Integrand size = 13, antiderivative size = 53 \[ \int \left (a-a \cos ^2(x)\right )^{5/2} \, dx=-\frac {8}{15} a^2 \cot (x) \sqrt {a \sin ^2(x)}-\frac {4}{15} a \cot (x) \left (a \sin ^2(x)\right )^{3/2}-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2} \] Output:
-8/15*a^2*cot(x)*(a*sin(x)^2)^(1/2)-4/15*a*cot(x)*(a*sin(x)^2)^(3/2)-1/5*c ot(x)*(a*sin(x)^2)^(5/2)
Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.68 \[ \int \left (a-a \cos ^2(x)\right )^{5/2} \, dx=-\frac {1}{240} a^2 (150 \cos (x)-25 \cos (3 x)+3 \cos (5 x)) \csc (x) \sqrt {a \sin ^2(x)} \] Input:
Integrate[(a - a*Cos[x]^2)^(5/2),x]
Output:
-1/240*(a^2*(150*Cos[x] - 25*Cos[3*x] + 3*Cos[5*x])*Csc[x]*Sqrt[a*Sin[x]^2 ])
Time = 0.39 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, 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.
\(\displaystyle \int \left (a-a \cos ^2(x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-a \sin \left (x+\frac {\pi }{2}\right )^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \left (a \sin ^2(x)\right )^{5/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \sin (x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 3682 |
\(\displaystyle \frac {4}{5} a \int \left (a \sin ^2(x)\right )^{3/2}dx-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} a \int \left (a \sin (x)^2\right )^{3/2}dx-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3682 |
\(\displaystyle \frac {4}{5} a \left (\frac {2}{3} a \int \sqrt {a \sin ^2(x)}dx-\frac {1}{3} \cot (x) \left (a \sin ^2(x)\right )^{3/2}\right )-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} a \left (\frac {2}{3} a \int \sqrt {a \sin (x)^2}dx-\frac {1}{3} \cot (x) \left (a \sin ^2(x)\right )^{3/2}\right )-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {4}{5} a \left (\frac {2}{3} a \csc (x) \sqrt {a \sin ^2(x)} \int \sin (x)dx-\frac {1}{3} \cot (x) \left (a \sin ^2(x)\right )^{3/2}\right )-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} a \left (\frac {2}{3} a \csc (x) \sqrt {a \sin ^2(x)} \int \sin (x)dx-\frac {1}{3} \cot (x) \left (a \sin ^2(x)\right )^{3/2}\right )-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3118 |
\(\displaystyle \frac {4}{5} a \left (-\frac {1}{3} \cot (x) \left (a \sin ^2(x)\right )^{3/2}-\frac {2}{3} a \cot (x) \sqrt {a \sin ^2(x)}\right )-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2}\) |
Input:
Int[(a - a*Cos[x]^2)^(5/2),x]
Output:
-1/5*(Cot[x]*(a*Sin[x]^2)^(5/2)) + (4*a*((-2*a*Cot[x]*Sqrt[a*Sin[x]^2])/3 - (Cot[x]*(a*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 = 0.48 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.60
method | result | size |
default | \(-\frac {\sin \left (x \right ) a^{3} \cos \left (x \right ) \left (3 \sin \left (x \right )^{4}+4 \sin \left (x \right )^{2}+8\right )}{15 \sqrt {a \sin \left (x \right )^{2}}}\) | \(32\) |
risch | \(-\frac {i a^{2} {\mathrm e}^{6 i x} \sqrt {-a \left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}{160 \left ({\mathrm e}^{2 i x}-1\right )}-\frac {5 i a^{2} {\mathrm e}^{2 i x} \sqrt {-a \left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}{16 \left ({\mathrm e}^{2 i x}-1\right )}-\frac {5 i \sqrt {-a \left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}\, a^{2}}{16 \left ({\mathrm e}^{2 i x}-1\right )}+\frac {5 i a^{2} {\mathrm e}^{-2 i x} \sqrt {-a \left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}{96 \left ({\mathrm e}^{2 i x}-1\right )}+\frac {11 i \sqrt {-a \left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}\, a^{2} \cos \left (4 x \right )}{240 \left ({\mathrm e}^{2 i x}-1\right )}-\frac {7 \sqrt {-a \left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}\, a^{2} \sin \left (4 x \right )}{120 \left ({\mathrm e}^{2 i x}-1\right )}\) | \(228\) |
Input:
int((a-a*cos(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/15*sin(x)*a^3*cos(x)*(3*sin(x)^4+4*sin(x)^2+8)/(a*sin(x)^2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \left (a-a \cos ^2(x)\right )^{5/2} \, dx=-\frac {{\left (3 \, a^{2} \cos \left (x\right )^{5} - 10 \, a^{2} \cos \left (x\right )^{3} + 15 \, a^{2} \cos \left (x\right )\right )} \sqrt {-a \cos \left (x\right )^{2} + a}}{15 \, \sin \left (x\right )} \] Input:
integrate((a-a*cos(x)^2)^(5/2),x, algorithm="fricas")
Output:
-1/15*(3*a^2*cos(x)^5 - 10*a^2*cos(x)^3 + 15*a^2*cos(x))*sqrt(-a*cos(x)^2 + a)/sin(x)
Timed out. \[ \int \left (a-a \cos ^2(x)\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((a-a*cos(x)**2)**(5/2),x)
Output:
Timed out
\[ \int \left (a-a \cos ^2(x)\right )^{5/2} \, dx=\int { {\left (-a \cos \left (x\right )^{2} + a\right )}^{\frac {5}{2}} \,d x } \] Input:
integrate((a-a*cos(x)^2)^(5/2),x, algorithm="maxima")
Output:
integrate((-a*cos(x)^2 + a)^(5/2), x)
Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.42 \[ \int \left (a-a \cos ^2(x)\right )^{5/2} \, dx=-\frac {16 \, {\left (10 \, a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{4} + 5 \, a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2} + a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )\right )}}{15 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{5}} \] Input:
integrate((a-a*cos(x)^2)^(5/2),x, algorithm="giac")
Output:
-16/15*(10*a^(5/2)*sgn(tan(1/2*x)^3 + tan(1/2*x))*tan(1/2*x)^4 + 5*a^(5/2) *sgn(tan(1/2*x)^3 + tan(1/2*x))*tan(1/2*x)^2 + a^(5/2)*sgn(tan(1/2*x)^3 + tan(1/2*x)))/(tan(1/2*x)^2 + 1)^5
Timed out. \[ \int \left (a-a \cos ^2(x)\right )^{5/2} \, dx=\int {\left (a-a\,{\cos \left (x\right )}^2\right )}^{5/2} \,d x \] Input:
int((a - a*cos(x)^2)^(5/2),x)
Output:
int((a - a*cos(x)^2)^(5/2), x)
\[ \int \left (a-a \cos ^2(x)\right )^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {-\cos \left (x \right )^{2}+1}d x +\int \sqrt {-\cos \left (x \right )^{2}+1}\, \cos \left (x \right )^{4}d x -2 \left (\int \sqrt {-\cos \left (x \right )^{2}+1}\, \cos \left (x \right )^{2}d x \right )\right ) \] Input:
int((a-a*cos(x)^2)^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt( - cos(x)**2 + 1),x) + int(sqrt( - cos(x)**2 + 1)*c os(x)**4,x) - 2*int(sqrt( - cos(x)**2 + 1)*cos(x)**2,x))