\(\int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx\) [344]

Optimal result

Integrand size = 18, antiderivative size = 88 \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=-\frac {20 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right )}{147 b^2}+\frac {20 \cos (a+b x) \sqrt {\sin (a+b x)}}{147 b^2}+\frac {4 \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x)}{49 b^2}+\frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b} \]

[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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3524, 2715, 2720} \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=-\frac {20 \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{147 b^2}+\frac {4 \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x)}{49 b^2}+\frac {20 \sqrt {\sin (a+b x)} \cos (a+b x)}{147 b^2}+\frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b} \]

[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 2715

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] && Integ
erQ[2*n]

Rule 2720

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

Rule 3524

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*} \text {integral}& = \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 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\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*}

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\frac {40 \operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right )+\sqrt {\sin (a+b x)} \left (46 \cos (a+b x)-6 \cos (3 (a+b x))+84 b x \sin ^3(a+b x)\right )}{294 b^2} \]

[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)

Maple [F]

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

[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)

Fricas [F(-2)]

Exception generated. \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\text {Exception raised: TypeError} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\int x\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^{5/2} \,d x \]

[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)