3.349 \(\int \frac {x \cos (a+b x)}{\sin ^{\frac {5}{2}}(a+b x)} \, dx\)

Optimal. Leaf size=65 \[ -\frac {4 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{3 b^2}-\frac {4 \cos (a+b x)}{3 b^2 \sqrt {\sin (a+b x)}}-\frac {2 x}{3 b \sin ^{\frac {3}{2}}(a+b x)} \]

[Out]

4/3*(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))
/b^2-2/3*x/b/sin(b*x+a)^(3/2)-4/3*cos(b*x+a)/b^2/sin(b*x+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3443, 2636, 2639} \[ -\frac {4 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{3 b^2}-\frac {4 \cos (a+b x)}{3 b^2 \sqrt {\sin (a+b x)}}-\frac {2 x}{3 b \sin ^{\frac {3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(x*Cos[a + b*x])/Sin[a + b*x]^(5/2),x]

[Out]

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

Rule 2636

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

Rule 2639

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

Rule 3443

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

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 56, normalized size = 0.86 \[ -\frac {2 \left (\sin (2 (a+b x))-2 \sin ^{\frac {3}{2}}(a+b x) E\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )+b x\right )}{3 b^2 \sin ^{\frac {3}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Cos[a + b*x])/Sin[a + b*x]^(5/2),x]

[Out]

(-2*(b*x - 2*EllipticE[(-2*a + Pi - 2*b*x)/4, 2]*Sin[a + b*x]^(3/2) + Sin[2*(a + b*x)]))/(3*b^2*Sin[a + b*x]^(
3/2))

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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)/sin(b*x+a)^(5/2),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 \[ \int \frac {x \cos \left (b x + a\right )}{\sin \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(b*x+a)/sin(b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(x*cos(b*x + a)/sin(b*x + a)^(5/2), x)

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {x \cos \left (b x +a \right )}{\sin \left (b x +a \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(b*x+a)/sin(b*x+a)^(5/2),x)

[Out]

int(x*cos(b*x+a)/sin(b*x+a)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \cos \left (b x + a\right )}{\sin \left (b x + a\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(b*x+a)/sin(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(x*cos(b*x + a)/sin(b*x + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x\,\cos \left (a+b\,x\right )}{{\sin \left (a+b\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*cos(a + b*x))/sin(a + b*x)^(5/2),x)

[Out]

int((x*cos(a + b*x))/sin(a + b*x)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \cos {\left (a + b x \right )}}{\sin ^{\frac {5}{2}}{\left (a + b x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(b*x+a)/sin(b*x+a)**(5/2),x)

[Out]

Integral(x*cos(a + b*x)/sin(a + b*x)**(5/2), x)

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