\(\int \sin ^{\frac {3}{2}}(a+b x) \, dx\) [19]

Optimal result

Integrand size = 10, antiderivative size = 47 \[ \int \sin ^{\frac {3}{2}}(a+b x) \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right )}{3 b}-\frac {2 \cos (a+b x) \sqrt {\sin (a+b x)}}{3 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2715, 2720} \[ \int \sin ^{\frac {3}{2}}(a+b x) \, dx=\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{3 b}-\frac {2 \sqrt {\sin (a+b x)} \cos (a+b x)}{3 b} \]

[In]

Int[Sin[a + b*x]^(3/2),x]

[Out]

(2*EllipticF[(a - Pi/2 + b*x)/2, 2])/(3*b) - (2*Cos[a + b*x]*Sqrt[Sin[a + b*x]])/(3*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 2720

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos (a+b x) \sqrt {\sin (a+b x)}}{3 b}+\frac {1}{3} \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx \\ & = \frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right )}{3 b}-\frac {2 \cos (a+b x) \sqrt {\sin (a+b x)}}{3 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.85 \[ \int \sin ^{\frac {3}{2}}(a+b x) \, dx=-\frac {2 \left (\operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right )+\cos (a+b x) \sqrt {\sin (a+b x)}\right )}{3 b} \]

[In]

Integrate[Sin[a + b*x]^(3/2),x]

[Out]

(-2*(EllipticF[(-2*a + Pi - 2*b*x)/4, 2] + Cos[a + b*x]*Sqrt[Sin[a + b*x]]))/(3*b)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.87

method	result	size
default	\(\frac {\frac {\sqrt {\sin \left (b x +a \right )+1}\, \sqrt {-2 \sin \left (b x +a \right )+2}\, \sqrt {-\sin \left (b x +a \right )}\, F\left (\sqrt {\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )}{3}-\frac {2 \left (\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}}{\cos \left (b x +a \right ) \sqrt {\sin \left (b x +a \right )}\, b}\)	\(88\)

[In]

int(sin(b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

(1/3*(sin(b*x+a)+1)^(1/2)*(-2*sin(b*x+a)+2)^(1/2)*(-sin(b*x+a))^(1/2)*EllipticF((sin(b*x+a)+1)^(1/2),1/2*2^(1/
2))-2/3*cos(b*x+a)^2*sin(b*x+a))/cos(b*x+a)/sin(b*x+a)^(1/2)/b

Fricas [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.53 \[ \int \sin ^{\frac {3}{2}}(a+b x) \, dx=\frac {\sqrt {2} \sqrt {-i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + \sqrt {2} \sqrt {i} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - 2 \, \cos \left (b x + a\right ) \sqrt {\sin \left (b x + a\right )}}{3 \, b} \]

[In]

integrate(sin(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \sin ^{\frac {3}{2}}(a+b x) \, dx=\int \sin ^{\frac {3}{2}}{\left (a + b x \right )}\, dx \]

[In]

integrate(sin(b*x+a)**(3/2),x)

[Out]

Integral(sin(a + b*x)**(3/2), x)

Maxima [F]

\[ \int \sin ^{\frac {3}{2}}(a+b x) \, dx=\int { \sin \left (b x + a\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(sin(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^(3/2), x)

Giac [F]

\[ \int \sin ^{\frac {3}{2}}(a+b x) \, dx=\int { \sin \left (b x + a\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(sin(b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.89 \[ \int \sin ^{\frac {3}{2}}(a+b x) \, dx=-\frac {\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{2};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{5/4}} \]

[In]

int(sin(a + b*x)^(3/2),x)

[Out]

-(cos(a + b*x)*sin(a + b*x)^(5/2)*hypergeom([-1/4, 1/2], 3/2, cos(a + b*x)^2))/(b*(sin(a + b*x)^2)^(5/4))