3.1.18 \(\int \sin ^{\frac {5}{2}}(a+b x) \, dx\) [18]

Optimal. Leaf size=47 \[ \frac {6 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right )}{5 b}-\frac {2 \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x)}{5 b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2715, 2719} \begin {gather*} \frac {6 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{5 b}-\frac {2 \sin ^{\frac {3}{2}}(a+b x) \cos (a+b x)}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^(5/2),x]

[Out]

(6*EllipticE[(a - Pi/2 + b*x)/2, 2])/(5*b) - (2*Cos[a + b*x]*Sin[a + 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*} \int \sin ^{\frac {5}{2}}(a+b x) \, dx &=-\frac {2 \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x)}{5 b}+\frac {3}{5} \int \sqrt {\sin (a+b x)} \, dx\\ &=\frac {6 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right )}{5 b}-\frac {2 \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x)}{5 b}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 44, normalized size = 0.94 \begin {gather*} -\frac {6 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right )+\sqrt {\sin (a+b x)} \sin (2 (a+b x))}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^(5/2),x]

[Out]

-1/5*(6*EllipticE[(-2*a + Pi - 2*b*x)/4, 2] + Sqrt[Sin[a + b*x]]*Sin[2*(a + b*x)])/b

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Maple [A]
time = 0.06, size = 142, normalized size = 3.02

method	result	size
default	\(\frac {\frac {2 \left (\sin ^{4}\left (b x +a \right )\right )}{5}-\frac {2 \left (\sin ^{2}\left (b x +a \right )\right )}{5}-\frac {6 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \EllipticE \left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}+\frac {3 \sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, \EllipticF \left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{5}}{\cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b}\)	\(142\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2/5*sin(b*x+a)^4-2/5*sin(b*x+a)^2-6/5*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Ellipt
icE((sin(b*x+a)+1)^(1/2),1/2*2^(1/2))+3/5*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*Ell
ipticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/2)))/cos(b*x+a)/sin(b*x+a)^(1/2)/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 80, normalized size = 1.70 \begin {gather*} -\frac {2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right )^{\frac {3}{2}} - 3 i \, \sqrt {2} \sqrt {-i} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 i \, \sqrt {2} \sqrt {i} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}{5 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{\frac {5}{2}}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(5/2), x)

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Mupad [B]
time = 0.45, size = 42, normalized size = 0.89 \begin {gather*} -\frac {\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^{7/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{7/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^(5/2),x)

[Out]

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

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