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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4303, 4292} \[ \frac {2 \sin (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}-\frac {\cos (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 43, normalized size = 0.81 \[ \frac {\sqrt {\sin (2 (a+b x))} \left (\frac {1}{4} \sec (a+b x)-\frac {1}{12} \cot (a+b x) \csc (a+b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/12*(Cot[a + b*x]*Csc[a + b*x]) + Sec[a + b*x]/4)*Sqrt[Sin[2*(a + b*x)]])/b

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fricas [A]  time = 0.59, size = 74, normalized size = 1.40 \[ \frac {4 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 4 \, \cos \left (b x + a\right )}{12 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*cos(b*x + a)^3 + sqrt(2)*(4*cos(b*x + a)^2 - 3)*sqrt(cos(b*x + a)*sin(b*x + a)) - 4*cos(b*x + a))/(b*c
os(b*x + a)^3 - b*cos(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 33.70, size = 194, normalized size = 3.66 \[ -\frac {\sqrt {-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1}}\, \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \left (2 \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+2}\, \sqrt {-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {b x}{2}+\frac {a}{2}\right )-\left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+1\right )}{24 b \tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}\, \sqrt {\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/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)*(2*(t
an(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.17, size = 104, normalized size = 1.96 \[ -\frac {2\,\sqrt {\sin \left (2\,a+2\,b\,x\right )}\,\left (3\,\cos \left (a+b\,x\right )-6\,\cos \left (3\,a+3\,b\,x\right )+4\,\cos \left (5\,a+5\,b\,x\right )-\cos \left (7\,a+7\,b\,x\right )\right )}{3\,b\,\left (4\,\cos \left (2\,a+2\,b\,x\right )+4\,\cos \left (4\,a+4\,b\,x\right )-4\,\cos \left (6\,a+6\,b\,x\right )+\cos \left (8\,a+8\,b\,x\right )-5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*sin(2*a + 2*b*x)^(1/2)*(3*cos(a + b*x) - 6*cos(3*a + 3*b*x) + 4*cos(5*a + 5*b*x) - cos(7*a + 7*b*x)))/(3*b
*(4*cos(2*a + 2*b*x) + 4*cos(4*a + 4*b*x) - 4*cos(6*a + 6*b*x) + cos(8*a + 8*b*x) - 5))

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

[Out]

Timed out

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