^{3/2}dx-\frac {1}{5} \sin ^2(x)^{5/2} \cot (x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} \int \left (\sin (x)^2\right )^{3/2}dx-\frac {1}{5} \sin ^2(x)^{5/2} \cot (x)\) |
\(\Big \downarrow \) 3682 |
\(\displaystyle \frac {4}{5} \left (\frac {2}{3} \int \sqrt {\sin ^2(x)}dx-\frac {1}{3} \sin ^2(x)^{3/2} \cot (x)\right )-\frac {1}{5} \sin ^2(x)^{5/2} \cot (x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} \left (\frac {2}{3} \int \sqrt {\sin (x)^2}dx-\frac {1}{3} \sin ^2(x)^{3/2} \cot (x)\right )-\frac {1}{5} \sin ^2(x)^{5/2} \cot (x)\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {4}{5} \left (\frac {2}{3} \sqrt {\sin ^2(x)} \csc (x) \int \sin (x)dx-\frac {1}{3} \sin ^2(x)^{3/2} \cot (x)\right )-\frac {1}{5} \sin ^2(x)^{5/2} \cot (x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} \left (\frac {2}{3} \sqrt {\sin ^2(x)} \csc (x) \int \sin (x)dx-\frac {1}{3} \sin ^2(x)^{3/2} \cot (x)\right )-\frac {1}{5} \sin ^2(x)^{5/2} \cot (x)\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {4}{5} \left (-\frac {1}{3} \sin ^2(x)^{3/2} \cot (x)-\frac {2}{3} \sqrt {\sin ^2(x)} \cot (x)\right )-\frac {1}{5} \sin ^2(x)^{5/2} \cot (x)\) |
Input:
Int[(1 - Cos[x]^2)^(5/2),x]
Output:
-1/5*(Cot[x]*(Sin[x]^2)^(5/2)) + (4*((-2*Cot[x]*Sqrt[Sin[x]^2])/3 - (Cot[x ]*(Sin[x]^2)^(3/2))/3))/5
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/(2*f*p)), x] + Simp[b*((2*p - 1)/(2*p)) Int[(b*S in[e + f*x]^2)^(p - 1), x], x] /; FreeQ[{b, e, f}, x] && !IntegerQ[p] && G tQ[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]])
Time = 0.49 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.63
method | result | size |
default | \(-\frac {2 \sin \left (x \right ) \cos \left (x \right ) \left (3 \sin \left (x \right )^{4}+4 \sin \left (x \right )^{2}+8\right )}{15 \sqrt {2-2 \cos \left (2 x \right )}}\) | \(27\) |
risch | \(-\frac {i {\mathrm e}^{6 i x} \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}{160 \left ({\mathrm e}^{2 i x}-1\right )}-\frac {5 i {\mathrm e}^{2 i x} \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}{16 \left ({\mathrm e}^{2 i x}-1\right )}-\frac {5 i \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}{16 \left ({\mathrm e}^{2 i x}-1\right )}+\frac {5 i {\mathrm e}^{-2 i x} \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}{96 \left ({\mathrm e}^{2 i x}-1\right )}+\frac {11 i \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}\, \cos \left (4 x \right )}{240 \left ({\mathrm e}^{2 i x}-1\right )}-\frac {7 \sqrt {-\left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}\, \sin \left (4 x \right )}{120 \left ({\mathrm e}^{2 i x}-1\right )}\) | \(204\) |
Input:
int((1-cos(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/15*sin(x)*cos(x)*(3*sin(x)^4+4*sin(x)^2+8)/(sin(x)^2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.40 \[ \int \left (1-\cos ^2(x)\right )^{5/2} \, dx=-\frac {1}{5} \, \cos \left (x\right )^{5} + \frac {2}{3} \, \cos \left (x\right )^{3} - \cos \left (x\right ) \] Input:
integrate((1-cos(x)^2)^(5/2),x, algorithm="fricas")
Output:
-1/5*cos(x)^5 + 2/3*cos(x)^3 - cos(x)
Timed out. \[ \int \left (1-\cos ^2(x)\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((1-cos(x)**2)**(5/2),x)
Output:
Timed out
Time = 0.16 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.40 \[ \int \left (1-\cos ^2(x)\right )^{5/2} \, dx=\frac {1}{80} \, \cos \left (5 \, x\right ) - \frac {5}{48} \, \cos \left (3 \, x\right ) + \frac {5}{8} \, \cos \left (x\right ) \] Input:
integrate((1-cos(x)^2)^(5/2),x, algorithm="maxima")
Output:
1/80*cos(5*x) - 5/48*cos(3*x) + 5/8*cos(x)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (31) = 62\).
Time = 0.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.51 \[ \int \left (1-\cos ^2(x)\right )^{5/2} \, dx=-\frac {16 \, {\left (10 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{4} + 5 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2} + \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )\right )}}{15 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{5}} \] Input:
integrate((1-cos(x)^2)^(5/2),x, algorithm="giac")
Output:
-16/15*(10*sgn(tan(1/2*x)^3 + tan(1/2*x))*tan(1/2*x)^4 + 5*sgn(tan(1/2*x)^ 3 + tan(1/2*x))*tan(1/2*x)^2 + sgn(tan(1/2*x)^3 + tan(1/2*x)))/(tan(1/2*x) ^2 + 1)^5
Timed out. \[ \int \left (1-\cos ^2(x)\right )^{5/2} \, dx=\int {\left (1-{\cos \left (x\right )}^2\right )}^{5/2} \,d x \] Input:
int((1 - cos(x)^2)^(5/2),x)
Output:
int((1 - cos(x)^2)^(5/2), x)
\[ \int \left (1-\cos ^2(x)\right )^{5/2} \, dx=\int \sqrt {-\cos \left (x \right )^{2}+1}d x +\int \sqrt {-\cos \left (x \right )^{2}+1}\, \cos \left (x \right )^{4}d x -2 \left (\int \sqrt {-\cos \left (x \right )^{2}+1}\, \cos \left (x \right )^{2}d x \right ) \] Input:
int((1-cos(x)^2)^(5/2),x)
Output:
int(sqrt( - cos(x)**2 + 1),x) + int(sqrt( - cos(x)**2 + 1)*cos(x)**4,x) - 2*int(sqrt( - cos(x)**2 + 1)*cos(x)**2,x)