3.80 \(\int \frac{\sin (a+b x)}{\sin ^{\frac{9}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=105 \[ \frac{8 \sin (a+b x)}{35 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{\sin (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}-\frac{6 \cos (a+b x)}{35 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{16 \cos (a+b x)}{35 b \sqrt{\sin (2 a+2 b x)}} \]

[Out]

Sin[a + b*x]/(7*b*Sin[2*a + 2*b*x]^(7/2)) - (6*Cos[a + b*x])/(35*b*Sin[2*a + 2*b*x]^(5/2)) + (8*Sin[a + b*x])/
(35*b*Sin[2*a + 2*b*x]^(3/2)) - (16*Cos[a + b*x])/(35*b*Sqrt[Sin[2*a + 2*b*x]])

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Rubi [A]  time = 0.0833013, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4304, 4303, 4291} \[ \frac{8 \sin (a+b x)}{35 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{\sin (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}-\frac{6 \cos (a+b x)}{35 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{16 \cos (a+b x)}{35 b \sqrt{\sin (2 a+2 b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]/Sin[2*a + 2*b*x]^(9/2),x]

[Out]

Sin[a + b*x]/(7*b*Sin[2*a + 2*b*x]^(7/2)) - (6*Cos[a + b*x])/(35*b*Sin[2*a + 2*b*x]^(5/2)) + (8*Sin[a + b*x])/
(35*b*Sin[2*a + 2*b*x]^(3/2)) - (16*Cos[a + b*x])/(35*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 4303

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(Cos[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[Sin[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] && Integ
erQ[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{9}{2}}(2 a+2 b x)} \, dx &=\frac{\sin (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}+\frac{6}{7} \int \frac{\cos (a+b x)}{\sin ^{\frac{7}{2}}(2 a+2 b x)} \, dx\\ &=\frac{\sin (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}-\frac{6 \cos (a+b x)}{35 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{24}{35} \int \frac{\sin (a+b x)}{\sin ^{\frac{5}{2}}(2 a+2 b x)} \, dx\\ &=\frac{\sin (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}-\frac{6 \cos (a+b x)}{35 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{8 \sin (a+b x)}{35 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{16}{35} \int \frac{\cos (a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\\ &=\frac{\sin (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}-\frac{6 \cos (a+b x)}{35 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{8 \sin (a+b x)}{35 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{16 \cos (a+b x)}{35 b \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.146025, size = 67, normalized size = 0.64 \[ \frac{\sqrt{\sin (2 (a+b x))} (-10 \cos (2 (a+b x))+4 \cos (4 (a+b x))+4 \cos (6 (a+b x))-5) \csc ^3(a+b x) \sec ^4(a+b x)}{560 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]/Sin[2*a + 2*b*x]^(9/2),x]

[Out]

((-5 - 10*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)] + 4*Cos[6*(a + b*x)])*Csc[a + b*x]^3*Sec[a + b*x]^4*Sqrt[Sin[2
*(a + b*x)]])/(560*b)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{\sin \left ( bx+a \right ) \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{-{\frac{9}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)/sin(2*b*x+2*a)^(9/2),x)

[Out]

int(sin(b*x+a)/sin(2*b*x+2*a)^(9/2),x)

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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{9}{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)^(9/2),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)/sin(2*b*x + 2*a)^(9/2), x)

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Fricas [A]  time = 0.525527, size = 300, normalized size = 2.86 \begin{align*} -\frac{\sqrt{2}{\left (128 \, \cos \left (b x + a\right )^{6} - 160 \, \cos \left (b x + a\right )^{4} + 20 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 128 \,{\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \sin \left (b x + a\right )}{560 \,{\left (b \cos \left (b x + a\right )^{6} - b \cos \left (b x + a\right )^{4}\right )} \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)^(9/2),x, algorithm="fricas")

[Out]

-1/560*(sqrt(2)*(128*cos(b*x + a)^6 - 160*cos(b*x + a)^4 + 20*cos(b*x + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x +
a)) + 128*(cos(b*x + a)^6 - cos(b*x + a)^4)*sin(b*x + a))/((b*cos(b*x + a)^6 - b*cos(b*x + a)^4)*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)**(9/2),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)^(9/2),x, algorithm="giac")

[Out]

Timed out