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

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

[Out]

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

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Rubi [A]  time = 0.0164747, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2635, 2641} \[ \frac{2 F\left (\left .\frac{1}{2} \left (a+b x-\frac{\pi }{2}\right )\right |2\right )}{3 b}-\frac{2 \sqrt{\sin (a+b x)} \cos (a+b x)}{3 b} \]

Antiderivative was successfully verified.

[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 2635

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] && Integer
Q[2*n]

Rule 2641

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

Rubi steps

\begin{align*} \int \sin ^{\frac{3}{2}}(a+b x) \, dx &=-\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 F\left (\left .\frac{1}{2} \left (a-\frac{\pi }{2}+b x\right )\right |2\right )}{3 b}-\frac{2 \cos (a+b x) \sqrt{\sin (a+b x)}}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.0378158, size = 40, normalized size = 0.85 \[ -\frac{2 \left (F\left (\left .\frac{1}{4} (-2 a-2 b x+\pi )\right |2\right )+\sqrt{\sin (a+b x)} \cos (a+b x)\right )}{3 b} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.026, size = 88, normalized size = 1.9 \begin{align*}{\frac{1}{b\cos \left ( bx+a \right ) } \left ({\frac{1}{3}\sqrt{\sin \left ( bx+a \right ) +1}\sqrt{-2\,\sin \left ( bx+a \right ) +2}\sqrt{-\sin \left ( bx+a \right ) }{\it EllipticF} \left ( \sqrt{\sin \left ( bx+a \right ) +1},{\frac{\sqrt{2}}{2}} \right ) }-{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sin \left ( bx+a \right ) }{3}} \right ){\frac{1}{\sqrt{\sin \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (b x + a\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sin \left (b x + a\right )^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sin(b*x + a)^(3/2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin ^{\frac{3}{2}}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (b x + a\right )^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(3/2), x)