\(\int \frac {x \cos (a+b x)}{\csc ^{\frac {3}{2}}(a+b x)} \, dx\) [358]

Optimal result

Integrand size = 18, antiderivative size = 85 \[ \int \frac {x \cos (a+b x)}{\csc ^{\frac {3}{2}}(a+b x)} \, dx=\frac {2 x}{5 b \csc ^{\frac {5}{2}}(a+b x)}+\frac {4 \cos (a+b x)}{25 b^2 \csc ^{\frac {3}{2}}(a+b x)}-\frac {12 \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{25 b^2} \]

[Out]

2/5*x/b/csc(b*x+a)^(5/2)+4/25*cos(b*x+a)/b^2/csc(b*x+a)^(3/2)+12/25*(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)*EllipticE(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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4298, 3854, 3856, 2719} \[ \int \frac {x \cos (a+b x)}{\csc ^{\frac {3}{2}}(a+b x)} \, dx=\frac {4 \cos (a+b x)}{25 b^2 \csc ^{\frac {3}{2}}(a+b x)}-\frac {12 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{25 b^2}+\frac {2 x}{5 b \csc ^{\frac {5}{2}}(a+b x)} \]

[In]

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

[Out]

(2*x)/(5*b*Csc[a + b*x]^(5/2)) + (4*Cos[a + b*x])/(25*b^2*Csc[a + b*x]^(3/2)) - (12*Sqrt[Csc[a + b*x]]*Ellipti
cE[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/(25*b^2)

Rule 2719

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

Rule 3854

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 3856

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 4298

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*} \text {integral}& = \frac {2 x}{5 b \csc ^{\frac {5}{2}}(a+b x)}-\frac {2 \int \frac {1}{\csc ^{\frac {5}{2}}(a+b x)} \, dx}{5 b} \\ & = \frac {2 x}{5 b \csc ^{\frac {5}{2}}(a+b x)}+\frac {4 \cos (a+b x)}{25 b^2 \csc ^{\frac {3}{2}}(a+b x)}-\frac {6 \int \frac {1}{\sqrt {\csc (a+b x)}} \, dx}{25 b} \\ & = \frac {2 x}{5 b \csc ^{\frac {5}{2}}(a+b x)}+\frac {4 \cos (a+b x)}{25 b^2 \csc ^{\frac {3}{2}}(a+b x)}-\frac {\left (6 \sqrt {\csc (a+b x)} \sqrt {\sin (a+b x)}\right ) \int \sqrt {\sin (a+b x)} \, dx}{25 b} \\ & = \frac {2 x}{5 b \csc ^{\frac {5}{2}}(a+b x)}+\frac {4 \cos (a+b x)}{25 b^2 \csc ^{\frac {3}{2}}(a+b x)}-\frac {12 \sqrt {\csc (a+b x)} E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {\sin (a+b x)}}{25 b^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.34 \[ \int \frac {x \cos (a+b x)}{\csc ^{\frac {3}{2}}(a+b x)} \, dx=\frac {\left (-10+4 \cos (a+b x)+2 \cos (2 (a+b x))+4 \sqrt {2} \sqrt {\frac {1}{1+\cos (a+b x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right )+10 b x \sin (a+b x)+5 b x \sin (2 (a+b x))\right ) \tan \left (\frac {1}{2} (a+b x)\right )}{25 b^2 \sqrt {\csc (a+b x)}} \]

[In]

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

[Out]

((-10 + 4*Cos[a + b*x] + 2*Cos[2*(a + b*x)] + 4*Sqrt[2]*Sqrt[(1 + Cos[a + b*x])^(-1)]*Hypergeometric2F1[1/2, 3
/4, 7/4, -Tan[(a + b*x)/2]^2] + 10*b*x*Sin[a + b*x] + 5*b*x*Sin[2*(a + b*x)])*Tan[(a + b*x)/2])/(25*b^2*Sqrt[C
sc[a + b*x]])

Maple [F]

\[\int \frac {x \cos \left (x b +a \right )}{\csc \left (x b +a \right )^{\frac {3}{2}}}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {x \cos (a+b x)}{\csc ^{\frac {3}{2}}(a+b x)} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {x \cos (a+b x)}{\csc ^{\frac {3}{2}}(a+b x)} \, dx=\int \frac {x \cos {\left (a + b x \right )}}{\csc ^{\frac {3}{2}}{\left (a + b x \right )}}\, dx \]

[In]

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

[Out]

Integral(x*cos(a + b*x)/csc(a + b*x)**(3/2), x)

Maxima [F]

\[ \int \frac {x \cos (a+b x)}{\csc ^{\frac {3}{2}}(a+b x)} \, dx=\int { \frac {x \cos \left (b x + a\right )}{\csc \left (b x + a\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x \cos (a+b x)}{\csc ^{\frac {3}{2}}(a+b x)} \, dx=\int { \frac {x \cos \left (b x + a\right )}{\csc \left (b x + a\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x \cos (a+b x)}{\csc ^{\frac {3}{2}}(a+b x)} \, dx=\int \frac {x\,\cos \left (a+b\,x\right )}{{\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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