\(\int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx\) [345]

Optimal result

Integrand size = 18, antiderivative size = 65 \[ \int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx=-\frac {12 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right )}{25 b^2}+\frac {4 \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x)}{25 b^2}+\frac {2 x \sin ^{\frac {5}{2}}(a+b x)}{5 b} \]

[Out]

12/25*(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+4/25*cos(b*x+a)*sin(b*x+a)^(3/2)/b^2+2/5*x*sin(b*x+a)^(5/2)/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3524, 2715, 2719} \[ \int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx=-\frac {12 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{25 b^2}+\frac {4 \sin ^{\frac {3}{2}}(a+b x) \cos (a+b x)}{25 b^2}+\frac {2 x \sin ^{\frac {5}{2}}(a+b x)}{5 b} \]

[In]

Int[x*Cos[a + b*x]*Sin[a + b*x]^(3/2),x]

[Out]

(-12*EllipticE[(a - Pi/2 + b*x)/2, 2])/(25*b^2) + (4*Cos[a + b*x]*Sin[a + b*x]^(3/2))/(25*b^2) + (2*x*Sin[a +
b*x]^(5/2))/(5*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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 {5}{2}}(a+b x)}{5 b}-\frac {2 \int \sin ^{\frac {5}{2}}(a+b x) \, dx}{5 b} \\ & = \frac {4 \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x)}{25 b^2}+\frac {2 x \sin ^{\frac {5}{2}}(a+b x)}{5 b}-\frac {6 \int \sqrt {\sin (a+b x)} \, dx}{25 b} \\ & = -\frac {12 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right )}{25 b^2}+\frac {4 \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x)}{25 b^2}+\frac {2 x \sin ^{\frac {5}{2}}(a+b x)}{5 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 2.16 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.66 \[ \int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx=\frac {\sqrt {\sin (a+b x)} \left (5 b x-5 b x \cos (2 (a+b x))+2 \sin (2 (a+b x))-12 \tan \left (\frac {1}{2} (a+b x)\right )+4 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\tan ^2\left (\frac {1}{2} (a+b x)\right )\right ) \sqrt {\sec ^2\left (\frac {1}{2} (a+b x)\right )} \tan \left (\frac {1}{2} (a+b x)\right )\right )}{25 b^2} \]

[In]

Integrate[x*Cos[a + b*x]*Sin[a + b*x]^(3/2),x]

[Out]

(Sqrt[Sin[a + b*x]]*(5*b*x - 5*b*x*Cos[2*(a + b*x)] + 2*Sin[2*(a + b*x)] - 12*Tan[(a + b*x)/2] + 4*Hypergeomet
ric2F1[1/2, 3/4, 7/4, -Tan[(a + b*x)/2]^2]*Sqrt[Sec[(a + b*x)/2]^2]*Tan[(a + b*x)/2]))/(25*b^2)

Maple [F]

\[\int x \cos \left (x b +a \right ) \sin \left (x b +a \right )^{\frac {3}{2}}d x\]

[In]

int(x*cos(b*x+a)*sin(b*x+a)^(3/2),x)

[Out]

int(x*cos(b*x+a)*sin(b*x+a)^(3/2),x)

Fricas [F(-2)]

Exception generated. \[ \int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x*cos(b*x+a)*sin(b*x+a)^(3/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx=\int x \sin ^{\frac {3}{2}}{\left (a + b x \right )} \cos {\left (a + b x \right )}\, dx \]

[In]

integrate(x*cos(b*x+a)*sin(b*x+a)**(3/2),x)

[Out]

Integral(x*sin(a + b*x)**(3/2)*cos(a + b*x), x)

Maxima [F]

\[ \int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx=\int { x \cos \left (b x + a\right ) \sin \left (b x + a\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(x*cos(b*x+a)*sin(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(x*cos(b*x + a)*sin(b*x + a)^(3/2), x)

Giac [F]

\[ \int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx=\int { x \cos \left (b x + a\right ) \sin \left (b x + a\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate(x*cos(b*x+a)*sin(b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate(x*cos(b*x + a)*sin(b*x + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int x \cos (a+b x) \sin ^{\frac {3}{2}}(a+b x) \, dx=\int x\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^{3/2} \,d x \]

[In]

int(x*cos(a + b*x)*sin(a + b*x)^(3/2),x)

[Out]

int(x*cos(a + b*x)*sin(a + b*x)^(3/2), x)