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

Optimal. Leaf size=24 \[ -\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {4291} \[ -\frac {\cos (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.96 \[ -\frac {\cos (a+b x)}{b \sqrt {\sin (2 (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 39, normalized size = 1.62 \[ -\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + \sin \left (b x + a\right )}{2 \, b \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 10.74, size = 55902198, normalized size = 2329258.25 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.19, size = 24, normalized size = 1.00 \[ -\frac {\sqrt {\sin \left (2\,a+2\,b\,x\right )}}{2\,b\,\sin \left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**(3/2),x)

[Out]

Timed out

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