\(\int \sin ^{\frac {5}{2}}(b x) \, dx\) [10]

Optimal result

Integrand size = 8, antiderivative size = 41 \[ \int \sin ^{\frac {5}{2}}(b x) \, dx=-\frac {6 E\left (\left .\frac {\pi }{4}-\frac {b x}{2}\right |2\right )}{5 b}-\frac {2 \cos (b x) \sin ^{\frac {3}{2}}(b x)}{5 b} \]

[Out]

-6/5*(sin(1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/4*Pi+1/2*b*x)*EllipticE(cos(1/4*Pi+1/2*b*x),2^(1/2))/b-2/5*cos(b*x)*s
in(b*x)^(3/2)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 2719} \[ \int \sin ^{\frac {5}{2}}(b x) \, dx=-\frac {6 E\left (\left .\frac {\pi }{4}-\frac {b x}{2}\right |2\right )}{5 b}-\frac {2 \sin ^{\frac {3}{2}}(b x) \cos (b x)}{5 b} \]

[In]

Int[Sin[b*x]^(5/2),x]

[Out]

(-6*EllipticE[Pi/4 - (b*x)/2, 2])/(5*b) - (2*Cos[b*x]*Sin[b*x]^(3/2))/(5*b)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos (b x) \sin ^{\frac {3}{2}}(b x)}{5 b}+\frac {3}{5} \int \sqrt {\sin (b x)} \, dx \\ & = -\frac {6 E\left (\left .\frac {\pi }{4}-\frac {b x}{2}\right |2\right )}{5 b}-\frac {2 \cos (b x) \sin ^{\frac {3}{2}}(b x)}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \sin ^{\frac {5}{2}}(b x) \, dx=-\frac {2 \left (3 E\left (\left .\frac {1}{4} (\pi -2 b x)\right |2\right )+\cos (b x) \sin ^{\frac {3}{2}}(b x)\right )}{5 b} \]

[In]

Integrate[Sin[b*x]^(5/2),x]

[Out]

(-2*(3*EllipticE[(Pi - 2*b*x)/4, 2] + Cos[b*x]*Sin[b*x]^(3/2)))/(5*b)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(117\) vs. \(2(58)=116\).

Time = 0.11 (sec) , antiderivative size = 118, normalized size of antiderivative = 2.88

method	result	size
default	\(\frac {\frac {2 \left (\sin ^{4}\left (b x \right )\right )}{5}-\frac {2 \left (\sin ^{2}\left (b x \right )\right )}{5}-\frac {6 \sqrt {\sin \left (b x \right )+1}\, \sqrt {-2 \sin \left (b x \right )+2}\, \sqrt {-\sin \left (b x \right )}\, E\left (\sqrt {\sin \left (b x \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {3 \sqrt {\sin \left (b x \right )+1}\, \sqrt {-2 \sin \left (b x \right )+2}\, \sqrt {-\sin \left (b x \right )}\, F\left (\sqrt {\sin \left (b x \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}}{\cos \left (b x \right ) \sqrt {\sin \left (b x \right )}\, b}\)	\(118\)

[In]

int(sin(b*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

(2/5*sin(b*x)^4-2/5*sin(b*x)^2-6/5*(sin(b*x)+1)^(1/2)*(-2*sin(b*x)+2)^(1/2)*(-sin(b*x))^(1/2)*EllipticE((sin(b
*x)+1)^(1/2),1/2*2^(1/2))+3/5*(sin(b*x)+1)^(1/2)*(-2*sin(b*x)+2)^(1/2)*(-sin(b*x))^(1/2)*EllipticF((sin(b*x)+1
)^(1/2),1/2*2^(1/2)))/cos(b*x)/sin(b*x)^(1/2)/b

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.66 \[ \int \sin ^{\frac {5}{2}}(b x) \, dx=-\frac {2 \, \cos \left (b x\right ) \sin \left (b x\right )^{\frac {3}{2}} - 3 i \, \sqrt {2} \sqrt {-i} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x\right ) + i \, \sin \left (b x\right )\right )\right ) + 3 i \, \sqrt {2} \sqrt {i} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x\right ) - i \, \sin \left (b x\right )\right )\right )}{5 \, b} \]

[In]

integrate(sin(b*x)^(5/2),x, algorithm="fricas")

[Out]

-1/5*(2*cos(b*x)*sin(b*x)^(3/2) - 3*I*sqrt(2)*sqrt(-I)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x
) + I*sin(b*x))) + 3*I*sqrt(2)*sqrt(I)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(b*x) - I*sin(b*x)))
)/b

Sympy [F]

\[ \int \sin ^{\frac {5}{2}}(b x) \, dx=\int \sin ^{\frac {5}{2}}{\left (b x \right )}\, dx \]

[In]

integrate(sin(b*x)**(5/2),x)

[Out]

Integral(sin(b*x)**(5/2), x)

Maxima [F]

\[ \int \sin ^{\frac {5}{2}}(b x) \, dx=\int { \sin \left (b x\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(sin(b*x)^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x)^(5/2), x)

Giac [F]

\[ \int \sin ^{\frac {5}{2}}(b x) \, dx=\int { \sin \left (b x\right )^{\frac {5}{2}} \,d x } \]

[In]

integrate(sin(b*x)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(b*x)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \sin ^{\frac {5}{2}}(b x) \, dx=-\frac {\cos \left (b\,x\right )\,{\sin \left (b\,x\right )}^{7/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (b\,x\right )}^2\right )}{b\,{\left ({\sin \left (b\,x\right )}^2\right )}^{7/4}} \]

[In]

int(sin(b*x)^(5/2),x)

[Out]

-(cos(b*x)*sin(b*x)^(7/2)*hypergeom([-3/4, 1/2], 3/2, cos(b*x)^2))/(b*(sin(b*x)^2)^(7/4))