3.184 \(\int \frac{\cos ^3(a+b x)}{\sin ^{\frac{11}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=107 \[ \frac{4 \sin (a+b x)}{45 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{\cos ^3(a+b x)}{9 b \sin ^{\frac{9}{2}}(2 a+2 b x)}-\frac{\cos (a+b x)}{15 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{8 \cos (a+b x)}{45 b \sqrt{\sin (2 a+2 b x)}} \]

[Out]

-Cos[a + b*x]^3/(9*b*Sin[2*a + 2*b*x]^(9/2)) - Cos[a + b*x]/(15*b*Sin[2*a + 2*b*x]^(5/2)) + (4*Sin[a + b*x])/(
45*b*Sin[2*a + 2*b*x]^(3/2)) - (8*Cos[a + b*x])/(45*b*Sqrt[Sin[2*a + 2*b*x]])

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Rubi [A]  time = 0.0904905, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {4295, 4303, 4304, 4291} \[ \frac{4 \sin (a+b x)}{45 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{\cos ^3(a+b x)}{9 b \sin ^{\frac{9}{2}}(2 a+2 b x)}-\frac{\cos (a+b x)}{15 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{8 \cos (a+b x)}{45 b \sqrt{\sin (2 a+2 b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(11/2),x]

[Out]

-Cos[a + b*x]^3/(9*b*Sin[2*a + 2*b*x]^(9/2)) - Cos[a + b*x]/(15*b*Sin[2*a + 2*b*x]^(5/2)) + (4*Sin[a + b*x])/(
45*b*Sin[2*a + 2*b*x]^(3/2)) - (8*Cos[a + b*x])/(45*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4295

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))/(2*b*g*(p + 1)), x] + Dist[(e^2*(m + 2*p + 2))/(4*g^2*(p + 1)), Int[(e*Cos[a
+ b*x])^(m - 2)*(g*Sin[c + d*x])^(p + 2), x], x] /; FreeQ[{a, b, c, d, e, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d
/b, 2] &&  !IntegerQ[p] && GtQ[m, 1] && LtQ[p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && Inte
gersQ[2*m, 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 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{\cos ^3(a+b x)}{\sin ^{\frac{11}{2}}(2 a+2 b x)} \, dx &=-\frac{\cos ^3(a+b x)}{9 b \sin ^{\frac{9}{2}}(2 a+2 b x)}+\frac{1}{3} \int \frac{\cos (a+b x)}{\sin ^{\frac{7}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{\cos ^3(a+b x)}{9 b \sin ^{\frac{9}{2}}(2 a+2 b x)}-\frac{\cos (a+b x)}{15 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{4}{15} \int \frac{\sin (a+b x)}{\sin ^{\frac{5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{\cos ^3(a+b x)}{9 b \sin ^{\frac{9}{2}}(2 a+2 b x)}-\frac{\cos (a+b x)}{15 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{4 \sin (a+b x)}{45 b \sin ^{\frac{3}{2}}(2 a+2 b x)}+\frac{8}{45} \int \frac{\cos (a+b x)}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{\cos ^3(a+b x)}{9 b \sin ^{\frac{9}{2}}(2 a+2 b x)}-\frac{\cos (a+b x)}{15 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{4 \sin (a+b x)}{45 b \sin ^{\frac{3}{2}}(2 a+2 b x)}-\frac{8 \cos (a+b x)}{45 b \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.0935069, size = 62, normalized size = 0.58 \[ -\frac{\sqrt{\sin (2 (a+b x))} \left (5 \csc ^5(a+b x)+17 \csc ^3(a+b x)+113 \csc (a+b x)-15 \tan (a+b x) \sec (a+b x)\right )}{1440 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(11/2),x]

[Out]

-(Sqrt[Sin[2*(a + b*x)]]*(113*Csc[a + b*x] + 17*Csc[a + b*x]^3 + 5*Csc[a + b*x]^5 - 15*Sec[a + b*x]*Tan[a + b*
x]))/(1440*b)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( bx+a \right ) \right ) ^{3} \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{-{\frac{11}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x)

[Out]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(11/2), x)

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Fricas [A]  time = 0.525805, size = 358, normalized size = 3.35 \begin{align*} -\frac{\sqrt{2}{\left (128 \, \cos \left (b x + a\right )^{6} - 288 \, \cos \left (b x + a\right )^{4} + 180 \, \cos \left (b x + a\right )^{2} - 15\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 128 \,{\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{1440 \,{\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x, algorithm="fricas")

[Out]

-1/1440*(sqrt(2)*(128*cos(b*x + a)^6 - 288*cos(b*x + a)^4 + 180*cos(b*x + a)^2 - 15)*sqrt(cos(b*x + a)*sin(b*x
+ a)) + 128*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*sin(b*x + a))/((b*cos(b*x + a)^6 - 2*b*cos(b
*x + a)^4 + 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(cos(b*x+a)**3/sin(2*b*x+2*a)**(11/2),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 )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(11/2), x)