3.3.62 \(\int \frac {\cos (a+b x)}{\sqrt {\csc (a+b x)}} \, dx\) [262]

Optimal. Leaf size=17 \[ \frac {2}{3 b \csc ^{\frac {3}{2}}(a+b x)} \]

[Out]

2/3/b/csc(b*x+a)^(3/2)

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Rubi [A]
time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2701, 30} \begin {gather*} \frac {2}{3 b \csc ^{\frac {3}{2}}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]/Sqrt[Csc[a + b*x]],x]

[Out]

2/(3*b*Csc[a + b*x]^(3/2))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2701

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int \frac {\cos (a+b x)}{\sqrt {\csc (a+b x)}} \, dx &=-\frac {\text {Subst}\left (\int \frac {1}{x^{5/2}} \, dx,x,\csc (a+b x)\right )}{b}\\ &=\frac {2}{3 b \csc ^{\frac {3}{2}}(a+b x)}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 17, normalized size = 1.00 \begin {gather*} \frac {2}{3 b \csc ^{\frac {3}{2}}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]/Sqrt[Csc[a + b*x]],x]

[Out]

2/(3*b*Csc[a + b*x]^(3/2))

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Maple [A]
time = 0.04, size = 14, normalized size = 0.82

method	result	size
derivativedivides	\(\frac {2}{3 b \csc \left (b x +a \right )^{\frac {3}{2}}}\)	\(14\)
default	\(\frac {2}{3 b \csc \left (b x +a \right )^{\frac {3}{2}}}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)/csc(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/3/b/csc(b*x+a)^(3/2)

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Maxima [A]
time = 0.28, size = 13, normalized size = 0.76 \begin {gather*} \frac {2 \, \sin \left (b x + a\right )^{\frac {3}{2}}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/csc(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

2/3*sin(b*x + a)^(3/2)/b

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Fricas [A]
time = 0.36, size = 23, normalized size = 1.35 \begin {gather*} -\frac {2 \, {\left (\cos \left (b x + a\right )^{2} - 1\right )}}{3 \, b \sqrt {\sin \left (b x + a\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/csc(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos {\left (a + b x \right )}}{\sqrt {\csc {\left (a + b x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/csc(b*x+a)**(1/2),x)

[Out]

Integral(cos(a + b*x)/sqrt(csc(a + b*x)), x)

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Giac [A]
time = 0.47, size = 13, normalized size = 0.76 \begin {gather*} \frac {2 \, \sin \left (b x + a\right )^{\frac {3}{2}}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/csc(b*x+a)^(1/2),x, algorithm="giac")

[Out]

2/3*sin(b*x + a)^(3/2)/b

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Mupad [B]
time = 0.21, size = 15, normalized size = 0.88 \begin {gather*} \frac {2}{3\,b\,{\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(3*b*(1/sin(a + b*x))^(3/2))

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