3.353 \(\int x \cos (a+b x) \csc ^{\frac {7}{2}}(a+b x) \, dx\)

Optimal. Leaf size=85 \[ -\frac {4 \cos (a+b x) \csc ^{\frac {3}{2}}(a+b x)}{15 b^2}+\frac {4 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{15 b^2}-\frac {2 x \csc ^{\frac {5}{2}}(a+b x)}{5 b} \]

[Out]

-4/15*cos(b*x+a)*csc(b*x+a)^(3/2)/b^2-2/5*x*csc(b*x+a)^(5/2)/b-4/15*(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, 3768, 3771, 2641} \[ -\frac {4 \cos (a+b x) \csc ^{\frac {3}{2}}(a+b x)}{15 b^2}+\frac {4 \sqrt {\sin (a+b x)} \sqrt {\csc (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{15 b^2}-\frac {2 x \csc ^{\frac {5}{2}}(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Cos[a + b*x]*Csc[a + b*x]^(7/2),x]

[Out]

(-4*Cos[a + b*x]*Csc[a + b*x]^(3/2))/(15*b^2) - (2*x*Csc[a + b*x]^(5/2))/(5*b) + (4*Sqrt[Csc[a + b*x]]*Ellipti
cF[(a - Pi/2 + b*x)/2, 2]*Sqrt[Sin[a + b*x]])/(15*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[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 x \cos (a+b x) \csc ^{\frac {7}{2}}(a+b x) \, dx &=-\frac {2 x \csc ^{\frac {5}{2}}(a+b x)}{5 b}+\frac {2 \int \csc ^{\frac {5}{2}}(a+b x) \, dx}{5 b}\\ &=-\frac {4 \cos (a+b x) \csc ^{\frac {3}{2}}(a+b x)}{15 b^2}-\frac {2 x \csc ^{\frac {5}{2}}(a+b x)}{5 b}+\frac {2 \int \sqrt {\csc (a+b x)} \, dx}{15 b}\\ &=-\frac {4 \cos (a+b x) \csc ^{\frac {3}{2}}(a+b x)}{15 b^2}-\frac {2 x \csc ^{\frac {5}{2}}(a+b x)}{5 b}+\frac {\left (2 \sqrt {\csc (a+b x)} \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx}{15 b}\\ &=-\frac {4 \cos (a+b x) \csc ^{\frac {3}{2}}(a+b x)}{15 b^2}-\frac {2 x \csc ^{\frac {5}{2}}(a+b x)}{5 b}+\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)}}{15 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 65, normalized size = 0.76 \[ -\frac {2 \sqrt {\csc (a+b x)} \left (2 \cot (a+b x)+3 b x \csc ^2(a+b x)+2 \sqrt {\sin (a+b x)} F\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )\right )}{15 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x]*Csc[a + b*x]^(7/2),x]

[Out]

(-2*Sqrt[Csc[a + b*x]]*(2*Cot[a + b*x] + 3*b*x*Csc[a + b*x]^2 + 2*EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[Sin
[a + b*x]]))/(15*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)^(7/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cos \left (b x + a\right ) \csc \left (b x + a\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(b*x+a)*csc(b*x+a)^(7/2),x, algorithm="giac")

[Out]

integrate(x*cos(b*x + a)*csc(b*x + a)^(7/2), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int x \cos \left (b x +a \right ) \left (\csc ^{\frac {7}{2}}\left (b x +a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(b*x+a)*csc(b*x+a)^(7/2),x)

[Out]

int(x*cos(b*x+a)*csc(b*x+a)^(7/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cos \left (b x + a\right ) \csc \left (b x + a\right )^{\frac {7}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(b*x+a)*csc(b*x+a)^(7/2),x, algorithm="maxima")

[Out]

integrate(x*cos(b*x + a)*csc(b*x + a)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\cos \left (a+b\,x\right )\,{\left (\frac {1}{\sin \left (a+b\,x\right )}\right )}^{7/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(a + b*x)*(1/sin(a + b*x))^(7/2),x)

[Out]

int(x*cos(a + b*x)*(1/sin(a + b*x))^(7/2), 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(x*cos(b*x+a)*csc(b*x+a)**(7/2),x)

[Out]

Timed out

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