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

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

[Out]

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

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Rubi [A]  time = 0.0799988, 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 = {4303, 4304, 4292} \[ \frac{6 \sin (a+b x)}{35 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{16 \sin (a+b x)}{35 b \sqrt{\sin (2 a+2 b x)}}-\frac{8 \cos (a+b x)}{35 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{\cos (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 4292

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[((e*Sin[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] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps

\begin{align*} \int \frac{\cos (a+b x)}{\sin ^{\frac{9}{2}}(2 a+2 b x)} \, dx &=-\frac{\cos (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}+\frac{6}{7} \int \frac{\sin (a+b x)}{\sin ^{\frac{7}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{\cos (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}+\frac{6 \sin (a+b x)}{35 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{24}{35} \int \frac{\cos (a+b x)}{\sin ^{\frac{5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{\cos (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}+\frac{6 \sin (a+b x)}{35 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{8 \cos (a+b x)}{35 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{16}{35} \int \frac{\sin (a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{\cos (a+b x)}{7 b \sin ^{\frac{7}{2}}(2 a+2 b x)}+\frac{6 \sin (a+b x)}{35 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{8 \cos (a+b x)}{35 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{16 \sin (a+b x)}{35 b \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.14338, size = 67, normalized size = 0.64 \[ \frac{\sqrt{\sin (2 (a+b x))} (-10 \cos (2 (a+b x))-4 \cos (4 (a+b x))+4 \cos (6 (a+b x))+5) \csc ^4(a+b x) \sec ^3(a+b x)}{560 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[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]^4*Sec[a + b*x]^3*Sqrt[Sin[2*
(a + b*x)]])/(560*b)

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Maple [C]  time = 183.381, size = 222, normalized size = 2.1 \begin{align*}{\frac{1}{2688\,b}\sqrt{-{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) ^{-1}}} \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) \left ( 3\, \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{8}+40\,\sqrt{\tan \left ( 1/2\,bx+a/2 \right ) +1}\sqrt{-2\,\tan \left ( 1/2\,bx+a/2 \right ) +2}\sqrt{-\tan \left ( 1/2\,bx+a/2 \right ) }{\it EllipticF} \left ( \sqrt{\tan \left ( 1/2\,bx+a/2 \right ) +1},1/2\,\sqrt{2} \right ) \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{3}-26\, \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{6}+26\, \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{2}-3 \right ) \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) }}}{\frac{1}{\sqrt{ \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{3}-\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2688/b*(-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)^2-1)/tan(1/2*b*x+1/2*a)^3*(3
*tan(1/2*b*x+1/2*a)^8+40*(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)^3-26*tan(1/2*b*x+1/2*a)^6+26*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)^3-tan(1/2*b*x+1/2*a)
)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \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(cos(b*x+a)/sin(2*b*x+2*a)^(9/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.534884, size = 320, normalized size = 3.05 \begin{align*} \frac{128 \, \cos \left (b x + a\right )^{7} - 256 \, \cos \left (b x + a\right )^{5} + 128 \, \cos \left (b x + a\right )^{3} + \sqrt{2}{\left (128 \, \cos \left (b x + a\right )^{6} - 224 \, \cos \left (b x + a\right )^{4} + 84 \, \cos \left (b x + a\right )^{2} + 7\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{560 \,{\left (b \cos \left (b x + a\right )^{7} - 2 \, b \cos \left (b x + a\right )^{5} + b \cos \left (b x + a\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/sin(2*b*x+2*a)^(9/2),x, algorithm="fricas")

[Out]

1/560*(128*cos(b*x + a)^7 - 256*cos(b*x + a)^5 + 128*cos(b*x + a)^3 + sqrt(2)*(128*cos(b*x + a)^6 - 224*cos(b*
x + a)^4 + 84*cos(b*x + a)^2 + 7)*sqrt(cos(b*x + a)*sin(b*x + a)))/(b*cos(b*x + a)^7 - 2*b*cos(b*x + a)^5 + b*
cos(b*x + a)^3)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \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(cos(b*x+a)/sin(2*b*x+2*a)^(9/2),x, algorithm="giac")

[Out]

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