3.78 \(\int \frac{\sin (a+b x)}{\sin ^{\frac{5}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=53 \[ \frac{\sin (a+b x)}{3 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{2 \cos (a+b x)}{3 b \sqrt{\sin (2 a+2 b x)}} \]

[Out]

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

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Rubi [A]  time = 0.0395403, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {4304, 4291} \[ \frac{\sin (a+b x)}{3 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{2 \cos (a+b x)}{3 b \sqrt{\sin (2 a+2 b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4304

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[(Sin[a + b*x]*(g*Sin[c +
d*x])^(p + 1))/(2*b*g*(p + 1)), x] + Dist[(2*p + 3)/(2*g*(p + 1)), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p + 1),
x], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && LtQ[p, -1] && Inte
gerQ[2*p]

Rule 4291

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> -Simp[((e*Cos[a +
b*x])^m*(g*Sin[c + d*x])^(p + 1))/(b*g*m), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && E
qQ[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps

\begin{align*} \int \frac{\sin (a+b x)}{\sin ^{\frac{5}{2}}(2 a+2 b x)} \, dx &=\frac{\sin (a+b x)}{3 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{2}{3} \int \frac{\cos (a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\\ &=\frac{\sin (a+b x)}{3 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{2 \cos (a+b x)}{3 b \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.102788, size = 43, normalized size = 0.81 \[ \frac{\sqrt{\sin (2 (a+b x))} \left (\frac{1}{12} \tan (a+b x) \sec (a+b x)-\frac{1}{4} \csc (a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Sin[2*(a + b*x)]]*(-Csc[a + b*x]/4 + (Sec[a + b*x]*Tan[a + b*x])/12))/b

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Maple [C]  time = 18.474, size = 597, normalized size = 11.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/b*(-tan(1/2*b*x+1/2*a)/(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*(6*(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1
/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticE((tan(1/
2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^2-3*(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*
(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*
x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^2+6*(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*(tan
(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticE((tan(1/2*b*x+1/
2*a)+1)^(1/2),1/2*2^(1/2))-3*(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*
(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2)
)+2*(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*tan(1/2*b*x+1/2*a)^4+2*(tan(1/2*b*x+1/2*a)^3-tan(1/2*b
*x+1/2*a))^(1/2)*tan(1/2*b*x+1/2*a)^4-2*(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*tan(1/2*b*x+1/2*a)
^2-2*(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2))/tan(1/2*b*x+1/2*a)/(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2
*a))^(1/2)/(tan(1/2*b*x+1/2*a)^2+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.509805, size = 184, normalized size = 3.47 \begin{align*} -\frac{4 \, \cos \left (b x + a\right )^{2} \sin \left (b x + a\right ) + \sqrt{2}{\left (4 \, \cos \left (b x + a\right )^{2} - 1\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{12 \, b \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*cos(b*x + a)^2*sin(b*x + a) + sqrt(2)*(4*cos(b*x + a)^2 - 1)*sqrt(cos(b*x + a)*sin(b*x + a)))/(b*cos(
b*x + a)^2*sin(b*x + a))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out