3.357 \(\int \frac {x \cos (a+b x)}{\sqrt {\csc (a+b x)}} \, dx\)

Optimal. Leaf size=85 \[ \frac {4 \cos (a+b x)}{9 b^2 \sqrt {\csc (a+b x)}}-\frac {4 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{9 b^2}+\frac {2 x}{3 b \csc ^{\frac {3}{2}}(a+b x)} \]

[Out]

2/3*x/b/csc(b*x+a)^(3/2)+4/9*cos(b*x+a)/b^2/csc(b*x+a)^(1/2)+4/9*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a
+1/4*Pi+1/2*b*x)*EllipticF(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*csc(b*x+a)^(1/2)*sin(b*x+a)^(1/2)/b^2

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Rubi [A]  time = 0.04, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4213, 3769, 3771, 2641} \[ \frac {4 \cos (a+b x)}{9 b^2 \sqrt {\csc (a+b x)}}-\frac {4 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{9 b^2}+\frac {2 x}{3 b \csc ^{\frac {3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x)/(3*b*Csc[a + b*x]^(3/2)) + (4*Cos[a + b*x])/(9*b^2*Sqrt[Csc[a + b*x]]) - (4*Sqrt[Csc[a + b*x]]*EllipticF
[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/(9*b^2)

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 4213

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*Csc[(a_.) + (b_.)*(x_)^(n_.)]^(p_)*(x_)^(m_.), x_Symbol] :> -Simp[(x^(m - n
+ 1)*Csc[a + b*x^n]^(p - 1))/(b*n*(p - 1)), x] + Dist[(m - n + 1)/(b*n*(p - 1)), Int[x^(m - n)*Csc[a + b*x^n]^
(p - 1), x], x] /; FreeQ[{a, b, p}, x] && IntegerQ[n] && GeQ[m - n, 0] && NeQ[p, 1]

Rubi steps

\begin {align*} \int \frac {x \cos (a+b x)}{\sqrt {\csc (a+b x)}} \, dx &=\frac {2 x}{3 b \csc ^{\frac {3}{2}}(a+b x)}-\frac {2 \int \frac {1}{\csc ^{\frac {3}{2}}(a+b x)} \, dx}{3 b}\\ &=\frac {2 x}{3 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {4 \cos (a+b x)}{9 b^2 \sqrt {\csc (a+b x)}}-\frac {2 \int \sqrt {\csc (a+b x)} \, dx}{9 b}\\ &=\frac {2 x}{3 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {4 \cos (a+b x)}{9 b^2 \sqrt {\csc (a+b x)}}-\frac {\left (2 \sqrt {\csc (a+b x)} \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx}{9 b}\\ &=\frac {2 x}{3 b \csc ^{\frac {3}{2}}(a+b x)}+\frac {4 \cos (a+b x)}{9 b^2 \sqrt {\csc (a+b x)}}-\frac {4 \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{9 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 65, normalized size = 0.76 \[ \frac {2 \sqrt {\csc (a+b x)} \left (3 b x \sin ^2(a+b x)+\sin (2 (a+b x))+2 \sqrt {\sin (a+b x)} F\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )\right )}{9 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[Csc[a + b*x]]*(2*EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[Sin[a + b*x]] + 3*b*x*Sin[a + b*x]^2 + Sin[2
*(a + b*x)]))/(9*b^2)

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {x \cos \left (b x +a \right )}{\sqrt {\csc \left (b x +a \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\cos \left (a+b\,x\right )}{\sqrt {\frac {1}{\sin \left (a+b\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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