Integrand size = 18, antiderivative size = 88 \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=-\frac {20 \operatorname {EllipticF}\left (\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right ),2\right )}{147 b^2}+\frac {20 \cos (a+b x) \sqrt {\sin (a+b x)}}{147 b^2}+\frac {4 \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x)}{49 b^2}+\frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b} \] Output:
-20/147*InverseJacobiAM(1/2*a-1/4*Pi+1/2*b*x,2^(1/2))/b^2+20/147*cos(b*x+a )*sin(b*x+a)^(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
Time = 0.52 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76 \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\frac {40 \operatorname {EllipticF}\left (\frac {1}{4} (-2 a+\pi -2 b x),2\right )+\sqrt {\sin (a+b x)} \left (46 \cos (a+b x)-6 \cos (3 (a+b x))+84 b x \sin ^3(a+b x)\right )}{294 b^2} \] Input:
Integrate[x*Cos[a + b*x]*Sin[a + b*x]^(5/2),x]
Output:
(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*Sin[a + b*x]^3))/(294*b^2)
Time = 0.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3924, 3042, 3115, 3042, 3115, 3042, 3120}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x) \, dx\) |
\(\Big \downarrow \) 3924 |
\(\displaystyle \frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {2 \int \sin ^{\frac {7}{2}}(a+b x)dx}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {2 \int \sin (a+b x)^{7/2}dx}{7 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {2 \left (\frac {5}{7} \int \sin ^{\frac {3}{2}}(a+b x)dx-\frac {2 \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x)}{7 b}\right )}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {2 \left (\frac {5}{7} \int \sin (a+b x)^{3/2}dx-\frac {2 \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x)}{7 b}\right )}{7 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {2 \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (a+b x)}}dx-\frac {2 \sqrt {\sin (a+b x)} \cos (a+b x)}{3 b}\right )-\frac {2 \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x)}{7 b}\right )}{7 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {2 \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (a+b x)}}dx-\frac {2 \sqrt {\sin (a+b x)} \cos (a+b x)}{3 b}\right )-\frac {2 \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x)}{7 b}\right )}{7 b}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 x \sin ^{\frac {7}{2}}(a+b x)}{7 b}-\frac {2 \left (\frac {5}{7} \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right ),2\right )}{3 b}-\frac {2 \sqrt {\sin (a+b x)} \cos (a+b x)}{3 b}\right )-\frac {2 \sin ^{\frac {5}{2}}(a+b x) \cos (a+b x)}{7 b}\right )}{7 b}\) |
Input:
Int[x*Cos[a + b*x]*Sin[a + b*x]^(5/2),x]
Output:
(2*x*Sin[a + b*x]^(7/2))/(7*b) - (2*((5*((2*EllipticF[(a - Pi/2 + b*x)/2, 2])/(3*b) - (2*Cos[a + b*x]*Sqrt[Sin[a + b*x]])/(3*b)))/7 - (2*Cos[a + b*x ]*Sin[a + b*x]^(5/2))/(7*b)))/(7*b)
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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]
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] - Simp[(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]
\[\int x \cos \left (b x +a \right ) \sin \left (b x +a \right )^{\frac {5}{2}}d x\]
Input:
int(x*cos(b*x+a)*sin(b*x+a)^(5/2),x)
Output:
int(x*cos(b*x+a)*sin(b*x+a)^(5/2),x)
Exception generated. \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x*cos(b*x+a)*sin(b*x+a)^(5/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
Timed out. \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\text {Timed out} \] Input:
integrate(x*cos(b*x+a)*sin(b*x+a)**(5/2),x)
Output:
Timed out
\[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\int { x \cos \left (b x + a\right ) \sin \left (b x + a\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(x*cos(b*x+a)*sin(b*x+a)^(5/2),x, algorithm="maxima")
Output:
integrate(x*cos(b*x + a)*sin(b*x + a)^(5/2), x)
\[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\int { x \cos \left (b x + a\right ) \sin \left (b x + a\right )^{\frac {5}{2}} \,d x } \] Input:
integrate(x*cos(b*x+a)*sin(b*x+a)^(5/2),x, algorithm="giac")
Output:
integrate(x*cos(b*x + a)*sin(b*x + a)^(5/2), x)
Timed out. \[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\int x\,\cos \left (a+b\,x\right )\,{\sin \left (a+b\,x\right )}^{5/2} \,d x \] Input:
int(x*cos(a + b*x)*sin(a + b*x)^(5/2),x)
Output:
int(x*cos(a + b*x)*sin(a + b*x)^(5/2), x)
\[ \int x \cos (a+b x) \sin ^{\frac {5}{2}}(a+b x) \, dx=\int \sqrt {\sin \left (b x +a \right )}\, \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2} x d x \] Input:
int(x*cos(b*x+a)*sin(b*x+a)^(5/2),x)
Output:
int(sqrt(sin(a + b*x))*cos(a + b*x)*sin(a + b*x)**2*x,x)