3.266 \(\int \cos ^3(a+b x) \sqrt {\csc (a+b x)} \, dx\)

Optimal. Leaf size=33 \[ \frac {2}{b \sqrt {\csc (a+b x)}}-\frac {2}{5 b \csc ^{\frac {5}{2}}(a+b x)} \]

[Out]

-2/5/b/csc(b*x+a)^(5/2)+2/b/csc(b*x+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2621, 14} \[ \frac {2}{b \sqrt {\csc (a+b x)}}-\frac {2}{5 b \csc ^{\frac {5}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Sqrt[Csc[a + b*x]],x]

[Out]

-2/(5*b*Csc[a + b*x]^(5/2)) + 2/(b*Sqrt[Csc[a + b*x]])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2621

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 \cos ^3(a+b x) \sqrt {\csc (a+b x)} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {-1+x^2}{x^{7/2}} \, dx,x,\csc (a+b x)\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int \left (-\frac {1}{x^{7/2}}+\frac {1}{x^{3/2}}\right ) \, dx,x,\csc (a+b x)\right )}{b}\\ &=-\frac {2}{5 b \csc ^{\frac {5}{2}}(a+b x)}+\frac {2}{b \sqrt {\csc (a+b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 27, normalized size = 0.82 \[ \frac {\cos (2 (a+b x))+9}{5 b \sqrt {\csc (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Sqrt[Csc[a + b*x]],x]

[Out]

(9 + Cos[2*(a + b*x)])/(5*b*Sqrt[Csc[a + b*x]])

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fricas [A]  time = 0.58, size = 23, normalized size = 0.70 \[ \frac {2 \, {\left (\cos \left (b x + a\right )^{2} + 4\right )} \sqrt {\sin \left (b x + a\right )}}{5 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*csc(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

2/5*(cos(b*x + a)^2 + 4)*sqrt(sin(b*x + a))/b

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giac [A]  time = 1.09, size = 24, normalized size = 0.73 \[ -\frac {2 \, {\left (\sin \left (b x + a\right )^{\frac {5}{2}} - 5 \, \sqrt {\sin \left (b x + a\right )}\right )}}{5 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*csc(b*x+a)^(1/2),x, algorithm="giac")

[Out]

-2/5*(sin(b*x + a)^(5/2) - 5*sqrt(sin(b*x + a)))/b

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maple [A]  time = 0.09, size = 26, normalized size = 0.79 \[ \frac {-\frac {2 \left (\sin ^{\frac {5}{2}}\left (b x +a \right )\right )}{5}+2 \left (\sqrt {\sin }\left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3*csc(b*x+a)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.70, size = 25, normalized size = 0.76 \[ \frac {2 \, {\left (\frac {5}{\sin \left (b x + a\right )^{2}} - 1\right )} \sin \left (b x + a\right )^{\frac {5}{2}}}{5 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*csc(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\cos \left (a+b\,x\right )}^3\,\sqrt {\frac {1}{\sin \left (a+b\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(a + b*x)^3*(1/sin(a + b*x))^(1/2), 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)**3*csc(b*x+a)**(1/2),x)

[Out]

Timed out

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