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

Optimal. Leaf size=88 \[ -\frac {20 F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{147 b^2}+\frac {4 \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x)}{49 b^2}+\frac {20 \sqrt {\sin (a+b x)} \cos (a+b x)}{147 b^2}+\frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b} \]

[Out]

20/147*(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))/b^2+4/49*cos(b*x+a)*sin(b*x+a)^(5/2)/b^2+2/7*x*sin(b*x+a)^(7/2)/b+20/147*cos(b*x+a)*sin(b*x+a)^(1/2)/b^2

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3443, 2635, 2641} \[ -\frac {20 F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{147 b^2}+\frac {4 \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x)}{49 b^2}+\frac {20 \sqrt {\sin (a+b x)} \cos (a+b x)}{147 b^2}+\frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-20*EllipticF[(a - Pi/2 + b*x)/2, 2])/(147*b^2) + (20*Cos[a + b*x]*Sqrt[Sin[a + b*x]])/(147*b^2) + (4*Cos[a +
b*x]*Sin[a + b*x]^(5/2))/(49*b^2) + (2*x*Sin[a + b*x]^(7/2))/(7*b)

Rule 2635

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

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 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 x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx &=\frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {2 \int \sin ^{\frac {7}{2}}(a+b x) \, dx}{7 b}\\ &=\frac {4 \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x)}{49 b^2}+\frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {10 \int \sin ^{\frac {3}{2}}(a+b x) \, dx}{49 b}\\ &=\frac {20 \cos (a+b x) \sqrt {\sin (a+b x)}}{147 b^2}+\frac {4 \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x)}{49 b^2}+\frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {10 \int \frac {1}{\sqrt {\sin (a+b x)}} \, dx}{147 b}\\ &=-\frac {20 F\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right )}{147 b^2}+\frac {20 \cos (a+b x) \sqrt {\sin (a+b x)}}{147 b^2}+\frac {4 \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x)}{49 b^2}+\frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}\\ \end {align*}

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Mathematica [A]  time = 0.54, size = 67, normalized size = 0.76 \[ \frac {40 F\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )+\sqrt {\sin (a+b x)} \left (84 b x \sin ^3(a+b x)+46 \cos (a+b x)-6 \cos (3 (a+b x))\right )}{294 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(40*EllipticF[(-2*a + Pi - 2*b*x)/4, 2] + Sqrt[Sin[a + b*x]]*(46*Cos[a + b*x] - 6*Cos[3*(a + b*x)] + 84*b*x*Si
n[a + b*x]^3))/(294*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)*sin(b*x+a)^(5/2),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 \[ \int x \cos \left (b x + a\right ) \sin \left (b x + a\right )^{\frac {5}{2}}\,{d x} \]

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 = 0.06, size = 0, normalized size = 0.00 \[ \int x \cos \left (b x +a \right ) \left (\sin ^{\frac {5}{2}}\left (b x +a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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 = -0.01 \[ \int x\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^{5/2} \,d x \]

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

[Out]

Timed out

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