\(\int x \cos ^{\frac {5}{2}}(a+b x) \sin (a+b x) \, dx\) [328]

Optimal result

Integrand size = 18, antiderivative size = 83 \[ \int x \cos ^{\frac {5}{2}}(a+b x) \sin (a+b x) \, dx=-\frac {2 x \cos ^{\frac {7}{2}}(a+b x)}{7 b}+\frac {20 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{147 b^2}+\frac {20 \sqrt {\cos (a+b x)} \sin (a+b x)}{147 b^2}+\frac {4 \cos ^{\frac {5}{2}}(a+b x) \sin (a+b x)}{49 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, 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 = {3525, 2715, 2720} \[ \int x \cos ^{\frac {5}{2}}(a+b x) \sin (a+b x) \, dx=\frac {20 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{147 b^2}+\frac {4 \sin (a+b x) \cos ^{\frac {5}{2}}(a+b x)}{49 b^2}+\frac {20 \sin (a+b x) \sqrt {\cos (a+b x)}}{147 b^2}-\frac {2 x \cos ^{\frac {7}{2}}(a+b x)}{7 b} \]

[In]

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

[Out]

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

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 3525

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Simp[(-x^(m - n
+ 1))*(Cos[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] + Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Cos[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 \cos ^{\frac {7}{2}}(a+b x)}{7 b}+\frac {2 \int \cos ^{\frac {7}{2}}(a+b x) \, dx}{7 b} \\ & = -\frac {2 x \cos ^{\frac {7}{2}}(a+b x)}{7 b}+\frac {4 \cos ^{\frac {5}{2}}(a+b x) \sin (a+b x)}{49 b^2}+\frac {10 \int \cos ^{\frac {3}{2}}(a+b x) \, dx}{49 b} \\ & = -\frac {2 x \cos ^{\frac {7}{2}}(a+b x)}{7 b}+\frac {20 \sqrt {\cos (a+b x)} \sin (a+b x)}{147 b^2}+\frac {4 \cos ^{\frac {5}{2}}(a+b x) \sin (a+b x)}{49 b^2}+\frac {10 \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx}{147 b} \\ & = -\frac {2 x \cos ^{\frac {7}{2}}(a+b x)}{7 b}+\frac {20 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )}{147 b^2}+\frac {20 \sqrt {\cos (a+b x)} \sin (a+b x)}{147 b^2}+\frac {4 \cos ^{\frac {5}{2}}(a+b x) \sin (a+b x)}{49 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int x \cos ^{\frac {5}{2}}(a+b x) \sin (a+b x) \, dx=\frac {40 \operatorname {EllipticF}\left (\frac {1}{2} (a+b x),2\right )+\sqrt {\cos (a+b x)} (-63 b x \cos (a+b x)-21 b x \cos (3 (a+b x))+46 \sin (a+b x)+6 \sin (3 (a+b x)))}{294 b^2} \]

[In]

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

[Out]

(40*EllipticF[(a + b*x)/2, 2] + Sqrt[Cos[a + b*x]]*(-63*b*x*Cos[a + b*x] - 21*b*x*Cos[3*(a + b*x)] + 46*Sin[a
+ b*x] + 6*Sin[3*(a + b*x)]))/(294*b^2)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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