3.181 \(\int \frac {\cos ^3(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=28 \[ -\frac {\cos ^3(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[Out]

-1/3*cos(b*x+a)^3/b/sin(2*b*x+2*a)^(3/2)

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Rubi [A]  time = 0.03, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {4291} \[ -\frac {\cos ^3(a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 27, normalized size = 0.96 \[ -\frac {\sin ^{\frac {3}{2}}(2 (a+b x)) \csc ^3(a+b x)}{24 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

-1/24*(Csc[a + b*x]^3*Sin[2*(a + b*x)]^(3/2))/b

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fricas [B]  time = 0.64, size = 53, normalized size = 1.89 \[ \frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \cos \left (b x + a\right ) + \cos \left (b x + a\right )^{2} - 1}{12 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(5/2), x)

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maple [C]  time = 167.45, size = 192, normalized size = 6.86 \[ \frac {\sqrt {-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1}}\, \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \left (4 \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+2}\, \sqrt {-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{24 \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}\, \sqrt {\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x)

[Out]

1/24*(-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)*(4*(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)+tan(1/2*b*x+1/2*a)^4-1)/(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)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(5/2), x)

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mupad [B]  time = 1.17, size = 94, normalized size = 3.36 \[ \frac {\sqrt {\sin \left (2\,a+2\,b\,x\right )}\,\left (\frac {2\,{\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2}{3}-{\sin \left (\frac {3\,a}{2}+\frac {3\,b\,x}{2}\right )}^2+\frac {{\sin \left (\frac {5\,a}{2}+\frac {5\,b\,x}{2}\right )}^2}{3}\right )}{b\,\left (30\,{\sin \left (a+b\,x\right )}^2-12\,{\sin \left (2\,a+2\,b\,x\right )}^2+2\,{\sin \left (3\,a+3\,b\,x\right )}^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3/sin(2*a + 2*b*x)^(5/2),x)

[Out]

(sin(2*a + 2*b*x)^(1/2)*((2*sin(a/2 + (b*x)/2)^2)/3 - sin((3*a)/2 + (3*b*x)/2)^2 + sin((5*a)/2 + (5*b*x)/2)^2/
3))/(b*(2*sin(3*a + 3*b*x)^2 - 12*sin(2*a + 2*b*x)^2 + 30*sin(a + b*x)^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3/sin(2*b*x+2*a)**(5/2),x)

[Out]

Timed out

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