Integrand size = 13, antiderivative size = 53 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\frac {8}{15} a^2 \sqrt {a \cos ^2(x)} \tan (x)+\frac {4}{15} a \left (a \cos ^2(x)\right )^{3/2} \tan (x)+\frac {1}{5} \left (a \cos ^2(x)\right )^{5/2} \tan (x) \] Output:
8/15*a^2*(a*cos(x)^2)^(1/2)*tan(x)+4/15*a*(a*cos(x)^2)^(3/2)*tan(x)+1/5*(a *cos(x)^2)^(5/2)*tan(x)
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.62 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\frac {1}{15} a^2 \sqrt {a \cos ^2(x)} \left (15-10 \sin ^2(x)+3 \sin ^4(x)\right ) \tan (x) \] Input:
Integrate[(a - a*Sin[x]^2)^(5/2),x]
Output:
(a^2*Sqrt[a*Cos[x]^2]*(15 - 10*Sin[x]^2 + 3*Sin[x]^4)*Tan[x])/15
Time = 0.40 (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, 3117}
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.
\(\displaystyle \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a-a \sin (x)^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \left (a \cos ^2(x)\right )^{5/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \sin \left (x+\frac {\pi }{2}\right )^2\right )^{5/2}dx\) |
\(\Big \downarrow \) 3682 |
\(\displaystyle \frac {4}{5} a \int \left (a \cos ^2(x)\right )^{3/2}dx+\frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} a \int \left (a \sin \left (x+\frac {\pi }{2}\right )^2\right )^{3/2}dx+\frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3682 |
\(\displaystyle \frac {4}{5} a \left (\frac {2}{3} a \int \sqrt {a \cos ^2(x)}dx+\frac {1}{3} \tan (x) \left (a \cos ^2(x)\right )^{3/2}\right )+\frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} a \left (\frac {2}{3} a \int \sqrt {a \sin \left (x+\frac {\pi }{2}\right )^2}dx+\frac {1}{3} \tan (x) \left (a \cos ^2(x)\right )^{3/2}\right )+\frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {4}{5} a \left (\frac {2}{3} a \sec (x) \sqrt {a \cos ^2(x)} \int \cos (x)dx+\frac {1}{3} \tan (x) \left (a \cos ^2(x)\right )^{3/2}\right )+\frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{5} a \left (\frac {2}{3} a \sec (x) \sqrt {a \cos ^2(x)} \int \sin \left (x+\frac {\pi }{2}\right )dx+\frac {1}{3} \tan (x) \left (a \cos ^2(x)\right )^{3/2}\right )+\frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}\) |
\(\Big \downarrow \) 3117 |
\(\displaystyle \frac {1}{5} \tan (x) \left (a \cos ^2(x)\right )^{5/2}+\frac {4}{5} a \left (\frac {1}{3} \tan (x) \left (a \cos ^2(x)\right )^{3/2}+\frac {2}{3} a \tan (x) \sqrt {a \cos ^2(x)}\right )\) |
Input:
Int[(a - a*Sin[x]^2)^(5/2),x]
Output:
((a*Cos[x]^2)^(5/2)*Tan[x])/5 + (4*a*((2*a*Sqrt[a*Cos[x]^2]*Tan[x])/3 + (( a*Cos[x]^2)^(3/2)*Tan[x])/3))/5
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
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 = 0.60 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.60
method | result | size |
default | \(\frac {\cos \left (x \right ) a^{3} \sin \left (x \right ) \left (3 \cos \left (x \right )^{4}+4 \cos \left (x \right )^{2}+8\right )}{15 \sqrt {a \cos \left (x \right )^{2}}}\) | \(32\) |
risch | \(-\frac {i a^{2} {\mathrm e}^{6 i x} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}{160 \left ({\mathrm e}^{2 i x}+1\right )}-\frac {5 i a^{2} {\mathrm e}^{2 i x} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}{16 \left ({\mathrm e}^{2 i x}+1\right )}+\frac {5 i a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}{16 \left ({\mathrm e}^{2 i x}+1\right )}+\frac {5 i a^{2} {\mathrm e}^{-2 i x} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}{96 \left ({\mathrm e}^{2 i x}+1\right )}-\frac {11 i a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}\, \cos \left (4 x \right )}{240 \left ({\mathrm e}^{2 i x}+1\right )}+\frac {7 a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}\, \sin \left (4 x \right )}{120 \left ({\mathrm e}^{2 i x}+1\right )}\) | \(222\) |
Input:
int((a-a*sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/15*cos(x)*a^3*sin(x)*(3*cos(x)^4+4*cos(x)^2+8)/(a*cos(x)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.75 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\frac {{\left (3 \, a^{2} \cos \left (x\right )^{4} + 4 \, a^{2} \cos \left (x\right )^{2} + 8 \, a^{2}\right )} \sqrt {a \cos \left (x\right )^{2}} \sin \left (x\right )}{15 \, \cos \left (x\right )} \] Input:
integrate((a-a*sin(x)^2)^(5/2),x, algorithm="fricas")
Output:
1/15*(3*a^2*cos(x)^4 + 4*a^2*cos(x)^2 + 8*a^2)*sqrt(a*cos(x)^2)*sin(x)/cos (x)
Timed out. \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\text {Timed out} \] Input:
integrate((a-a*sin(x)**2)**(5/2),x)
Output:
Timed out
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.58 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\frac {1}{240} \, {\left (3 \, a^{2} \sin \left (5 \, x\right ) + 25 \, a^{2} \sin \left (3 \, x\right ) + 150 \, a^{2} \sin \left (x\right )\right )} \sqrt {a} \] Input:
integrate((a-a*sin(x)^2)^(5/2),x, algorithm="maxima")
Output:
1/240*(3*a^2*sin(5*x) + 25*a^2*sin(3*x) + 150*a^2*sin(x))*sqrt(a)
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (41) = 82\).
Time = 0.56 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.58 \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=-\frac {2 \, {\left (15 \, a^{\frac {5}{2}} {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{4} - 1\right ) - 40 \, a^{\frac {5}{2}} {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{4} - 1\right ) + 48 \, a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{4} - 1\right )\right )}}{15 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{5}} \] Input:
integrate((a-a*sin(x)^2)^(5/2),x, algorithm="giac")
Output:
-2/15*(15*a^(5/2)*(1/tan(1/2*x) + tan(1/2*x))^4*sgn(tan(1/2*x)^4 - 1) - 40 *a^(5/2)*(1/tan(1/2*x) + tan(1/2*x))^2*sgn(tan(1/2*x)^4 - 1) + 48*a^(5/2)* sgn(tan(1/2*x)^4 - 1))/(1/tan(1/2*x) + tan(1/2*x))^5
Timed out. \[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\int {\left (a-a\,{\sin \left (x\right )}^2\right )}^{5/2} \,d x \] Input:
int((a - a*sin(x)^2)^(5/2),x)
Output:
int((a - a*sin(x)^2)^(5/2), x)
\[ \int \left (a-a \sin ^2(x)\right )^{5/2} \, dx=\sqrt {a}\, a^{2} \left (\int \sqrt {-\sin \left (x \right )^{2}+1}d x +\int \sqrt {-\sin \left (x \right )^{2}+1}\, \sin \left (x \right )^{4}d x -2 \left (\int \sqrt {-\sin \left (x \right )^{2}+1}\, \sin \left (x \right )^{2}d x \right )\right ) \] Input:
int((a-a*sin(x)^2)^(5/2),x)
Output:
sqrt(a)*a**2*(int(sqrt( - sin(x)**2 + 1),x) + int(sqrt( - sin(x)**2 + 1)*s in(x)**4,x) - 2*int(sqrt( - sin(x)**2 + 1)*sin(x)**2,x))