Integrand size = 10, antiderivative size = 53 \[ \int \left (a \sin ^2(x)\right )^{5/2} \, dx=-\frac {8}{15} a^2 \cot (x) \sqrt {a \sin ^2(x)}-\frac {4}{15} a \cot (x) \left (a \sin ^2(x)\right )^{3/2}-\frac {1}{5} \cot (x) \left (a \sin ^2(x)\right )^{5/2} \] 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 = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.68 \[ \int \left (a \sin ^2(x)\right )^{5/2} \, dx=-\frac {1}{240} a^2 (150 \cos (x)-25 \cos (3 x)+3 \cos (5 x)) \csc (x) \sqrt {a \sin ^2(x)} \] Input:
Integrate[(a*Sin[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 = 0.36 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {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.
\(\displaystyle \int \left (a \sin ^2(x)\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 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*Sin[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[((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 = 0.67 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.60
method | result | size |
default | \(-\frac {a^{3} \sin \left (x \right ) \cos \left (x \right ) \left (3 \sin \left (x \right )^{4}+4 \sin \left (x \right )^{2}+8\right )}{15 \sqrt {a \sin \left (x \right )^{2}}}\) | \(32\) |
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*sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/15*a^3*sin(x)*cos(x)*(3*sin(x)^4+4*sin(x)^2+8)/(a*sin(x)^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.81 \[ \int \left (a \sin ^2(x)\right )^{5/2} \, dx=-\frac {{\left (3 \, a^{2} \cos \left (x\right )^{5} - 10 \, a^{2} \cos \left (x\right )^{3} + 15 \, a^{2} \cos \left (x\right )\right )} \sqrt {-a \cos \left (x\right )^{2} + a}}{15 \, \sin \left (x\right )} \] Input:
integrate((a*sin(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)
\[ \int \left (a \sin ^2(x)\right )^{5/2} \, dx=\int \left (a \sin ^{2}{\left (x \right )}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((a*sin(x)**2)**(5/2),x)
Output:
Integral((a*sin(x)**2)**(5/2), x)
\[ \int \left (a \sin ^2(x)\right )^{5/2} \, dx=\int { \left (a \sin \left (x\right )^{2}\right )^{\frac {5}{2}} \,d x } \] Input:
integrate((a*sin(x)^2)^(5/2),x, algorithm="maxima")
Output:
integrate((a*sin(x)^2)^(5/2), x)
Time = 0.43 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.60 \[ \int \left (a \sin ^2(x)\right )^{5/2} \, dx=-\frac {1}{15} \, {\left ({\left (3 \, \cos \left (x\right )^{5} - 10 \, \cos \left (x\right )^{3} + 15 \, \cos \left (x\right )\right )} \mathrm {sgn}\left (\sin \left (x\right )\right ) - 8 \, \mathrm {sgn}\left (\sin \left (x\right )\right )\right )} a^{\frac {5}{2}} \] Input:
integrate((a*sin(x)^2)^(5/2),x, algorithm="giac")
Output:
-1/15*((3*cos(x)^5 - 10*cos(x)^3 + 15*cos(x))*sgn(sin(x)) - 8*sgn(sin(x))) *a^(5/2)
Timed out. \[ \int \left (a \sin ^2(x)\right )^{5/2} \, dx=\int {\left (a\,{\sin \left (x\right )}^2\right )}^{5/2} \,d x \] Input:
int((a*sin(x)^2)^(5/2),x)
Output:
int((a*sin(x)^2)^(5/2), x)
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.55 \[ \int \left (a \sin ^2(x)\right )^{5/2} \, dx=\frac {\sqrt {a}\, a^{2} \left (-3 \cos \left (x \right ) \sin \left (x \right )^{4}-4 \cos \left (x \right ) \sin \left (x \right )^{2}-8 \cos \left (x \right )+8\right )}{15} \] Input:
int((a*sin(x)^2)^(5/2),x)
Output:
(sqrt(a)*a**2*( - 3*cos(x)*sin(x)**4 - 4*cos(x)*sin(x)**2 - 8*cos(x) + 8)) /15